PROBABILISTIC ANALYSIS OF THE (1 + 1)-EVOLUTIONARY ALGORITHM

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1 /59 PROBABILISTIC ANALYSIS OF THE (1 + 1)-EVOLUTIONARY ALGORITHM Hsien-Kuei Hwang (joint with Alois Panholzer, Nicolas Rolin, Tsung-Hsi Tsai, Wei-Mei Chen) October 13, 2017

2 /59 MIT: EVOLUTONARY COMPUTATION

3 3/59 WE ARE ALWAYS SEARCHING & RESEARCHING Life Algorithms Searching Searching Life = Algorithms?

4 4/59 OPTIMIZATION PROBLEMS EVERYWHERE Conflicting criteria Impasse Intractable Life = NP-Hard? Dilemmas, Quandaries Mess, Obstacles Puzzles

5 5/59 SEARCH ALGORITHMS Backtracking Branch-and-bound Greedy Dynamic programming Simulated annealing Evolutionary algorithms Ant colony optimization Particle swarm Tabu search GRASP... Meta-heuristics

6 6/59 EVOLUTIONARY ALGORITHMS

7 EVOLUTIONARY ALGORITHM The use of Darwinian principles for automated problem solving originated in the 1950s. Darwin s theory of evolution: survival of the fittest stochastic evolution on computers cultivating problem solutions instead of calculating them randomized search heuristics generate and test (or trial and error) useful for global optimization, if the problem is too complex to be handled by an exact method or no exact method is available 7/59 Pioneers: John Holland, Lawrence J. Fogel, Ingo Rechenberg,...

8 8/59 EVOLUTIONARY ALGORITHMS (EAs) Anne-Wil Harzing s s/w Publish or Perish Most popular: Multiobjective optimization problems

9 ELITISM IN MULTIOBJECTIVE EVOLUTIONARY ALGORITHM 9/59 Our motivation: from maxima (skylines) to elites to EA

10 10/59 COMPONTENTS OF EAs Representation Coding of solutions Initialization Parent selection Evaluation: Fitness function Initialize Population Randomly vary individuals Evaluate fitness Apply selection Survivor selection Offspring Reproduction Genetic operators Termination condition Stop? yes Output no

11 11/59 TYPICAL PROGRESS OF AN EA Initialization Halfway Termination

12 PROS AND CONS OF EAs Disadvantages Large convergence time Difficult adjustment of parameters Heuristic principle No guarantee of global max Advantages Reasonably good solutions quickly Suitable for complex search spaces Easy to parallelize 12/59 Scalable to higher dimensional problems

13 13/59 DIFFICULTY OF ANALYSIS OF EA A typical EA comprises several ingredients coding of solution population of individuals selection for reproduction operations for breeding new individuals fitness function to evaluate the new individual... Mathematical description of the dynamics of the algorithms or the asymptotics of the complexity = challenging

14 14/59 Droste et al. (2002): Theory is far behind experimental knowledge... rigorous research is hard to find. Applications Theory Algorithms

15 15/59 SIMPLEST VERSION: 1 PARENT, 1 CHILD, AND MUTATION ONLY Algorithm (1 + 1)-EA 1 Choose an initial string x {0, 1} n uniformly at random 2 Repeat until a terminating condition (mutation) Create y by flipping each bit of x independently with probability p Replace x by y iff f (y) f (x) f : fitness (or objective) function

16 ANALYSIS OF (1 + 1)-EA UNDER ONEMAX Known results for ONEMAX f (x) = x x n X n := # steps used by the (1 + 1)-EA to reach the optimum state f (x) = n when the mutation rate is 1 n Bäck (1992): transition probabilities Mühlenbein (1992): E(X n ) n log n Droste et al. (1998, 2002): E(X n ) n log n ONEMAX function Linear functions w i x i Doerr et al. lower bound Jagerskupper upper bound (2010) (1 o(1))en log(n) (2011) 2.02en log(n) Sudholt lower bound Doerr et al. upper bound 2010 en log(n) 2n log log(n) (2010) 1.39en log(n) Doerr et al. (2011) en log(n) Θ(n) Witt (2013) upper bound en log(n) + O(n) 16/59 Approaches used: Markov chain, martingale, coupon collection,...

17 17/59 KNOWN BOUNDS FOR E(X n ) UNDER ONEMAX ONEMAX Droste et al. (2002) en(log n + γ) Doerr et al. (2011) en log n n Our result = e(n ) log n c 1n + O(1) c 1 = Lehre & Witt (2014) en log n n O(log n) Doerr et al. (2011) Sudholt (2010) Doerr et al. (2010) Droste et al. (2002) en log n Θ(n) en log n 2n log log n (1 o(1))en log n 0.196n log n

18 18/59 GARNIER, KALLEL & SCHOENAUER (1999) Rigorous hitting times for binary mutations Strongest results obtained so far but proof incomplete (probabilistic arguments) E(X n ) = en log n + c 1 n + o(n), where c X n en log n c d 1 log Exp(1) (double-exponential) their results had remained obscure in the EA-literature

19 19/59 OUR RESULTS ( ) log n E(X n ) = en log n + c 1 n + 1e log n + c O n c 1 = e ( ( log 2 γ φ 1 )) , where γ is Euler s constant, z ( 1 φ 1 (z) := 0 S 1 (t) 1 ) dt, t S 1 (z) := z l (1 z)j (l j). l! j! l 1 0 j<l Indeed E(X n ) n k 0 c k log n + c k n k

20 LIMIT GUMBEL DISTRIBUTION ( ) Xn P en log n + log 2 φ 1( 1) x e e x 2 φ 1 ( 1 ) /59 left: n = right: e x e x

21 21/59 OUR APPROACH RECURRENCE ASYMPTOTICS

22 THE RANDOM VARIABLE X n,m X n := 0 m n ( ) n p m q n m X n,m m X n,m := # steps used by the (1 + 1)-EA to reach f (x) = n when starting from f (x) = n m Let Q n,m (t) := E(t Xn,m ). Then Q n,0 (t) = 1 and t λ n,n m,l Q n,m l (t) Q n,m (t) = 1 1 l m ( 1 1 l m λ n,n m,l ) t 22/59 for 1 m n, where (P(m 1 s m + l 1 s)) ( λ n,m,l := 1 1 ) n ( )( ) m n m (n 1) l 2j n j j + l 0 j min{m,n m l}

23 23/59 λ n,m,l := P(m 1 s m + l 1 s) 0 j min{m,n m l} ( ) ( ) j ( m ) m j ( ) ( ) j+l ( n m ) n m j l j n n j + l n n }{{}}{{} λ n,m,l 1 l m 1 λ n,m,2 λ n,m,m... optimum state X n,m n m 1s λ n,m,1 X n,m 1 n m + 1 1s X n,m 2 n m + 2 1s... X n,0 n 1s

24 24/59 Q: How to solve this recurrence? t λ n,n m,l Q n,m l (t) 1 l m Q n,m (t) = ( 1 1 ) λ n,n m,l t 1 l m m = O(1)

25 25/59 SIMPLEST CASE: m = 1 λ n,n 1,1 = 1 n m = 1 (n 1 1s = n 1s) ( ( 1 n) 1 n 1 ( 1 = Geometric 1 1 ) ) n 1 n n Q n,1 (t) = ( 1 n 1 1 n 1 n) t ( ( ) ) n 1 1 n 1. t n = X n,1 en P d Exp(1) ( ) Xn,1 en x 1 e x

26 26/59 EXPONENTIAL DISTRIBUTION

27 27/59 BY INDUCTION m = O(1) X n,m en P d Exp(1) + + Exp(m) ) en x ( 1 e x) m ( Xn,m E(X n,m ) eh m n and V(X n,m ) e 2 H (2) m n 2 m = 2, 4, 6, 8 & n = 5,..., 50 An LLT also holds

28 28/59 m = 2; n = 5,..., 50

29 29/59 SKETCH OF PROOF λ n,n m,l = = ( 1 1 ) n n ( m l 0 j min{n m,m l} )e 1 n l ( 1 + O ( m j + l ( m l n(l + 1) + l n )( n m j )) m = O(1) = j = 1 is dominant ) (n 1) l 2j

30 29/59 SKETCH OF PROOF λ n,n m,l = = ( 1 1 ) n n ( m l 0 j min{n m,m l} )e 1 n l ( 1 + O ( m j + l ( m l n(l + 1) + l n )( n m j )) m = O(1) = j = 1 is dominant Q n,m (t) = t 1 = Q n,m (t) 1 l m ( λ n,n m,l Q n,m l (t) 1 1 l m m t en 1 ( 1 m en λ n,n m,l ) ) (n 1) l 2j t ) t Q n,m 1 (t)

31 30/59 SKETCH OF PROOF: BY INDUCTION Q n,m (t) 1 r m r t en 1 ( 1 en) r t

32 30/59 SKETCH OF PROOF: BY INDUCTION Q n,m (t) 1 r m = Q n,m ( e s/(en) ) 1 r m r t en 1 ( 1 en) r t 1 1 s r Fails when m = 1 r m Exp(r)

33 30/59 SKETCH OF PROOF: BY INDUCTION Q n,m (t) 1 r m = Q n,m ( e s/(en) ) 1 r m r t en 1 ( 1 en) r t 1 1 s r Fails when m = 1 r m Exp(r) Let Y m := 1 r m Exp(r). Then as m E ( e (Ym Hm)iθ) = 1 r m e iθ r 1 iθ r e iθ r 1 iθ r 1 r = e γiθ Γ(1 iθ)

34 30/59 SKETCH OF PROOF: BY INDUCTION Q n,m (t) 1 r m = Q n,m ( e s/(en) ) 1 r m r t en 1 ( 1 en) r t 1 1 s r Fails when m = 1 r m Exp(r) Let Y m := 1 r m Exp(r). Then as m E ( e (Ym Hm)iθ) = 1 r m e iθ r 1 iθ r e iθ r 1 iθ r 1 r = e γiθ Γ(1 iθ) = P (Y m log m x) e e x (x R)

35 31/59 m

36 EXPECTED VALUES E(X n,m ) µ n,m = µ n,m := E(X n,m ) = Q n,m(1) (µ n,0 = 0) 1 + λ n,n m,l µ n,m l 1 l m 1 l m λ n,n m,l µ n,m l (1 m n) = E(X n ) = 2 n 0 m n ( ) n µ n,m m 32/59 Let e n := ( ) 1 1 n+1 n+1 and µ n,m := en µ n n+1,m ( ) λ n,m,l µ n,m µ 1 n,m l = n 1 l m λ n,m,l := λ n+1,n+1 m,l e n = 0 j min{n+1 m,m l} ( n + 1 m j )( ) m j + l n l 2j

37 33/59 µ n,1 = 1 & 1 l m λ n,m,l ( µ n,m µ n,m l) = 1 n µ n,2 = 3 n2 +n 1 2 n 2 +2 n 1 µ n,3 = 22 n6 +40 n 5 19 n 4 42 n n n 6 (2 n 2 +2 n 1)(6 n n 3 7 n 2 9 n+6)

38 33/59 µ n,1 = 1 & 1 l m λ n,m,l ( µ n,m µ n,m l) = 1 n µ n,2 = 3 n2 +n 1 2 n 2 +2 n 1 µ n,3 = 22 n6 +40 n 5 19 n 4 42 n n n 6 µ n,4 = µ n,5 = (2 n 2 +2 n 1)(6 n n 3 7 n 2 9 n+6) 600 n n n n n n n n n n n n ( 2 n n 1 ) ( 6 n n 3 7 n 2 9 n + 6 ) ( 24 n n 5 48 n n n n 60 ) n n n n n n n n n n n n n n n n n n n n ( 2 n n 1 ) ( 6 n n 3 7 n 2 9 n + 6 ) ( 24 n n 5 48 n n n n 60 ) (120 n n n n n n n n + 840)

39 34/59 µ n,1 = 1 ASYMPTOTICS OF µ n,m ( ) 1 l m λ n,m,l µ n,m µ n,m l = 1 n µ n,2 = 3 2 n n n n n 5 + µ n,3 = n n n n 4 + µ n,4 = n n n n 4 + µ n,5 = n n n n 4 + µ n,6 = n n n n 4 +

40 ASYMPTOTICS OF µ n,m ( ) 1 l m λ n,m,l µ n,m µ n,m l = 1 n 34/59 µ n,1 = 1 µ n,2 = 3 2 n n n n n 5 + µ n,3 = n n n n 4 + µ n,4 = n n n n 4 + µ n,5 = n n n n 4 + µ n,6 = n n n n 4 + H m = 1 j m 1 { j = 1, 3 2, 11 6, 25 12, , 49 } 20,...

41 35/59 HEURISTICS An Ansatz approximation: µ n,m k 0 d 0 (m) = H m (m 0) d k (m) n k d 1 (m) = H m m (m 1) d 2 (m) = 2 3 H m m m2 (m 2) d 3 (m) = 1 2 H m m m m3 (m 2) d 4 (m) = 5 18 H m m m m m4 (m 4)

42 35/59 HEURISTICS An Ansatz approximation: µ n,m k 0 d 0 (m) = H m (m 0) d k (m) n k d 1 (m) = H m m (m 1) d 2 (m) = 2 3 H m m m2 (m 2) d 3 (m) = 1 2 H m m m m3 (m 2) d 4 (m) = 5 18 H m m m m m4 (m 4) Complication: d k (m) holds for m 2 k 2

43 35/59 HEURISTICS An Ansatz approximation: µ n,m k 0 d 0 (m) = H m (m 0) d k (m) n k d 1 (m) = H m m (m 1) d 2 (m) = 2 3 H m m m2 (m 2) d 3 (m) = 1 2 H m m m m3 (m 2) d 4 (m) = 5 18 H m m m m m4 (m 4) Complication: d k (m) holds for m 2 k 2 General pattern: µ n,m ( k 0 n k b k H m + ) 0 j k ϖ k,jm j

44 35/59 HEURISTICS An Ansatz approximation: µ n,m k 0 d 0 (m) = H m (m 0) d k (m) n k d 1 (m) = H m m (m 1) d 2 (m) = 2 3 H m m m2 (m 2) d 3 (m) = 1 2 H m m m m3 (m 2) d 4 (m) = 5 18 H m m m m m4 (m 4) Complication: d k (m) holds for m 2 k 2 General pattern: µ n,m ( k 0 n k b k H m + ) 0 j k ϖ k,jm j α := m n = µ n,m H m + φ(α) for 1 m n

45 6/59 A MORE GENERAL ANSATZ µ n,m H m + φ 1 (α) + b 1H m+φ 2 (α) n + b 2H m+φ 3 (α) n 2 + µ n,m µ n,m H m µ n,m (H m + φ 1 (α)) µ n,m ( H m + φ 1 (α) + Hm+φ ) 2(α) n

46 φ 1 (z) := φ 2 (z) = 1 2 S r (z) := l 1 z 0 z 0 z l l! ( 1 S 1 (t) 1 ) dt t ( S2 (t)s 1 (t) 2S 1 (t) 3 0 j<l r (1 z)j (l j) j! S 0(t) S 1 (t) 2 1 2S 1 (t) 1 2t ) dt t (analytic in z 1) 37/59 S 1 (x) & S 2 (x) φ 1 (x) φ 2 (x)

47 38/59 A GENERATING FUNCTION APPROACH? λ n,m,l = 0 j min{n+1 m,m l} f n (z) := m 1 ( n + 1 m j µ n,mz m )( ) m j + l n l 2j ( ) λ n,m,m l µ n,m µ 1 n,l = n 0 l<m = 1 ( ( ) ) 1 2πi 1 t z t + 1 n t ( 1 + t z ( )) t + 1 n n ( 1 + t ) ( n+1 ( )) z t + 1 n f n n t ( ) dt = 1 + t n z n(1 z)

48 9/59 A HEURISTIC 1 z f n (w)φ n (z, w) dw = 2πi n(1 z) ( ( Φ n (z, w) := 1 + τ ) ( ) ) n+1 1 n 1 τ z τ + 1 n τ ( 1 + τ n z ( )) τ + 1 n z(w 1) n(w z) 2 + Assume f n (w) φ(w). dτ dw z φ(w)(w 1) 2πin (w z) 2 dw = z n (φ(z) (1 z)φ (z)) = z 1 n 1 z = RHS Then (φ(0) = 0) φ(z) (1 z)φ (z) = 1 1 z = µ n,m H m. = φ(z) = 1 1 z log 1 1 z.

49 HOW TO GUESS φ 1 (α)? Assume µ n,m H m + φ(α) (α := m n ) H m H m l = l l(l 1) + + m 2m 2 ( m ) ( ) m l φ φ = φ (α) lm ( ) l 2 n n + O m 2 40/59 1 n = Matched asymptotics 1 l m ( ) λ n,m,l µ n,m µ n,m l ( l λ n,m,l m + φ (α) l ) n 1 l m 1 ( ) 1 n α + φ (α) 1 l m lλ n,m,l,

50 41/59 1 l m lλ n,m,l = j 1 1 n 1 ( ) 1 n α + φ (α) lλ n,m,l 1 l m j 1 = S 1 (α) ( n + 1 m j ) n j (1 α) j l αl j! l! l>j Then we see that φ must satisfy j<l m ( ) m l n l l φ (x) = 1 S 1 (x) 1 x = x = φ = φ 1 The justification relies on a careful error analysis

51 TOOLS NEEDED Lemma 1. Asymptotics of A n,m := 1 l m a lλ n,m,l Assume that A(z) = l 1 a lz l 1 has a nonzero radius of convergence in the z-plane. Then where Ã 0 (α) := l 1 Ã 1 (α) := l 1 α l l! α l l! A n,m = Ã0(α) Ã1(α) 2n + O ( n 2), 0 j<l 0 j<l (1 α) j a l j j! ( ) (1 α)j+2 (1 α)j 1 (1 α)j 2 a l j α 2 + (1 α) (j + 2)! (j 1)! (j 2)! 2/59 A n,m = 1 ( A(z) ) m ( 1 + z n+1 m dz 2πi z =c nz n)

52 43/59 TOOLS NEEDED Lemma 2. (Asymptotic tranfer) λ n,m,l(a n,m a n,m l ) = b n,m 1 l m If b n,m c/n, uniformly for 1 m n and n 1, where c > 0, then a n,m ch m (1 m n). In particular, µ n,m H m Λ n,m := 1 l m λ n,m,l m n (1 m n) a n,m b n,m Λ n,m + a n,m 1 c n n m + ch m 1 = ch m, Useful for error analysis

53 44/59 TOOLS NEEDED Lemma 3 If φ C 2 [0, 1] and φ (x) 0 for x [0, 1], then ( ( m ) ( )) m l λ n,m,l φ φ n n 1 l m = φ (α) n 1 l m = φ (α) S 1 (α) n uniformly for 1 m n. lλ n,m,l + O ( n 2) + O ( n 2) Bootstrapping & induction = µ n,m k 0 b k H m+φ k+1 (α) n k

54 45/59 INITIAL BITS ARE BERNOULLI(p): m ( n m) p m q n m µ n,m E(X n ) en = log qn + γ + φ 1(q) + 1 ( log qn + γ + 3 φ 1 (q) 2n + 2qφ (q) + pqφ 1(q) + 2φ 2 (q) + O ( n 2 log n ) ) p = 1 2 E(X n ) en = log n log 2 + γ + φ 1( 1) + 1 ( log n log 2 + γ 2 2n ) + 3 φ 1 ( 1) + 2 φ ( 1) φ 1( 1) + 2φ 2 2( 1) + 2

55 VARIANCE OF X n,m Uniformly for 1 m n V(X n,m ) = e 2 H (2) m n 2 e(2e + 1) ( n + 1 2) Hm + ψ 1 (α)n + ψ 2 (α) + O ( n 1 H m ) The two dominating terms independent of p 46/59 V(X n ) = e2 π 2 6 n2 e(2e + 1) ( ) n log n + c 1n + c 2 + O ( n 1 log n ) ψ 1 (α) = ( ) α S2 (x) 0 S dx (x) 3 x 2 x ψ 2 (α) = 7 12 ( α 5S 1 (x)s 2(x) 2 0 2S 1 (x) 5 S 0(x) S (x) 3 S (x) 2 x 3 x 2 2x 2S 1 (x)s 3(x)+S 2 (x)s 2 (x)+6s 0(x)S 2 (x) 2S 1 (x) 4 ) dx

56 V n,m := (1 1 n+1 )n+1 n 2 (V(X n+1,m ) + E(X n+1,m )) = H (2) m + 1 k<k r k H m + s k H (2) m + t k n k + O ( H m n K ) V n,m V n,m H (2) m 47/59 K = 2 K = 3

57 48/59 LIMIT GUMBEL DISTRIBUTION OF X n,m P ( 1 r m Exp(r) log m x ) e e x If m with n and m n, then ( ) Xn,m P en log m φ 1( m) x e e x n By induction ) ( E (e Xn,ms/(en) (Hm+φ 1( m n ))s = 1 + O ( Hm n )) 1 r m uniformly for 1 m n (proof long and messy). e s/r 1 s, r

58 49/59 X n = 0 m n ( ) n p m q n m X n,m m ( ) Xn P en log pn φ 1(ρ) x e e x (x R) P ( X n en log n 2 φ 1( 1 2 ) x) convergence rates

59 MAIN STEPS F n,m (s) = F n,m (s) := E ( e ) Xn,ms/(en) e φ( m n )s 1 l m 1 r m 1 1 s r = Q ( ) n,m e s/(en) e Hms φ( m n )s. 1 r m e s/r 1 s r λ n,n m,l F n,m l (s)e ( φ( m n ) φ( m l n ) ) s e s/(en) ( 1 1 l m λ n,n m,l ) m l+1 r m ( ) 1 s r 50/59 Prove F n,m (s) 1 Cn 1 H m when φ = φ 1

60 51/59 AN AUXILIARY FUNCTION G n,m (s) := 1 l m λ n,n m,l e ( φ( m n ) φ( m l n ) ) s e s/(en) ( 1 1 l m m l+1 r m λ n,n m,l ) e s/(en) ( ) 1 s r If φ C 2 [0, 1], then G n,m (s) = 1 s (1 + m αφ (α)) S 1(α) + O ( ) 1 S(α) mn 1 s α + O ( ), 1 m S(α) mn If φ = φ 1 then G n,m (s) = 1 + O((mn) 1 )

61 52/59 A SUMMARY OF THE APPROACHES Recurrence Ansatz Error analysis

62 53/59 (1 + 1)-EA FOR LEADINGONES f (x) = x j ; Y n := optimization time 1 k n 1 j k Rudolph (1997): introduced LEADINGONES Droste et al. (2002): E(Y n ) n 2 Ladret (2005): CLT(c 1 n 2, c 2 n 3 ) Böttcher et al. (2010): re-derived mean many other papers Prove Ladret s results by a direct analytic approach

63 4/59 TIME TO OPTIMUM STATE UNDER LEADINGONES Y n (starting with n random bits (each being 1 with probability 1 2 ) E ( e Yns) := 2 n + 1 m n 2 m n 1 Q n,m (s) where the conditional moment generating function Q n,m (s) satisfies the recurrence relation ( 1 (1 pq n m )e s) Q n,m (s) = pq n m e s 2 1 m + Q n,l (s) 2 m l, for 1 m n, where q = 1 p. p n 1 1 l<m

64 55/59 CLOSED-FORM SOLUTION FOR Q n,m Q n,m (s) = e s pq n m 1 j<m 1 1 e s 2pq n j 1 1 e s pq n j Y n,m d = Z [0] n,m }{{} geom + Z [m 1] n,m ) R m (t) := E (t Z [0] n,m = + + Z [m 1] n,m } {{ } geom ) E (t Z [j] n,m = (1 2pqn j )t 1 (1 pq n j )t (j = 1,..., m 1) pq n m t 1 (1 pq n m )t = R j(t) 2

65 Y n E(Y n ) V(Yn ) N (0, 1) E(Y n,m ) = 1 ( 1 q m 1 pq n 1 2p E(Y n ) = 1 m n p= c n = n2 2c 2 + q m 1 ) ( e c e c(1 α) + O ( c(c + 1) 2 n+m 1 E(Y n,m ) = q 2p 2 ( q n 1 ) n )). = ec 1 2c 2 n 2 + (c 2)ec + 2 n + cec (3c 4) + 4c 48 56/59 V(Y n ) = 3q 2 ( q 2n 4p 3 1 ) µ n (1 + q) p= c n = e2c 1 8c 3 n 3 + 3e2c (2c 3) 8e c c 2 n 2 + (6c2 10c + 3)e 2c 8(c 2)e c 19 32c n + O(1)

66 57/59 Properties ONEMAX (X n) LEADINGONES (Y n) Mean Variance Limit law en log n + c 1 n π 2 6 (en)2 (2e + 1)en log n Gumbel distribution P ( X n log n φ en 2 1( 1 ) x) 2 e e x e 1 2 n 2 e n 3 Gaussian distribution P e 1 Yn n 2 2 e 2 1 n 8 3 x 1 x t2 2π e 2 dt Approach Ansatz & error analysis Analytic combinatorics

67 58/59 THANK YOU (1 + 1)-EA = 232

A Gentle Introduction to the Time Complexity Analysis of Evolutionary Algorithms

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