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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
A Gentle Introduction to the Time Complexity Analysis of
Evolutionary Algorithms
Pietro S. Oliveto and Xin Yao
CERCIA, School of Computer Science, University of Birmingham, UK
WCCI 2012
Brisbane, Australia, 10 June 2012
P. S. Oliveto & X. Yao
(University of Birmingham)
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Aims and Goals of this Tutorial
This tutorial will provide an overview of
the goals of time complexity analysis of Evolutionary Algorithms (EAs)
the most common and effective techniques
You should attend if you wish to
theoretically understand the behaviour and performance of the search algorithms you
design
familiarise with the techniques used in the time complexity analysis of EAs
pursue research in the area
enable you or enhance your ability to
1
2
3
4
5
P. S. Oliveto & X. Yao
understand theoretically the behaviour of EAs on different problems
perform time complexity analysis of simple EAs on common toy problems
read and understand research papers on the computational complexity of EAs
have the basic skills to start independent research in the area
follow the time complexity of EAs for combinatorial optimization Tutorial later on today
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Outline I
1
Motivation
Introduction to the theory of EAs
Convergence analysis of EAs
Computational complexity of EAs
2
Basic Probability Theory
Probability space
Union bound
Random variables and expectations
Law of total probability
3
Evolutionary Algorithms
General EAs
(1+1)-EA and RLS
General properties
General upper bound
4
Tail Inequalities
Markov’s inequality
Chernoff bounds
5
Artificial Fitness Levels
Coupon collector’s problem
AFL method for upper bounds
AFL method for parent populations
AFL for non-elitist EAs
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Outline II
AFL for lower bounds
6
Drift Analysis
Additive Drift Theorem
Multiplicative Drift Theorem
Simplified Negative Drift Theorem
7
Typical Run Investigations
8
Conclusions
Overview
State-of-the-art
Further reading
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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)
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Introduction to the theory of EAs
Brief history
Theoretical studies of Evolutionary Algorithms (EAs), albeit few, have always existed
since the seventies [Goldberg, 1989];
Early studies were concerned with explaining the behaviour rather than analysing
their 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.
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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!
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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 any
other point
2
The best found solution is never removed from the population (elitism)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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 any
other point
2
The best found solution is never removed from the population (elitism)
Canonical GAs using mutation, crossover and proportional selection Do Not
converge!
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?
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Computational complexity of EAs
Computational Complexity of EAs
P. K. Lehre, 2011
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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 algorithm
Polynomial runtime: “Efficient” algorithm
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Computational complexity of EAs
Computational Complexity of EAs
P. K. Lehre, 2011
However (EAs):
1
2
In practice the time for a fitness function evaluation is much higher than the rest;
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 .
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Computational complexity of EAs
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)
f (n) ∈ o(g(n)) ⇐⇒ lim
=0
n→∞ g(n)
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Computational complexity of EAs
Exercise 1: Asymptotic Notation
[Lehre, Tutorial]
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Computational complexity of EAs
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;
Next
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 not
always sufficient)
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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: P r : = → R (eg. P r(E) = 2/6 = 1/3)
Probability Axioms
For any event E, 0 ≤ P r(E) ≤ 1
P r(Ω) = 1
For any countably finite sequence of pairwise mutually disjoint events
E1 , E2 , E3 , . . .
´ X
`[
P r(Ei )
Ei =
Pr
i≥1
P. S. Oliveto & X. Yao
(University of Birmingham)
i≥1
Runtime Analysis of EAs
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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: P r : = → R (eg. P r(E) = 2/6 = 1/3)
Probability Axioms
For any event E, 0 ≤ P r(E) ≤ 1
P r(Ω) = 1
For any countably finite sequence of pairwise mutually disjoint events
E1 , E2 , E3 , . . .
´ X
`[
P r(Ei )
Ei =
Pr
i≥1
i≥1
1 Die (Ei the event that i shows up)
Ω
1
2
3
4
5
6
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
12 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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: P r : = → R (eg. P r(E) = 2/6 = 1/3)
Probability Axioms
For any event E, 0 ≤ P r(E) ≤ 1
P r(Ω) = 1
For any countably finite sequence of pairwise mutually disjoint events
E1 , E2 , E3 , . . .
´ X
`[
P r(Ei )
Ei =
Pr
i≥1
i≥1
1 Die (Ei the event that i shows up)
Ω P r(E1 ) = P (E2 ) = · · · = P (E6 ) = 1/6
1
2
3
P r(E1 ∪ E2 ) = P r(E1 ) + P r(E2 ) = 2/6 = 1/3
P r(E1 ∪ E2 ∪ E3 ) = P r(E1 ) + P r(E2 ) + P r(E3 ) = 3/6 = 1/2
4
5
P. S. Oliveto & X. Yao
6
P r(E1 ∪ E2 ∪ · · · ∪ E6 ) = P r(E1 ) + . . . P r(E6 ) = 6/6 = 1 = P r(Ω)
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : 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?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : 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? P r(E1 ) =
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : 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? P r(E1 ) =
What is the probability that the second die is 1?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
6
36
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
What is the probability that the second die is 1? P r(E2 ) =
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
6
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WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
What is the probability that the second die is 1? P r(E2 ) =
What is the probability that both dice are 1?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
6
36
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
6
What is the probability that the second die is 1? P r(E2 ) = 36
What is the probability that both dice are 1? P r(E3 ) = P r(E1 ∩ E2 ) =
(independent events)
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
1
36
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
6
What is the probability that the second die is 1? P r(E2 ) = 36
What is the probability that both dice are 1? P r(E3 ) = P r(E1 ∩ E2 ) =
(independent events)
1
36
Definition
Two events E and F are independent if and only if P r(E ∩ F ) = P r(E) · P r(F )
What is the probability that second die is 1 given that the first die is 1?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
6
What is the probability that the second die is 1? P r(E2 ) = 36
What is the probability that both dice are 1? P r(E3 ) = P r(E1 ∩ E2 ) =
(independent events)
1
36
Definition
Two events E and F are independent if and only if P r(E ∩ F ) = P r(E) · P r(F )
What is the probability that second die is 1 given that the first die is 1?
P r(E2 ∩E1 )
1/36
P r(E2 |E1 ) = P r(E
) = 1/6 = 1/6 (conditional probabilities)
)
1
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
6
What is the probability that the second die is 1? P r(E2 ) = 36
What is the probability that both dice are 1? P r(E3 ) = P r(E1 ∩ E2 ) =
(independent events)
1
36
Definition
Two events E and F are independent if and only if P r(E ∩ F ) = P r(E) · P r(F )
What is the probability that second die is 1 given that the first die is 1?
P r(E2 ∩E1 )
1/36
P r(E2 |E1 ) = P r(E
) = 1/6 = 1/6 (conditional probabilities)
)
1
What about at least one of the two dice is a 1?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
6
What is the probability that the second die is 1? P r(E2 ) = 36
What is the probability that both dice are 1? P r(E3 ) = P r(E1 ∩ E2 ) =
(independent events)
1
36
Definition
Two events E and F are independent if and only if P r(E ∩ F ) = P r(E) · P r(F )
What is the probability that second die is 1 given that the first die is 1?
P r(E2 ∩E1 )
1/36
P r(E2 |E1 ) = P r(E
) = 1/6 = 1/6 (conditional probabilities)
)
1
What about at least one of the two dice is a 1?P r(E1 ∪ E2 ) =
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
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36
WCCI 2012
13 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Probability space
Independent Events and Conditional Probabilities
Two dice
E1 : Event that the first die is a 1;
E2 : Event that the second die is a 1;
6
What is the probability that the first die is 1? P r(E1 ) = 36
6
What is the probability that the second die is 1? P r(E2 ) = 36
What is the probability that both dice are 1? P r(E3 ) = P r(E1 ∩ E2 ) =
(independent events)
1
36
Definition
Two events E and F are independent if and only if P r(E ∩ F ) = P r(E) · P r(F )
What is the probability that second die is 1 given that the first die is 1?
P r(E2 ∩E1 )
1/36
P r(E2 |E1 ) = P r(E
) = 1/6 = 1/6 (conditional probabilities)
)
1
What about at least one of the two dice is a 1?P r(E1 ∪ E2 ) =
11
36
E1 and E2 are not mutually disjoint events!
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union bound
Ω
1,1 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
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union bound
Ω
1,1 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
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
P r(E1 ) = 6/36 = 1/6
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union bound
Ω
1,1 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
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
P r(E1 ) = 6/36 = 1/6
P r(E2 ) = 6/36 = 1/6
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union bound
Ω
1,1 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
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
P r(E1 ) = 6/36 = 1/6
P r(E2 ) = 6/36 = 1/6
P r(E2 |E1 ) = 1/6
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union bound
Ω
1,1 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
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
P r(E1 ) = 6/36 = 1/6
P r(E2 ) = 6/36 = 1/6
P r(E2 |E1 ) = 1/6
P r(E1 ∪ E2 ) = 6/36 + 5/36 = 11/36 ≤ 6/36 + 6/36 = P r(E1 ) + P r(E2 )
P. S. Oliveto & X. Yao
(University of Birmingham)
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union bound
Ω
1,1 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
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
P r(E1 ) = 6/36 = 1/6
P r(E2 ) = 6/36 = 1/6
P r(E2 |E1 ) = 1/6
P r(E1 ∪ E2 ) = 6/36 + 5/36 = 11/36 ≤ 6/36 + 6/36 = P r(E1 ) + P r(E2 )
E1 and E2 are not mutually disjoint events!
P. S. Oliveto & X. Yao
(University of Birmingham)
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Union bound
Union Bound
Theorem (Union bound)
For any finite or countably finite sequence of events E1 , E2 , . . .
`[
´ X
Pr
Ei ≤
P r(Ei )
i≥1
P. S. Oliveto & X. Yao
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i≥1
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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. A
discrete random variable takes only a finite or countably finite number of values.
Probability:
P r(X = a) =
P
s∈Ω;X(s)=a
P r(s)
Expectation:
E(X) =
P. S. Oliveto & X. Yao
(University of Birmingham)
P
i
i · P r(X = i)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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. A
discrete random variable takes only a finite or countably finite number of values.
Probability:
P r(X = a) =
P
s∈Ω;X(s)=a
P r(s)
Expectation:
E(X) =
P
i
i · P r(X = i)
Example 1 (one die):Let X be the value of one die;
1
6
1
1
1
1
7
E(X) = · 1 + · 2 + · 3 + . . . · 6 =
6
6
6
6
2
P r(X = 6) =
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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. A
discrete random variable takes only a finite or countably finite number of values.
Probability:
P r(X = a) =
P
s∈Ω;X(s)=a
P r(s)
Expectation:
E(X) =
P
i
i · P r(X = i)
Example 1 (two dice): Let X be the value of the sum of two dice;
P r(X = 4) = P r(1, 3) + P r(3, 1) + P r(2, 2) =
E(X) =
P. S. Oliveto & X. Yao
(University of Birmingham)
3
1
=
36
12
1
2
3
1
·2+
·3+
· 4 + ...
· 12 = 7
36
36
36
36
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Random variables and expectations
Linearity of Expectation
Definition
For any collection of discrete random variables X1 , X2 , . . . , Xn with finite
expectations,
n
n
ˆX
˜ X
E
Xi =
E(Xi )
i=1
i=1
Example 1 (two dice): Let X be the value of the sum of two dice, X1 the value of die
1 and X2 the value of die 2
7
2
E(X) = E(X1 ) + E(X2 ) = 7
E(Xi ) =
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Random variables and expectations
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 the
number of successes in n independent experiments each of which succeds with
probability p.
Probability:
P r(X = j) =
`n´ j
p (1 − p)n−j
j
Expectation: Let Xi = 1 if the ith trial is successful; (linearity of expectation)
ˆ Pn
˜ Pn
E(X) = E
i Xi =
i=1 E(Xi ) = np
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Random variables and expectations
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 the
number of successes in n independent experiments each of which succeds with
probability p.
Probability:
P r(X = j) =
`n´ j
p (1 − p)n−j
j
Expectation: Let Xi = 1 if the ith trial is successful; (linearity of expectation)
ˆ Pn
˜ Pn
E(X) = E
i Xi =
i=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
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Random variables and expectations
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 until
the first success.
Probability:
P r(X = n) = (1 − p)n−1 p
Expectation: E(X) = 1/p (waiting time argument)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Random variables and expectations
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 until
the first success.
Probability:
P r(X = n) = (1 − p)n−1 p
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 i
flips with mutation rate pm = 1/n
P r(X = n ·
P. S. Oliveto & X. Yao
√
n + 1) = (1 −
(University of Birmingham)
p)n
E(X) = 1/p = n
„
n+1−1 p = 1 −
√
Runtime Analysis of EAs
1
n
«n√n
1
n
„ «√n
≤
1
e
1
n
√
≤ e−
n
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Law of total probability
Law of Total Probability
Let E, F be mutually disjoint events and E ∪ F = Ω.
Theorem
P r(E) = P r(E|F ) · P r(F ) + P r(E|F̄ ) · P r(F̄ )
E(X) = E(X|F ) · P r(F ) + E(X|F̄ ) · P r(F̄ )
Immediate Consequence:
P r(E) ≥ P r(E|F ) · P r(F )
E(X) ≥ E(X|F ) · P r(F )
Often used to derive lower bounds on the expected time!
We will use this to show that expected values may not be sufficient → success
probabilities!
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General EAs
Evolutionary Algorithms
Algorithm ((µ+λ)-EA)
1 Let t = 0;
2
3
Initialize P0 with µ individuals chosen uniformly at random;
Repeat
Create λ new individuals:
1
2
choose x ∈ Pt uniformly at random;
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.
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General EAs
Evolutionary Algorithms
Algorithm ((µ+λ)-EA)
1 Let t = 0;
2
3
Initialize P0 with µ individuals chosen uniformly at random;
Repeat
Create λ new individuals:
1
2
choose x ∈ Pt uniformly at random;
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;
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General EAs
Evolutionary Algorithms
Algorithm ((µ+λ)-EA)
1 Let t = 0;
2
3
Initialize P0 with µ individuals chosen uniformly at random;
Repeat
Create λ new individuals:
1
2
choose x ∈ Pt uniformly at random;
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 [Bäck, 1993, Droste et al., 1998];
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General EAs
Evolutionary Algorithms
Algorithm ((µ+λ)-EA)
1 Let t = 0;
2
3
Initialize P0 with µ individuals chosen uniformly at random;
Repeat
Create λ new individuals:
1
2
choose x ∈ Pt uniformly at random;
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 [Bäck, 1993, Droste et al., 1998];
By introducing stochastic selection and crossover we obtain a Genetic
Algorithm(GA)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
(1+1)-EA and RLS
1+1-EA
Algorithm ((1+1)-EA)
Initialise P0 with x ∈ {1, 0}n by flipping each bit with p = 1/2 ;
Repeat
Create x0 by flipping each bit in x with p = 1/n;
If f (x0 ) ≥ f (x) Then x0 ∈ 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?
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
(1+1)-EA and RLS
1+1-EA
Algorithm ((1+1)-EA)
Initialise P0 with x ∈ {1, 0}n by flipping each bit with p = 1/2 ;
Repeat
Create x0 by flipping each bit in x with p = 1/n;
If f (x0 ) ≥ f (x) Then x0 ∈ 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?
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
(1+1)-EA and RLS
1+1-EA
Algorithm ((1+1)-EA)
Initialise P0 with x ∈ {1, 0}n by flipping each bit with p = 1/2 ;
Repeat
Create x0 by flipping each bit in x with p = 1/n;
If f (x0 ) ≥ f (x) Then x0 ∈ 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 ] =
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
(1+1)-EA and RLS
1+1-EA
Algorithm ((1+1)-EA)
Initialise P0 with x ∈ {1, 0}n by flipping each bit with p = 1/2 ;
Repeat
Create x0 by flipping each bit in x with p = 1/n;
If f (x0 ) ≥ f (x) Then x0 ∈ 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
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(University of Birmingham)
Runtime Analysis of EAs
«
E(X) = np
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
(1+1)-EA and RLS
1+1-EA
Algorithm ((1+1)-EA)
Initialise P0 with x ∈ {1, 0}n by flipping each bit with p = 1/2 ;
Repeat
Create x0 by flipping each bit in x with p = 1/n;
If f (x0 ) ≥ f (x) Then x0 ∈ 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
=
n
X
«
E(X) = np
1 · 1/n = n/n = 1
i=1
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General properties
1+1-EA: 2
How likely is it that exactly one bit flips?
P. S. Oliveto & X. Yao
(University of Birmingham)
„
«
` ´ j
n−j
P r(X = j) = n
p
(1
−
p)
j
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General properties
1+1-EA: 2
How likely is it that exactly one bit flips?
„
«
` ´ j
n−j
P r(X = j) = n
p
(1
−
p)
j
What is the probability of exactly one bit flipping?
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General properties
1+1-EA: 2
How likely is it that exactly one bit flips?
„
«
` ´ j
n−j
P r(X = j) = n
p
(1
−
p)
j
What is the probability of exactly one bit flipping?
“n”
P r(X = 1) =
· 1/n · (1 − 1/n)n−1 = (1 − 1/n)n−1 ≥ 1/e ≈ 0.37
1
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General properties
1+1-EA: 2
How likely is it that exactly one bit flips?
„
«
` ´ j
n−j
P r(X = j) = n
p
(1
−
p)
j
What is the probability of exactly one bit flipping?
“n”
P r(X = 1) =
· 1/n · (1 − 1/n)n−1 = (1 − 1/n)n−1 ≥ 1/e ≈ 0.37
1
Is it more likely that 2 bits flip or none?
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General properties
1+1-EA: 2
How likely is it that exactly one bit flips?
„
«
` ´ j
n−j
P r(X = j) = n
p
(1
−
p)
j
What is the probability of exactly one bit flipping?
“n”
P r(X = 1) =
· 1/n · (1 − 1/n)n−1 = (1 − 1/n)n−1 ≥ 1/e ≈ 0.37
1
Is it more likely that 2 bits flip or none?
“n”
P r(X = 2) =
· 1/n2 · (1 − 1/n)n−2 =
2
=
n · (n − 1)
1/n2 · (1 − 1/n)n−2 =
2
= 1/2 · (1 − 1/n)n−1 ≈ 1/(2e)
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General properties
1+1-EA: 2
How likely is it that exactly one bit flips?
„
«
` ´ j
n−j
P r(X = j) = n
p
(1
−
p)
j
What is the probability of exactly one bit flipping?
“n”
P r(X = 1) =
· 1/n · (1 − 1/n)n−1 = (1 − 1/n)n−1 ≥ 1/e ≈ 0.37
1
Is it more likely that 2 bits flip or none?
“n”
P r(X = 2) =
· 1/n2 · (1 − 1/n)n−2 =
2
=
n · (n − 1)
1/n2 · (1 − 1/n)n−2 =
2
= 1/2 · (1 − 1/n)n−1 ≈ 1/(2e)
While
P r(X = 0) =
P. S. Oliveto & X. Yao
(University of Birmingham)
“n”
(1/n)0 · (1 − 1/n)n ≈ 1/e
0
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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, 1}n is
O(nn )
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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, 1}n is
O(nn )
Proof
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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, 1}n is
O(nn )
Proof
1
Let i be the number of bit positions in which the current solution x and the
global optimum x∗ differ;
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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, 1}n is
O(nn )
Proof
1
Let i be the number of bit positions in which the current solution x and the
global optimum x∗ differ;
2
Each bit flips with probability 1/n, hence does not flip with probability (1 − 1/n);
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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, 1}n is
O(nn )
Proof
1
Let i be the number of bit positions in which the current solution x and the
global 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 and
leave the n − i bits unchanged;
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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, 1}n is
O(nn )
Proof
1
Let i be the number of bit positions in which the current solution x and the
global 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 and
leave the n − i bits unchanged;
4
Then:
p(x∗ |x) =
P. S. Oliveto & X. Yao
(University of Birmingham)
„ «i „
«
„ «n
`
´
1
1 n−i
1
1−
≥
= n−n p = n−n
n
n
n
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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, 1}n is
O(nn )
Proof
1
Let i be the number of bit positions in which the current solution x and the
global 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 and
leave the n − i bits unchanged;
4
Then:
p(x∗ |x) =
5
„ «i „
«
„ «n
`
´
1
1 n−i
1
1−
≥
= n−n p = n−n
n
n
n
it implies an upper bound on the expected runtime of O(nn )
(E(X) = 1/p = nn ) (waiting time argument).
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
General Upper bound Exercises
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = 1/2 for an
arbitrary function defined in {0, 1}n is O(2n )
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General upper bound
General Upper bound Exercises
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = 1/2 for an
arbitrary function defined in {0, 1}n is O(2n )
Proof Left as Exercise.
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = χ/n for an
arbitrary function defined in {0, 1}n is O((n/χ)n )
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General upper bound
General Upper bound Exercises
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = 1/2 for an
arbitrary function defined in {0, 1}n is O(2n )
Proof Left as Exercise.
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = χ/n for an
arbitrary function defined in {0, 1}n is O((n/χ)n )
Proof Left as Exercise.
Theorem
The expected runtime of RLS for an arbitrary function defined in {0, 1}n is infinite.
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(University of Birmingham)
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General upper bound
General Upper bound Exercises
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = 1/2 for an
arbitrary function defined in {0, 1}n is O(2n )
Proof Left as Exercise.
Theorem
The expected runtime of the (1+1)-EA with mutation probability p = χ/n for an
arbitrary function defined in {0, 1}n is O((n/χ)n )
Proof Left as Exercise.
Theorem
The expected runtime of RLS for an arbitrary function defined in {0, 1}n is infinite.
Proof Left as Exercise.
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
P. S. Oliveto & X. Yao
(University of Birmingham)
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
What about RLS?
P. S. Oliveto & X. Yao
(University of Birmingham)
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
What about RLS?
Expectation: E[X] = 1
P. S. Oliveto & X. Yao
(University of Birmingham)
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
What about RLS?
Expectation: E[X] = 1
P(1-bitflip) = 1
P. S. Oliveto & X. Yao
(University of Birmingham)
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
What about RLS?
Expectation: E[X] = 1
P(1-bitflip) = 1
What about initialisation?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
26 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
What about RLS?
Expectation: E[X] = 1
P(1-bitflip) = 1
What about initialisation?
How many one-bits in expectation after initialisation?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
26 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
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?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
26 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
General upper bound
1+1-EA: Conclusions & Exercises
In general:
P (i − bitf lip) =
«
„
«
“n” 1 „
1
1
1 n−i
1 n−i
≤
≈
1
−
1
−
i ni
n
i!
n
i!e
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?
P r(X = n/2) =
«
„
«
“ n ” 1 „
1 n/2
1
−
n
=
100,
P
r(X
=
50)
≈
0.0796
n/2 nn/2
n
Tail Inequalities help us to deal with these kind of problems.
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
26 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Tail Inequalities
Given a random variable X it may assume values that are considerably larger or lower
than its expectation;
Tail inequalities:
Estimate the probability that X deviates from the expectation by a defined
amount δ;
For many intermediate results, expected values are useless;
May turn expected runtimes into bounds that hold with overwhelming probability.
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Markov’s inequality
Markov Inequality
The fundamental inequality from which many others are derived.
P. S. Oliveto & X. Yao
(University of Birmingham)
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WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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] its
expectation. Then for all t ∈ R+ ,
P r[X ≥ t] ≤
P. S. Oliveto & X. Yao
(University of Birmingham)
E[X]
.
t
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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] its
expectation. Then for all t ∈ R+ ,
P r[X ≥ t] ≤
E[X] = 1; then: P r[X ≥ 2] ≤
P. S. Oliveto & X. Yao
(University of Birmingham)
E[X]
2
≤
1
2
E[X]
.
t
(Number of bits that flip)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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] its
expectation. Then for all t ∈ R+ ,
P r[X ≥ t] ≤
E[X] = 1; then: P r[X ≥ 2] ≤
E[X]
2
E[X]
.
t
1
(Number of bits that
2
E[X]
n/2
= (2/3)n ≤ 43
(2/3)n
≤
flip)
E[X] = n/2; then P r[X ≥ (2/3)n] ≤
(Number of one-bits after initialisation)
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
28 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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] its
expectation. Then for all t ∈ R+ ,
P r[X ≥ t] ≤
E[X] = 1; then: P r[X ≥ 2] ≤
E[X]
2
E[X]
.
t
1
(Number of bits that
2
E[X]
n/2
= (2/3)n ≤ 43
(2/3)n
≤
flip)
E[X] = n/2; then P r[X ≥ (2/3)n] ≤
(Number of one-bits after initialisation)
Usually Markov’s inequality is used iteratively to obtain stronger bounds!
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bounds
Let X1 , XP
2 , . . . Xn be independent Poisson trials
Pneach with probability pi ;
For X = n
i=1 pi .
i=1 Xi the expectation is E(X) =
Definition (Chernoff Bounds)
−E[X]δ 2
1
2
2
.
for 0 ≤ δ ≤ 1, P r(X ≤ (1 − δ)E[X]) ≤ e
–E[X]
»
δ
e
.
for δ > 0, P r(X > (1 + δ)E[X]) ≤ (1+δ)
1+δ
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bounds
Let X1 , XP
2 , . . . Xn be independent Poisson trials
Pneach with probability pi ;
For X = n
i=1 pi .
i=1 Xi the expectation is E(X) =
Definition (Chernoff Bounds)
−E[X]δ 2
1
2
2
.
for 0 ≤ δ ≤ 1, P r(X ≤ (1 − δ)E[X]) ≤ e
–E[X]
»
δ
e
.
for δ > 0, P r(X > (1 + δ)E[X]) ≤ (1+δ)
1+δ
What is the probability that we have more than (2/3)n one-bits at initialisation?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
29 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bounds
Let X1 , XP
2 , . . . Xn be independent Poisson trials
Pneach with probability pi ;
For X = n
i=1 pi .
i=1 Xi the expectation is E(X) =
Definition (Chernoff Bounds)
−E[X]δ 2
1
2
2
.
for 0 ≤ δ ≤ 1, P r(X ≤ (1 − δ)E[X]) ≤ e
–E[X]
»
δ
e
.
for δ > 0, P r(X > (1 + δ)E[X]) ≤ (1+δ)
1+δ
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,
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
29 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bounds
Let X1 , XP
2 , . . . Xn be independent Poisson trials
Pneach with probability pi ;
For X = n
i=1 pi .
i=1 Xi the expectation is E(X) =
Definition (Chernoff Bounds)
−E[X]δ 2
1
2
2
.
for 0 ≤ δ ≤ 1, P r(X ≤ (1 − δ)E[X]) ≤ e
–E[X]
»
δ
e
.
for δ > 0, P r(X > (1 + δ)E[X]) ≤ (1+δ)
1+δ
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:
„
«n/2
e1/3
P r[X > (2/3)n] ≤
= c−n/2
4/3
(4/3)
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bound Simple Application
Bitstring of length n = 100
P r(Xi ) = 1/2 and E(X) = np = 100/2 = 50.
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bound Simple Application
Bitstring of length n = 100
P r(Xi ) = 1/2 and E(X) = np = 100/2 = 50.
What is the probability to have at least 75 1-bits?
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
30 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
Chernoff Bound Simple Application
Bitstring of length n = 100
P r(Xi ) = 1/2 and E(X) = np = 100/2 = 50.
What is the probability to have at least 75 1-bits?
Markov: P r(X ≥ 75) ≤
50
75
=
2
3
„
Chernoff: P r(X ≥ (1 + 1/2)50) ≤
Truth: P r(X ≥ 75) =
P. S. Oliveto & X. Yao
(University of Birmingham)
P100 `100´
i=75
i
√
e
(3/2)3/2
«50
< 0.0045
2−100 < 0.000000282
Runtime Analysis of EAs
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Chernoff bounds
OneMax
P
OneMax (x)= n
i=1 x[i])
f(x)
n
2
1
n
1 2
P. S. Oliveto & X. Yao
(University of Birmingham)
Runtime Analysis of EAs
ones(x)
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
0
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
p0 =
Runtime Analysis of EAs
6
6
E(T0 ) =
6
6
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
0
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
p0 =
Runtime Analysis of EAs
6
6
E(T0 ) =
6
6
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
p0 =
Runtime Analysis of EAs
6
6
E(T0 ) =
6
6
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
0
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p0 =
6
6
E(T0 ) =
6
6
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
0
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p1 =
5
6
E(T1 ) =
6
5
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p2 =
4
6
E(T2 ) =
6
4
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T0 ) =
6
3
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
0
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
WCCI 2012
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p2 =
4
6
E(T2 ) =
6
4
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p3 =
3
6
E(T3 ) =
6
3
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p4 =
2
6
E(T4 ) =
6
2
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p4 =
2
6
E(T4 ) =
6
2
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p4 =
2
6
E(T4 ) =
6
2
p4 =
2
6
E(T4 ) =
6
2
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p4 =
2
6
E(T4 ) =
6
2
p5 =
1
6
E(T5 ) =
6
1
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
P. S. Oliveto & X. Yao
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
1
1
1
1
1
1
0
1
2
3
4
5
(University of Birmingham)
Pn
i=1 x[i])
Runtime Analysis of EAs
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p4 =
2
6
E(T4 ) =
6
2
p5 =
1
6
E(T5 ) =
6
1
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
0
1
2
3
4
5
0
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
0
1
0
1
2
3
4
5
1
0
1
0
1
1
0
1
2
3
4
5
1
1
1
0
1
1
0
1
2
3
4
5
1
1
1
1
1
1
0
1
2
3
4
5
Pn
i=1 x[i])
p0 =
6
6
E(T0 ) =
6
6
p1 =
5
6
E(T1 ) =
6
5
p2 =
4
6
E(T2 ) =
6
4
p3 =
3
6
E(T3 ) =
6
3
p4 =
2
6
E(T4 ) =
6
2
p5 =
1
6
E(T5 ) =
6
1
E(T ) = E(T0 ) + E(T1 ) + · · · + E(T5 ) = 1/p0 + 1/p1 + · · · + 1/p5 =
=
P. S. Oliveto & X. Yao
5
5
6
X
X
X
1
6
1
=
=6
= 6 · 2.45 = 14.7
p
i
i
i
i=0
i=0
i=1
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
32 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
P. S. Oliveto & X. Yao
0
(University of Birmingham)
0
0
0
Pn
i=1 x[i])
0
0
n
Runtime Analysis of EAs
: Generalisation
p0 =
n
n
E(T0 ) =
n
n
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
P. S. Oliveto & X. Yao
0
(University of Birmingham)
0
5
0
0
Pn
i=1 x[i])
0
0
n
Runtime Analysis of EAs
: Generalisation
p0 =
n
n
E(T0 ) =
n
n
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
P. S. Oliveto & X. Yao
0
(University of Birmingham)
1
0
0
Pn
i=1 x[i])
0
0
n
Runtime Analysis of EAs
: Generalisation
p0 =
n
n
E(T0 ) =
n
n
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
0
0
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
E(T0 ) =
n
n
0
1
0
0
0
0
n
p0 =
n
n
E(T0 ) =
n
n
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
0
0
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
(University of Birmingham)
Runtime Analysis of EAs
E(T0 ) =
E(T1 ) =
n
n
n
n−1
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
(University of Birmingham)
Runtime Analysis of EAs
E(T0 ) =
E(T1 ) =
n
n
n
n−1
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
(University of Birmingham)
Runtime Analysis of EAs
E(T0 ) =
n
n
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
0
n
p2 =
n−2
n
E(T2 ) =
n
n−2
(University of Birmingham)
Runtime Analysis of EAs
E(T0 ) =
n
n
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
1
n
p2 =
n−2
n
E(T2 ) =
n
n−2
(University of Birmingham)
Runtime Analysis of EAs
E(T0 ) =
n
n
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
1
1
1
1
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
1
n
p2 =
n−2
n
E(T2 ) =
n
n−2
0
(University of Birmingham)
1
1
1
1
1
n
Runtime Analysis of EAs
pn−1 =
1
n
E(T0 ) =
n
n
E(Tn−1 ) =
n
1
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
1
1
1
1
0
0
1
2
3
4
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
1
n
p2 =
n−2
n
E(T2 ) =
n
n−2
(University of Birmingham)
1
1
1
1
1
n
Runtime Analysis of EAs
pn−1 =
1
n
E(T0 ) =
n
n
E(Tn−1 ) =
n
1
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
1
1
1
1
0
1
2
3
P. S. Oliveto & X. Yao
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
1
n
p2 =
n−2
n
E(T2 ) =
n
n−2
1
(University of Birmingham)
1
1
1
1
1
n
Runtime Analysis of EAs
pn−1 =
1
n
E(T0 ) =
n
n
E(Tn−1 ) =
n
1
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
RLS for OneMax ( OneMax (x)=
1
0
0
0
0
0
0
1
2
3
0
0
1
0
0
1
2
3
0
0
1
0
0
1
2
3
1
1
1
1
2
3
Pn
i=1 x[i])
: Generalisation
0
1
0
0
0
0
n
p0 =
n
n
0
1
0
0
0
0
n
p1 =
n−1
n
E(T1 ) =
n
n−1
0
1
0
0
0
1
n
p2 =
n−2
n
E(T2 ) =
n
n−2
1
1
1
1
1
1
n
pn−1 =
1
n
E(T0 ) =
n
n
E(Tn−1 ) =
n
1
E(T ) = E(T0 ) + E(T1 ) + · · · + E(Tn−1 ) = 1/p1 + 1/p2 + · · · + 1/pn−1 =
=
n−1
X
i=0
P. S. Oliveto & X. Yao
n
n
X
X
1
n
1
=
=n
= n · H(n) = n log n + Θ(n) = O(n log n)
pi
i
i
i=1
i=1
(University of Birmingham)
Runtime Analysis of EAs
WCCI 2012
33 / 92
Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Coupon collector’s problem
Coupon collector’s problem
The Coupon collector’s problem
There are n types of coupons and at each trial one coupon is chosen at random. Each
coupon has the same probability of being extracted. The goal is to find the exact
number 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−1
X
1
i=0
pi+1
=
n−1
X
i=0
n−1
X 1
n
=n
=
n−i
i
i=0
= n(log n + Θ(1)) = n log n + O(n).
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Coupon collector’s problem
Coupon collector’s problem: Upper bound on time
What is the probability that the time to collect n coupons is greater than
n ln n + O(n)?
Theorem (Coupon collector upper bound on time)
Let T be the time for all the n coupons to be collected. Then
P r(T ≥ (1 + )n ln n) ≤ n−
Proof
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Coupon collector’s problem
Coupon collector’s problem: Upper bound on time
What is the probability that the time to collect n coupons is greater than
n ln n + O(n)?
Theorem (Coupon collector upper bound on time)
Let T be the time for all the n coupons to be collected. Then
P r(T ≥ (1 + )n ln n) ≤ n−
Proof
1
n
1
1− n
„
«t
1
1− n
Probability of choosing a given coupon
Probability of not choosing a given coupon
Probability 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
„
«t
1
n· 1− n
(Union Bound)
Hence, for t = cn ln n
`
P r(T ≥ cn ln n) ≤ n 1 − 1/n)cn ln n ≤ n · e−c ln n = n · n−c = n−c+1
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Coupon collector’s problem
Coupon collector’s problem: lower bound on time
What is the probability that the time to collect n coupons is less than n ln n + 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 > 0
P r(T < (1 − )(n − 1) ln n) ≤ exp(−n )
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Coupon collector’s problem
Coupon collector’s problem: lower bound on time
What is the probability that the time to collect n coupons is less than n ln n + 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 > 0
P r(T < (1 − )(n − 1) ln n) ≤ exp(−n )
Corollary
The expected time for RLS to optimise OneMax is Θ(n ln n). Furthermore,
P r(T ≥ (1 + )n ln n) ≤ n−
and
P r(T < (1 − )(n − 1) ln n) ≤ exp(−n )
What about the (1+1)-EA?
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AFL method for upper bounds
Artificial Fitness Levels [Droste et al., 2002]
Observation
Divide the search space |S| = 2n into m < 2n sets A1 , . . . Am such that:
Idea
1
2
3
Due to elitism, fitness is monotone increasing
∀i 6= j :
Ai ∩ Aj = ∅
Sm
n
i=0 Ai = {0, 1}
for all points a ∈ Ai and b ∈ Aj it happens that f (a) < f (b) if i < j.
requirement
P. S. Oliveto & X. Yao
Am contains only optimal search points.
(University of Birmingham)
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AFL method for upper bounds
Artificial Fitness Levels [Droste et al., 2002]
Observation
Divide the search space |S| = 2n into m < 2n sets A1 , . . . Am such that:
Idea
1
2
3
Due to elitism, fitness is monotone increasing
∀i 6= j :
Ai ∩ Aj = ∅
Sm
n
i=0 Ai = {0, 1}
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.
P
Expected time: E(T ) ≤ i p1
i
Very simple, yet often powerful method for upper bounds
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AFL method for upper bounds
Artificial Fitness Levels
D. Sudholt, Tutorial 2011
Let:
p(Ai ) be the probability that a random chosen point belongs to level Ai
si be the probability to leave level Ai for Aj with j > i
Then:
„
« „
« m−1
X
X 1
1
1
1
1
E(T ) ≤
p(Ai ) ·
+ ··· +
≤
+ ··· +
=
s
s
s
s
s
m−1
1
m−1
i
i=1 i
1≤i≤m−1
`
´
P
Inequality 1: Law of total probability E(T ) = i P r(F ) · E(T |F )
Inequality 2: Trivial!
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AFL method for upper bounds
(1+1)-EA for OneMax
Theorem
The expected runtime of the (1+1)-EA for OneMax is O(n ln n).
Proof
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AFL method for upper bounds
(1+1)-EA for OneMax
Theorem
The expected runtime of the (1+1)-EA for OneMax is O(n ln n).
Proof
The current solution is in level Ai if it has i zeroes (hence n − i ones)
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AFL method for upper bounds
(1+1)-EA for OneMax
Theorem
The expected runtime of the (1+1)-EA for OneMax is O(n ln n).
Proof
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 leave
the other bits unchanged
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AFL method for upper bounds
(1+1)-EA for OneMax
Theorem
The expected runtime of the (1+1)-EA for OneMax is O(n ln n).
Proof
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 leave
the other bits unchanged
„
«n−1
1
1
i
The probability is si ≥ i · n
1− n
≥ en
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AFL method for upper bounds
(1+1)-EA for OneMax
Theorem
The expected runtime of the (1+1)-EA for OneMax is O(n ln n).
Proof
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 leave
the other bits unchanged
„
«n−1
1
1
i
The probability is si ≥ i · n
1− n
≥ en
Hence,
P. S. Oliveto & X. Yao
1
si
≤
en
i
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
AFL method for upper bounds
(1+1)-EA for OneMax
Theorem
The expected runtime of the (1+1)-EA for OneMax is O(n ln n).
Proof
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 leave
the other bits unchanged
„
«n−1
1
1
i
The probability is si ≥ i · n
1− n
≥ en
Hence, s1 ≤ en
i
i
`
´
Then Artificial Fitness Levels :
E(T ) ≤
m−1
X
s−1
≤
i
i=1
m−1
n
X 1
X
en
≤e·n
≤ e · n · (ln n + 1) = O(n ln n)
i
i
i=1
i=1
Is the (1+1)-EA quicker than n ln n?
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AFL method for upper bounds
(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 ln n).
Proof Idea
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AFL method for upper bounds
(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 ln n).
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).
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AFL method for upper bounds
(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 ln n).
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 ln n steps one of the n/2 remaining
zero-bits does not flip.
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
AFL method for upper bounds
Lower bound for OneMax
Theorem (Droste,Jansen,Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof of 2.
1 − 1/n
P. S. Oliveto & X. Yao
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a given bit does not flip
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AFL method for upper bounds
Lower bound for OneMax
Theorem (Droste,Jansen,Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof of 2.
1 − 1/n
(1 − 1/n)t
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a given bit does not flip
a given bit does not flip in t steps
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AFL method for upper bounds
Lower bound for OneMax
Theorem (Droste,Jansen,Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof of 2.
1 − 1/n
(1 − 1/n)t
1 − (1 − 1/n)t
P. S. Oliveto & X. Yao
(University of Birmingham)
a given bit does not flip
a given bit does not flip in t steps
it flips at least once in t steps
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AFL method for upper bounds
Lower bound for OneMax
Theorem (Droste,Jansen,Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof of 2.
1 − 1/n
(1 − 1/n)t
1 − (1 − 1/n)t
(1 − (1 − 1/n)t )n/2
P. S. Oliveto & X. Yao
(University of Birmingham)
a given bit does not flip
a given bit does not flip in t steps
it flips at least once in t steps
n/2 bits flip at least once in t steps
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AFL method for upper bounds
Lower bound for OneMax
Theorem (Droste,Jansen,Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof of 2.
1 − 1/n
(1 − 1/n)t
1 − (1 − 1/n)t
(1 − (1 − 1/n)t )n/2
1 − [1 − (1 − 1/n)t ]n/2
P. S. Oliveto & X. Yao
(University of Birmingham)
a given bit does not flip
a given bit does not flip in t steps
it flips at least once in t steps
n/2 bits flip at least once in t steps
at least one of the n/2 bits does not flip in t steps
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AFL method for upper bounds
Lower bound for OneMax
Theorem (Droste,Jansen,Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof of 2.
1 − 1/n
(1 − 1/n)t
1 − (1 − 1/n)t
(1 − (1 − 1/n)t )n/2
1 − [1 − (1 − 1/n)t ]n/2
a given bit does not flip
a given bit does not flip in t steps
it flips at least once in t steps
n/2 bits flip at least once in t steps
at least one of the n/2 bits does not flip in t steps
Set
t = (n − 1) log n. Then:
1 − [1 − (1 − 1/n)t ]n/2 = 1 − [1 − (1 − 1/n)(n−1) log n ]n/2 ≥
≥ 1 − [1 − (1/e)log n ]n/2 = 1 − [1 − 1/n]n/2 =
= 1 − [1 − 1/n]n·1/2 ≥ 1 − (2e)−1/2 = c
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AFL method for upper bounds
Lower bound for OneMax (2)
Theorem (Droste, Jansen, Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof
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 log n steps one of the n/2 remaining
zero-bits does not flip.
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AFL method for upper bounds
Lower bound for OneMax (2)
Theorem (Droste, Jansen, Wegener, 2002)
The expected runtime of the (1+1)-EA for OneMax is Ω(n log n).
Proof
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 log n steps one of the n/2 remaining
zero-bits does not flip.
The Expected runtime is:
E[T ] =
∞
X
t · p(t) ≥ [(n − 1) log n] · p[t = (n − 1) log n] ≥
t=1
≥ [(n − 1) log n] · [(1/2) · (1 − (2e)−1/2 ) = Ω(n log n)
First inequality: law of total probability
The upper bound given by artificial fitness levels is indeed tight!
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AFL method for upper bounds
Pn Qi
Artificial Fitness Levels Exercises: LeadingOnes(x) = i=1 j=1 x[j]
Theorem
The expected runtime of RLS for LeadingOnes is O(n2 ).
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AFL method for upper bounds
Pn Qi
Artificial Fitness Levels Exercises: LeadingOnes(x) = i=1 j=1 x[j]
Theorem
The expected runtime of RLS for LeadingOnes is O(n2 ).
Proof
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
and s−1
=n
i
P
P
−1
2
E(T ) ≤ n−1
s
= n
i=1 n = O(n )
i=1 i
si =
P. S. Oliveto & X. Yao
1
n
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AFL method for upper bounds
Pn Qi
Artificial Fitness Levels Exercises: LeadingOnes(x) = i=1 j=1 x[j]
Theorem
The expected runtime of RLS for LeadingOnes is O(n2 ).
Proof
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
and s−1
=n
i
P
P
−1
2
E(T ) ≤ n−1
s
= n
i=1 n = O(n )
i=1 i
si =
1
n
Theorem
The expected runtime of the (1+1)-EA for LeadingOnes is O(n2 ).
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AFL method for upper bounds
Pn Qi
Artificial Fitness Levels Exercises: LeadingOnes(x) = i=1 j=1 x[j]
Theorem
The expected runtime of RLS for LeadingOnes is O(n2 ).
Proof
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
and s−1
=n
i
P
P
−1
2
E(T ) ≤ n−1
s
= n
i=1 n = O(n )
i=1 i
si =
1
n
Theorem
The expected runtime of the (1+1)-EA for LeadingOnes is O(n2 ).
Proof Left as Exercise.
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AFL method for upper bounds
Fitness Levels Advanced Exercises (Populations)
Theorem
The expected runtime of (1+λ)-EA for LeadingOnes is O(λn + n2 )
[Jansen et al., 2005].
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AFL method for upper bounds
Fitness Levels Advanced Exercises (Populations)
Theorem
The expected runtime of (1+λ)-EA for LeadingOnes is O(λn + n2 )
[Jansen et al., 2005].
Proof
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
„
«λ
1
si = 1 − 1 − en
≥ 1 − e−λ/(en)
1
2
si ≥ 1 −
λ
si ≥ 2en
1
e
Case 1: λ ≥ en
Case 2: λ < en
„„
« „
Pn 1
Pn
P
−1
E(T ) ≤ λ · n−1
s
≤
λ
+
i=1 c
i=1
i=1 i
„ „
««
n2
O λ· n+ λ
= O(λ · n + n2 )
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2en
λ
««
=
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AFL method for upper bounds
Fitness Levels Advanced Exercises (Populations)
Theorem
The expected runtime of the (µ+1)-EA for LeadingOnes is O(µ · n2 ).
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AFL method for upper bounds
Fitness Levels Advanced Exercises (Populations)
Theorem
The expected runtime of the (µ+1)-EA for LeadingOnes is O(µ · n2 ).
Proof Left as Exercise.
Theorem
The expected runtime of the (µ+1)-EA for OneMax is O(µ · n log n).
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AFL method for upper bounds
Fitness Levels Advanced Exercises (Populations)
Theorem
The expected runtime of the (µ+1)-EA for LeadingOnes is O(µ · n2 ).
Proof Left as Exercise.
Theorem
The expected runtime of the (µ+1)-EA for OneMax is O(µ · n log n).
Proof Left as Exercise.
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AFL method for parent populations
Artificial Fitness Levels for Populations
D. Sudholt, Tutorial 2011
Let:
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
Then:
E(T ) ≤
m−1
X „
i=1
P. S. Oliveto & X. Yao
(University of Birmingham)
1
+ To
si
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«
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
Let partition Ai contain search points with exactly i leading ones
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
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 of the best individual
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
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 of the best individual
We set χ(i) = n/ ln n
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
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 of the best individual
We set χ(i) = n/ ln n
Given
„ j copies
«n of the best individual another replica is created with probability
j
j
1
1
−
≥ 2eµ
µ
n
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
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 of the best individual
We set χ(i) = n/ ln n
Given
„ j copies
«n of the best individual another replica is created with probability
j
j
1
1
−
≥ 2eµ
µ
n
To ≤
P. S. Oliveto & X. Yao
Pn/ ln n
j=1
2eµ
j
= 2eµ ln n
(University of Birmingham)
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
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 of the best individual
We set χ(i) = n/ ln n
Given
„ j copies
«n of the best individual another replica is created with probability
j
j
1
1
−
≥ 2eµ
µ
n
To ≤
P. S. Oliveto & X. Yao
Pn/ ln n
j=1
1
si ≥
2
si ≥
2eµ
j
n/ ln n
µ
n/ ln n
µ
·
·
= 2eµ ln n
1
en
1
en
(University of Birmingham)
=
≥
1
eµ ln n
1
en
n
ln n
n
ln n
Case 1: µ >
Case 2: µ ≤
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AFL method for parent populations
Applications to (µ+1)-EA
Theorem
The expected runtime of (µ+1)-EA for LeadingOnes is O(µn log n + n2 )
[Witt, 2006].
Proof
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 of the best individual
We set χ(i) = n/ ln n
Given
„ j copies
«n of the best individual another replica is created with probability
j
j
1
1
−
≥ 2eµ
µ
n
To ≤
Pn/ ln n
j=1
1
si ≥
2
si ≥
2eµ
j
n/ ln n
µ
n/ ln n
µ
·
·
= 2eµ ln n
1
en
1
en
=
≥
1
eµ ln n
1
en
n
ln n
n
ln n
Case 1: µ >
Case 2: µ ≤
„
«
`
´
E(T ) ≤ i=1 (To + s−1
=
i=1 2eµ ln n + en + eµ ln n
i )≤
„
«
`
´
n · 2eµ ln n + en + eµ ln n
= O(nµ ln n + n2 )
Pn−1
P. S. Oliveto & X. Yao
(University of Birmingham)
Pn
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AFL method for parent populations
Populations Fitness Levels: Exercise
Theorem
The expected runtime of the (µ+1)-EA for OneMax is O(µn + n log n).
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AFL method for parent populations
Populations Fitness Levels: Exercise
Theorem
The expected runtime of the (µ+1)-EA for OneMax is O(µn + n log n).
Proof Left as Exercise.
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AFL for non-elitist EAs
Advanced: Fitness Levels for non-Elitist Populations [Lehre, 2011]
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AFL for lower bounds
Advanced: Fitness Levels for Lower Bounds [Sudholt, 2010]
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AFL for lower bounds
Artificial Fitness Levels: Conclusions
It’s a powerful general method to obtain (often) tight upper bounds on the
runtime of simple EAs;
For offspring populations tight bounds can often be achieved with the general
method;
For parent populations takeover times have to be introduced;
Recent methods have been presented to deal with non-elitism and for lower
bounds.
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
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
Question
How many steps does Peter need to reach his hotel?
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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
Question
How many steps does Peter need to reach his hotel?
n steps
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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 the
goal;
(
0, if Xt = 0,
d(Xt ) − d(Xt+1 ) =
1, if Xt ∈ {1, . . . , n}
Time
Then the expected time to reach the hotel (goal) is:
E(T ) =
P. S. Oliveto & X. Yao
(University of Birmingham)
n
maximum distance
=
=n
drif t
1
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Drift Analysis: Example 2
Friday night dinner at the restaurant.
Peter walks back from the restaurant to the hotel but had a few drinks.
The restaurant is n meters away from the hotel;
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.
Question
How many steps does Peter need to reach his hotel?
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Drift Analysis: Example 2
Friday night dinner at the restaurant.
Peter walks back from the restaurant to the hotel but had a few drinks.
The restaurant is n meters away from the hotel;
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.
Question
How many steps does Peter need to reach his hotel?
5n steps
Let us calculate this through drift analysis.
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Drift Analysis (2): Formalisation
Define the same distance function d(x) as before to measure the distance from
the 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 the
goal;
8
>
<0, if Xt = 0,
d(Xt ) − d(Xt+1 ) = 1, if Xt ∈ {1, . . . , n}with probability 0.6
>
:
−1, if Xt ∈ {1, . . . , n}with 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
Time
Then the expected time to reach the hotel (goal) is:
E(T ) =
P. S. Oliveto & X. Yao
(University of Birmingham)
maximum distance
n
=
= 5n
drif t
0.2
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Additive Drift Theorem
Additive Drift Theorem
Theorem (Additive Drift Theorem for Upper Bounds [He and Yao, 2001])
Let {Xt }t≥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 the
optimum be T := min{t ≥ 0 : d(Xt ) = 0}. If there exists δ > 0 such that at any time
step 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
then
E(T | X0 > 0) ≤
and
E(T ) ≤
P. S. Oliveto & X. Yao
(University of Birmingham)
d(X0 )
δ
E(d(X0 ))
.
δ
Runtime Analysis of EAs
(1)
(2)
(3)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Drift Analysis for Leading Ones
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2 )
Proof
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Additive Drift Theorem
Drift Analysis for Leading Ones
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2 )
Proof
1
Let g(Xt ) = i where i is the number of missing leading ones;
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Drift Analysis for Leading Ones
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2 )
Proof
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 solution
the new point will not be accepted;
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Drift Analysis for Leading Ones
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2 )
Proof
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 solution
the 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 and
the leading ones are remained unchanged:
E(∆+ (t)) =
n−i
X
`
k · (p(∆+ (t)) = k) ≥ 1 · 1/n · 1 − 1/n)n−1 ≥ 1/(en)
k=1
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Drift Analysis for Leading Ones
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2 )
Proof
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 solution
the 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 and
the leading ones are remained unchanged:
E(∆+ (t)) =
n−i
X
`
k · (p(∆+ (t)) = k) ≥ 1 · 1/n · 1 − 1/n)n−1 ≥ 1/(en)
k=1
4
Hence, E[∆(t)] = E(∆+ (t)) − E(∆− (t)) ≥ 1/(en) = δ
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Drift Analysis for Leading Ones
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is O(n2 )
Proof
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 solution
the 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 and
the leading ones are remained unchanged:
E(∆+ (t)) =
n−i
X
`
k · (p(∆+ (t)) = k) ≥ 1 · 1/n · 1 − 1/n)n−1 ≥ 1/(en)
k=1
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) ≤
P. S. Oliveto & X. Yao
(University of Birmingham)
g(X0 )
n
≤
= e · n2 = O(n2 )
δ
1/(en)
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Additive Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise LeadingOnes is O(n2 )
Proof
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Additive Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise LeadingOnes is O(n2 )
Proof Left as exercise.
Theorem
Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes is
O(λn)
Proof
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise LeadingOnes is O(n2 )
Proof Left as exercise.
Theorem
Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes is
O(λn)
Proof Left as exercise.
Theorem
Let λ < en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes is
O(n2 )
Proof
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise LeadingOnes is O(n2 )
Proof Left as exercise.
Theorem
Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes is
O(λn)
Proof Left as exercise.
Theorem
Let λ < en. Then the expected time for the (1+λ)-EA to optimise LeadingOnes is
O(n2 )
Proof Left as exercise.
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
(1,λ)-EA Analysis for LeadingOnes
Theorem
Let λ = n. Then the expected time for the (1,λ)-EA to optimise LeadingOnes is
O(n2 ) (Jansen and Neumann, Tutorial)
Proof
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
(1,λ)-EA Analysis for LeadingOnes
Theorem
Let λ = n. Then the expected time for the (1,λ)-EA to optimise LeadingOnes is
O(n2 ) (Jansen and Neumann, Tutorial)
Proof
Distance: let d(x) = n − i where i is the number of leading ones;
Drift:
E[d(Xt ) − d(Xt+1 )]
„
„
« «
„
„
« «
1 n
1 n n
≥1· 1− 1−
−n· 1− 1−
en
n
= c1 − n · cn
2 = Ω(1)
Hence,
E(generations) ≤
max
n
distance
=
= O(n)
drif t
Ω(1)
and,
E(T ) ≤ n · E(generations) = O(n2 )
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Additive Drift Theorem
Theorem (Additive Drift Theorem for Lower Bounds [He and Yao, 2004])
Let {Xt }t≥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 the
optimum be T := min{t ≥ 0 : d(Xt ) = 0}. If there exists δ > 0 such that at any time
step 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
then
E(T | X0 > 0) ≥
and
E(T ) ≥
P. S. Oliveto & X. Yao
(University of Birmingham)
d(X0 )
δ
E(d(X0 ))
.
δ
Runtime Analysis of EAs
(4)
(5)
(6)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2 ).
P. S. Oliveto & X. Yao
(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
1
Let the current solution have n − i leading ones (i.e. 1n−i 0∗).
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
1
Let the current solution have n − i leading ones (i.e. 1n−i 0∗).
2
We define the distance function as the number of missing leading ones, i.e.
g(X) = i.
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
1
Let the current solution have n − i leading ones (i.e. 1n−i 0∗).
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;
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
1
Let the current solution have n − i leading ones (i.e. 1n−i 0∗).
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 free
riders).
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
1
Let the current solution have n − i leading ones (i.e. 1n−i 0∗).
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 free
riders).
5
Such i − 1 bits are uniformely distributed at initialisation and still are!
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
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).
Proof
1
Let the current solution have n − i leading ones (i.e. 1n−i 0∗).
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 free
riders).
5
Such i − 1 bits are uniformely distributed at initialisation and still are!
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(University of Birmingham)
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Motivation Basic Probability Theory Evolutionary Algorithms Tail Inequalities Artificial Fitness Levels Drift Analysis Typical Run Investigations Conclusio
Additive Drift Theorem
Drift Theorem for LeadingOnes (lower bound)
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2 ).
The expected number of free riders is:
E[Y ] =
i−1
X
k · P r(Y = k) =
k=1
P. S. Oliveto & X. Yao
(University of Birmingham)
i−1
X
P r(Y ≥ k) =
k=1
Runtime Analysis of EAs
i−1
X
(1/2)k ≤ 1
k=1
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Additive Drift Theorem
Drift Theorem for LeadingOnes (lower bound)
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2 ).
The expected number of free riders is:
E[Y ] =
i−1
X
k · P r(Y = k) =
k=1
i−1
X
P r(Y ≥ k) =
k=1
i−1
X
(1/2)k ≤ 1
k=1
The negative drift is 0;
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Additive Drift Theorem
Drift Theorem for LeadingOnes (lower bound)
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2 ).
The expected number of free riders is:
E[Y ] =
i−1
X
k · P r(Y = k) =
k=1
i−1
X
P r(Y ≥ k) =
k=1
i−1
X
(1/2)k ≤ 1
k=1
The negative drift is 0;
Let p(A) be the probability that the first zero-bit flips into a one-bit.
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Additive Drift Theorem
Drift Theorem for LeadingOnes (lower bound)
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2 ).
The expected number of free riders is:
E[Y ] =
i−1
X
k · P r(Y = k) =
k=1
i−1
X
P r(Y ≥ k) =
k=1
i−1
X
(1/2)k ≤ 1
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
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Additive Drift Theorem
Drift Theorem for LeadingOnes (lower bound)
Theorem
The expected time for the (1+1)-EA to optimise LeadingOnes is Ω(n2 ).
The expected number of free riders is:
E[Y ] =
i−1
X
k · P r(Y = k) =
k=1
i−1
X
P r(Y ≥ k) =
k=1
i−1
X
(1/2)k ≤ 1
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, it
follows that E[d(X0 )] ≥ n − 2,
By applying the Drift Theorem we get
E(T ) ≥
P. S. Oliveto & X. Yao
(University of Birmingham)
E(d(X0 )
n−2
=
= Ω(n2 )
δ
2/n
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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;
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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;
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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 remain
unchanged:
E(∆+ (t)) = E[d(Xt ) − d(Xt+1 )] ≥ 1 ·
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„
«
i
i
1 n−1
1
1−
≥
≥
n
n
en
en
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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 remain
unchanged:
E(∆+ (t)) = E[d(Xt ) − d(Xt+1 )] ≥ 1 ·
4
„
«
i
i
1 n−1
1
1−
≥
≥
n
n
en
en
Hence, E[∆(t)] = E(∆+ (t)) − E(∆− (t)) ≥ 1/(en) = δ
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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 remain
unchanged:
E(∆+ (t)) = E[d(Xt ) − d(Xt+1 )] ≥ 1 ·
„
«
i
i
1 n−1
1
1−
≥
≥
n
n
en
en
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 )]
n/2
≤
= e/2 · n2 = O(n2 )
δ
1/(en)
We need a different distance function!
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Multiplicative Drift Theorem
Drift Analysis for OneMax
1
Let g(Xt ) = ln(i + 1) where i is the number of zeroes in the bitstring;
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Multiplicative Drift Theorem
Drift Analysis for OneMax
1
Let g(Xt ) = ln(i + 1) 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;
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Multiplicative Drift Theorem
Drift Analysis for OneMax
1
Let g(Xt ) = ln(i + 1) 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 remain
unchanged:
E(∆+ (t)) = E[d(Xt )−d(Xt+1 )] ≥ ln(i+1)·
P. S. Oliveto & X. Yao
(University of Birmingham)
„
«
i
1 n−1
ln(i + 1)
ln(2)
1−
≥
≥
n
n
en
en
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Multiplicative Drift Theorem
Drift Analysis for OneMax
1
Let g(Xt ) = ln(i + 1) 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 remain
unchanged:
E(∆+ (t)) = E[d(Xt )−d(Xt+1 )] ≥ ln(i+1)·
4
„
«
i
1 n−1
ln(i + 1)
ln(2)
1−
≥
≥
n
n
en
en
Hence, ∆(t) = E(∆+ (t)) − E(∆− (t)) ≥ ln(2)/(en) = δ
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Multiplicative Drift Theorem
Drift Analysis for OneMax
1
Let g(Xt ) = ln(i + 1) 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 remain
unchanged:
E(∆+ (t)) = E[d(Xt )−d(Xt+1 )] ≥ ln(i+1)·
„
«
i
1 n−1
ln(i + 1)
ln(2)
1−
≥
≥
n
n
en
en
4
Hence, ∆(t) = E(∆+ (t)) − E(∆− (t)) ≥ ln(2)/(en) = δ
5
The distance is d(X0 ) ≤ ln(n + 1)
The expected runtime is (i.e. Eq. (6)):
E(T | d(X0 ) > 0) ≤
ln(n + 1)
d(X0 )
≤
= O(n ln n)
δ
ln(2)/(en)
If the amount of progress depends on the distance from the optimum we need to use a
logarithmic distance!
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Multiplicative Drift Theorem
Multiplicative Drift Theorem
Theorem (Multiplicative Drift, [Doerr et al., 2010])
Let {Xt }t∈N0 be random variables describing a Markov process over a finite state
space S ⊆ R. Let T be the random variable that denotes the earliest point in time
t ∈ 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 ] ≤
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„
«
2
cmax
· ln 1 +
δ
cmin
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Multiplicative Drift Theorem
(1+1)-EA Analysis for OneMax
Theorem
The expected time for the (1+1)-EA to optimise OneMax is O(n ln n)
Proof
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Multiplicative Drift Theorem
(1+1)-EA Analysis for OneMax
Theorem
The expected time for the (1+1)-EA to optimise OneMax is O(n ln n)
Proof
Distance: let i be the number of zeroes;
„
«
„
«
1
1
i
= i · 1 − en
= Xt · 1 − en
E[Xt+1 |Xt = i] ≥ i − 1 · en
„
«
1
1
1
E[Xt − Xt+1 |Xt = i] ≤ Xt − Xt · 1 − en
= en
Xt (δ = en
)
1 = cmin ≤ Xt ≤ cmax = n
Hence,
E[T ] ≤
P. S. Oliveto & X. Yao
„
«
cmax
2
· ln 1 +
= 2en ln(1 + n) = O(n ln n)
δ
cmin
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Multiplicative Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise OneMax is O(n log n)
Proof
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Multiplicative Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise OneMax is O(n log n)
Proof Left as exercise.
Theorem
Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise OneMax is O(λn)
Proof
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Multiplicative Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise OneMax is O(n log n)
Proof Left as exercise.
Theorem
Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise OneMax is O(λn)
Proof Left as exercise.
Theorem
Let λ < en. Then the expected time for the (1+λ)-EA to optimise OneMax is
O(n log n)
Proof
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Multiplicative Drift Theorem
Exercises
Theorem
The expected time for RLS to optimise OneMax is O(n log n)
Proof Left as exercise.
Theorem
Let λ ≥ en. Then the expected time for the (1+λ)-EA to optimise OneMax is O(λn)
Proof Left as exercise.
Theorem
Let λ < en. Then the expected time for the (1+λ)-EA to optimise OneMax is
O(n log n)
Proof Left as exercise.
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Simplified Negative Drift Theorem
Drift Analysis: Example 3
Friday night dinner at the restaurant.
Peter walks back from the restaurant to the hotel but had too many drinks.
The restaurant is n meters away from the hotel;
Peter moves towards the hotel of 1 meter in each step with probability 0.4.
Peter moves away from the hotel of 1 meter in each step with probability 0.6.
Question
How many steps does Peter need to reach his hotel?
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Simplified Negative Drift Theorem
Drift Analysis: Example 3
Friday night dinner at the restaurant.
Peter walks back from the restaurant to the hotel but had too many drinks.
The restaurant is n meters away from the hotel;
Peter moves towards the hotel of 1 meter in each step with probability 0.4.
Peter moves away from the hotel of 1 meter in each step with probability 0.6.
Question
How many steps does Peter need to reach his hotel?
at least 2cn steps with overwhelming probability (exponential time)
We need Negative-Drift Analysis.
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Simplified Negative Drift Theorem
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
[a, b]
T∗
(Interval in State space S)
:= min{t ≥ 0 : Xt ≤ a | X0 ≥ b}
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(Time to reach a)
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Simplified Negative Drift Theorem
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
[a, b]
T∗
(Drift)
(Interval in State space S)
:= min{t ≥ 0 : Xt ≤ a | X0 ≥ b}
(Time to reach a)
Theorem (Simplified Negative-Drift Theorem, [Oliveto and Witt, 2011])
Suppose there exist three constants δ,,r such that for all t ≥ 0:
1
E(∆t (i)) ≥ for a < i < b,
2
Prob(∆t (i) = −j) ≤
Then
1
(1+δ)j−r
for i > a and j ≥ 1.
∗
Prob(T ∗ ≤ 2c
(b−a)
) = 2−Ω(b−a)
.
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Simplified Negative Drift Theorem
Simplified Drift Theorem
1
E(∆t (i)) ≥ for a < i < b,
2
Prob(∆t (i) = −j) ≤
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1
(1+δ)j−r
for i > a and j ≥ 1.
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Simplified Negative Drift Theorem
Negative-Drift Analysis: Example (3)
Define the same distance function d(x) = x, x ∈ {0, . . . , n} (metres from the
hotel) (b=n-1, a=1).
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Simplified Negative Drift Theorem
Negative-Drift Analysis: Example (3)
Define the same distance function d(x) = x, x ∈ {0, . . . , n} (metres from the
hotel) (b=n-1, a=1).
Estimate the increase in distance from the goal (negative drift);
8
>
<0, if Xt = 0,
d(Xt ) − d(Xt+1 ) = 1, if Xt ∈ {1, . . . , n}with probability 0.6
>
:
−1, if Xt ∈ {1, . . . , n}with probability 0.4
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Simplified Negative Drift Theorem
Negative-Drift Analysis: Example (3)
Define the same distance function d(x) = x, x ∈ {0, . . . , n} (metres from the
hotel) (b=n-1, a=1).
Estimate the increase in distance from the goal (negative drift);
8
>
<0, if Xt = 0,
d(Xt ) − d(Xt+1 ) = 1, if Xt ∈ {1, . . . , n}with probability 0.6
>
:
−1, if Xt ∈ {1, . . . , n}with probability 0.4
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
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Simplified Negative Drift Theorem
Negative-Drift Analysis: Example (3)
Define the same distance function d(x) = x, x ∈ {0, . . . , n} (metres from the
hotel) (b=n-1, a=1).
Estimate the increase in distance from the goal (negative drift);
8
>
<0, if Xt = 0,
d(Xt ) − d(Xt+1 ) = 1, if Xt ∈ {1, . . . , n}with probability 0.6
>
:
−1, if Xt ∈ {1, . . . , n}with probability 0.4
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):
(
P r(∆t (i) = −j) =
P. S. Oliveto & X. Yao
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0 < (1/2)j−1 , if j > 1,
0.6 < (1/2)0 = 1, if j = 1
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Simplified Negative Drift Theorem
Negative-Drift Analysis: Example (3)
Define the same distance function d(x) = x, x ∈ {0, . . . , n} (metres from the
hotel) (b=n-1, a=1).
Estimate the increase in distance from the goal (negative drift);
8
>
<0, if Xt = 0,
d(Xt ) − d(Xt+1 ) = 1, if Xt ∈ {1, . . . , n}with probability 0.6
>
:
−1, if Xt ∈ {1, . . . , n}with probability 0.4
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):
(
P r(∆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:
P r(T ≤ 2c(b−a) ) = P r(T ≤ 2c(n−2) ) = 2−Ω(n)
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Simplified Negative Drift Theorem
Needle in a Haystack
Theorem
Let η > 0 be constant. Then there is a constant c > 0 such that with probability
1 − 2−Ω(n) the (1+1)-EA on Needle creates only search points with at most
n/2 + ηn ones in 2cn steps.
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Simplified Negative Drift Theorem
Needle in a Haystack
Theorem
Let η > 0 be constant. Then there is a constant c > 0 such that with probability
1 − 2−Ω(n) the (1+1)-EA on Needle creates only search points with at most
n/2 + ηn ones in 2cn steps.
Proof Idea
1
By Chernoff bounds the probability that the initial bit string has less than
n/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 theorem
hold.
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Simplified Negative Drift Theorem
Needle in a Haystack (2)
Proof of Condition 1
Condition 1 holds if E(∆(i)) ≥ for some constant > 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 expected
number of zero-bits flipped into one-bits is i/n and the viceversa is (n − i)/n.
3
Hence,
E(∆(i)) =
n−i
i
n − 2i
−
=
≥ 2γ = n
n
n
and Condition 1 holds.
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Simplified Negative Drift Theorem
Needle in a Haystack (3)
Proof of Condition 2
Condition 2 is: P rob(∆(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, the
probability of a drift of length j can be bounded as follows:
P rob(∆(i) ≤ −j) ≤
“n” „ 1 «j
j
n
≤
1
≤
j!
„ «j−1
1
2
This proves Condition 2 by setting δ = r = 1.
Applying the Simplified Drift Theorem
∗
For a constant c∗ > 0 the global optimum is found in 2c (b−a) = 2cn steps with
−Ω(b−a)
−Ω(n)
probability at most 2
=2
.
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Simplified Negative Drift Theorem
Exercise: Trap Functions

Trap(x) =
n+1
OneMax(x)
if x = 0n
otherwise.
f(x)
n+1
n
3
2
1
0123
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n
ones(x)
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Simplified Negative Drift Theorem
Exercise: Trap Functions

Trap(x) =
n+1
OneMax(x)
if x = 0n
otherwise.
f(x)
n+1
n
3
2
1
0123
n
ones(x)
Theorem
With overwhelming probability at least 1 − 2−Ω(n) the (1+1)-EA requires 2Ω(n) steps
to optimise Trap .
Proof Left as exercise.
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Simplified Negative Drift Theorem
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];
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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;
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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 even
happens 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;
P
The failure probability is 1 − ki=1 pi ;
4
5
The runtime is lower than the sum of each phase time.
Generally Chernoff bounds are used to obtain the failure probabilities!
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(1+1)-EA for Trap functions
Theorem
With overwhelming probability at least 1 − e−Ω(n) the (1+1)-EA requires nΩ(n) steps
to optimise Trap .
The deceptive trap function is the following:

n+1
Trap(x) =
OneMax(x)
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if x = 0n
otherwise.
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(1+1)-EA for Trap functions
Theorem
With overwhelming probability at least 1 − e−Ω(n) the (1+1)-EA requires nΩ(n) steps
to optimise Trap .
The deceptive trap function is the following:

n+1
Trap(x) =
OneMax(x)
if x = 0n
otherwise.
Proof Idea
We use the Typical run investigations method and split the analysis into the following
three phases:
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(1+1)-EA for Trap functions
Theorem
With overwhelming probability at least 1 − e−Ω(n) the (1+1)-EA requires nΩ(n) steps
to optimise Trap .
The deceptive trap function is the following:

n+1
Trap(x) =
OneMax(x)
if x = 0n
otherwise.
Proof Idea
We use the Typical run investigations method and split the analysis into the following
three phases:
1
This phase ends when at least n/3 one-bits are in the current solution.
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(1+1)-EA for Trap functions
Theorem
With overwhelming probability at least 1 − e−Ω(n) the (1+1)-EA requires nΩ(n) steps
to optimise Trap .
The deceptive trap function is the following:

n+1
Trap(x) =
OneMax(x)
if x = 0n
otherwise.
Proof Idea
We use the Typical run investigations method and split the analysis into the following
three 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.
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(1+1)-EA for Trap functions
Theorem
With overwhelming probability at least 1 − e−Ω(n) the (1+1)-EA requires nΩ(n) steps
to optimise Trap .
The deceptive trap function is the following:

n+1
Trap(x) =
OneMax(x)
if x = 0n
otherwise.
Proof Idea
We use the Typical run investigations method and split the analysis into the following
three 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.
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(1+1)-EA for Trap functions (2)
We need to show that the runtime required to complete the three phases is nΩ(n) and
that 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 current
solution.
Proof
1
By using Chernoff bounds we calculate the probability that the algorithm is
initialised with less than n/3 one-bits.
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(1+1)-EA for Trap functions (2)
We need to show that the runtime required to complete the three phases is nΩ(n) and
that 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 current
solution.
Proof
1
By using Chernoff bounds we calculate the probability that the algorithm is
initialised with less than n/3 one-bits.
2
n/2 expected one-bits after initialisation (i.e. E(X) = n · p = n/2).
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(1+1)-EA for Trap functions (2)
We need to show that the runtime required to complete the three phases is nΩ(n) and
that 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 current
solution.
Proof
1
By using Chernoff bounds we calculate the probability that the algorithm is
initialised 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)δ
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2
/2
= e−(n/2)·(1/9)·(1/2) = e−n/36 = e−Ω(n)
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(1+1)-EA for Trap functions (3)
Phase 2 This phase ends when the 1n bitstring is the current solution.
Proof
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(1+1)-EA for Trap functions (3)
Phase 2 This phase ends when the 1n bitstring is the current solution.
Proof
1
The probability of reaching the 0n bitstring in one step is less than n−n/3 ;
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(1+1)-EA for Trap functions (3)
Phase 2 This phase ends when the 1n bitstring is the current solution.
Proof
1
The probability of reaching the 0n bitstring in one step is less than n−n/3 ;
2
in expected time c · n log n the algorithm climbs up the OneMax .
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(1+1)-EA for Trap functions (3)
Phase 2 This phase ends when the 1n bitstring is the current solution.
Proof
1
The probability of reaching the 0n bitstring in one step is less than n−n/3 ;
2
in expected time c · n log n the algorithm climbs up the OneMax .
3
By Markov’s inequality,
`
´
P X ≥ e · c · n log n ≤
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c · n log n
= 1/e
e · c · n log n
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(1+1)-EA for Trap functions (3)
Phase 2 This phase ends when the 1n bitstring is the current solution.
Proof
1
The probability of reaching the 0n bitstring in one step is less than n−n/3 ;
2
in expected time c · n log n the algorithm climbs up the OneMax .
3
By Markov’s inequality,
`
´
P X ≥ e · c · n log n ≤
4
c · n log n
= 1/e
e · c · n log n
`
´
By applying it iteratively: P X ≥ e · c · n2 log n ≤ e−n
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(1+1)-EA for Trap functions (3)
Phase 2 This phase ends when the 1n bitstring is the current solution.
Proof
1
The probability of reaching the 0n bitstring in one step is less than n−n/3 ;
2
in expected time c · n log n the algorithm climbs up the OneMax .
3
By Markov’s inequality,
`
´
P X ≥ e · c · n log n ≤
4
5
c · n log n
= 1/e
e · c · n log n
`
´
By applying it iteratively: P X ≥ e · c · n2 log n ≤ e−n
The probability that in c · n2 log n steps n/3 precise bits flip is
c · n2 log n · n−n/3 = n−Ω(n) . (union bound)
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(1+1)-EA for Trap functions (4)
Phase 3
This phase ends when the global optimum is found.
Proof
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(1+1)-EA for Trap functions (4)
Phase 3
This phase ends when the global optimum is found.
Proof
1
The expected time to reach the global optimum is nn because all the bits have to
be flipped at the same time;
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(1+1)-EA for Trap functions (4)
Phase 3
This phase ends when the global optimum is found.
Proof
1
The expected time to reach the global optimum is nn because all the bits have to
be 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−n
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n
·1/4·1/2
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= e−n
n
·1/8
= e−Ω(n)
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(1+1)-EA for Trap functions (4)
Phase 3
This phase ends when the global optimum is found.
Proof
1
The expected time to reach the global optimum is nn because all the bits have to
be 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−n
n
·1/4·1/2
= e−n
n
·1/8
= e−Ω(n)
Summing up
1
the algorithm requires exponential runtime 1 + c · n2 log n + (1/2)nn = nΩ(n)
2
With probability at least 1 − e−Ω(n) .
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ExpectationTrap

ExpectationTrap(x) =
n − 1/2
OneMax(x)
if x = 0n
otherwise.
Theorem
`
´
The expected time for the (1+1)-EA to optimise ExpectationTrap is Ω (n/2)n .
[Jansen and Neumann, Tutorial]
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ExpectationTrap

ExpectationTrap(x) =
n − 1/2
OneMax(x)
if x = 0n
otherwise.
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)
Proof
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ExpectationTrap

ExpectationTrap(x) =
n − 1/2
OneMax(x)
if x = 0n
otherwise.
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)
Proof
E(T ) ≥ E(T |X0 = 0n ) = nn · 2−n = (n/2)n
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(law
of
total
probability)
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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]
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State-of-the-art
State of the Art in Computational Complexity of RSHs
OneMax
Linear Functions
Max. Matching
Sorting
SS Shortest Path
(1+1) EA
(1+λ) EA
(µ+1) EA
1-ANT
(µ+1) IA
(1+1) EA
cGA
Max. Clique
(1+1) EA
(1+1) EA
(1+1) EA
MO (1+1) EA
(1+1) EA
(1+λ) EA
1-ANT
(1+1) EA
(rand. planar)
Eulerian Cycle
Partition
(16n+1) RLS
(1+1) EA
(1+1) EA
Vertex Cover
(1+1) EA
Set Cover
(1+1) EA
SEMO
(1+1) EA
MST
Intersection of
p ≥ 3 matroids
UIO/FSM conf.
(1+1) EA
O(n log n)
O(λn + n log n)
O(µn + n log n)
O(n2 ) w.h.p.
O(µn + n log n)
Θ(n log n)
Θ(n2+ε ), ε > 0 const.
eΩ(n) , PRAS
Θ(n2 log n)
O(n3 log(nwmax ))
O(n3 )
Θ(m2 log(nwmax ))
O(n log(nwmax ))
O(mn log(nwmax ))
Θ(n5 )
Θ(n5/3 )
Θ(m2 log m)
4/3 approx., competitive avg.
eΩ(n) , arb. bad approx.
eΩ(n) , arb. bad approx.
Pol. O(log n)-approx.
1/p-approximation in
O(|E|p+2 log(|E|wmax ))
eΩ(n)
See [Oliveto et al., 2007] for an overview.
P. K. Lehre, 2008
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Further reading
Further Reading
[Auger and Doerr, 2011, Neumann and Witt, 2010]
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Further reading
Coming up Conference
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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.
Bäck, 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.
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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 Artificial
Intelligence.
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.
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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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