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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusions

A Gentle Introduction to the Time Complexity Analysis ofEvolutionary Algorithms

Pietro S. Oliveto and Xin Yao

CERCIA, School of Computer Science, University of Birmingham, UK

WCCI 2012Brisbane, Australia, 10 June 2012

P. S. Oliveto & X. Yao (University of Birmingham) Runtime Analysis of EAs WCCI 2012 1 / 92

Aims and Goals of this Tutorial

This tutorial will provide an overview ofthe goals of time complexity analysis of Evolutionary Algorithms (EAs)the most common and effective techniques

You should attend if you wish totheoretically understand the behaviour and performance of the search algorithms youdesignfamiliarise with the techniques used in the time complexity analysis of EAspursue research in the area

enable you or enhance your ability to1 understand theoretically the behaviour of EAs on different problems2 perform time complexity analysis of simple EAs on common toy problems3 read and understand research papers on the computational complexity of EAs4 have the basic skills to start independent research in the area5 follow the time complexity of EAs for combinatorial optimization Tutorial later on today

Outline I

1 MotivationIntroduction to the theory of EAsConvergence analysis of EAsComputational complexity of EAs

2 Basic Probability TheoryProbability spaceUnion boundRandom variables and expectationsLaw of total probability

3 Evolutionary AlgorithmsGeneral EAs(1+1)-EA and RLSGeneral propertiesGeneral upper bound

4 Tail InequalitiesMarkov’s inequalityChernoff bounds

5 Artificial Fitness LevelsCoupon collector’s problemAFL method for upper boundsAFL method for parent populationsAFL for non-elitist EAs

Outline II

AFL for lower bounds

6 Drift AnalysisAdditive Drift TheoremMultiplicative Drift TheoremSimplified Negative Drift Theorem

7 Typical Run Investigations

8 ConclusionsOverviewState-of-the-artFurther reading

Introduction to the theory of EAs

Evolutionary Algorithms and Computer Science

Goals of design and analysis of algorithms

1 correctness“does the algorithm always output the correct solution?”

2 computational complexity“how many computational resources are required?”

For Evolutionary Algorithms (General purpose)

1 convergence“Does the EA find the solution in finite time?”

2 time complexity“how long does it take to find the optimum?”(time = n. of fitness function evaluations)

Introduction to the theory of EAs

Brief history

Theoretical studies of Evolutionary Algorithms (EAs), albeit few, have always existedsince the seventies [Goldberg, 1989];

Early studies were concerned with explaining the behaviour rather than analysingtheir performance.

Schema Theory was considered fundamental;First proposed to understand the behaviour of the simple GA [Holland, 1992];It cannot explain the performance or limit behaviour of EAs;Building Block Hypothesis was wrong [Reeves and Rowe, 2002];

Convergence results appeared in the nineties [Rudolph, 1998];Related to the time limit behaviour of EAs.

Convergence analysis of EAs

Convergence

Definition

Ideally the EA should find the solution in finite steps with probability 1(visit the global optimum in finite time);

If the solution is held forever after, then the algorithm converges to the optimum!

Conditions for Convergence ([Rudolph, 1998])

1 There is a positive probability to reach any point in the search space from anyother point

2 The best found solution is never removed from the population (elitism)

Canonical GAs using mutation, crossover and proportional selection Do Notconverge!

Elitist variants Do converge!

In practice, is it interesting that an algorithm converges to the optimum?

Most EAs visit the global optimum in finite time (RLS does not!)

How much time?

Convergence

Definition

How much time?

Convergence

Definition

How much time?

Computational complexity of EAs

Computational Complexity of EAs

P. K. Lehre, 2011

Generally means predicting the resources the algorithm requires:

Usually the computational time: the number of primitive steps;

Usually grows with size of the input;

Usually expressed in asymptotic notation;

Exponential runtime: Inefficient algorithmPolynomial runtime: “Efficient” algorithm

P. K. Lehre, 2011

However (EAs):

1 In practice the time for a fitness function evaluation is much higher than the rest;2 EAs are randomised algorithms

They do not perform the same operations even if the input is the same!They do not output the same result if run twice!

Hence, the runtime of an EA is a random variable Tf .We are interested in:

1 Estimating E(Tf ), the expected runtime of the EA for f ;

2 Estimating p(Tf ≤ t), the success probability of the EA in t steps for f .

Asymptotic notation

f(n) ∈ O(g(n)) ⇐⇒ ∃ constants c, n0 > 0 st. 0 ≤ f(n)≤cg(n) ∀n ≥ n0

f(n) ∈ Ω(g(n)) ⇐⇒ ∃ constants c, n0 > 0 st. 0 ≤ cg(n)≤f(n) ∀n ≥ n0

f(n) ∈ Θ(g(n)) ⇐⇒ f(n) ∈ O(g(n)) and f(n) ∈ Ω(g(n))

f(n) ∈ o(g(n)) ⇐⇒ limn→∞

g(n)= 0

Exercise 1: Asymptotic Notation

[Lehre, Tutorial]

Motivation Overview

Overview

Goal: Analyze the correctness and performance of EAs;

Difficulties: General purpose, randomised;

EAs find the solution in finite time; (convergence analysis)

How much time? → Derive the expected runtime and the success probability;

Basic Probability Theory: probability space, random variables, expectations(expected runtime)

Randomised Algorithm Tools: Tail inequalities (success probabilities)

Along the way

Understand that the analysis cannot be done over all functions

Understand why the success probability is important (expected runtime notalways sufficient)

Probability space

Probability Axioms

Probability space

Sample space Ω (eg. 1 die: 1, 2, 3, 4, 5, 6 ∈ Ω)

Allowable events: = ⊆ Ω (eg. E ∈ = := 2 or 3 is the outcome of 1 die)

Probability function: Pr : = → R (eg. Pr(E) = 2/6 = 1/3)

Probability Axioms

For any event E, 0 ≤ Pr(E) ≤ 1

Pr(Ω) = 1

For any countably finite sequence of pairwise mutually disjoint eventsE1, E2, E3, . . .

P r` [i≥1

=Xi≥1

Pr(Ei)

1 Die (Ei the event that i shows up)

Pr(E1) = P (E2) = · · · = P (E6) = 1/6

Pr(E1 ∪ E2) = Pr(E1) + Pr(E2) = 2/6 = 1/3

Pr(E1 ∪ E2 ∪ E3) = Pr(E1) + Pr(E2) + Pr(E3) = 3/6 = 1/2

Pr(E1 ∪ E2 ∪ · · · ∪ E6) = Pr(E1) + . . . P r(E6) = 6/6 = 1 = Pr(Ω)

Probability space

Probability Axioms

Probability space

Probability Axioms

Pr(Ω) = 1

P r` [i≥1

=Xi≥1

Pr(Ei)

Pr(E1) = P (E2) = · · · = P (E6) = 1/6

Pr(E1 ∪ E2) = Pr(E1) + Pr(E2) = 2/6 = 1/3

Pr(E1 ∪ E2 ∪ E3) = Pr(E1) + Pr(E2) + Pr(E3) = 3/6 = 1/2

Pr(E1 ∪ E2 ∪ · · · ∪ E6) = Pr(E1) + . . . P r(E6) = 6/6 = 1 = Pr(Ω)

Probability space

Probability Axioms

Probability space

Probability Axioms

Pr(Ω) = 1

P r` [i≥1

=Xi≥1

Pr(Ei)

Pr(E1) = P (E2) = · · · = P (E6) = 1/6

Pr(E1 ∪ E2) = Pr(E1) + Pr(E2) = 2/6 = 1/3

Pr(E1 ∪ E2 ∪ E3) = Pr(E1) + Pr(E2) + Pr(E3) = 3/6 = 1/2

Pr(E1 ∪ E2 ∪ · · · ∪ E6) = Pr(E1) + . . . P r(E6) = 6/6 = 1 = Pr(Ω)

Probability space

Independent Events and Conditional Probabilities

Two diceE1: Event that the first die is a 1;E2: Event that the second die is a 1;

What is the probability that the first die is 1?

Pr(E1) = 636

What is the probability that the second die is 1? Pr(E2) = 636

What is the probability that both dice are 1? Pr(E3) = Pr(E1 ∩ E2) = 136

(independent events)

Definition

Two events E and F are independent if and only if Pr(E ∩ F ) = Pr(E) · Pr(F )

What is the probability that second die is 1 given that the first die is 1?

Pr(E2|E1) =Pr(E2∩E1)Pr(E1)

) =1/361/6

= 1/6 (conditional probabilities)

What about at least one of the two dice is a 1?Pr(E1 ∪ E2) = 1136

E1 and E2 are not mutually disjoint events!

Probability space

What is the probability that the first die is 1? Pr(E1) = 636

Definition

) =1/361/6

Probability space

What is the probability that the second die is 1?

Pr(E2) = 636

Definition

) =1/361/6

Probability space

Definition

) =1/361/6

Probability space

What is the probability that both dice are 1?

Pr(E3) = Pr(E1 ∩ E2) = 136

Definition

) =1/361/6

Probability space

Definition

) =1/361/6

Probability space

Definition

) =1/361/6

Probability space

Definition

) =1/361/6

Probability space

Definition

) =1/361/6

What about at least one of the two dice is a 1?

Pr(E1 ∪ E2) = 1136

Probability space

Definition

) =1/361/6

Probability space

Definition

) =1/361/6

Union bound

Ω1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

6,1 6,2 6,3 6,4 6,5 6,6

5,1 5,2 5,3 5,4 5,5 5,6

Union bound

Ω1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

6,1 6,2 6,3 6,4 6,5 6,6

5,1 5,2 5,3 5,4 5,5 5,6

Pr(E1) = 6/36 = 1/6

Union bound

Ω1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

6,1 6,2 6,3 6,4 6,5 6,6

5,1 5,2 5,3 5,4 5,5 5,6

Pr(E1) = 6/36 = 1/6Pr(E2) = 6/36 = 1/6

Union bound

Ω1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

6,1 6,2 6,3 6,4 6,5 6,6

5,1 5,2 5,3 5,4 5,5 5,6

Pr(E1) = 6/36 = 1/6Pr(E2) = 6/36 = 1/6Pr(E2|E1) = 1/6

Union bound

Ω1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

6,1 6,2 6,3 6,4 6,5 6,6

5,1 5,2 5,3 5,4 5,5 5,6

Pr(E1) = 6/36 = 1/6Pr(E2) = 6/36 = 1/6Pr(E2|E1) = 1/6Pr(E1 ∪ E2) = 6/36 + 5/36 = 11/36 ≤ 6/36 + 6/36 = Pr(E1) + Pr(E2)

Union bound

Ω1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

6,1 6,2 6,3 6,4 6,5 6,6

5,1 5,2 5,3 5,4 5,5 5,6

Pr(E1) = 6/36 = 1/6Pr(E2) = 6/36 = 1/6Pr(E2|E1) = 1/6Pr(E1 ∪ E2) = 6/36 + 5/36 = 11/36 ≤ 6/36 + 6/36 = Pr(E1) + Pr(E2)E1 and E2 are not mutually disjoint events!

Union bound

Union Bound

Theorem (Union bound)

For any finite or countably finite sequence of events E1, E2, . . .

P r` [i≥1

Ei´≤Xi≥1

Pr(Ei)

Random variables and expectations

Random Variables

Definition (Random Variable)

A random variable X on a sample space Ω is a real-valued function X : Ω→ R. Adiscrete random variable takes only a finite or countably finite number of values.

Probability:

Pr(X = a) =Ps∈Ω;X(s)=a Pr(s)

Expectation:

E(X) =Pi i · Pr(X = i)

Random Variables

Probability:

Expectation:

Example 1 (one die):Let X be the value of one die;

Pr(X = 6) =1

E(X) =1

6· 1 +

6· 2 +

6· 3 + . . .

6· 6 =

Random Variables

Probability:

Expectation:

Example 1 (two dice): Let X be the value of the sum of two dice;

Pr(X = 4) = Pr(1, 3) + Pr(3, 1) + Pr(2, 2) =3

E(X) =1

36· 2 +

36· 3 +

36· 4 + . . .

36· 12 = 7

Linearity of Expectation

Definition

For any collection of discrete random variables X1, X2, . . . , Xn with finiteexpectations,

Eˆ nXi=1

=nXi=1

Example 1 (two dice): Let X be the value of the sum of two dice, X1 the value of die1 and X2 the value of die 2

E(Xi) =7

E(X) = E(X1) + E(X2) = 7

Binomial Random Variables

We run an experiment that succeeds with probability p and fails 1− p (Eg. coin flips)Consider n trials.

Definition

A binomial random variable X ∼ B(n, p) with parameters n and p represents thenumber of successes in n independent experiments each of which succeds withprobability p.

Probability:

Pr(X = j) =`nj

´pj(1− p)n−j

Expectation: Let Xi = 1 if the ith trial is successful; (linearity of expectation)

E(X) = EˆPn

i Xi˜

=Pni=1 E(Xi) = np

Example 1 (Initialisation of EA): Let X be the number of ones in the initial bit-string(p= 1/2, bit-string length =n)

E(X) = np = n/2

Binomial Random Variables

We run an experiment that succeeds with probability p and fails 1− p (Eg. coin flips)Consider n trials.

Definition

A binomial random variable X ∼ B(n, p) with parameters n and p represents thenumber of successes in n independent experiments each of which succeds withprobability p.

Probability:

Pr(X = j) =`nj

´pj(1− p)n−j

Expectation: Let Xi = 1 if the ith trial is successful; (linearity of expectation)

E(X) = EˆPn

i Xi˜

=Pni=1 E(Xi) = np

Example 1 (Initialisation of EA): Let X be the number of ones in the initial bit-string(p= 1/2, bit-string length =n)

E(X) = np = n/2

Geometric Random Variables

We run an experiment that succeeds with probability p and fails 1− p (Eg. coin flips)What is the number of trials until we get a success? (eg. heads?)

Definition

A geometric random variable X with parameter p represents the number of trials untilthe first success.

Probability:

Pr(X = n) = (1− p)n−1p

Expectation: E(X) = 1/p (waiting time argument)

Example (expected time for bit i to flip): Let X be the number of steps until bit iflips with mutation rate pm = 1/n

E(X) = 1/p = n

Pr(X = n ·√n+ 1) = (1− p)n

√n+1−1p =

„1− 1

«n√n1n≤„

«√n1n≤ e−

Geometric Random Variables

We run an experiment that succeeds with probability p and fails 1− p (Eg. coin flips)What is the number of trials until we get a success? (eg. heads?)

Definition

A geometric random variable X with parameter p represents the number of trials untilthe first success.

Probability:

Pr(X = n) = (1− p)n−1p

Expectation: E(X) = 1/p (waiting time argument)

Example (expected time for bit i to flip): Let X be the number of steps until bit iflips with mutation rate pm = 1/n

E(X) = 1/p = n

Pr(X = n ·√n+ 1) = (1− p)n

√n+1−1p =

„1− 1

«n√n1n≤„

«√n1n≤ e−

Law of total probability

Law of Total Probability

Let E,F be mutually disjoint events and E ∪ F = Ω.

Theorem

Pr(E) = Pr(E|F ) · Pr(F ) + Pr(E|F ) · Pr(F )

E(X) = E(X|F ) · Pr(F ) + E(X|F ) · Pr(F )

Immediate Consequence:

Pr(E) ≥ Pr(E|F ) · Pr(F )

E(X) ≥ E(X|F ) · Pr(F )

Often used to derive lower bounds on the expected time!

We will use this to show that expected values may not be sufficient → successprobabilities!

General EAs

Evolutionary Algorithms

Algorithm ((µ+λ)-EA)

1 Let t = 0;

2 Initialize P0 with µ individuals chosen uniformly at random;Repeat

3 Create λ new individuals:1 choose x ∈ Pt uniformly at random;2 flip each bit in x with probability p;

4 Create the new population Pt+1 by choosing the best µ individuals out of µ+ λ;

5 Let t = t+ 1.Until a stopping condition is fulfilled.

if µ = λ = 1 we get a (1+1)-EA;

p = 1/n is generally considered as best choice [Back, 1993, Droste et al., 1998];

By introducing stochastic selection and crossover we obtain a GeneticAlgorithm(GA)

General EAs

1 Let t = 0;

if µ = λ = 1 we get a (1+1)-EA;

General EAs

1 Let t = 0;

if µ = λ = 1 we get a (1+1)-EA;

General EAs

1 Let t = 0;

if µ = λ = 1 we get a (1+1)-EA;

(1+1)-EA and RLS

1+1-EA

Algorithm ((1+1)-EA)

Initialise P0 with x ∈ 1, 0n by flipping each bit with p = 1/2 ;Repeat

Create x′ by flipping each bit in x with p = 1/n;

If f(x′) ≥ f(x) Then x′ ∈ Pt+1 Else x ∈ Pt+1;

Let t = t+ 1; Until stopping condition.

If only one bit is flipped per iteration: Random Local Search (RLS).How does it work?

Given x, how many bits will flip in expectation?

E[X] = E[X1 +X2 + · · ·+Xn] = E[X1] + E[X2] + · · ·+ E[Xn] =

„E[Xi] = 1 · 1/n+ 0 · (1− 1/n) = 1 · 1/n = 1/n E(X) = np

1 · 1/n = n/n = 1

(1+1)-EA and RLS

1+1-EA

E[X] = E[X1 +X2 + · · ·+Xn] = E[X1] + E[X2] + · · ·+ E[Xn] =

„E[Xi] = 1 · 1/n+ 0 · (1− 1/n) = 1 · 1/n = 1/n E(X) = np

1 · 1/n = n/n = 1

(1+1)-EA and RLS

1+1-EA

E[X] = E[X1 +X2 + · · ·+Xn] = E[X1] + E[X2] + · · ·+ E[Xn] =

„E[Xi] = 1 · 1/n+ 0 · (1− 1/n) = 1 · 1/n = 1/n E(X) = np

1 · 1/n = n/n = 1

(1+1)-EA and RLS

1+1-EA

E[X] = E[X1 +X2 + · · ·+Xn] = E[X1] + E[X2] + · · ·+ E[Xn] =

„E[Xi] = 1 · 1/n+ 0 · (1− 1/n) = 1 · 1/n = 1/n E(X) = np

1 · 1/n = n/n = 1

(1+1)-EA and RLS

1+1-EA

E[X] = E[X1 +X2 + · · ·+Xn] = E[X1] + E[X2] + · · ·+ E[Xn] =

„E[Xi] = 1 · 1/n+ 0 · (1− 1/n) = 1 · 1/n = 1/n E(X) = np

=nXi=1

1 · 1/n = n/n = 1

General properties

1+1-EA: 2

How likely is it that exactly one bit flips?

„Pr(X = j) =

´pj(1− p)n−j

What is the probability of exactly one bit flipping?

Pr(X = 1) =“n

”· 1/n · (1− 1/n)n−1 = (1− 1/n)n−1 ≥ 1/e ≈ 0.37

Is it more likely that 2 bits flip or none?

Pr(X = 2) =“n

”· 1/n2 · (1− 1/n)n−2 =

=n · (n− 1)

21/n2 · (1− 1/n)n−2 =

= 1/2 · (1− 1/n)n−1 ≈ 1/(2e)

WhilePr(X = 0) =

”(1/n)0 · (1− 1/n)n ≈ 1/e

General properties

1+1-EA: 2

„Pr(X = j) =

´pj(1− p)n−j

«What is the probability of exactly one bit flipping?

Pr(X = 1) =“n

”· 1/n · (1− 1/n)n−1 = (1− 1/n)n−1 ≥ 1/e ≈ 0.37

Pr(X = 2) =“n

”· 1/n2 · (1− 1/n)n−2 =

=n · (n− 1)

21/n2 · (1− 1/n)n−2 =

= 1/2 · (1− 1/n)n−1 ≈ 1/(2e)

WhilePr(X = 0) =

”(1/n)0 · (1− 1/n)n ≈ 1/e

General properties

1+1-EA: 2

„Pr(X = j) =

´pj(1− p)n−j

Pr(X = 1) =“n

”· 1/n · (1− 1/n)n−1 = (1− 1/n)n−1 ≥ 1/e ≈ 0.37

Pr(X = 2) =“n

”· 1/n2 · (1− 1/n)n−2 =

=n · (n− 1)

21/n2 · (1− 1/n)n−2 =

= 1/2 · (1− 1/n)n−1 ≈ 1/(2e)

WhilePr(X = 0) =

”(1/n)0 · (1− 1/n)n ≈ 1/e

General properties

1+1-EA: 2

„Pr(X = j) =

´pj(1− p)n−j

Pr(X = 1) =“n

”· 1/n · (1− 1/n)n−1 = (1− 1/n)n−1 ≥ 1/e ≈ 0.37

Pr(X = 2) =“n

”· 1/n2 · (1− 1/n)n−2 =

=n · (n− 1)

21/n2 · (1− 1/n)n−2 =

= 1/2 · (1− 1/n)n−1 ≈ 1/(2e)

WhilePr(X = 0) =

”(1/n)0 · (1− 1/n)n ≈ 1/e

General properties

1+1-EA: 2

„Pr(X = j) =

´pj(1− p)n−j

Pr(X = 1) =“n

”· 1/n · (1− 1/n)n−1 = (1− 1/n)n−1 ≥ 1/e ≈ 0.37

Pr(X = 2) =“n

”· 1/n2 · (1− 1/n)n−2 =

=n · (n− 1)

21/n2 · (1− 1/n)n−2 =

= 1/2 · (1− 1/n)n−1 ≈ 1/(2e)

WhilePr(X = 0) =

”(1/n)0 · (1− 1/n)n ≈ 1/e

General properties

1+1-EA: 2

„Pr(X = j) =

´pj(1− p)n−j

Pr(X = 1) =“n

”· 1/n · (1− 1/n)n−1 = (1− 1/n)n−1 ≥ 1/e ≈ 0.37

Pr(X = 2) =“n

”· 1/n2 · (1− 1/n)n−2 =

=n · (n− 1)

21/n2 · (1− 1/n)n−2 =

= 1/2 · (1− 1/n)n−1 ≈ 1/(2e)

WhilePr(X = 0) =

”(1/n)0 · (1− 1/n)n ≈ 1/e

General upper bound

1+1-EA: General Upper bound

Theorem ([Droste et al., 2002])

The expected runtime of the (1+1)-EA for an arbitrary function defined in 0, 1n isO(nn)

1 Let i be the number of bit positions in which the current solution x and theglobal optimum x∗ differ;

2 Each bit flips with probability 1/n, hence does not flip with probability (1− 1/n);

3 In order to reach the global optimum the algorithm has to mutate the i bits andleave the n− i bits unchanged;

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

´5 it implies an upper bound on the expected runtime of O(nn)

(E(X) = 1/p = nn) (waiting time argument).

General upper bound

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

General upper bound

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

General upper bound

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

General upper bound

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

General upper bound

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

5 it implies an upper bound on the expected runtime of O(nn)(E(X) = 1/p = nn) (waiting time argument).

General upper bound

4 Then:

p(x∗|x) =

«i„1−

«n−i≥„

«n= n−n

`p = n−n

General upper bound

General Upper bound Exercises

Theorem

The expected runtime of the (1+1)-EA with mutation probability p = 1/2 for anarbitrary function defined in 0, 1n is O(2n)

Proof Left as Exercise.

Theorem

The expected runtime of the (1+1)-EA with mutation probability p = χ/n for anarbitrary function defined in 0, 1n is O((n/χ)n)

Theorem

The expected runtime of RLS for an arbitrary function defined in 0, 1n is infinite.

General upper bound

Theorem

General upper bound

Theorem

General upper bound

Theorem

General upper bound

1+1-EA: Conclusions & Exercises

In general:

P (i− bitflip) =“ni

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

Expectation: E[X] = 1

P(1-bitflip) = 1

What about initialisation?

How many one-bits in expectation after initialisation?

E[X] = n · 1/2 = n/2

How likely is it that we get exactly n/2 one-bits?

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

Tail Inequalities help us to deal with these kind of problems.

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

General upper bound

In general:

„1−

«n−i≤

„1−

«n−i≈

What about RLS?

P(1-bitflip) = 1

E[X] = n · 1/2 = n/2

Pr(X = n/2) =“ n

„1−

«n/2„n = 100, P r(X = 50) ≈ 0.0796

Tail Inequalities

Given a random variable X it may assume values that are considerably larger or lowerthan its expectation;

Tail inequalities:

Estimate the probability that X deviates from the expectation by a definedamount δ;

For many intermediate results, expected values are useless;

May turn expected runtimes into bounds that hold with overwhelming probability.

Markov’s inequality

Markov Inequality

The fundamental inequality from which many others are derived.

Definition (Markov’s Inequality)

Let X be a random variable assuming only non-negative values, and E[X] itsexpectation. Then for all t ∈ R+,

Pr[X ≥ t] ≤E[X]

E[X] = 1; then: Pr[X ≥ 2] ≤ E[X]2≤ 1

2(Number of bits that flip)

E[X] = n/2; then Pr[X ≥ (2/3)n] ≤ E[X](2/3)n

(2/3)n≤ 3

(Number of one-bits after initialisation)

Usually Markov’s inequality is used iteratively to obtain stronger bounds!

Markov Inequality

Pr[X ≥ t] ≤E[X]

E[X] = 1; then: Pr[X ≥ 2] ≤ E[X]2≤ 1

E[X] = n/2; then Pr[X ≥ (2/3)n] ≤ E[X](2/3)n

(2/3)n≤ 3

Markov Inequality

Pr[X ≥ t] ≤E[X]

E[X] = 1; then: Pr[X ≥ 2] ≤ E[X]2≤ 1

E[X] = n/2; then Pr[X ≥ (2/3)n] ≤ E[X](2/3)n

(2/3)n≤ 3

Markov Inequality

Pr[X ≥ t] ≤E[X]

E[X] = 1; then: Pr[X ≥ 2] ≤ E[X]2≤ 1

E[X] = n/2; then Pr[X ≥ (2/3)n] ≤ E[X](2/3)n

(2/3)n≤ 3

Markov Inequality

Pr[X ≥ t] ≤E[X]

E[X] = 1; then: Pr[X ≥ 2] ≤ E[X]2≤ 1

E[X] = n/2; then Pr[X ≥ (2/3)n] ≤ E[X](2/3)n

(2/3)n≤ 3

Chernoff bounds

Chernoff Bounds

Let X1, X2, . . . Xn be independent Poisson trials each with probability pi;For X =

Pni=1 Xi the expectation is E(X) =

Pni=1 pi.

Definition (Chernoff Bounds)

1 for 0 ≤ δ ≤ 1, Pr(X ≤ (1− δ)E[X]) ≤ e−E[X]δ2

2 for δ > 0, Pr(X > (1 + δ)E[X]) ≤»

(1+δ)1+δ

–E[X]

What is the probability that we have more than (2/3)n one-bits at initialisation?

pi = 1/2, E[X] = n · 1/2 = n/2,(we fix δ = 1/3→ (1 + δ)E[X] = (2/3)n); then:

Pr[X > (2/3)n] ≤„

(4/3)4/3

«n/2= c−n/2

Chernoff bounds

Chernoff Bounds

Pni=1 pi.

1 for 0 ≤ δ ≤ 1, Pr(X ≤ (1− δ)E[X]) ≤ e−E[X]δ2

2 for δ > 0, Pr(X > (1 + δ)E[X]) ≤»

(1+δ)1+δ

–E[X]

Pr[X > (2/3)n] ≤„

(4/3)4/3

«n/2= c−n/2

Chernoff bounds

Chernoff Bounds

Pni=1 pi.

1 for 0 ≤ δ ≤ 1, Pr(X ≤ (1− δ)E[X]) ≤ e−E[X]δ2

2 for δ > 0, Pr(X > (1 + δ)E[X]) ≤»

(1+δ)1+δ

–E[X]

pi = 1/2, E[X] = n · 1/2 = n/2,

(we fix δ = 1/3→ (1 + δ)E[X] = (2/3)n); then:

Pr[X > (2/3)n] ≤„

(4/3)4/3

«n/2= c−n/2

Chernoff bounds

Chernoff Bounds

Pni=1 pi.

1 for 0 ≤ δ ≤ 1, Pr(X ≤ (1− δ)E[X]) ≤ e−E[X]δ2

2 for δ > 0, Pr(X > (1 + δ)E[X]) ≤»

(1+δ)1+δ

–E[X]

Pr[X > (2/3)n] ≤„

(4/3)4/3

«n/2= c−n/2

Chernoff bounds

Chernoff Bound Simple Application

Bitstring of length n = 100

Pr(Xi) = 1/2 and E(X) = np = 100/2 = 50.

What is the probability to have at least 75 1-bits?

Markov: Pr(X ≥ 75) ≤ 5075

Chernoff: Pr(X ≥ (1 + 1/2)50) ≤„ √

(3/2)3/2

< 0.0045

Truth: Pr(X ≥ 75) =P100i=75

´2−100 < 0.000000282

Chernoff bounds

Pr(Xi) = 1/2 and E(X) = np = 100/2 = 50.What is the probability to have at least 75 1-bits?

Markov: Pr(X ≥ 75) ≤ 5075

Chernoff: Pr(X ≥ (1 + 1/2)50) ≤„ √

(3/2)3/2

< 0.0045

Truth: Pr(X ≥ 75) =P100i=75

´2−100 < 0.000000282

Chernoff bounds

Pr(Xi) = 1/2 and E(X) = np = 100/2 = 50.What is the probability to have at least 75 1-bits?

Markov: Pr(X ≥ 75) ≤ 5075

Chernoff: Pr(X ≥ (1 + 1/2)50) ≤„ √

(3/2)3/2

< 0.0045

Truth: Pr(X ≥ 75) =P100i=75

´2−100 < 0.000000282

Chernoff bounds

OneMax

OneMax (x)=Pni=1 x[i])

ones(x)

RLS for OneMax ( OneMax (x)=∑n

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 0 p0 = 66

E(T0) = 66

i=1 x[i])

50 1 2 3 4 5

0 0 0 0 0 0 p0 = 66

E(T0) = 66

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

20 1 2 3 4 5

0 0 0 0 0 1 p1 = 56

E(T1) = 65

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

00 1 2 3 4 5

0 0 1 0 0 1 p2 = 46

E(T2) = 64

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

50 1 2 3 4 5

1 0 1 0 0 1 p3 = 36

E(T0) = 63

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 0 0 p3 = 36

E(T3) = 63

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

40 1 2 3 4 5

1 0 1 0 0 1 p3 = 36

E(T3) = 63

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

10 1 2 3 4 5

1 0 1 0 1 1 p4 = 26

E(T4) = 62

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

0 1 2 3 4 5

1 1 1 0 1 1 p4 = 26

E(T4) = 62

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

0 1 2 3 4 5

1 1 1 0 1 1 p4 = 26

E(T4) = 62

0 1 2 3 4 5

1 1 1 0 1 1 p4 = 26

E(T4) = 62

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

0 1 2 3 4 5

1 1 1 0 1 1 p4 = 26

E(T4) = 62

30 1 2 3 4 5

1 1 1 0 1 1 p5 = 16

E(T5) = 61

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

0 1 2 3 4 5

1 1 1 0 1 1 p4 = 26

E(T4) = 62

0 1 2 3 4 5

1 1 1 1 1 1 p5 = 16

E(T5) = 61

i=1 x[i])

0 1 2 3 4 5

0 0 0 0 0 1 p0 = 66

E(T0) = 66

0 1 2 3 4 5

0 0 1 0 0 1 p1 = 56

E(T1) = 65

0 1 2 3 4 5

1 0 1 0 0 1 p2 = 46

E(T2) = 64

0 1 2 3 4 5

1 0 1 0 1 1 p3 = 36

E(T3) = 63

0 1 2 3 4 5

1 1 1 0 1 1 p4 = 26

E(T4) = 62

0 1 2 3 4 5

1 1 1 1 1 1 p5 = 16

E(T5) = 61

E(T ) = E(T0) + E(T1) + · · ·+ E(T5) = 1/p0 + 1/p1 + · · ·+ 1/p5 =

i= 6 · 2.45 = 14.7

i=1 x[i]) : Generalisation

0 1 2 3 n

0 0 0 0 0 0 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 5 n

0 0 0 0 0 0 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

20 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

n0 1 2 3

0 0 1 0 0 1 0 0 0 0 p2 = n−2n

E(T2) = nn−2

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 1 0 0 1 0 0 0 1 p2 = n−2n

E(T2) = nn−2

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 1 0 0 1 0 0 0 1 p2 = n−2n

E(T2) = nn−2

0 1 2 3 n

1 1 1 1 0 1 1 1 1 1 pn−1 = 1n

E(Tn−1) = n1

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 1 0 0 1 0 0 0 1 p2 = n−2n

E(T2) = nn−2

40 1 2 3 n

1 1 1 1 0 1 1 1 1 1 pn−1 = 1n

E(Tn−1) = n1

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 1 0 0 1 0 0 0 1 p2 = n−2n

E(T2) = nn−2

0 1 2 3 n

1 1 1 1 1 1 1 1 1 1 pn−1 = 1n

E(Tn−1) = n1

0 1 2 3 n

0 0 0 0 0 1 0 0 0 0 p0 = nn

E(T0) = nn

0 1 2 3 n

0 0 1 0 0 1 0 0 0 0 p1 = n−1n

E(T1) = nn−1

0 1 2 3 n

0 0 1 0 0 1 0 0 0 1 p2 = n−2n

E(T2) = nn−2

0 1 2 3 n

1 1 1 1 1 1 1 1 1 1 pn−1 = 1n

E(Tn−1) = n1

E(T ) = E(T0) + E(T1) + · · ·+ E(Tn−1) = 1/p1 + 1/p2 + · · ·+ 1/pn−1 =

n−1Xi=0

i= n ·H(n) = n logn+ Θ(n) = O(n logn)

Coupon collector’s problem

The Coupon collector’s problemThere are n types of coupons and at each trial one coupon is chosen at random. Eachcoupon has the same probability of being extracted. The goal is to find the exactnumber of trials before the collector has obtained all the n coupons.

Theorem (The coupon collector’s Theorem)

Let T be the time for all the n coupons to be collected. Then

E(T ) =

n−1Xi=0

n− i= n

n−1Xi=0

= n(logn+ Θ(1)) = n logn+O(n).

Coupon collector’s problem: Upper bound on time

What is the probability that the time to collect n coupons is greater thann lnn+O(n)?

Theorem (Coupon collector upper bound on time)

Pr(T ≥ (1 + ε)n lnn) ≤ n−ε

Probability of choosing a given coupon

1− 1n

Probability of not choosing a given coupon„1− 1

«tProbability of not choosing a given coupon for t rounds

The probability that one of the n coupons is not chosen in t rounds is less than

n ·„

1− 1n

«t(Union Bound)

Hence, for t = cn lnn

Pr(T ≥ cn lnn) ≤ n`1− 1/n)cn lnn ≤ n · e−c lnn = n · n−c = n−c+1

Coupon collector’s problem: Upper bound on time

What is the probability that the time to collect n coupons is greater thann lnn+O(n)?

Theorem (Coupon collector upper bound on time)

Pr(T ≥ (1 + ε)n lnn) ≤ n−ε

Probability of choosing a given coupon

1− 1n

Probability of not choosing a given coupon„1− 1

«tProbability of not choosing a given coupon for t rounds

The probability that one of the n coupons is not chosen in t rounds is less than

n ·„

1− 1n

«t(Union Bound)

Hence, for t = cn lnn

Pr(T ≥ cn lnn) ≤ n`1− 1/n)cn lnn ≤ n · e−c lnn = n · n−c = n−c+1

Coupon collector’s problem: lower bound on time

What is the probability that the time to collect n coupons is less than n lnn+O(n)?

Theorem (Coupon collector lower bound on time (Doerr, 2011))

Let T be the time for all the n coupons to be collected. Then for all ε > 0

Pr(T < (1− ε)(n− 1) lnn) ≤ exp(−nε)

Corollary

The expected time for RLS to optimise OneMax is Θ(n lnn). Furthermore,

Pr(T ≥ (1 + ε)n lnn) ≤ n−ε

andPr(T < (1− ε)(n− 1) lnn) ≤ exp(−nε)

What about the (1+1)-EA?

Coupon collector’s problem: lower bound on time

What is the probability that the time to collect n coupons is less than n lnn+O(n)?

Theorem (Coupon collector lower bound on time (Doerr, 2011))

Let T be the time for all the n coupons to be collected. Then for all ε > 0

Pr(T < (1− ε)(n− 1) lnn) ≤ exp(−nε)

Corollary

The expected time for RLS to optimise OneMax is Θ(n lnn). Furthermore,

Pr(T ≥ (1 + ε)n lnn) ≤ n−ε

andPr(T < (1− ε)(n− 1) lnn) ≤ exp(−nε)

What about the (1+1)-EA?

AFL method for upper bounds

Artificial Fitness Levels [Droste et al., 2002]

Observation Due to elitism, fitness is monotone increasing

Idea Divide the search space |S| = 2n into m < 2n sets A1, . . . Am such that:

1 ∀i 6= j : Ai ∩Aj = ∅2Smi=0 Ai = 0, 1n

3 for all points a ∈ Ai and b ∈ Aj it happens that f(a) < f(b) if i < j.

requirement Am contains only optimal search points.

Then:pi probability that point in Ai is mutated to a point in Aj with j > i.

Expected time: E(T ) ≤Pi

Very simple, yet often powerful method for upper bounds

Artificial Fitness Levels [Droste et al., 2002]

Observation Due to elitism, fitness is monotone increasing

Idea Divide the search space |S| = 2n into m < 2n sets A1, . . . Am such that:

1 ∀i 6= j : Ai ∩Aj = ∅2Smi=0 Ai = 0, 1n

3 for all points a ∈ Ai and b ∈ Aj it happens that f(a) < f(b) if i < j.

requirement Am contains only optimal search points.

Then:pi probability that point in Ai is mutated to a point in Aj with j > i.

Expected time: E(T ) ≤Pi

Very simple, yet often powerful method for upper bounds

Artificial Fitness Levels

D. Sudholt, Tutorial 2011

p(Ai) be the probability that a random chosen point belongs to level Aisi be the probability to leave level Ai for Aj with j > i

E(T ) ≤X

1≤i≤m−1

p(Ai) ·„

si+ · · ·+

sm−1

«≤„

s1+ · · ·+

sm−1

m−1Xi=1

Inequality 1: Law of total probability`E(T ) =

Pi Pr(F ) · E(T |F )

´Inequality 2: Trivial!

(1+1)-EA for OneMax

Theorem

The expected runtime of the (1+1)-EA for OneMax is O(n lnn).

The current solution is in level Ai if it has i zeroes (hence n− i ones)

To reach a higher fitness level it is sufficient to flip a zero into a one and leavethe other bits unchanged

The probability is si ≥ i · 1n

„1− 1

«n−1

≥ ien

Hence, 1si≤ en

Artificial Fitness Levels´:

E(T ) ≤m−1Xi=1

s−1i ≤

i≤ e · n

m−1Xi=1

i≤ e · n · (lnn+ 1) = O(n lnn)

Is the (1+1)-EA quicker than n lnn?

(1+1)-EA for OneMax

Theorem

„1− 1

«n−1

≥ ien

Hence, 1si≤ en

E(T ) ≤m−1Xi=1

s−1i ≤

i≤ e · n

m−1Xi=1

i≤ e · n · (lnn+ 1) = O(n lnn)

(1+1)-EA for OneMax

Theorem

„1− 1

«n−1

≥ ien

Hence, 1si≤ en

E(T ) ≤m−1Xi=1

s−1i ≤

i≤ e · n

m−1Xi=1

i≤ e · n · (lnn+ 1) = O(n lnn)

(1+1)-EA for OneMax

Theorem

„1− 1

«n−1

≥ ien

Hence, 1si≤ en

E(T ) ≤m−1Xi=1

s−1i ≤

i≤ e · n

m−1Xi=1

i≤ e · n · (lnn+ 1) = O(n lnn)

(1+1)-EA for OneMax

Theorem

„1− 1

«n−1

≥ ien

Hence, 1si≤ en

E(T ) ≤m−1Xi=1

s−1i ≤

i≤ e · n

m−1Xi=1

i≤ e · n · (lnn+ 1) = O(n lnn)

(1+1)-EA for OneMax

Theorem

„1− 1

«n−1

≥ ien

Hence, 1si≤ en

E(T ) ≤m−1Xi=1

s−1i ≤

i≤ e · n

m−1Xi=1

i≤ e · n · (lnn+ 1) = O(n lnn)

(1+1)-EA lower bound for OneMax

Theorem (Droste,Jansen,Wegener, 2002)

The expected runtime of the (1+1)-EA for OneMax is Ω(n lnn).

Proof Idea

1 At most n/2 one-bits are created during initialisation with probability at least 1/2(By symmetry of the binomial distribution).

2 There is a constant probability that in cn lnn steps one of the n/2 remainingzero-bits does not flip.

Proof Idea

Lower bound for OneMax

The expected runtime of the (1+1)-EA for OneMax is Ω(n logn).

Proof of 2.1− 1/n a given bit does not flip

(1− 1/n)t a given bit does not flip in t steps1− (1− 1/n)t it flips at least once in t steps

(1− (1− 1/n)t)n/2 n/2 bits flip at least once in t steps

1− [1− (1− 1/n)t]n/2 at least one of the n/2 bits does not flip in t steps

t = (n− 1) logn. Then:

1− [1− (1− 1/n)t]n/2 = 1− [1− (1− 1/n)(n−1) logn]n/2 ≥

≥ 1− [1− (1/e)logn]n/2 = 1− [1− 1/n]n/2 =

= 1− [1− 1/n]n·1/2 ≥ 1− (2e)−1/2 = c

Proof of 2.1− 1/n a given bit does not flip(1− 1/n)t a given bit does not flip in t steps

1− (1− 1/n)t it flips at least once in t steps

1− [1− (1− 1/n)t]n/2 = 1− [1− (1− 1/n)(n−1) logn]n/2 ≥

≥ 1− [1− (1/e)logn]n/2 = 1− [1− 1/n]n/2 =

= 1− [1− 1/n]n·1/2 ≥ 1− (2e)−1/2 = c

Proof of 2.1− 1/n a given bit does not flip(1− 1/n)t a given bit does not flip in t steps1− (1− 1/n)t it flips at least once in t steps

1− [1− (1− 1/n)t]n/2 = 1− [1− (1− 1/n)(n−1) logn]n/2 ≥

≥ 1− [1− (1/e)logn]n/2 = 1− [1− 1/n]n/2 =

= 1− [1− 1/n]n·1/2 ≥ 1− (2e)−1/2 = c

1− [1− (1− 1/n)t]n/2 = 1− [1− (1− 1/n)(n−1) logn]n/2 ≥

≥ 1− [1− (1/e)logn]n/2 = 1− [1− 1/n]n/2 =

= 1− [1− 1/n]n·1/2 ≥ 1− (2e)−1/2 = c

1− [1− (1− 1/n)t]n/2 = 1− [1− (1− 1/n)(n−1) logn]n/2 ≥

≥ 1− [1− (1/e)logn]n/2 = 1− [1− 1/n]n/2 =

= 1− [1− 1/n]n·1/2 ≥ 1− (2e)−1/2 = c

1− [1− (1− 1/n)t]n/2 = 1− [1− (1− 1/n)(n−1) logn]n/2 ≥

≥ 1− [1− (1/e)logn]n/2 = 1− [1− 1/n]n/2 =

= 1− [1− 1/n]n·1/2 ≥ 1− (2e)−1/2 = c

Lower bound for OneMax (2)

Theorem (Droste, Jansen, Wegener, 2002)

2 There is a constant probability that in cn logn steps one of the n/2 remainingzero-bits does not flip.

The Expected runtime is:

E[T ] =

∞Xt=1

t · p(t) ≥ [(n− 1) logn] · p[t = (n− 1) logn] ≥

≥ [(n− 1) logn] · [(1/2) · (1− (2e)−1/2) = Ω(n logn)

First inequality: law of total probability

The upper bound given by artificial fitness levels is indeed tight!

Lower bound for OneMax (2)

Theorem (Droste, Jansen, Wegener, 2002)

2 There is a constant probability that in cn logn steps one of the n/2 remainingzero-bits does not flip.

The Expected runtime is:

E[T ] =

∞Xt=1

t · p(t) ≥ [(n− 1) logn] · p[t = (n− 1) logn] ≥

≥ [(n− 1) logn] · [(1/2) · (1− (2e)−1/2) = Ω(n logn)

First inequality: law of total probability

The upper bound given by artificial fitness levels is indeed tight!

Artificial Fitness Levels Exercises:

(LeadingOnes(x) =

∑ni=1

∏ij=1 x[j]

Theorem

The expected runtime of RLS for LeadingOnes is O(n2).

Let partition Ai contain search points with exactly i leading ones

To leave level Ai it suffices to flip the zero at position i+ 1

si = 1n

and s−1i = n

E(T ) ≤Pn−1i=1 s−1

i =Pni=1 n = O(n2)

Theorem

The expected runtime of the (1+1)-EA for LeadingOnes is O(n2).

(LeadingOnes(x) =

∑ni=1

∏ij=1 x[j]

Theorem

si = 1n

and s−1i = n

E(T ) ≤Pn−1i=1 s−1

i =Pni=1 n = O(n2)

Theorem

(LeadingOnes(x) =

∑ni=1

∏ij=1 x[j]

Theorem

si = 1n

and s−1i = n

E(T ) ≤Pn−1i=1 s−1

i =Pni=1 n = O(n2)

Theorem

(LeadingOnes(x) =

∑ni=1

∏ij=1 x[j]

Theorem

si = 1n

and s−1i = n

E(T ) ≤Pn−1i=1 s−1

i =Pni=1 n = O(n2)

Theorem

Fitness Levels Advanced Exercises (Populations)

Theorem

The expected runtime of (1+λ)-EA for LeadingOnes is O(λn+ n2)[Jansen et al., 2005].

si = 1−„

1− 1en

«λ≥ 1− e−λ/(en)

1 si ≥ 1− 1e Case 1: λ ≥ en

2 si ≥ λ2en Case 2: λ < en

E(T ) ≤ λ ·Pn−1i=1 s−1

i ≤ λ„„Pn

„Pni=1

„λ ·„n+ n2

««= O(λ · n+ n2)

Theorem

The expected runtime of (1+λ)-EA for LeadingOnes is O(λn+ n2)[Jansen et al., 2005].

si = 1−„

1− 1en

«λ≥ 1− e−λ/(en)

1 si ≥ 1− 1e Case 1: λ ≥ en

2 si ≥ λ2en Case 2: λ < en

E(T ) ≤ λ ·Pn−1i=1 s−1

i ≤ λ„„Pn

„Pni=1

„λ ·„n+ n2

««= O(λ · n+ n2)

Theorem

The expected runtime of the (µ+1)-EA for LeadingOnes is O(µ · n2).

Theorem

The expected runtime of the (µ+1)-EA for OneMax is O(µ · n logn).

Theorem

AFL method for parent populations

Artificial Fitness Levels for Populations

D. Sudholt, Tutorial 2011

To be the expected time for a fraction χ(i) of the population to be in level Ai

si be the probability to leave level Ai for Aj with j > i given χ(i) in level Ai

E(T ) ≤m−1Xi=1

si+ To

Applications to (µ+1)-EA

Theorem

The expected runtime of (µ+1)-EA for LeadingOnes is O(µn logn+ n2)[Witt, 2006].

To leave level Ai it suffices to flip the zero at position i+ 1 of the best individual

We set χ(i) = n/ lnn

Given j copies of the best individual another replica is created with probability

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

en = 1eµ lnn Case 1: µ > n

2 si ≥ n/ lnnµ · 1

en ≥1en Case 2: µ ≤ n

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

2eµ lnn+`en+ eµ lnn

´«= O(nµ lnn+ n2)

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Theorem

„1− 1

«n≥ j

To ≤Pn/ lnnj=1

= 2eµ lnn

1 si ≥ n/ lnnµ · 1

2 si ≥ n/ lnnµ · 1

E(T ) ≤Pn−1i=1 (To + s−1

i ) ≤Pni=1

„2eµ lnn+

`en+ eµ lnn

n ·„

Populations Fitness Levels: Exercise

Theorem

The expected runtime of the (µ+1)-EA for OneMax is O(µn+ n logn).

Populations Fitness Levels: Exercise

Theorem

The expected runtime of the (µ+1)-EA for OneMax is O(µn+ n logn).

AFL for non-elitist EAs

Advanced: Fitness Levels for non-Elitist Populations [Lehre, 2011]

Advanced: Fitness Levels for Lower Bounds [Sudholt, 2010]

Artificial Fitness Levels: Conclusions

It’s a powerful general method to obtain (often) tight upper bounds on theruntime of simple EAs;

For offspring populations tight bounds can often be achieved with the generalmethod;

For parent populations takeover times have to be introduced;

Recent methods have been presented to deal with non-elitism and for lowerbounds.

Drift Analysis: Example 1

Friday night dinner at the restaurant.Peter walks back from the restaurant to the hotel.

The restaurant is n meters away from the hotel;

Peter moves towards the hotel of 1 meter in each step

QuestionHow many steps does Peter need to reach his hotel?

n steps

Friday night dinner at the restaurant.Peter walks back from the restaurant to the hotel.

Peter moves towards the hotel of 1 meter in each step

QuestionHow many steps does Peter need to reach his hotel?n steps

Drift Analysis: Formalisation

Define a distance function d(x) to measure the distance from the hotel;

d(x) = x, x ∈ 0, . . . , n

(In our case the distance is simply the number of metres from the hotel).

Estimate the expected “speed” (drift), the expected decrease in distance from thegoal;

d(Xt)− d(Xt+1) =

(0, if Xt = 0,

1, if Xt ∈ 1, . . . , n

TimeThen the expected time to reach the hotel (goal) is:

E(T ) =maximum distance

drift=n

Friday night dinner at the restaurant.Peter walks back from the restaurant to the hotel but had a few drinks.

Peter moves towards the hotel of 1 meter in each step with probability 0.6.

Peter moves away from the hotel of 1 meter in each step with probability 0.4.

5n stepsLet us calculate this through drift analysis.

Friday night dinner at the restaurant.Peter walks back from the restaurant to the hotel but had a few drinks.

QuestionHow many steps does Peter need to reach his hotel?5n stepsLet us calculate this through drift analysis.

Drift Analysis (2): Formalisation

Define the same distance function d(x) as before to measure the distance fromthe hotel;

d(x) = x, x ∈ 0, . . . , n

(simply the number of metres from the hotel).

Estimate the expected “speed” (drift), the expected decrease in distance from thegoal;

d(Xt)− d(Xt+1) =

8><>:0, if Xt = 0,

1, if Xt ∈ 1, . . . , nwith probability 0.6

−1, if Xt ∈ 1, . . . , nwith probability 0.4

The expected dicrease in distance (drift) is:

E[d(Xt)− d(Xt+1)] = 0.6 · 1 + 0.4 · (−1) = 0.6− 0.4 = 0.2

TimeThen the expected time to reach the hotel (goal) is:

E(T ) =maximum distance

drift=

0.2= 5n

Additive Drift Theorem

Theorem (Additive Drift Theorem for Upper Bounds [He and Yao, 2001])

Let Xtt≥0 be a Markov process over a set of states S, and d : S → R+0 a function

that assigns a non-negative real number to every state. Let the time to reach theoptimum be T := mint ≥ 0 : d(Xt) = 0. If there exists δ > 0 such that at any timestep t ≥ 0 and at any state Xt > 0 the following condition holds:

E(∆(t)|d(Xt) > 0) = E(d(Xt)− d(Xt+1) | d(Xt) > 0) ≥ δ (1)

E(T | X0 > 0) ≤d(X0)

E(T ) ≤E(d(X0))

δ. (3)

Drift Analysis for Leading Ones

Theorem

The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2)

1 Let g(Xt) = i where i is the number of missing leading ones;

2 The negative drift is 0 since if a leading one is removed from the current solutionthe new point will not be accepted;

3 A positive drift (i.e. of length 1) is achieved as long as the first 0 is flipped andthe leading ones are remained unchanged:

E(∆+(t)) =

n−iXk=1

k · (p(∆+(t)) = k) ≥ 1 · 1/n ·`1− 1/n)n−1 ≥ 1/(en)

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

5 The expected runtime is (i.e. Eq. (6)):

E(T | d(X0) > 0) ≤g(X0)

1/(en)= e · n2 = O(n2)

Theorem

E(∆+(t)) =

n−iXk=1

k · (p(∆+(t)) = k) ≥ 1 · 1/n ·`1− 1/n)n−1 ≥ 1/(en)

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤g(X0)

1/(en)= e · n2 = O(n2)

Theorem

E(∆+(t)) =

n−iXk=1

k · (p(∆+(t)) = k) ≥ 1 · 1/n ·`1− 1/n)n−1 ≥ 1/(en)

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤g(X0)

1/(en)= e · n2 = O(n2)

Theorem

E(∆+(t)) =

n−iXk=1

k · (p(∆+(t)) = k) ≥ 1 · 1/n ·`1− 1/n)n−1 ≥ 1/(en)

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤g(X0)

1/(en)= e · n2 = O(n2)

Theorem

E(∆+(t)) =

n−iXk=1

k · (p(∆+(t)) = k) ≥ 1 · 1/n ·`1− 1/n)n−1 ≥ 1/(en)

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤g(X0)

1/(en)= e · n2 = O(n2)

Theorem

E(∆+(t)) =

n−iXk=1

k · (p(∆+(t)) = k) ≥ 1 · 1/n ·`1− 1/n)n−1 ≥ 1/(en)

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤g(X0)

1/(en)= e · n2 = O(n2)

Exercises

Theorem

The expected time for RLS to optimise LeadingOnes is O(n2)

Left as exercise.

Theorem

Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes isO(λn)

Proof Left as exercise.

Theorem

Let λ < en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes isO(n2)

Exercises

Theorem

Left as exercise.

Theorem

Exercises

Theorem

Left as exercise.

Exercises

Theorem

(1,λ)-EA Analysis for LeadingOnes

Theorem

Let λ = n. Then the expected time for the (1,λ)-EA to optimise LeadingOnes isO(n2) (Jansen and Neumann, Tutorial)

Distance: let d(x) = n− i where i is the number of leading ones;

Drift:

E[d(Xt)− d(Xt+1)]

≥ 1 ·„

1−„

«n«− n ·

„1−

«n«n= c1 − n · cn2 = Ω(1)

Hence,

E(generations) ≤max distance

drift=

Ω(1)= O(n)

and,E(T ) ≤ n · E(generations) = O(n2)

(1,λ)-EA Analysis for LeadingOnes

Theorem

Let λ = n. Then the expected time for the (1,λ)-EA to optimise LeadingOnes isO(n2) (Jansen and Neumann, Tutorial)

Distance: let d(x) = n− i where i is the number of leading ones;

Drift:

E[d(Xt)− d(Xt+1)]

≥ 1 ·„

1−„

«n«− n ·

„1−

«n«n= c1 − n · cn2 = Ω(1)

Hence,

E(generations) ≤max distance

drift=

Ω(1)= O(n)

and,E(T ) ≤ n · E(generations) = O(n2)

Theorem (Additive Drift Theorem for Lower Bounds [He and Yao, 2004])

Let Xtt≥0 be a Markov process over a set of states S, and d : S → R+0 a function

that assigns a non-negative real number to every state. Let the time to reach theoptimum be T := mint ≥ 0 : d(Xt) = 0. If there exists δ > 0 such that at any timestep t ≥ 0 and at any state Xt > 0 the following condition holds:

E(∆(t)|d(Xt) > 0) = E(d(Xt)− d(Xt+1) | d(Xt) > 0) ≤ δ (4)

E(T | X0 > 0) ≥d(X0)

E(T ) ≥E(d(X0))

δ. (6)

Theorem

The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2).

Sources of progress

1 Flipping the leftmost zero-bit;

2 Bits to right of the leftmost zero-bit that are one-bits (free riders).

1 Let the current solution have n− i leading ones (i.e. 1n−i0∗).

2 We define the distance function as the number of missing leading ones, i.e.g(X) = i.

3 The n− i+ 1 bit is a zero;

4 let E[Y ] be the expected number of one-bits after the first zero (i.e. the freeriders).

5 Such i− 1 bits are uniformely distributed at initialisation and still are!

Theorem

Sources of progress

Theorem

Sources of progress

Theorem

Sources of progress

Theorem

Sources of progress

Theorem

Sources of progress

Theorem

Sources of progress

Theorem

Sources of progress

Drift Theorem for LeadingOnes (lower bound)

Theorem

The expected number of free riders is:

E[Y ] =

i−1Xk=1

k · Pr(Y = k) =

i−1Xk=1

Pr(Y ≥ k) =

i−1Xk=1

(1/2)k ≤ 1

The negative drift is 0;

Let p(A) be the probability that the first zero-bit flips into a one-bit.

The positive drift (i.e. the decrease in distance) is bounded as follows:

E(∆+(t)) ≤ p(A) · E[∆+(t)|A] = 1/n · (1 + E[Y ]) ≤ 2/n = δ

Since, also at initialisation the expected number of free riders is less than 2, itfollows that E[d(X0)] ≥ n− 2,

By applying the Drift Theorem we get

E(T ) ≥E(d(X0)

δ=n− 2

2/n= Ω(n2)

Theorem

E[Y ] =

i−1Xk=1

k · Pr(Y = k) =

i−1Xk=1

Pr(Y ≥ k) =

i−1Xk=1

(1/2)k ≤ 1

E(∆+(t)) ≤ p(A) · E[∆+(t)|A] = 1/n · (1 + E[Y ]) ≤ 2/n = δ

E(T ) ≥E(d(X0)

δ=n− 2

2/n= Ω(n2)

Theorem

E[Y ] =

i−1Xk=1

k · Pr(Y = k) =

i−1Xk=1

Pr(Y ≥ k) =

i−1Xk=1

(1/2)k ≤ 1

E(∆+(t)) ≤ p(A) · E[∆+(t)|A] = 1/n · (1 + E[Y ]) ≤ 2/n = δ

E(T ) ≥E(d(X0)

δ=n− 2

2/n= Ω(n2)

Theorem

E[Y ] =

i−1Xk=1

k · Pr(Y = k) =

i−1Xk=1

Pr(Y ≥ k) =

i−1Xk=1

(1/2)k ≤ 1

E(∆+(t)) ≤ p(A) · E[∆+(t)|A] = 1/n · (1 + E[Y ]) ≤ 2/n = δ

E(T ) ≥E(d(X0)

δ=n− 2

2/n= Ω(n2)

Theorem

E[Y ] =

i−1Xk=1

k · Pr(Y = k) =

i−1Xk=1

Pr(Y ≥ k) =

i−1Xk=1

(1/2)k ≤ 1

E(∆+(t)) ≤ p(A) · E[∆+(t)|A] = 1/n · (1 + E[Y ]) ≤ 2/n = δ

E(T ) ≥E(d(X0)

δ=n− 2

2/n= Ω(n2)

Multiplicative Drift Theorem

Drift Analysis for OneMax

Lets calculate the runtime of the (1+1)-EA using the additive Drift Theorem.

1 Let g(Xt) = i where i is the number of zeroes in the bitstring;

2 The negative drift is 0 since solution with less one-bits will not be accepted;

3 A positive drift is achieved as long as a 0 is flipped and the ones remainunchanged:

E(∆+(t)) = E[d(Xt)− d(Xt+1)] ≥ 1 ·i

„1−

«n−1

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

5 The expected initial distance is E(d(X0)) = n/2

The expected runtime is (i.e. Eq. (6)):

E(T | d(X0) > 0) ≤E[(d(X0)]

1/(en)= e/2 · n2 = O(n2)

We need a different distance function!

E(∆+(t)) = E[d(Xt)− d(Xt+1)] ≥ 1 ·i

„1−

«n−1

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤E[(d(X0)]

1/(en)= e/2 · n2 = O(n2)

E(∆+(t)) = E[d(Xt)− d(Xt+1)] ≥ 1 ·i

„1−

«n−1

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤E[(d(X0)]

1/(en)= e/2 · n2 = O(n2)

E(∆+(t)) = E[d(Xt)− d(Xt+1)] ≥ 1 ·i

„1−

«n−1

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤E[(d(X0)]

1/(en)= e/2 · n2 = O(n2)

E(∆+(t)) = E[d(Xt)− d(Xt+1)] ≥ 1 ·i

„1−

«n−1

4 Hence, E[∆(t)] = E(∆+(t))− E(∆−(t)) ≥ 1/(en) = δ

E(T | d(X0) > 0) ≤E[(d(X0)]

1/(en)= e/2 · n2 = O(n2)

1 Let g(Xt) = ln(i+ 1) where i is the number of zeroes in the bitstring;

E(∆+(t)) = E[d(Xt)−d(Xt+1)] ≥ ln(i+1)·i

„1−

«n−1

≥ln(i+ 1)

4 Hence, ∆(t) = E(∆+(t))− E(∆−(t)) ≥ ln(2)/(en) = δ

5 The distance is d(X0) ≤ ln(n+ 1)

E(T | d(X0) > 0) ≤d(X0)

ln(n+ 1)

ln(2)/(en)= O(n lnn)

If the amount of progress depends on the distance from the optimum we need to use alogarithmic distance!

E(∆+(t)) = E[d(Xt)−d(Xt+1)] ≥ ln(i+1)·i

„1−

«n−1

≥ln(i+ 1)

E(T | d(X0) > 0) ≤d(X0)

ln(n+ 1)

E(∆+(t)) = E[d(Xt)−d(Xt+1)] ≥ ln(i+1)·i

„1−

«n−1

≥ln(i+ 1)

E(T | d(X0) > 0) ≤d(X0)

ln(n+ 1)

E(∆+(t)) = E[d(Xt)−d(Xt+1)] ≥ ln(i+1)·i

„1−

«n−1

≥ln(i+ 1)

E(T | d(X0) > 0) ≤d(X0)

ln(n+ 1)

E(∆+(t)) = E[d(Xt)−d(Xt+1)] ≥ ln(i+1)·i

„1−

«n−1

≥ln(i+ 1)

E(T | d(X0) > 0) ≤d(X0)

ln(n+ 1)

Theorem (Multiplicative Drift, [Doerr et al., 2010])

Let Xtt∈N0 be random variables describing a Markov process over a finite statespace S ⊆ R. Let T be the random variable that denotes the earliest point in timet ∈ N0 such that Xt = 0.If there exist δ, cmin, cmax > 0 such that

1 E[Xt −Xt+1 | Xt] ≥ δXt and

2 cmin ≤ Xt ≤ cmax,

for all t < T , then

E[T ] ≤2

δ· ln„

1 +cmax

(1+1)-EA Analysis for OneMax

Theorem

The expected time for the (1+1)-EA to optimise OneMax is O(n lnn)

Distance: let i be the number of zeroes;

E[Xt+1|Xt = i] ≥ i− 1 · ien

= i ·„

1− 1en

«= Xt ·

„1− 1

«E[Xt −Xt+1|Xt = i] ≤ Xt −Xt ·

„1− 1

enXt (δ = 1

1 = cmin ≤ Xt ≤ cmax = n

Hence,

E[T ] ≤2

δ· ln„

1 +cmax

«= 2en ln(1 + n) = O(n lnn)

(1+1)-EA Analysis for OneMax

Theorem

The expected time for the (1+1)-EA to optimise OneMax is O(n lnn)

Distance: let i be the number of zeroes;

E[Xt+1|Xt = i] ≥ i− 1 · ien

= i ·„

1− 1en

«= Xt ·

„1− 1

«E[Xt −Xt+1|Xt = i] ≤ Xt −Xt ·

„1− 1

enXt (δ = 1

1 = cmin ≤ Xt ≤ cmax = n

Hence,

E[T ] ≤2

δ· ln„

1 +cmax

«= 2en ln(1 + n) = O(n lnn)

Exercises

Theorem

The expected time for RLS to optimise OneMax is O(n logn)

Left as exercise.

Theorem

Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise OneMax is O(λn)

Theorem

Let λ < en. Then the expected time for the (1+λ)-EA to optimise OneMax isO(n logn)

Exercises

Theorem

Left as exercise.

Theorem

Exercises

Theorem

Left as exercise.

Exercises

Theorem

Simplified Negative Drift Theorem

Friday night dinner at the restaurant.Peter walks back from the restaurant to the hotel but had too many drinks.

at least 2cn steps with overwhelming probability (exponential time)We need Negative-Drift Analysis.

Friday night dinner at the restaurant.Peter walks back from the restaurant to the hotel but had too many drinks.

QuestionHow many steps does Peter need to reach his hotel?at least 2cn steps with overwhelming probability (exponential time)We need Negative-Drift Analysis.

Simplified Drift Theorem

Xt, t ≥ 0, (Markov process over S ⊆ [0, N ])

∆t(i) := (Xt+1 −Xt | Xt = i) for i ∈ S and t ≥ 0 (Drift)

[a, b] (Interval in State space S)

T ∗ := mint ≥ 0: Xt ≤ a | X0 ≥ b (Time to reach a)

Theorem (Simplified Negative-Drift Theorem, [Oliveto and Witt, 2011])

Suppose there exist three constants δ,ε,r such that for all t ≥ 0:

1 E(∆t(i)) ≥ ε for a < i < b,

2 Prob(∆t(i) = −j) ≤ 1(1+δ)j−r

for i > a and j ≥ 1.

ThenProb(T ∗ ≤ 2c

∗(b−a)) = 2−Ω(b−a)

Xt, t ≥ 0, (Markov process over S ⊆ [0, N ])

∆t(i) := (Xt+1 −Xt | Xt = i) for i ∈ S and t ≥ 0 (Drift)

[a, b] (Interval in State space S)

T ∗ := mint ≥ 0: Xt ≤ a | X0 ≥ b (Time to reach a)

Theorem (Simplified Negative-Drift Theorem, [Oliveto and Witt, 2011])

Suppose there exist three constants δ,ε,r such that for all t ≥ 0:

1 E(∆t(i)) ≥ ε for a < i < b,

2 Prob(∆t(i) = −j) ≤ 1(1+δ)j−r

ThenProb(T ∗ ≤ 2c

∗(b−a)) = 2−Ω(b−a)

1 E(∆t(i)) ≥ ε for a < i < b,

2 Prob(∆t(i) = −j) ≤ 1(1+δ)j−r

Negative-Drift Analysis: Example (3)

Define the same distance function d(x) = x, x ∈ 0, . . . , n (metres from thehotel) (b=n-1, a=1).

Estimate the increase in distance from the goal (negative drift);

d(Xt)− d(Xt+1) =

8><>:0, if Xt = 0,

The expected increase in distance (negative drift) is: (Condition 1)

E[d(Xt)− d(Xt+1)] = 0.6 · 1 + 0.4 · (−1) = 0.6− 0.4 = 0.2

Probability of jumps (i.e. Prob(∆t(i) = −j) ≤ 1(1+δ)j−r

) (set δ = r = 1)

(Condition 2):

Pr(∆t(i) = −j) =

(0 < (1/2)j−1, if j > 1,

0.6 < (1/2)0 = 1, if j = 1

Then the expected time to reach the hotel (goal) is:

Pr(T ≤ 2c(b−a)) = Pr(T ≤ 2c(n−2)) = 2−Ω(n)

d(Xt)− d(Xt+1) =

8><>:0, if Xt = 0,

E[d(Xt)− d(Xt+1)] = 0.6 · 1 + 0.4 · (−1) = 0.6− 0.4 = 0.2

) (set δ = r = 1)

(Condition 2):

Pr(∆t(i) = −j) =

(0 < (1/2)j−1, if j > 1,

0.6 < (1/2)0 = 1, if j = 1

Pr(T ≤ 2c(b−a)) = Pr(T ≤ 2c(n−2)) = 2−Ω(n)

d(Xt)− d(Xt+1) =

8><>:0, if Xt = 0,

E[d(Xt)− d(Xt+1)] = 0.6 · 1 + 0.4 · (−1) = 0.6− 0.4 = 0.2

) (set δ = r = 1)

(Condition 2):

Pr(∆t(i) = −j) =

(0 < (1/2)j−1, if j > 1,

0.6 < (1/2)0 = 1, if j = 1

Pr(T ≤ 2c(b−a)) = Pr(T ≤ 2c(n−2)) = 2−Ω(n)

d(Xt)− d(Xt+1) =

8><>:0, if Xt = 0,

E[d(Xt)− d(Xt+1)] = 0.6 · 1 + 0.4 · (−1) = 0.6− 0.4 = 0.2

) (set δ = r = 1)

(Condition 2):

Pr(∆t(i) = −j) =

(0 < (1/2)j−1, if j > 1,

0.6 < (1/2)0 = 1, if j = 1

Pr(T ≤ 2c(b−a)) = Pr(T ≤ 2c(n−2)) = 2−Ω(n)

d(Xt)− d(Xt+1) =

8><>:0, if Xt = 0,

E[d(Xt)− d(Xt+1)] = 0.6 · 1 + 0.4 · (−1) = 0.6− 0.4 = 0.2

) (set δ = r = 1)

(Condition 2):

Pr(∆t(i) = −j) =

(0 < (1/2)j−1, if j > 1,

0.6 < (1/2)0 = 1, if j = 1

Pr(T ≤ 2c(b−a)) = Pr(T ≤ 2c(n−2)) = 2−Ω(n)

Needle in a Haystack

Theorem

Let η > 0 be constant. Then there is a constant c > 0 such that with probability1− 2−Ω(n) the (1+1)-EA on Needle creates only search points with at mostn/2 + ηn ones in 2cn steps.

Proof Idea

1 By Chernoff bounds the probability that the initial bit string has less thann/2− γn zeroes is e−Ω(n).

2 we set b := n/2− γn and a := n/2− 2γn where γ := η/2;

3 Let Xt denote the number of zeroes in the bit string at time step t.;

4 Let ∆(i) denote the random increase of the number of zeroes (i.e. the drift);

5 Now we have to check that the two conditions of the Simplified drift theoremhold.

Needle in a Haystack

Theorem

Let η > 0 be constant. Then there is a constant c > 0 such that with probability1− 2−Ω(n) the (1+1)-EA on Needle creates only search points with at mostn/2 + ηn ones in 2cn steps.

Proof Idea

1 By Chernoff bounds the probability that the initial bit string has less thann/2− γn zeroes is e−Ω(n).

2 we set b := n/2− γn and a := n/2− 2γn where γ := η/2;

3 Let Xt denote the number of zeroes in the bit string at time step t.;

4 Let ∆(i) denote the random increase of the number of zeroes (i.e. the drift);

5 Now we have to check that the two conditions of the Simplified drift theoremhold.

Needle in a Haystack (2)

Proof of Condition 1Condition 1 holds if E(∆(i)) ≥ ε for some constant ε > 0.

1 The (1+1)-EA flips 0-bits and 1-bits independently with probability 1/n.

2 Since there are i zero-bits and n− i one-bits in the current string, the expectednumber of zero-bits flipped into one-bits is i/n and the viceversa is (n− i)/n.

3 Hence,

E(∆(i)) =n− in−i

n− 2i

n≥ 2γ = ε

and Condition 1 holds.

Needle in a Haystack (3)

Proof of Condition 2Condition 2 is: Prob(∆(i) = −j) ≤ 1/(1 + δ)j−r for j ∈ N0.

1 Since to reach state i− j or less from state i, at least j bits have to flip, theprobability of a drift of length j can be bounded as follows:

Prob(∆(i) ≤ −j) ≤“nj

”„ 1

«j≤

j!≤„

«j−1

This proves Condition 2 by setting δ = r = 1.

Applying the Simplified Drift TheoremFor a constant c∗ > 0 the global optimum is found in 2c

∗(b−a) = 2cn steps withprobability at most 2−Ω(b−a) = 2−Ω(n).

Exercise: Trap Functions

Trap(x) =

n+ 1 if x = 0n

OneMax(x) otherwise.

0 ones(x)

Theorem

With overwhelming probability at least 1− 2−Ω(n) the (1+1)-EA requires 2Ω(n) stepsto optimise Trap .

Exercise: Trap Functions

Trap(x) =

n+ 1 if x = 0n

0 ones(x)

Theorem

With overwhelming probability at least 1− 2−Ω(n) the (1+1)-EA requires 2Ω(n) stepsto optimise Trap .

Proof Left as exercise.P. S. Oliveto & X. Yao (University of Birmingham) Runtime Analysis of EAs WCCI 2012 75 / 92

Drift Analysis Conclusion

Overview

Additive Drift Analysis (upper and lower bounds);

Multiplicative Drift Analysis;

Simplified Negative-Drift Theorem;

Advanced Lower bound Drift Techniques

Drift Analysis for Stochastic Populations (mutation) [Lehre, 2010];

Simplified Drift Theorem combined with bandwidth analysis (mutation +crossover stochastic populations = GAs) [Oliveto and Witt, 2012];

Typical run investigations

Typical Runs have been introduced by considering that

the “global behaviour of a process” is predictable with high probability;

the “local behaviour” is quite unpredictable;

Method

1 Divide the process into k phases which are long enough to assure that some evenhappens with probability pk;

2 Hence, does not happen with failure probability 1− pk;

3 The last phase should lead the EA to the global optimum;

4 The failure probability is 1−Pki=1 pi;

5 The runtime is lower than the sum of each phase time.

Generally Chernoff bounds are used to obtain the failure probabilities!

Typical run investigations

Typical Runs have been introduced by considering that

the “global behaviour of a process” is predictable with high probability;

the “local behaviour” is quite unpredictable;

Method

1 Divide the process into k phases which are long enough to assure that some evenhappens with probability pk;

2 Hence, does not happen with failure probability 1− pk;

3 The last phase should lead the EA to the global optimum;

4 The failure probability is 1−Pki=1 pi;

5 The runtime is lower than the sum of each phase time.

Generally Chernoff bounds are used to obtain the failure probabilities!

(1+1)-EA for Trap functions

Theorem

With overwhelming probability at least 1− e−Ω(n) the (1+1)-EA requires nΩ(n) stepsto optimise Trap .

The deceptive trap function is the following:

Trap(x) =

n+ 1 if x = 0n

Proof IdeaWe use the Typical run investigations method and split the analysis into the followingthree phases:

1 This phase ends when at least n/3 one-bits are in the current solution.

2 This phase ends when the 1n bitstring is the current solution.

3 This phase ends when the global optimum is found.

Theorem

Trap(x) =

n+ 1 if x = 0n

Theorem

Trap(x) =

n+ 1 if x = 0n

Theorem

Trap(x) =

n+ 1 if x = 0n

Theorem

Trap(x) =

n+ 1 if x = 0n

(1+1)-EA for Trap functions (2)

We need to show that the runtime required to complete the three phases is nΩ(n) andthat the failure probability of each phase is at most e−Ω(n).

Phase 1 This phase ends when at least n/3 one-bits are in the currentsolution.

1 By using Chernoff bounds we calculate the probability that the algorithm isinitialised with less than n/3 one-bits.

2 n/2 expected one-bits after initialisation (i.e. E(X) = n · p = n/2).

3 Setting δ = 1/3 we get

P (X ≤ n/3) ≤ e−E(X)δ2/2 = e−(n/2)·(1/9)·(1/2) = e−n/36 = e−Ω(n)

Phase 2 This phase ends when the 1n bitstring is the current solution.

1 The probability of reaching the 0n bitstring in one step is less than n−n/3;

2 in expected time c · n logn the algorithm climbs up the OneMax .

3 By Markov’s inequality,

P`X ≥ e · c · n logn

c · n logn

e · c · n logn= 1/e

4 By applying it iteratively: P`X ≥ e · c · n2 logn

´≤ e−n

5 The probability that in c · n2 logn steps n/3 precise bits flip isc · n2 logn · n−n/3 = n−Ω(n). (union bound)

c · n logn

´≤ e−n

c · n logn

´≤ e−n

c · n logn

´≤ e−n

c · n logn

´≤ e−n

c · n logn

´≤ e−n

Phase 3

This phase ends when the global optimum is found.

1 The expected time to reach the global optimum is nn because all the bits have tobe flipped at the same time;

2 By Chernoff bounds the probability that the time is less than (1/2)nn is (i.e.δ = 1/2):

P (T ≤ (1/2)nn) ≤ e−nn·1/4·1/2 = e−n

n·1/8 = e−Ω(n)

Summing up

1 the algorithm requires exponential runtime 1 + c · n2 logn+ (1/2)nn = nΩ(n)

2 With probability at least 1− e−Ω(n).

Phase 3

P (T ≤ (1/2)nn) ≤ e−nn·1/4·1/2 = e−n

n·1/8 = e−Ω(n)

Summing up

Phase 3

P (T ≤ (1/2)nn) ≤ e−nn·1/4·1/2 = e−n

n·1/8 = e−Ω(n)

Summing up

Phase 3

P (T ≤ (1/2)nn) ≤ e−nn·1/4·1/2 = e−n

n·1/8 = e−Ω(n)

Summing up

ExpectationTrap

ExpectationTrap(x) =

n− 1/2 if x = 0n

Theorem

The expected time for the (1+1)-EA to optimise ExpectationTrap is Ω`(n/2)n

[Jansen and Neumann, Tutorial]

Observation

Prob(Local) = 2−n (probability the algorithm is initialised with the 0n bitstring)

E(T |X0 = 0n) = nn (all bits need to be flipped)

E(T ) ≥ E(T |X0 = 0n) = nn · 2−n = (n/2)n (law of total probability)

ExpectationTrap

n− 1/2 if x = 0n

Theorem

Observation

ExpectationTrap

n− 1/2 if x = 0n

Theorem

Observation

Overview

Final Overview

Overview

Basic Probability Theory

Tail Inequalities

Artificial Fitness Levels

Drift Analysis

Typical Runs

Other Techniques (Not covered)

Family Trees [Witt, 2006]

Gambler’s Ruin & Martingales [Jansen and Wegener, 2001]

State-of-the-art

State of the Art in Computational Complexity of RSHs

OneMax (1+1) EA O(n logn)(1+λ) EA O(λn + n logn)(µ+1) EA O(µn + n logn)1-ANT O(n2) w.h.p.(µ+1) IA O(µn + n logn)

Linear Functions (1+1) EA Θ(n logn)cGA Θ(n2+ε), ε > 0 const.

Max. Matching (1+1) EA eΩ(n) , PRAS

Sorting (1+1) EA Θ(n2 logn)SS Shortest Path (1+1) EA O(n3 log(nwmax))

MO (1+1) EA O(n3)MST (1+1) EA Θ(m2 log(nwmax))

(1+λ) EA O(n log(nwmax))1-ANT O(mn log(nwmax))

Max. Clique (1+1) EA Θ(n5)(rand. planar) (16n+1) RLS Θ(n5/3)Eulerian Cycle (1+1) EA Θ(m2 logm)Partition (1+1) EA 4/3 approx., competitive avg.

Vertex Cover (1+1) EA eΩ(n) , arb. bad approx.

Set Cover (1+1) EA eΩ(n) , arb. bad approx.SEMO Pol. O(logn)-approx.

Intersection of (1+1) EA 1/p-approximation in

p ≥ 3 matroids O(|E|p+2 log(|E|wmax))UIO/FSM conf. (1+1) EA eΩ(n)

See [Oliveto et al., 2007] for an overview.

P. K. Lehre, 2008

Further reading

Coming up Conference

Further reading

References I

Auger, A. and Doerr, B. (2011).

Theory of Randomized Search Heuristics: Foundations and Recent Developments.World Scientific Publishing Co., Inc., River Edge, NJ, USA.

Back, T. (1993).

Optimal mutation rates in genetic search.In In Proceedings of the Fifth International Conference on Genetic Algorithms (ICGA), pages 2–8.

Doerr, B., Johannsen, D., and Winzen, C. (2010).

Multiplicative drift analysis.In Proceedings of the 12th annual conference on Genetic and evolutionary computation, GECCO ’10, pages 1449–1456. ACM.

Droste, S., Jansen, T., and Wegener, I. (1998).

A rigorous complexity analysis of the (1 + 1) evolutionary algorithm for separable functions with boolean inputs.Evolutionary Computation, 6(2):185–196.

Droste, S., Jansen, T., and Wegener, I. (2002).

On the analysis of the (1+1) evolutionary algorithm.Theoretical Computer Science, 276(1-2):51–81.

Goldberg, D. E. (1989).

Genetic Algorithms for Search, Optimization, and Machine Learning.Addison-Wesley.

He, J. and Yao, X. (2001).

Drift analysis and average time complexity of evolutionary algorithms.Artificial Intelligence, 127(1):57–85.

Further reading

References II

He, J. and Yao, X. (2004).

A study of drift analysis for estimating computation time of evolutionary algorithms.Natural Computing: an international journal, 3(1):21–35.

Holland, J. H. (1992).

Adaptation in Natural and Artificial Systems: An Introductory Analysis with Applications to Biology, Control, and ArtificialIntelligence.The MIT Press.

Jansen, T., Jong, K. A. D., and Wegener, I. A. (2005).

On the choice of the offspring population size in evolutionary algorithms.Evolutionary Computation, 13(4):413–440.

Jansen, T. and Wegener, I. (2001).

Evolutionary algorithms - how to cope with plateaus of constant fitness and when to reject strings of the same fitness.IEEE Trans. Evolutionary Computation, 5(6):589–599.

Lehre, P. K. (2010).

Negative drift in populations.In PPSN (1), pages 244–253.

Lehre, P. K. (2011).

Fitness-levels for non-elitist populations.In Proceedings of the 13th annual conference on Genetic and evolutionary computation, GECCO ’11, pages 2075–2082. ACM.

Neumann, F. and Witt, C. (2010).

Bioinspired Computation in Combinatorial Optimization: Algorithms and Their Computational Complexity.Springer-Verlag New York, Inc., New York, NY, USA, 1st edition.

Further reading

References III

Oliveto, P. and Witt, C. (2012).

On the analysis of the simple genetic algorithm (to appear).In Proceedings of the 12th annual conference on Genetic and evolutionary computation, GECCO ’12, pages –. ACM.

Oliveto, P. S., He, J., and Yao, X. (2007).

Time complexity of evolutionary algorithms for combinatorial optimization: A decade of results.International Journal of Automation and Computing, 4(3):281–293.

Oliveto, P. S. and Witt, C. (2011).

Simplified drift analysis for proving lower bounds inevolutionary computation.Algorithmica, 59(3):369–386.

Reeves, C. R. and Rowe, J. E. (2002).

Genetic Algorithms: Principles and Perspectives: A Guide to GA Theory.Kluwer Academic Publishers, Norwell, MA, USA.

Rudolph, G. (1998).

Finite Markov chain results in evolutionary computation: A tour d’horizon.Fundamenta Informaticae, 35(1–4):67–89.

Sudholt, D. (2010).

General lower bounds for the running time of evolutionary algorithms.In PPSN (1), pages 124–133.

Witt, C. (2006).

Runtime analysis of the (µ+1) ea on simple pseudo-boolean functions evolutionary computation.In GECCO ’06: Proceedings of the 8th annual conference on Genetic and evolutionary computation, pages 651–658, New York,NY, USA. ACM Press.

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