Crossover Can Provably Speed Up Evolutionary Computation Benjamin - - PowerPoint PPT Presentation

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Crossover Can Provably Speed Up Evolutionary Computation

Benjamin Doerr, Edda Happ, Christian Klein January 29, 2008

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State-of-the-Art

Praxis

Bio-inspired-ness: Use both mutation and crossover! Practise shows that often crossover improves runtime.

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State-of-the-Art

Praxis

Bio-inspired-ness: Use both mutation and crossover! Practise shows that often crossover improves runtime.

...and Theory

Little proof that crossover is useful. Most theoretical analyses: only mutation.a

aE.g.: Pietro S. Oliveto, J. He, X. Yao. Evolutionary Algorithms and the

Vertex Cover Problem. In Proceedings of the IEEE Congress on Evolutionary Computation (CEC’07), 2007.

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State-of-the-Art

Pseudo-boolean jump Function

jump function jm : {0, 1}n → R jm(x1, . . . , xn) =      m + xi if xi ≤ n − m m + n if xi = n n − xi else

• T. Jansen and I. Wegener. On the analysis of evolutionary algorithms - a proof that

crossover really can help. ESA 1999, Algorithmica 2002

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State-of-the-Art

Pseudo-boolean jump Function

jump function jm : {0, 1}n → R jm(x1, . . . , xn) =      m + xi if xi ≤ n − m m + n if xi = n n − xi else

Result

Mutation only: Expected optimization time Θ(nm).

• T. Jansen and I. Wegener. On the analysis of evolutionary algorithms - a proof that

crossover really can help. ESA 1999, Algorithmica 2002

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State-of-the-Art

Pseudo-boolean jump Function

jump function jm : {0, 1}n → R jm(x1, . . . , xn) =      m + xi if xi ≤ n − m m + n if xi = n n − xi else

Result

Mutation only: Expected optimization time Θ(nm). Mutation and crossover: Expected optimization time O(n2 log n).

• T. Jansen and I. Wegener. On the analysis of evolutionary algorithms - a proof that

crossover really can help. ESA 1999, Algorithmica 2002

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State-of-the-Art

Simplified Ising Problem

Given a tree T = (V , E). Assign xi ∈ {−1, +1} to the vertices maximizing

• (vi,vj)∈E

xixj.

• D. Sudholt. Crossover is provably essential for the ising model on trees. GECCO 2005

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State-of-the-Art

Simplified Ising Problem

Given a tree T = (V , E). Assign xi ∈ {−1, +1} to the vertices maximizing

• (vi,vj)∈E

xixj. ◮ Find a monochromatic coloring!

• D. Sudholt. Crossover is provably essential for the ising model on trees. GECCO 2005

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State-of-the-Art

Simplified Ising Problem

Given a tree T = (V , E). Assign xi ∈ {−1, +1} to the vertices maximizing

• (vi,vj)∈E

xixj. ◮ Find a monochromatic coloring!

Result

Only mutation: Optimization time exponential.

• D. Sudholt. Crossover is provably essential for the ising model on trees. GECCO 2005

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State-of-the-Art

Simplified Ising Problem

Given a tree T = (V , E). Assign xi ∈ {−1, +1} to the vertices maximizing

• (vi,vj)∈E

xixj. ◮ Find a monochromatic coloring!

Result

Only mutation: Optimization time exponential.

• D. Sudholt. Crossover is provably essential for the ising model on trees. GECCO 2005

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State-of-the-Art

Simplified Ising Problem

Given a tree T = (V , E). Assign xi ∈ {−1, +1} to the vertices maximizing

• (vi,vj)∈E

xixj. ◮ Find a monochromatic coloring!

Result

Only mutation: Optimization time exponential. Mutation, fitness sharing and “1”-point crossover: O(n3)

• ptimization time.
• D. Sudholt. Crossover is provably essential for the ising model on trees. GECCO 2005

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State-of-the-Art

Simplified Ising Problem

Given a tree T = (V , E). Assign xi ∈ {−1, +1} to the vertices maximizing

• (vi,vj)∈E

xixj. ◮ Find a monochromatic coloring!

Result

Only mutation: Optimization time exponential. Mutation, fitness sharing and “1”-point crossover: O(n3)

• ptimization time.
• D. Sudholt. Crossover is provably essential for the ising model on trees. GECCO 2005

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State-of-the-Art

Summary Existing Results

Some proofs that crossover helps

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State-of-the-Art

Summary Existing Results

Some proofs that crossover helps Artificial problems only

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State-of-the-Art

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

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State-of-the-Art

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation

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State-of-the-Art

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem.

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem. Mutation only: Expected optimization time Θ(n4)

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem. Mutation only: Expected optimization time Θ(n4) Mutation and crossover: Exp. opt. time O(n3.5√log n)

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem. Mutation only: Expected optimization time Θ(n4) Mutation and crossover: Exp. opt. time O(n3.5√log n)

Remainder of the talk

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem. Mutation only: Expected optimization time Θ(n4) Mutation and crossover: Exp. opt. time O(n3.5√log n)

Remainder of the talk

All Pairs Shortest Path problem

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem. Mutation only: Expected optimization time Θ(n4) Mutation and crossover: Exp. opt. time O(n3.5√log n)

Remainder of the talk

All Pairs Shortest Path problem A (µ + 1) evolutionary algorithm for the APSP

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State-of-the-Art, our Results

Summary Existing Results

Some proofs that crossover helps Artificial problems only Some additional critique

1 − o(1) fraction mutation need shared fitness

Our Result

Rigorous analysis of the All Pairs Shortest Path problem. Mutation only: Expected optimization time Θ(n4) Mutation and crossover: Exp. opt. time O(n3.5√log n)

Remainder of the talk

All Pairs Shortest Path problem A (µ + 1) evolutionary algorithm for the APSP Some proof ideas.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v. In total: µ = n(n − 1) non-trivial SPs.

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The All Pairs Shortest Path (APSP) Problem

Given

Directed graph G = (V , E). edge weights w : E → Z ∪ {∞} No negative cycles!

Problem

For all pairs (u, v) ∈ V × V , find a shortest path (SP) from u to v. In total: µ = n(n − 1) non-trivial SPs.

Classical Results (|V | = n, |E| = m)

Floyd-Warshall algorithm: O(n3). Johnson’s algorithm: O(n2 log n + nm).

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A (µ + 1) Evolutionary Algorithm for the APSP

Individuals and Fitness Function

Individuals are simple paths: I = (e1, . . . , ek) with ei ∈ E. Fitness of individual I: Length of path f (I) =

k

• i=1

w(ei)

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A (µ + 1) Evolutionary Algorithm for the APSP

Individuals and Fitness Function

Individuals are simple paths: I = (e1, . . . , ek) with ei ∈ E. Fitness of individual I: Length of path f (I) =

k

• i=1

w(ei)

Population and Selection

Initial population: I := {(e) | e ∈ E}. At most one SP for each vertex pair (diversity mechanism).

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A (µ + 1) Evolutionary Algorithm for the APSP

Individuals and Fitness Function

Individuals are simple paths: I = (e1, . . . , ek) with ei ∈ E. Fitness of individual I: Length of path f (I) =

k

• i=1

w(ei)

Population and Selection

Initial population: I := {(e) | e ∈ E}. At most one SP for each vertex pair (diversity mechanism).

◮ At most µ = n(n − 1) individuals.

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A (µ + 1) Evolutionary Algorithm for the APSP

Individuals and Fitness Function

Individuals are simple paths: I = (e1, . . . , ek) with ei ∈ E. Fitness of individual I: Length of path f (I) =

k

• i=1

w(ei)

Population and Selection

Initial population: I := {(e) | e ∈ E}. At most one SP for each vertex pair (diversity mechanism).

◮ At most µ = n(n − 1) individuals.

Selection: Replace an existing path I from u to v by a newly generated one I ′ if f (I ′) ≤ f (I).

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A (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E}

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A (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E} repeat forever

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A (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E} repeat generate I ′ either via crossover forever

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A (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E} repeat generate I ′ either via crossover

• r via mutation

forever

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A (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E} repeat generate I ′ either via crossover

• r via mutation

if there is an I ∈ I with same endpoints as I ′ and f (I ′) ≤ f (I) then replace I by I ′ in I forever

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Mutation Operator

Single Mutation of an Individual I

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Mutation Operator

Single Mutation of an Individual I

Pick an edge e u.a.r. from all edges incident with an endvertex of I

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Mutation Operator

Single Mutation of an Individual I

Pick an edge e u.a.r. from all edges incident with an endvertex of I

Add e to I, or

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Mutation Operator

Single Mutation of an Individual I

Pick an edge e u.a.r. from all edges incident with an endvertex of I

Add e to I, or delete e from I, if e is already on the path I.

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Mutation Operator

Single Mutation of an Individual I

Pick an edge e u.a.r. from all edges incident with an endvertex of I

Add e to I, or delete e from I, if e is already on the path I.

Mutation Operator

Pick I u.a.r. from I. Pick s at random according to Pois(λ = 1). Sequentially perform s + 1 single mutations.

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Crossover Operator

Crossover Operator 1

Pick I1, I2 u.a.r. from I and concatenate.

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Crossover Operator

Crossover Operator 1

Pick I1, I2 u.a.r. from I and concatenate.

Crossover Operator 2

Pick I1, I2 u.a.r. from I. Pick i u.a.r from [0..|I1|]. Concatenate the first i edges from I1 and I2.

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Crossover Operator

Crossover Operator 1

Pick I1, I2 u.a.r. from I and concatenate.

Crossover Operator 2

Pick I1, I2 u.a.r. from I. Pick i u.a.r from [0..|I1|]. Concatenate the first i edges from I1 and I2.

Crossover Operator 3

Pick I1, I2 u.a.r. from I. Pick i ∈ [0..|I1|], j ∈ [0..|I2|] u.a.r. Concatenate first i edges from I1 and last j edges from I2.

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Crossover Operator

Crossover Operator 1

Pick I1, I2 u.a.r. from I and concatenate.

Crossover Operator 2

Pick I1, I2 u.a.r. from I. Pick i u.a.r from [0..|I1|]. Concatenate the first i edges from I1 and I2.

Crossover Operator 3

Pick I1, I2 u.a.r. from I. Pick i ∈ [0..|I1|], j ∈ [0..|I2|] u.a.r. Concatenate first i edges from I1 and last j edges from I2.

Fitness:

If the new individual is not a path, it shall never enter the population!

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Summary: The (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E}

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Summary: The (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E} repeat either do crossover or do mutation if f (I ′) ≤ f (I) then replace I by I ′ in I forever

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Summary: The (µ + 1) Evolutionary Algorithm for the APSP

Algorithm

I := {(e) | e ∈ E} repeat either do crossover or do mutation

Mutation: Add/delete Pois(λ = 1) random edges to random individual. Crossover: Concatenate parts/all from two random individuals.

if f (I ′) ≤ f (I) then replace I by I ′ in I forever

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else.

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)).

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)). Finding I means going from a leaf to I (“trail in auxiliary graph”).

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)). Finding I means going from a leaf to I (“trail in auxiliary graph”).

◮ Fixed trail of length ℓ: Pr[go trail in ≤ 10n4 steps] ≤

1 5ℓ .

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)). Finding I means going from a leaf to I (“trail in auxiliary graph”).

◮ Fixed trail of length ℓ: Pr[go trail in ≤ 10n4 steps] ≤

1 5ℓ .

◮ At most 4ℓ trails of length ℓ.

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)). Finding I means going from a leaf to I (“trail in auxiliary graph”).

◮ Fixed trail of length ℓ: Pr[go trail in ≤ 10n4 steps] ≤

1 5ℓ .

◮ At most 4ℓ trails of length ℓ.

Summing over all ℓ ≥ n − 2:

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)). Finding I means going from a leaf to I (“trail in auxiliary graph”).

◮ Fixed trail of length ℓ: Pr[go trail in ≤ 10n4 steps] ≤

1 5ℓ .

◮ At most 4ℓ trails of length ℓ.

Summing over all ℓ ≥ n − 2:

◮ Pr[Find I in ≤ 10n4 steps] ≤ ∞

ℓ=n−2 4ℓ 5ℓ = 5( 4 5)n−2 = unlikely.

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Proof Outline

Only Mutation: Expected Optimization Time Ω(n4)

Consider Kn with w(vi, vj) =

• 1

if i + 1 = j, n else. Claim: One needs Ω(n4) time to find I = ((v1, v2), . . . , (vn−1, vn)). Finding I means going from a leaf to I (“trail in auxiliary graph”).

◮ Fixed trail of length ℓ: Pr[go trail in ≤ 10n4 steps] ≤

1 5ℓ .

◮ At most 4ℓ trails of length ℓ.

Summing over all ℓ ≥ n − 2:

◮ Pr[Find I in ≤ 10n4 steps] ≤ ∞

ℓ=n−2 4ℓ 5ℓ = 5( 4 5)n−2 = unlikely.

⇒ Ω(n4) optimization time

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Proof Outline

Why are we Faster with Crossover?

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Proof Outline

Why are we Faster with Crossover?

Plan: If all SPs having ≤ k edges are in the population, then all SPs having ≤ 3

2k edges can be found in O(n4 log n k

) crossover steps [independent of the crossover operator used].

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Proof Outline

Why are we Faster with Crossover?

Plan: If all SPs having ≤ k edges are in the population, then all SPs having ≤ 3

2k edges can be found in O(n4 log n k

) crossover steps [independent of the crossover operator used]. First O(n3.5√log n) steps: Mutation finds all SPs having <

• n log(n) edges.

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Proof Outline

Why are we Faster with Crossover?

Plan: If all SPs having ≤ k edges are in the population, then all SPs having ≤ 3

2k edges can be found in O(n4 log n k

) crossover steps [independent of the crossover operator used]. First O(n3.5√log n) steps: Mutation finds all SPs having <

• n log(n) edges.

Next O(n3.5√log n) steps: Crossover finds all shorstest paths.

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Proof Outline

Why are we Faster with Crossover?

Plan: If all SPs having ≤ k edges are in the population, then all SPs having ≤ 3

2k edges can be found in O(n4 log n k

) crossover steps [independent of the crossover operator used]. First O(n3.5√log n) steps: Mutation finds all SPs having <

• n log(n) edges.

Next O(n3.5√log n) steps: Crossover finds all shorstest paths. In total: O(n3.5√log n) optimization time.

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Proof Outline

“Plan” becomes “Theorem”:

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Proof Outline

“Plan” becomes “Theorem”:

Plan: If for all vertices u, v connected via a SP having at most k edges a SP connecting them is in the population, then O(n4 log n

k

) crossover steps suffice to find SPs connecting all vertices u, v such that there is a SP from u to v having at most (3/2)k edges.

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Proof Outline

“Plan” becomes “Theorem”:

Plan: If for all vertices u, v connected via a SP having at most k edges a SP connecting them is in the population, then O(n4 log n

k

) crossover steps suffice to find SPs connecting all vertices u, v such that there is a SP from u to v having at most (3/2)k edges.

Crossover 1 (concatenate two individuals): True!

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Proof Outline

“Plan” becomes “Theorem”:

Plan: If for all vertices u, v connected via a SP having at most k edges a SP connecting them is in the population, then O(n4 log n

k

) crossover steps suffice to find SPs connecting all vertices u, v such that there is a SP from u to v having at most (3/2)k edges.

Crossover 1 (concatenate two individuals): True! Crossover 2 (concatenate initial segment plus indiviual): True if fitness prefers solution with fewer edges in case of equal lengths.

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Proof Outline

“Plan” becomes “Theorem”:

Plan: If for all vertices u, v connected via a SP having at most k edges a SP connecting them is in the population, then O(n4 log n

k

) crossover steps suffice to find SPs connecting all vertices u, v such that there is a SP from u to v having at most (3/2)k edges.

Crossover 1 (concatenate two individuals): True! Crossover 2 (concatenate initial segment plus indiviual): True if fitness prefers solution with fewer edges in case of equal lengths. Crossover 3 (concatenate initial plus terminal segment): True if shortest paths are unique.

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Proof Outline

“Plan” now is “Theorem”:

Theorem: If for all vertices u, v connected via a SP having at most k edges a SP connecting them is in the population, then O(n4 log n

k

) crossover steps suffice to find SPs connecting all vertices u, v such that there is a SP from u to v having at most (3/2)k edges.

Crossover 1 (concatenate two individuals): True! Crossover 2 (concatenate initial segment plus indiviual): True if fitness prefers solution with fewer edges in case of equal lengths. Crossover 3 (concatenate initial plus terminal segment): True if shortest paths are unique.

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Proof Outline

“Plan” now is “Theorem”:

Theorem: If for all vertices u, v connected via a SP having at most k edges a SP connecting them is in the population, then O(n4 log n

k

) crossover steps suffice to find SPs connecting all vertices u, v such that there is a SP from u to v having at most (3/2)k edges.

Crossover 1 (concatenate two individuals): True! Crossover 2 (concatenate initial segment plus indiviual): True if fitness prefers solution with fewer edges in case of equal lengths. Crossover 3 (concatenate initial plus terminal segment): True if shortest paths are unique.

Do we need the Extras?

Yes, theorem is wrong without! a

aThanks to an unknown referee for finding this error in a previous version :-)!

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover.

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover. Our result: A natural (µ + 1)-EA for the APSP problem needs

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover. Our result: A natural (µ + 1)-EA for the APSP problem needs

Ω(n4) time using mutation only,

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover. Our result: A natural (µ + 1)-EA for the APSP problem needs

Ω(n4) time using mutation only, O(n3.5√log n) time using mutation and suitable crossover.

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover. Our result: A natural (µ + 1)-EA for the APSP problem needs

Ω(n4) time using mutation only, O(n3.5√log n) time using mutation and suitable crossover.

Crossover operators seem to be less robust.

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover. Our result: A natural (µ + 1)-EA for the APSP problem needs

Ω(n4) time using mutation only, O(n3.5√log n) time using mutation and suitable crossover.

Crossover operators seem to be less robust.

Open Problem:

Other problems with such gaps. Larger (superpoly?) gaps?

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Summary and Conclusion

Summary

So far: Only artifical examples show the use of crossover. Our result: A natural (µ + 1)-EA for the APSP problem needs

Ω(n4) time using mutation only, O(n3.5√log n) time using mutation and suitable crossover.

Crossover operators seem to be less robust.

Open Problem:

Other problems with such gaps. Larger (superpoly?) gaps?

Thanks!