discrete mathematics transparency no. 2-1 chapter 2. basic structures: sets, functions, sequences...

Discrete Mathematics

Transparency No. 2-1

Chapter 2. Basic structures:

Sets, Functions, Sequences and Sums

Cheng-Chia Chen

The Foundations


2.1 Sets

Basic structure upon which all other (discrete and continuous ) structures are built.

A set is a collection of objects. An object is anything of interest, maybe itself a

set.Definition 1.

A set is a collection of objects. The objects in a set are called the elements or

members of the set. If x is a member of a set S, we say S contains x. notation: x S vs x S

Ex: In 1,2,3,4,5, the collection of 1,3 5 is a set.

The Foundations


Set descriptionHow to describe a set:?1. List all its member.

the set of all positive odd integer < 10 = ? The set all decimal digits = ? the set of all upper case English letters = ? The set of all nonnegative integers = ?

2. Set builder notation: P(x) : a property (or a propositional function) about

objects. e.g., P(x) = “ x > 0 and x is odd” then {x | P(x) } is the set of objects satisfying

property P. P(3) is true => 3 {x | P(x)} P(2) is false => 2 {x | P(x)}

The Foundations


Conventions

N =def {x | x is a natural numbers } = { 0,1,2,3,...}

N+ =def {1,2,3,...}

Z =def {...,-3,-2,-1,0,1,2,3,...}

R =def the set of real numbers.

Problem:

The same set may have many different descriptions. {x | 0 < x <10 /\ x is odd } {1,3,5,7,9}, {5,3,1,9,7} {9,7,1,3,5}.

The Foundations


Set predicatesDefinition 2.

Two sets S1, S2 are equal iff they have the same elements

S1 = S2 iff x (x S1 <-> x S2)Ex: {1,3,5} = {1,5,3} = {1,1,3,3, 5}Graphical representation of sets:

Venn Diagrams: rectangle: Universal set: U circles: sets points: particular elements point x inside circle S => x S point x outside S => x S.

S

S2

x

z

w

y

U

The Foundations


Set predicates (cont’d)Null set ={} = =def the collection of no objects.Def 3’: [empty set] for-all x, x.Def 3. [subset]

A B iff all elements of A are elements of B. A B x (x A x B)).

Def 3’’: A B =def A B /\ A B.Exercise : Show that:

1. For all set A, (Pf: Let x be any element => x(By def) => x x ⇒

A holds. Hence x (x x ⇒ A) and 2. (A B /\ B A) (A = B) 3. A ⇒

Diagram representation of the set inclusion relationship.

The Foundations


Size(or cardinality) of a finite set

An important property about sets is the number of members each set has. Such property is called its size or cardinality.

Def. 4: A: a set. Then |A| = the size(or cardinality) of set A = number of distinct elements in A.

Ex: |{1,3,3,5}| = 3 |{}| = 0 the set of decimal digits } | =|{0,1,..,9}| = 10

But how about |A| if A is not finite ? |N| = ? ; |Z| = ? ; |{2i | i N} | = ? ; |R| = ?∈ See section 2.4

The Foundations


The power set

Def 6. If A is a set, then the collection of all subsets

of A is also a set, called the power set of A and is denoted as P(A) or 2A.

Ex: P({0,1,2}) = ? P({}) = ? |P({1,2,..., n})| = ?

Elements in a set are not ordered. But sometimes we need to distinguish between (1,3,4) and (3,4,1) --> ordered n-tuples

The Foundations


Cartesian Products

Def. 7 [n-tuple] If a1,a2,...,an (n > 0) are n objects, then

“(a1,a2,...,an)” is a new object, called an (ordered) n-tuple [ with ai as its ith element. ]

Any ordered 2-tuple is called a pair.

Def. Two tuples (a1,a2,...,am) = (b1,b2,...,bn) iff (1) m = n and (2) ai = bi for all 1 ≤ i ≤ m.

The Foundations


Cartesian product

Def. 8: [Cartesian product] A, B : two sets.

A x B =def {(a,b) | a A and b B } ∈ ∈

A1 x A2 x ...x An =def {(a1,...,an) | ai A∈ i for all 1 ≤ i ≤ n}.

Note: AxB can be simply written as AB.

Ex: A = {1,2}, B = {a,b,c} , C = {0,1}

1. A x B = ? ; 2. B x A = ?

3. A x {} = ? ; 4. A x B x C = ?

Def. 8.1: Any subset of AB is called a relation from A to B.

Problems: 1. when will A x B = B x A ?

2. |A x B | = ?

The Foundations


2.2 Set operations

Union, intersection,difference , complement,Definition. If A and B are two sets from a

universe U, then 1. A B = {x | x A or x B }∈ ∈ 2. A B = {x | x A and x B }∈ ∈ 3. If A B = {} => call A and B disjoint. 4. A - B = {x | x A but x ∈ ∉ B } 5. ~ A = U – A = {x| x U but x∈ ∉A} = {x| x∉A} Venn diagram representationsEx: U = {1,..,10}, A = { 1,2,3,5,8} B = {2,4,6,8,10}=> A B , A B , A - B , ~ A = ?

The Foundations


Set identities (page 124, 6th edition)

Identity laws: A {} = A ; A U ⋂ = ADomination law: U A = A ; {} A⋂ = {}Idempotent law: A ⋂A = A ; AA = Acomplementation: ~~A = A

commutative : A⋂B = B ⋂A ; AB=BAAssociative: A⋂(B⋂C) = (A⋂B)⋂C; ___?_____Distributive: A (B ⋂ C) = ? ; ____?_____De Morgan laws: ~(AUB) = _?_ ; ~(A ⋂ B)= ~A U ~BAbsorption law : A(A ⋂ B) = A; A ⋂ (AB) = AComplement law: A~A = U; A⋂~A = {}

The Foundations


Prove set equality1. Prove that ~(A B) = ~A ~B by showing that

1.1 ~(A B)~A ~B and 1.2 ~A ~B ~(A B)

Pf. of 1 by definition and logic reasoning. It suffices to show that x. x~(AUB) iff x(~A ~B). Let x be any element. Then

x ~(AUB)

iff ~(xAUB) -- def. of complement

iff ~(xA \/ xB) -- def. of Union

iff (x A /\ x B) -- de Morgan law

iff (x ~A /\ x ~B) – def. of comlpement

iff x ~A ~B . -- def of intersection

Note that the top-down reading of the above proves 1.1 while the buttom-up reading proves 1.2.

The Foundations


2. The previous set equality can also be proven

by using set builder and logical equivalence.

Pf:

~(AUB)

= {x | ~(xAUB) } -- def. of complement

= {x | ~(xA \/ xB) } -- def. of union

= {x | ~(xA) /\ ~(xB) } -- de Morgan law

= {x | (x~A) /\ (x~B)} -- def. of compement

= {x | x (~A~B) } -- def. of intersection

= ~A ~B. -- set builder notation :

{x|xC} = C

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Membership Table

3. Show distributive law : A U (B ⋂ C) = (AUB) ⋂ (AUC) by using membership table.

Let x be any element. Then we need only consider 8 cases as to whether x is a member of A or B or C. In all cases, we find that x A U (B C) iff x⋂ (AUB) (AUC) . Hence the equality holds.⋂

x A x B x C xAUB

xAUC

xB⋂C

xA U (B ⋂ C)

x(AUB)⋂(AUC)

0 0 0 0 0 0 0 0

0 0 1 0 1 0 0 0

0 1 0 1 0 0 0 0

0 1 1 1 1 1 1 1

1 0 0 1 1 0 1 1

1 0 1 1 1 0 1 1

1 1 0 1 1 0 1 1

1 1 1 1 1 1 1 1

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Set equality reasoning

4 show ~(A (B C)) = (~C ~B) ~A by set identities.

pf:

~(A (B C))

= ~A C ) -- de Morgan law

= C ) ~A -- commutative law

= (~B ~C) ~A -- de Morgan law

= (~C ~B) ~A -- commutative law

The Foundations


Generalized set operations

Def. 6: Let A1,A2,...An: n sets and B = {A1,A2,...An }

12n{i=1,..n} Ai =def {x|xA for some AB}.

12n{i=1,..n} Ai=def {x| xA for all AB }.

Note: If B = {}, then = {} and U (why ?)Venn diagram representation of 123

and123Example:

123 = ?123

The Foundations


Set representation (in computer)

Ordered (or Unordered) list/array union, intersection, ~: time consuming.

Balanced–tree representation; HashtableBit string:

U = {a1,...,an} is the universal set.

A: any subset of U A can be represented by the string:

s(A) = x1 x2 x3....xn , where

xi = (ai A ? 1 : 0) for all i = 1..n.∈ fast set operations suitable only if U is not large. constant size representation.

The Foundations


2.3 FunctionsDef. 1 [functions] A, B: two sets

1. a partial function f from A to B is a set of pairs (x, y) AxB∈ s.t., for each x A there is at most one y B s.t. (x,y) f.∈ ∈ ∈ 2. f is a (total) function if for for each x A there is exactly one∈ y B with (x,y) f. ∈ ∈ Note if not specified otherwise, all

functions are assumed to be total. 3. if (x,y) f, we write f(x) = y.∈ 4. f : A B means f is a total function (or mapping) from A to B.

Def. 2. If f:A B then 1. A is the domain of f; 2. B is the codomain of f. If f(a)=b => 3. b is the image of a 4. a is the preimage of b 5. range(f) = f(A) = { f(x) | x A }.

The Foundations


Types of functionsDef 4. f : A B ; S: a subset of A,

T: a subset of B

1. f(S) =def { y B | x S with f(x) = y }

2. f-1(T) =def { x A | y T with f(x) = y }

Def 5,7,8. [1-1(injection), onto(surjection), bijection]

f: A -> B. f is 1-1 (an injection) iff xyA. (xyf(x) f(y) ). f is onto (surjective, a surjection) iff yBxA f(x) = y. If f is 1-1 & onto, then f is said to be bijective (a

bijection, 1-1 correspondence)

The Foundations


Properties of functions

1. If there is an onto mapping from A to B, then there is a 1-1 mapping from B to A.

2. If there is a 1-1 mapping from A to B then there is an onto mapping from B to A.

3. The onto (≥ ) and the 1-1(≤ ) relations between sets are reflexive and transitive (i.e., a preorder).

4. If there is a 1-1 mapping from A to B and a 1-1 mapping from B to A, then there is a 1-1 and onto mapping from A to B.

pf: (1) Let f:AB an onto. For each b B, let g(b) = {x A | f(x) = b} ≠ ∈ ∈∅ the function h:B⇒ A defined by h(b) = any x g(b) is 1-1. ∈

(2) f: A B a 1-1 mapping, b: any element in A. Then define g:B A with g(y) = (y = f(x) for some x A)? x : b). g is an onto. ∈

(3) is easy to verify. (4) is hard but understandable (reference).

The Foundations


Real valued functionsF:A B is a real valued function iff A and B are subsets

of R.f1,f2: real valued functions

=> 1. (f1+f2) (x) = f1(x) + f2(x)

2. (f1• f2) (x) = f1(x) x f2(x).

3. f is increasing iff ?

x y implies f(x) f(y)

4. f is strictly increasing iff ?

x > y implies f(x) > f(y) ceiling(.), floor(.): R Z.

x = least integer x; y = greatest integer y Ex: 3.4 = 4; -3.4 = -3 ; 2 = 2 ; ; -3.4 = -4

The Foundations


Operations on functions

A, B, C: any sets ; f: A B; g: B C, then

1. [identity function]

idA : A A s.t. idA(x) = x for all x in A.

2. [composition of g and f]

gf: A C s.t. (gf)(x) = g(f(x)) for x in A.

3. [inverse] If f is a bijection, then

f-1: B A s.t. f-1(y) = x iff f(x) = y for y in B. Graphical representations

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Properties of operations on functionsf: A B. g: B C; h: C D, then

1. f idA = f = idB f

2. h(g f) = (h g)f --- assoc.

3. (g f)-1 = f-1 g-1

If f: A B is a bijection then

4. f-1 f = idA and f f-1 = idB .

The Foundations


2.4 Sequences and Summations

Sequences: a discrete structure used to represent an ordered list.

Let [n] = [1,n] = {1,2,…,n}. A sequence 1,2,… n of length n with value from a set B is called a finite B-sequence (of length n). Formally, a finite B-sequence (of length n) is a function : [n] B.

A(k) or ak where k [n] is called a term( 項 ) of the sequence. We use the notation {an} where an is a function (or

expression) of n to represent a sequence.A B-sequence is a function : N B.A general C indexed B-sequence is a function: f:CB.

The Foundations


Summations

If {ak} is a sequence of length l, then we use

mjn ai or jm..n ai

to represent am + am+1 + … + an

If m(lower limit) = 1 and n(upper limit) = l =>ai =def a1 + a2+...al+ (...)

Summation rules :a + (a+d) + (a+2d)+... = ?a + ar +arr + ... = ?

The Foundations


Summation rules

1. a + (a+d) + (a+2d)+... (a + (n-1) d ) = ?2. a + ar +arr + ... + a rn-1 = ?Sol 1: Let S = a + (a+d) + (a+2d)+... (a + (n-1)d).Then S = (a + (n-1)d) + (a+(n-2)d) +… (a+d)+a 2S = (2a + (n-1)d) + …. (2a+ nd) = (2a+ nd) x n.

=> S = [a + (n-1)d]n/2 = (a1 + an)n/2.Sol 2: Let S = a + ar + ar2+... +a r (n-1).Then rS = ar + ar2 + ar3 +… arn

(1-r)S = a - arn. => S = a(1-rn)/(1-r). If |r| < 1, then limitn S = a/(1-r) Some facts: (proofs using induction left as exercises!)

k=1..n k = n(n+1) / 2 ; k=1..nk2 = n(n+1)(2n+1)/6 k=1..n k3 = n2(n+1)2 /4

The Foundations


cardinality of finite or infinite sets

Def. 4: |A| = the cardinality of set A = number of distinct elements in A.

Ex: |{1,3,3,5}| = 3 |{}| = 0But how about |A| if A is not finite ?

Ex: |N| = ? ; |Z| = ? ; | {2i | i N} | = ? |R| = ?∈Def. 4’: Two sets are said to have the same cardinality iff

there is a 1-1 & onto mapping between them. The collection of all sets which have the same cardinality is

said to be a cardinal number. Every non-negative integer n can be viewed as an

equivalence to the collection of all sets which has size n.

Def. 5: A set A is said to be finite iff |A| is a natural number; o/w it is infinite.

The Foundations


countability of sets

Def. A set A is said to be denumerable (or coutably

infinite) iff |A| = |N| ( = aleph0). A set is countable iff either |A| = n for some n in

N or |A| = |N|.Exercises: Show that

1. |N| = |Z | = | Q+ | = |{x | x N and x >3 }| 2. |R| = | (-1, 1) | = |(0,1)| 3. |(0,1)| is uncountable

By exercise 1,2,3, R is not countable. Q+ and Z is countable.

p/q 1 2 3 4 …

1 1/1 1/2 1/3 1/4 …

2 2/1 2/2 2/3 2/4 …

3 3/1 3/2 3/3 3/4 …

4 4/1 4/2 4/3 4/4 …

… … … … … 5/5

How to enumerate Q+

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Properties of countable sets A is countable iff there is a onto mapping from N to A.Pf. “=>” If A is finite, obviously there is a onto mapping from N to A.

If A is denumerable, by definition, there is also a onto (&1-1)

mapping from N to A.

“<=“ If A is finite, it is surely countable.

If A is infinite, let f:N A be the surjection from N to A.

we can define a bijection g: N A from f as follows:

g(0) = f(0),

g(k) = f(j) where j = least {m | f(m){g(0), g(1),…,g(k-1)} for k >0.

Ex: If f: N A is then

g: N A is

As a result, |A| = |N| and is countable.Corollary: A is countable iff there is a 1-1 mapping from A to N.

5 3 5 1 3 2 1 1 4 …

5 3 1 2 4 …

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Properties of countable setsSuppose A B. Then

1. If B is countable, then A is countable.

2. If A is uncountable, then B is uncountable.

Pf: Note (2) is logically equivalent to (1) and hence is implied by (1).

(Pf of 1) Obviously A is countable if B is finite or A is finite.

If both B and A are infinite. Since B is denumerable, there is a 1-1 mapping f: B N. now define f’: A N with f’(x) = f(x) for all x A. Then f’ is also a 1-1 from A to N. Hence A is countable.

Corollaries: Every subset of a countable set is countable.

Since Q is countable (because |Q+| = |Q| why? ), all its subsets (Z, N, Even, {x | x > 10} …) are countable.

Every superset of a uncountable set is uncountable. Since (0,1) is uncountable, R is uncountable.

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Some uncoutable sets

0 1 2 3 4 ... f(0) 1 0 0 f(1) 0 0 f(2) 0 f(3) 0 0 1 0 0 f(4) 0 0 ... 0 0 0 1 1

Theorem: The set 2N is uncountable.

pf: Suppose 2N is denumerable. Thenthere is a bijection f: N -> 2N. I.e., 2N = {f(0),f(1),f(2),...} . Now consider the diagonal set

= { k N | k f(k) }. Since kf(k) iff k for all k, f(k) for all k.but since is a subset of N, by assumption, must equal to f(k) forsome k. This is a contradiction!

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|A| ≠ |2A|

Pf: (1) The case that A is finite is trivial since |2A| = 2|A| > |A| and there is no bijection b/t two finite sets with different sizes.

(2) Assume |A| = |2A|, i.e., there is a bijection f: A B, where B is 2A and f(A) = { f(x) | x A } = B.

Now define the so-called diagonal set in terms of f as follows:

= {x A | x ∈ f(x) }. Now (*) is a subset of A and (**) f(x) for any x A :∈ ( since x ∈ iff x f(x) for each x A. )∈ ==> is not a subset of A ( f is onto to 2∵ A), This is a contradiction to (*). Hence f must not exist!!

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Real set (0,1) is not countable

Theorem: The real set (0,1) is uncountable.pf: If n is a real in (0,1), then it can be represented as an infinite sequence n = 0.n1n2n3… where each nk is one of {0,…,9}.Ex: 0 = 0.00000 ; 1 = 0.9999… ½ = 0.500000… = 0.49999…. 1/3 = 0.3333… -3 = 0.14159….For the representation to be unique, we exclude any sequence of the form: 0.d1d2…dk9999…. and say any sequence not in such form a normal sequence. Now suppose the set (0,1) is denumerable. Then there must exist a bijection f: N (0,1).

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We apply diagonalization principle to define a number d from f as follows:

let d = 0.d0d1d2…. where

dk = (s == 0)? 8 : 9 – s,

where s=f(k)k is the kth digit of the fraction part of f(k)

digit f(n) 0 1 2 3 4 ... f(0)= .1 0 5 0 0 0 f(1)= . 0 2 0 f(2)= . 5 7 f(3)= . 0 0 7 0 0 …

d= .8 8 4 2 . .

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Now it can be shown that d is a normal sequence representing a real number in (0,1)

since it contains no 9. d f(n) for all n since by construction dn f(n)n, where

f(n)k is the kth digit of the fraction part of f(n).

This contradicts the facts that d (0,1) and f(N) = ∈(0,1).

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The Halting Problem (Section 3.1, 6th edition; p176)L : any of your favorite programming languages (C,

C++, Java, BASIC, etc. )Problem: write an L-program HALT(P,X), which takes

another L-program P(-) and string X as input, and

HALT(P,X) returns true if P(X) halts and

returns false if P(X) does not halt.

P(X) Halt ?

yes true

nofalse

Pr(x:String) {…..

}

P

“It’s a test …”

X

HALT(P,X)

Program you are asked to

design

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How hard is it to predict if a program will halt ?

Consider the 3n+1 programint f( int n ) {

if ( n == 1 ) return 1 ;

if (n % 2 == 0 ) { // n is even n/2

f(n % 2) ;

} else { // n is odd 3n+1

f( n*3 + 1) ;

}}

It is conjectured that f(n) halts (returns 1) for all n ≥ 1.

Ex: f(10) 5 16 8 4 2 1 (halts)

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Halt(P,X) does not exist Ideas leading to the proof:Problem1 : What about the truth value of the sentence:

L: L is false Problem 2 : Let S = {X | X X}. Then does S belong to S or not ?The analysis: S S => S S; S S => S S.

Problem 3 : 矛盾說 : 　 1. 我的矛無盾不穿　 2. 我的盾可抵擋所有茅　　　結論 : 1. 2. 不可同時為真。

Problem 4 : 萬能上帝 : 　萬能上帝無所不能 => 可創造一個不服從他的子民 => 萬能上帝無法使所有子民服從 => 萬能上帝不是萬能 . 　　　　　　結論 : 萬能上帝不存在。

Conclusion: 1. S is not a set!! 2. If a language is too powerful, it may produce expressions that is

meaningless or can not be realized. Question: If HALT(P,X) can be programmed, will it incur any absurd result

like the case of S?Ans: yes!!

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H(P) : a program that HALT(-,-) cannot predict

Notes: 1. H(P) is simply { L : if(HALT(P,P)) then goto L } 2. H uses HALT(-,-) as a subroutine. 3. H(P) halts iff HALT(P,P) returns false iff P(P) does not halt. 4. Let input P be H H(H) halts iff H(H) does not halt. HALT is not a correct implementation!

H(P)

Pr(x:String) {…..

}

P

P(P) Halt ?

yestrue

no false

HALT(P,X) Loop

haltP

P

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The growth of functions ( Section 3.2, 6th edition)

If f is a positive valued function, then what do the following terms mean? O(f) means ? (f) means ? (f) means ? useful in the analysis of the efficiency of

computer algorithms.

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Problem, algorithm and Complexity

A problem is a general question: description of parameters [input] description of solution[output] description of how the outputs are related to the

inputs. An algorithm is a step by step procedure to get

the related output (i.e., answer) from the given input. a recipe a computer program

We want the most efficient algorithm fastest (mostly) most economical with memory (sometimes) expressed as a function of problem input size

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Parameters: Set of cities Inter-city distances

Example: Traveling Salesman Problem

c

bd

a

3

5

6

9

910

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Solution [output]

Solution: The shortest tour passing all cities Example: a,b,d,c,a has length 27

c

bd

a

3

5

6

9

910

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Problem Size

What is appropriate measure of problem size? m nodes? m(m+1)/2 distances?

Use an encoding of the problem alphabet of symbols: a,b,c,d,0-9, |. strings: abcd||10|5|9|6|9|3.

Measures Problem Size: length of encoding. Time Complexity: how long an algorithm takes,

as function of problem size.

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Time Complexity

What is tractable? A function f(n) is O(g(n)) whenever ∃ c > 0 n∃ 0 > 0 s.t. |f(n)| ≤ c ∙ |g(n)| for all n > n0.

A polynomial time algorithm is one whose time complexity is O(p(n)) for some polynomial p(n).

An exponential time algorithm is one whose time complexity cannot be bounded by a polynomial

(e.g., nlog n or 2n).

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Tractability Basic distinction:

polynomial time = tractable exponential time = intractable

10 20 30 40 50 60

n .00001s .00002s .00003s .00004s .00005s .00006s

n2 .0001s .0004s .0009s .0016s .0025s .0036s

n3 .001s .008s .027s .064s .125s .216s

n5 .1s 3.2s 24.3s 1.7 m 5.2 m 13.0 m

2n .001s 1.0s 17.9s 12.7d 35.7 y 366c

3n .059 s 58 m 6.5 y 3855 c 2∙108c 1.3∙10 13 c

execution time in microseconds(μs)

problem size

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Effect of Speed-ups for different order of functionsWait for faster hardware! Consider maximum problem size you can solve in an

hour. present 100 times faster 1000 times faster

n N1 100 N1 1000 N1

n2 N2 10N2 31.6 N2

n3 N3 4.64 N3 10 N3

n5 N4 2.5 N4 3.98 N4

2n N5 N5+6.64 N5+9.97

3n N6 N6+4.19 N6+2.29

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Asymptotic notations

Asymptotic analysis: care the value of f(n) only when n is very large. discard the difference of f and g when limit f/g is

bounded by two positive numbers (a,b).

((g(n)) = { f(n) | c and k > 0 s.t. f(n) cg(n) for all n > k }

((g(n)) = { f(n) | c2,c1 and k > 0 s.t. c2g(n) f(n) c1 g(n) for all n > k }

((g(n)) = { f(n) | c and k > 0 s.t. c g(n) f(n) for all n > k }

The Foundations


Examples

Ex1: Show that f(x) = x2 + 2x+ 1 is O(x2).

pf: Let c = 4, k = 1, It is easy to check that if x > k, then f(x) = x2 + 2x+ 1 ≤ x2 + 2 x2 + x2 ≤ c x2.

Hence by definition, f(x) = O(x2).

Ex2: Show that f(x) = x2 + 2x+ 1 is (x2).

pf: Let c = 1, k = 1. Obviously, x2+2x+1 ≥ x2 = cx2 for all x >1=k. Hence f(x) = (x2).

The Foundations


Ex2: Show that f(x) = x2 + 2x+ 1 is (x2).

pf: Let c1 = 1,c2=4, k = 1.

c1x2 ≤ x2+2x+1 ≤ c2 x2 for all x >1=k. Hence f(x) = (x2).

Proposition: f = (g) iff f = (g) and f = (g).pf: (=>:) direct from definition.(<=:) if f = O(g) = (g)

=> c∃ 1, k1 s.t f(x) ≤ c1 g(x) for all x > k1.

∃ c2, k2 s.t f(x) ≥ c2 g(x) for all x > k2.

let k = max(k1,k2) =>

c2 g(x) ≤ f(x) ≤ c1 g(x) for all x > k. Hence f(x) = (g).

The Foundations


Using the asymptotic notations

We use f(n) = (g(n)) to mean f(n) (g(n)), where = .

analogy: O (asymptotic upper bound): f(n) = O(g(n)) behaves like f(n) g(n) (asymptotic lower bound): f(n) =(g(n)) behaves like f(n) g(n)

(asymptotic tight bound : =

f(n) = (g(n)) behaves like f(n) = g(n)

The Foundations


Exercises

Let f and g be positive real function. Show that

1. f = (g) iff g = (f)

2. f= (g) iff limit n g/f is bounded above 0

discrete mathematics transparency no. 2-1 chapter 2. basic structures: sets, functions, sequences...

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