Probabilistic Method

Presented by Jordon LynCourse COMP 4804

Seminar Oulinebull Short talk about its origin and definitionbull Explain the various techniques used in performing this

method (Proofs will be used to show how it works)

Origin and definition

Created by Paul Erdos in 1947

Used to help prove the existence of certain objects without having to construct them explicitly

Achieved by the construction of a sample space with all possible candidates given that the probability of finding the desired object is positive

Examples

Probabilistic Method Succeds

bull Getting an even number on a fair 6-sided diebull Only 6 number on the die

bull (1 2 3 4 5 6)

bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist

Probabilistic Method Failsbull Finding 5 Aces in a shuffled

deck of cardsbull 52 cards in a shuffled deck

with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4

times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck

bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist

Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma

Basic Methodbull Best illustrated through many methods

bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a

subgraph) or an independent set (a set of vertices that induce edgeless subgraph)

bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored

with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem

bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero

bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Seminar Oulinebull Short talk about its origin and definitionbull Explain the various techniques used in performing this

method (Proofs will be used to show how it works)

Origin and definition

Created by Paul Erdos in 1947

Used to help prove the existence of certain objects without having to construct them explicitly

Achieved by the construction of a sample space with all possible candidates given that the probability of finding the desired object is positive

Examples

Probabilistic Method Succeds

bull Getting an even number on a fair 6-sided diebull Only 6 number on the die

bull (1 2 3 4 5 6)

bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist

Probabilistic Method Failsbull Finding 5 Aces in a shuffled

deck of cardsbull 52 cards in a shuffled deck

with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4

times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck

bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist

Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma

Basic Methodbull Best illustrated through many methods

bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a

subgraph) or an independent set (a set of vertices that induce edgeless subgraph)

bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored

with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem

bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero

bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Origin and definition

Created by Paul Erdos in 1947

Used to help prove the existence of certain objects without having to construct them explicitly

Achieved by the construction of a sample space with all possible candidates given that the probability of finding the desired object is positive

Examples

Probabilistic Method Succeds

bull Getting an even number on a fair 6-sided diebull Only 6 number on the die

bull (1 2 3 4 5 6)

bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist

Probabilistic Method Failsbull Finding 5 Aces in a shuffled

deck of cardsbull 52 cards in a shuffled deck

with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4

times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck

bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist

Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma

Basic Methodbull Best illustrated through many methods

bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a

subgraph) or an independent set (a set of vertices that induce edgeless subgraph)

bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored

with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem

bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero

bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Examples

Probabilistic Method Succeds

bull Getting an even number on a fair 6-sided diebull Only 6 number on the die

bull (1 2 3 4 5 6)

bull Pr(Even Number) = 36 = 12 gt 0 meaning an even number does exist

Probabilistic Method Failsbull Finding 5 Aces in a shuffled

deck of cardsbull 52 cards in a shuffled deck

with 13 different ranks and 4 suitsbull Each rank appears 5213 = 4

times (one for each suit) meaning there can only be a maximum of four Aces in a shuffled deck

bull Pr(5 Aces) = 0 meaning a fifth Ace does not exist

Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma

Basic Methodbull Best illustrated through many methods

bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a

subgraph) or an independent set (a set of vertices that induce edgeless subgraph)

bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored

with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem

bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero

bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Techniques bull Basic Methodbull Linearity of Expectationbull Alterationsbull The Second Momentbull Lovasz Local Lemma

Basic Methodbull Best illustrated through many methods

bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a

subgraph) or an independent set (a set of vertices that induce edgeless subgraph)

bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored

with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem

bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero

bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Basic Methodbull Best illustrated through many methods

bull Ramsey Numbersbull Any sufficiently large graph contains either a clique (a set of vertices that induce a

subgraph) or an independent set (a set of vertices that induce edgeless subgraph)

bull Hypergraph Coloringbull A hypergraph is c-colorable if the vertices of the graph can be colored

with c colors given that at least two colors appear in every edgebull Erdos-Ko-Rado Theorem

bull A family of sets F is intersecting if for every pair of sets within F the intersection set of both those sets is greater than zero

bull Overall goal prove a lower bound exists for each method when dealing with large integer valuesbull More sophisticated and simpler than counting arguments

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Basic Method (Hypergraph Coloring)bull A k-uniform hypergraph is a pair (X S) where X is the set

of vertices and S is the set of edgesbull S is a subset of X choose k where k is the amount of tuples of

vertices

bull Graphs are 2-colorablebull Generalized notion of graph coloring

bull m(k) = the smallest number of edges in a k-uniform hypergraph that are not 2 colorablebull m(2) = 3 -gt the smallest non-bipartite graph is a trianglebull Becomes difficult for k gt 3bull Hence the probabilistic method is used to retrieve a lower bound

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Hypergraph Coloring Proofbull Theorem For any k ge 2

m(k)ge 2^k-1bull Consider a hypergraph H that has less than

2^k-1 edges (Prove it is 2-colorable)bull each vertex gets painted red or blue independentlybull p = probability all edges are red or blue = 212^k bull p = 12^k-1

bull H has subset Sbull |S| lt 2^k-1 edgesbull probability there is an edge with only one color is

at most p|S| lt p2^k-1 bull p|S| lt 1 because p2^k-1 = 1

bull Therefore there is a probability greater than 0 that every edge in H has at least two colors

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Basic Method (Erdos-Ko-Rado Theorem)bull Created by Paul Erdos Chao

Ko and Richard Radobull Lemma X = 0 1n-1 with

modulo n and As = s s+1s+k-1 is a subset of Xbull 0 lt s lt nbull Then for n ge 2k any intersecting

family F contains at most k of the sets of As

bull Theorem |X| = n n ge 2k and F is an intersecting family of k-subsets of X bull |F| le n-1 choose k-1

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Erdos-Ko-Rado Theorem Proof

Lemma Proof

bull Assume the set Ai is an element of F bull Therefore any other set in As is

one of the followingbull Ai Ai+1 Ai+k-1

bull These 2k-2 remaining sets can be divided into k-1 pairs (As As+k)

bull Because n ge 2k the intersection between As and As+k is 0 and therefore only one set from each of the k

Theorem Proofbull Assume X = 0 1n-1 and

F (a subset of X choose k) is an intersecting familybull By following the lemma at

least k of the n sets are within Fbull Pr(As is an element of F) le knbull Pr(As is an element of F) =

|F|(n choose k)bull |F| = (n choose k) kn= n-1

choose k-1

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Linearity of Expectation bull Takes the expected values of individual indicators and uses

their summation as the expected value of a random variablebull Also knows as the first moment method

bull Proves how the expected value of a random variable is positivebull Easily done using Markovs Inequality

bull The probabilistic method plays an integral role in the linearity of expectationbull Can be used to help calculate the expected value of a random

variable

bull Such examplesbull Tournamentsbull Hamiltonian Paths

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Tournaments and Hamiltonian Paths

Tournaments

bull A graph where all undirected edges are replaced by directed edgesbull Graph must be completedbull Has a total of n vertices

and n choose 2 edges

Hamiltonian Paths

bull A path that visits every vertex in the graph exactly oncebull Named after William

Rowan Hamilton inventor of Hamiltons puzzle

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Hamiltonian Path Theorembull Theorem There is a tournament with n vertices that

has at least n2^n-1 Hamiltonian pathsbull Proofbull X = of Hamiltonian paths in the tournament

bull Each edge has 12 chance of being chosen independently

bull For a permutation p on 1 2 n consider the sequecne p(1)p(n) Xp is the indicator random variable for the event that all the edges (p(i) p(i+1)) are in the tournamentbull E(Xp) = Pr((p(i) p(i+1)) is in the tournament for i = 12n-1) =

12^n-1

bull E(X) = E(X1) + E(X2) + E(X3) + + E(Xn)E(X) = n2^n-1

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Alterationsbull Sometimes the first attempt at finding the desired

object fails but there does exist an object that nearly satisfies the desired conditionsbull It is possible to modify the object to have the desired

properties in a deterministic time

bull Such examplesbull High Girthbull High Chromatic Number

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

High Girth and High Chromatic Numberbull Part of a famous problem solved by Paul Erdosbull Could the non-existent of short cycles in a graph imply that

the graph can be colored with a small number of colorsbull Erdos proved this was false

bull There do exist some graphs that have large chromatic numbers despite having no short cycles

bull High Girth g(G) the length of the graphs shortest cyclebull High Chromatic Number x(G) the smallest integer k

such that the graph has proper k-coloring

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

High Girth and High Chromatic Example

High Girth High Chromatic

This graph has a girth of 6

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Second Momentbull Similar to the first moment method as it deals with showing how

a random variable has positive probability of being positivebull Accomplishes this with a random variables variance

bull Variance describes how much the variable fluctuates around its expectationbull Not a linear operator like expectation meaning that in order to calculate

the variance of the sum of random variables we need to understand the pairwise independence or covariance

bull Once the variance has been discovered we can apply to Chebyshev Inequality to estimate the probability that a random variable deviates from its expectation at least by a given number

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Equations and proofbull Variance equation

bull Var(X) = E((X - E(X))^2) = E(X^2) - (E(X))^2

bull Standard deviation of X = (Var(X))^12

bull If X1Xn are independent the variance is the sum of all Xi

bull Covariance equationbull Cov(XY) = E((X - E(X)) (Y-E(Y)))

= E(XY) = E(X) E(Y)bull If X1Xn are independent then

the covariance is 0bull Cov(XY) = 0 does not imply

independence of X and Y

bull Chebyshev Inequalitybull Pr((X-E(X)) ge t) le Var(X)tbull Proof

bull Var(X) = E((X - E(X)) ^2) ge t Pr((X-E(X)) ge t)

bull Contrast to empirical rule (68-95-997 rule)bull 75 of the values lie within

two standard deviations of the mean

bull 89 within three standard deviations of the mean

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Empirical Rule vs Chebyshevs Inequality

Empirical Rule Chebyshevs Inequality

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Lovasz Local Lemmabull Suppose you have an array of bad events (A1An)

and you want to avoid thembull Best approach is when they are independent of each otherbull Therefore their complements are independent

bull Still a positive probability none of the bad events happen

bull Proven by Paul Erdos and Laszlo Lovasz in 1975 bull Allows us to exclude all bad events given they have

relatively small probability and their dependency digraph does not have too many edges

bull Two Ways Symmetric and Generalbull Symmetric is the most often used

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

Symmetric Local Lemmabull Let A1An be events such that Pr(Ai) le p for all i and all

outdegrees in a dependency digraph of Ai are at most dbull Each Ai is independent of all but at most other d of the other Aj

bull If ep(d + 1) le 1 (e is the natural logarithm)bull Pr(A1An do not happen) gt 0

bull Proof bull If d = 0 the events are mutually independent and the result follows

easilybull Otherwise Xi = 1d+1 lt 1bull The dependency digraph the outdegree of any vertex is at most d so

bull xi the total product of (1-xj) ge 1d+1(1-1d+1)^d ge 1e(d+1) ge pbull With this we can apply the general local lemma

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

General Local Lemmabull Let A1A2An be events

and D = (VE) be the dependency digraph and Xi be an element in the set of real numbers between 0 and 1 in a way thatbull Pr(Ai) = Xi the total

product of (1-Xj)bull Pr(Complements of

A1An) ge the total product of (1-Xi) ge 0

bull Proof Prove that for any subset S 1n and i is not an element of Sbull Proceed with induction on

the size of Sbull If S = 0 the statement

follows directly from the lemmabull Pr(Ai) le the total product of

Xi (1-Xj)le Xi

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

General Local Lemma (cont)bull Suppose it holds for any S where |S| lt |S|

bull Set S1 = j is an element of S (i j) is an element of E and S2 =S S1bull Assume S1 gt 0 Ai is independent of all events Aj j is an element of S then we havebull Pr(Ai | intersection of all complement Aj in S ) = Pr(Ai and intersection of all

complement Aj in S1 | the intersection of all complement Al in S2) Pr(the intersection of all complement Aj in S1| the intersection of all complement Al in S2)

bull Ai is independent of the events Al l is an element of S2 bound the numeratorbull Pr(Ai and the intersection of all complement Aj in S1 | the intersection of all

complement Al in S2) le Pr(Ai |the intersection of all complement Al in S2) le Pr(Ai) le Xi the total product of (1-Xj)

bull Now bound the denominatorbull Pr(Aj1Ajr | the intersection of all complement of Al in S2) ge the total product of (1-Xj)

bull Once then Pr(Ai | the intersection of all complement Aj in S) le Xi bull Pr (All complements of Ai in S) ge the total product of (1 - Xi)

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem

References

For informationbull Matousek Jiri Jan Vondrak The

Probabilistic Method httpswwwcscmueduafscscmumatousek-vondrak-prob-lnpdf Department of Applied Mathematics Charles University March 2008 16 Mar 2015 PDF File

bull Li Jiayi The application of probabilistic method in graph theory httpswwwuni-ulmdefileadminwebsite_uni_ulmmawiinst110lehress10seminarjiayi_lipdf Ulm University 10 Oct 2006 18 Mar 2015 PDF File

For picturesbull httpenwikipediaorgwikiLovC3A1sz_local_le

mma

bull httpenwikipediaorgwikiTournament_(graph_theory)

bull httpenwikipediaorgwikiHamiltonian_path bull httpenwikipediaorgwikiHypergraph bull httpenwikipediaorgwikiRamsey27s_theorem bull httpenwikipediaorgwikiGirth_(graph_theory) bull httpmathworldwolframcomChromaticNumberht

ml

bull httpstatwikiucdaviseduUnder_ConstructionDescriptive_Statistics25_The_Empirical_Rule_and_Chebyshevs_Theorem