Mutation Rates of the (1+1)-EA on Pseudo-Boolean Functions of - PowerPoint PPT Presentation

Mutation Rates of the (1+1)-EA on Pseudo-Boolean Functions of Bounded Epistasis Andrew M Sutton, Darrell Whitley, Adele Howe Andrew M Sutton, Darrell Whitley, Adele Howe 0 / 19

Introduction Want to maximize f : { 0 , 1 } n → R + (1+1)-EA( ρ ) Choose x ∈ { 0 , 1 } n uniformly at random while stopping criteria not met do y ← x Flip each bit of y independently with prob. ρ if f ( y ) ≥ f ( x ) then x ← y Question: how do we choose the mutation rate ρ ? Andrew M Sutton, Darrell Whitley, Adele Howe 1 / 19

Motivation Early experimental evidence suggested 0 . 001 ≤ ρ ≤ 0 . 01 – De Jong (1975), Grefenstette (1986), Schaffer (1989) Droste et al. (1998) : linear functions ρ = 1 /n = ⇒ O ( n log n ) expected convergence Jansen and Wegener (2000) : PathToJump ρ = 1 /n = ⇒ superpolynomial runtime w.h.p. ρ = log n = ⇒ polytime convergence n Doerr et al. (2010) : monotone functions changing ρ by a constant factor = ⇒ exponential performance gap Andrew M Sutton, Darrell Whitley, Adele Howe 2 / 19

Expected offspring fitness under rate ρ We will study the class of functions f : { 0 , 1 } n → R + whose epistasis is bounded by a constant k . What is the “best” mutation rate? Maximizes probability of improvement (difficult to know in general) Maximizes the expected fitness of the offspring Let x ∈ { 0 , 1 } n . We define M x ( ρ ) - the expected fitness of the offspring of x under rate ρ . Andrew M Sutton, Darrell Whitley, Adele Howe 3 / 19

Expected offspring fitness under rate ρ How does one compute M x ( ρ ) ? Appeal to a basis function decomposition of the fitness function Standard and alternative bases e x e x f ϕ i f ϕ j e y e y ϕ k e x e x e y ( x ) = δ xy � � f ( x ) = v y e y ( x ) f ( x ) = a i ϕ i ( x ) y i Decomposition provides information about the relationship between the fitness function and the mutation operator . Andrew M Sutton, Darrell Whitley, Adele Howe 4 / 19

Appealing to Walsh decomposition We can write the fitness function as a Walsh polynomial � f ( x ) = w i ψ i ( x ) i Expected fitness of a point drawn uniformly at random at Hamming distance r from x (Sutton et al. 2011) � S r x = γ i,r w i ψ i ( x ) ( ∗ ) i From this we can obtain the expected fitness under rate ρ n � n � � ρ r (1 − ρ ) n − r S r M x ( ρ ) = x r r =0 M x ( ρ ) = A 0 + A 1 ρ + A 2 ρ 2 + · · · + A n ρ n where each A m is a linear combination of terms from ( ∗ ) . Andrew M Sutton, Darrell Whitley, Adele Howe 5 / 19

The expected offspring fitness polynomial For any pseudo-Boolean function, M x ( ρ ) is a degree at most n polynomial in ρ . opt M x ( ρ ) 0 ρ 1 Maximum in the inverval [0 , 1] gives the best ρ in terms of maximizing expected fitness. We are interested in exploring some properties of this polynomial. Andrew M Sutton, Darrell Whitley, Adele Howe 6 / 19

Degeneracy Let ρ ⋆ ∈ arg max M x ( ρ ) . ρ ∈ [0 , 1] ρ ⋆ � = 0 : there exists a mutation rate that produces an expected improvement over f ( x ) . ρ ⋆ = 0 : no mutation rate can produce an expected improvement over f ( x ) . Suppose we insist on flipping ℓ > 0 bits in expectation. Then ρ = ℓ/n . � ℓ � n � 1 − ℓ � f ( x ) ≤ M x < f ( x ) n n � ℓ � e − ℓ f ( x ) ≤ M x < f ( x ) . n We can conclude ρ = 1 /n minimizes expected loss in fitness. In this case, expected fitness of offspring is bounded below by f ( x ) e . Andrew M Sutton, Darrell Whitley, Adele Howe 7 / 19

Linear functions Linear functions Expressed as a sum of linear terms (epistasis is k = 1 ) Walsh coefficients of order higher than 1 vanish M x ( ρ ) = A 0 + A 1 ρ + A 2 ρ 2 + · · · + A n ρ n where A m is a linear combination of terms from Walsh series expansion of f . Proposition If f is a linear function, then  f ( x ) if m = 0 ; � ¯   � A m = 2 f − f ( x ) if m = 1 ;  0 otherwise.  Andrew M Sutton, Darrell Whitley, Adele Howe 8 / 19

Linear functions � ¯ � When f is linear, M x ( ρ ) = f ( x ) + 2 f − f ( x ) ρ f ( x ) < ¯ f ( x ) = ¯ f ( x ) > ¯ f f f M x ( ρ ) 0 ρ 1 0 ρ 1 0 ρ 1 Degenerate when f ( x ) > ¯ f , in this case 1 /n maximizes expected offspring s.t. ℓ > 0 bits flipping in expectation This illustrates a problem with using expectation: consider when f ( x ) = ¯ f . Andrew M Sutton, Darrell Whitley, Adele Howe 9 / 19

Epistatically bounded functions f : { 0 , 1 } n → R + where the epistasis is bounded by k = O (1) . Max- k -Sat Boolean formula over a set V of n variables and m clauses consisting of exactly k literals m � ( ℓ i, 1 ∨ ℓ i, 2 ∨ · · · ∨ ℓ i,k ) , where ℓ i,j ∈ { v, ¬ v : v ∈ V } i =1 f : { 0 , 1 } n → { 0 , . . . , m } counts clauses satisfied under x . NK-landscapes n f ( x ) = 1 � � , where g j : { 0 , 1 } K +1 → [0 , 1] x [ j ] , x [ b ( j ) 1 ] , . . . , x [ b ( j ) � g j K ] n j =1 Andrew M Sutton, Darrell Whitley, Adele Howe 10 / 19

Epistatically bounded functions Proposition When f is epistatically bounded by k , A m = 0 for m > k . Thus for any k -bounded pseudo-Boolean function, M x ( ρ ) is a degree- k polynomial in ρ . E.g., for Max- 2 -Sat , the mutation polynomial is quadratic . Furthermore, A m depends only on S r x for r ≤ m . Corollary If S r x < f ( x ) for all 0 < r ≤ k , then M x (0) is maximal. Andrew M Sutton, Darrell Whitley, Adele Howe 11 / 19

Numerical results Practically speaking, does solving M x ( ρ ) for the best mutation rate provide any insight? d dρ M x ( ρ ) = A 1 + 2 A 2 ρ + 3 A 3 ρ 2 + · · · + nA n ρ n − 1 , d 2 dρ 2 M x ( ρ ) = 2 A 2 + 6 A 3 ρ + 12 A 4 ρ 2 + · · · + n ( n − 1) A n ρ n − 2 . Numerically finding ρ ⋆ Find the stationary points of M x ( ρ ) by numerically solving for the d real roots of dρ M x ( ρ ) . Second derivative test for concavity ρ ⋆ is maximum of this set union { 0 , 1 } Andrew M Sutton, Darrell Whitley, Adele Howe 12 / 19

Numerical results Unrestricted NK-landscape, n = 100 , k = 2 0.500 optimal 1 /n 0 . 001 0.100 0.020 ρ 0.005 0.001 1 2 5 10 20 50 100 500 generation Andrew M Sutton, Darrell Whitley, Adele Howe 13 / 19

Numerical results Unrestricted NK-landscape, n = 100 , k = 2 optimal 1 /n 0 . 001 0.65 0.60 fitness 0.55 0.50 1 2 5 10 20 50 100 500 generation Andrew M Sutton, Darrell Whitley, Adele Howe 14 / 19

Numerical results Max- 3 -Sat , n = 100 0.500 optimal 1 /n 0 . 001 0.100 0.020 ρ 0.005 0.001 1 2 5 10 20 50 100 500 generation Andrew M Sutton, Darrell Whitley, Adele Howe 15 / 19

Numerical results Max- 3 -Sat , n = 100 420 optimal 1 /n 0 . 001 410 400 fitness 390 380 1 2 5 10 20 50 100 500 generation Andrew M Sutton, Darrell Whitley, Adele Howe 16 / 19

Numerical results (high epistasis) NK-landscape, N=10, K=9 0.75 0.500 optimal optimal standard standard hardwired hardwired 0.70 0.100 0.65 fitness rho 0.020 0.60 0.005 0.55 0.001 0.50 1 2 5 10 20 50 100 500 1 2 5 10 20 50 100 500 generation generation Andrew M Sutton, Darrell Whitley, Adele Howe 17 / 19

Research directions Analyzing w i for certain specific problems (e.g., Max- k -Sat ). Provide more precise statements about the expected fitness in general. Working with higher moments of the random variable distribution Connection to runtime analysis... might be possible if we can discover bounds for the higher moments of the distribution Generalization to broader problem classes (Fourier decomposition) Andrew M Sutton, Darrell Whitley, Adele Howe 18 / 19

Conclusion As long as epistasis is bounded by k , it is possible to efficiently compute the expected fitness of a mutation for any rate. it is possible to efficiently find the rate that results in the highest possible expected fitness. For strings with fitness higher than expectation in spheres up to radius k , 1 /n yields maximal expected fitness of the offspring while imposing the constraint that ℓ > 0 bits are flipped in expectation. Andrew M Sutton, Darrell Whitley, Adele Howe 19 / 19