Randomized algorithmsInge Li Gørtz

• Today• What are randomized algorithms?• Properties of randomized algorithms• Two examples:

• Median/Select.• Quick-sort

Randomized algorithms

Randomized Algorithms

• So far we dealt with deterministic algorithms: • Giving the same input to the algorithm repeatedly results in:

• The same running time. • The same output.

• Randomized algorithm:• Can make random choices (using a random number generator)

Randomized Algorithms

• A randomized algorithm has access to a random number generator rand(a, b), where a, b are integers, a < b.

• rand(a, b) returns a number x ∈ {a, (a + 1), . . . , (b − 1), b}.

• Each of the b − a + 1 numbers is equally likely.

• Calling rand(a, b) again results in a new, possibly different number.  

• rand(a, b) “has no memory”. The current number does not depend on the results of previous calls.

• Statistically speaking: rand(a, b) generates numbers independently according to uniform distribution.

Random number generator• The algorithm can make random decisions using rand

Randomized algorithms

Runningwhile (rand(0,1) = 0) do

print(“running”);end whileprint(“stopped”);

Julefrokostwhile (rand(1,4) = rand(1,4)) dohave another Christmas beer;

end whilego home;

• A randomized algorithm might never stop.

• Example: Find the position of 7 in a three-element array containing the numbers 4, 3 and 7.

Properties of randomized algorithms

Find7(A[1…3])goon := true;while (goon) doi := rand(1,3);if (A[i] = 7) then print(Bingo: 7 found at position i);goon := false;end if

end while

• A randomized algorithm might find the wrong solution.

• Example: Find the position of the minimum element in a three-element array.

• If A = [5,6,4] and i = 2 and j = 1 then the algorithm will wrongly return 5.

Properties of randomized algorithms

MinOfThree(A[1…3])i := rand(1,3);j := rand(1,3);return(min(A[i],A[j]);

• Both examples are not so bad as they look.

• It is highly unlikely that Find7 runs a long time.

• It is more likely that MinOfThree returns a correct result than a wrong one.

Properties of randomized algorithms• An event is something that can happen. • An atomic event is not composed of others. • Each atomic event x has an associated probability P[x] ∈ [0,1]. • The sum of the probabilities of all atomic events is 1.

• We consider only finite or countably infinite sets of atomic events.

Events

X

x is atomic event

P [x] = 1

• A random variable is an entity that can assume different values.

• The values are selected “randomly”; i.e., the process is governed by a probability distribution.

• Examples:

• The number shown by a die.

• The distance of two coins dropped on the floor.

• The running time of a randomized algorithm.

Random variables• Let X be a random variable with values in {x1,…xn}, where xi are

numbers.

• Let pi = P[X = xi] be the probability that X assumes value xi.

• The expected value (expectation) of X is defined as

• The expectation is the theoretical average.

Expected values

E[X] =nX

i=1

xi · pi

• Let X be the random variable “number shown by the die”.

• Atomic events: {1, 2, 3, 4, 5, 6}

• Probabilities: P[1] = ··· = P[6] = 1/6

• Composed event: Even = {2, 4, 6}

• Composed event: ge4 = {4, 5, 6}

• Expectation: E[X] = (1+2+3+4+5+6)/6 = 3.5

Example: a fair die• Let X be the random variable “number shown by the die”.

• Atomic events: 1, 2, 3, 4, 5, 6

• Probabilities: 0.1, 0.1, 0.1, 0.3, 0.2, 0.2

• Expectation

E[X] = 1 · 0.1 + 2 · 0.1 + 3 · 0.1 + 4 · 0.3 + 5 · 0.2 + 6 · 0.2 = 4.0

Example: a loaded die

• If we repeatedly perform independent trials of an experiment, each of which succeeds with probability p > 0, then the expected number of trails we need to perform until the first succes is 1/p.

• Linearity of expectation: For two random variables X and Y we haveE[X+Y] = E[X] + E[Y].

Expectation• Let A and B be events.

• If A and B are independent then

• P[A∧B] = P[A] · P[B]

• If A and B are disjoint then

• P[A∨B] = P[A] + P[B]

Rules for Probabilities

• A die is tossed twice.

• Let the random variable X1 be the result of first throw.

• Let the random variable X2 be the result of second throw.

• Events: A = (X1 = 2) and B = (X2 = 5)

• P[A] = 1/6 and P[B] = 1/6

• P[A∧B] = 1/36 = 1/6·1/6 = P[A] · P[B]

Independence example• Event A = Even = {2, 4, 6}: P[A] = 1/2

• Event B = ge4 = {4, 5, 6}: P[B] = 1/2

• A ∧ B = {4, 6}

• P[A∧B] = 1/3 ≠ 1/4 = P[A] · P[B]

Non-independent example

• Running time: O(1).• What is the chance that the answer of MinOfThree is correct? • 9 possible pairs all with probability 1/9:

• (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)• Assume wlog that A[1] is the minimum:

• 5 of the 9 pairs contain the index 1 and give the correct minimum. • P[correct] = 5/9 > 1/2.

Analysis of MinOfThreeMinOfThree(A[1…3])

i := rand(1,3);j := rand(1,3);return(min(A[i],A[j]);

• Running MinOfThree once gives the correct answer with probaility 5/9 and the incorrect one with probability 4/9.

• Idea: Run MinOfThree many times and pick the smallest value returned. • The probability that MinOfThree fails every time in k runs is:

• To be 99% sure to find the minimum, choose k such that

Analysis of MinOfThreeMinOfThree(A[1…3])

i := rand(1,3);j := rand(1,3);return(min(A[i],A[j]);

• Running time is variable. What is the expected running time?• Assume wlog that A[1] = 7.• If rand(1, 3) = 1 then the while-loop is terminated.

• P[rand(1,3) = 1] = 1/3 • If rand(1, 3) ≠ 1 then the while-loop is not terminated.

• P[rand(1,3) ≠ 1] = 2/3• The while-loop is terminated after one iteration with probability

P[rand(1,3) = 1] = 1/3.

Analysis of Find7Find7(A[1…3])

goon := true;while (goon) doi := rand(1,3);if (A[i] = 7) then print(Bingo: 7 found at position i);goon := false;end if

end while

• The while-loop stops after exactly two iterations if • rand(1, 3) ≠ 1 in the first iteration. • rand(1, 3) = 1 in the second iteration. • This happens with probability 2/3 and 1/3, resp.• The probability for both to happen is - by independence -

Analysis of Find7Find7(A[1…3])

goon := true;while (goon) doi := rand(1,3);if (A[i] = 7) then print(Bingo: 7 found at position i);goon := false;end if

end while

• The while-loop stops after exactly three iterations if • rand(1, 3) ≠ 1 in the first two iterations. • rand(1, 3) = 1 in the third iteration. • This happens with probability 2/3, 2/3 and 1/3, resp.• The probability for all three to happen is - by independence -

• The while-loop stops after exactly k iterations if • rand(1, 3) ≠ 1 in the first k-1 iterations. • rand(1, 3) = 1 in the kth iteration. • This happens with probability 2/3, 2/3 and 1/3, resp.• The probability for all three to happen is - by independence -

Analysis of Find7• We have

• The expected running time of Find7 is

• Idea: Stop Las Vegas algorithm if it runs “much longer” than the expected time and restart it.

Expected running time

• We have seen two kinds of algorithms: • Monte Carlo algorithms: stop after a fixed (polynomial) time and give

the correct answer with probability greater 50%.

• Las Vegas algorithms: have variable running time but always give the correct answer.

Types of randomized algorithms

Median/Select

• Given n numbers S = {a1, a2, …, an}. • Median: number that is in the middle position if in sorted order.• Select(S,k): Return the kth smallest number in S.

• Min(S) = Select(S,1), Max(S)= Select(S,n), Median = Select(S,n/2).

• Assume the numbers are distinct.

Select

Select(S, k) {

Choose a pivot s ∈ S uniformly at random.

For each element e in S if e < s put e in S’ if e > s put e in S’’

if |S’| = k-1 then return s

if |S’| ≥ k then call Select(S’, k)

if |S’| < k then call Select(S’’, k - |S’| - 1) }

• Worst case running time: • If there is at least an fraction of elements both larger and smaller than s:

• Limit number of bad pivots.• Intuition: A fairly large fraction of elements are “well-centered” => random pivot

likely to be good.

SelectSelect(S, k) {

Choose a pivot s ∈ S uniformly at random.

For each element e in S if e < s put e in S’ if e > s put e in S’’

if |S’| = k-1 then return s

if |S’| ≥ k then call Select(S’, k)

if |S’| < k then call Select(S’’, k - |S’| - 1) }

T (n) = cn+ c(n� 1) + c(n� 2) + · · · = ⇥(n2).

"

T (n) = cn+ (1� ")cn+ (1� ")2cn+ · · ·=

�1 + (1� ") + (1� ")2 + · · ·

�cn

cn/".

• Phase j: Size of set at most and at least .• Element is central if at least a quarter of the elements in the current call are smaller

and at least a quarter are larger.• At least half the elements are central.• Pivot central => size of set shrinks with by at least a factor 3/4 => current phase

ends.• Pr[s is central] = 1/2.• Expected number of iterations before a central pivot is found = 2 => expected number of iterations in phase j at most 2.

• X: random variable equal to number of steps taken by algorithm.• Xj: expected number of steps in phase j.• X = X1+ X2 + .…• Number of steps in one iteration in phase j is at most .• E[Xj] = .• Expected running time:

Selectn(3/4)j n(3/4)j+1

E[X] =X

j

E[Xj ] X

j

2cn

✓3

4

◆j

= 2cnX

j

✓3

4

◆j

8cn.

2cn(3/4)jcn(3/4)j

Quicksort

• Given n numbers S = {a1, a2, …, an} return the sorted list. • Assume the numbers are distinct.

Quicksort

Quicksort(A,p,r) {

if |S| ≤ 1 return S else

Choose a pivot s ∈ S uniformly at random.

For each element e in S if e < s put e in S’ if e > s put e in S’’

L = Quicksort(S’) R = Quicksort(S’’)

Return the sorted list L◦s◦R. }

• Worst case Quicksort requires Ω(n2) comparisons: if pivot is the smallest element in the list in each recursive call.

• If pivot aways is the median then T(n) = O(n log n).• for i < j: random variable

• X total number of comparisons:

• Expected number of comparisons:

Quicksort: Analysis

Xij =

(1 if ai and aj compared by algorithm

0 otherwise

X =n�1X

i=1

nX

j=i+1

Xij

E[X] = E[n�1X

i=1

nX

j=i+1

Xij ] =n�1X

i=1

nX

j=i+1

E[Xij ]

• Expected number of comparisons:

• Since Xij only takes values 0 and 1: • ai and aj compared iff ai or aj is the first pivot chosen from Zij = {ai,…,aj}.• Pivot chosen independently uniformly at random => all elements from Zij equally

likely to be chosen as first pivot from this set.

• We have • Thus

Quicksort: Analysis

E[X] = E[n�1X

i=1

nX

j=i+1

Xij ] =n�1X

i=1

nX

j=i+1

E[Xij ]

E[Xij ] = Pr[Xij = 1]

Pr[Xij = 1] = 2/(j � i+ 1)

E[X] =n�1X

i=1

nX

j=i+1

E[Xij ] =n�1X

i=1

nX

j=i+1

Pr[Xij ] =n�1X

i=1

nX

j=i+1

2

j � i+ 1

<n�1X

i=1

nX

k=1

2

k=

n�1X

i=1

O(log n) = O(n log n)=n�1X

i=1

n�i+1X

k=2

2

k

• Prize for best three teams.

• Can count as a mandatory assignment.

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