Drift analysis and average time complexity of evolutionary algorithms

Transcription

1 Artificial Itelligece 127 (2001) Drift aalysis ad average time complexity of evolutioary algorithms Ju He a,xiyao b, a Departmet of Computer Sciece, Norther Jiaotog Uiversity, Beijig , PR Chia b School of Computer Sciece, The Uiversity of Birmigham, Birmigham B15 2TT, UK Received 19 July 1999; received i revised form 15 August 2000 Abstract The computatioal time complexity is a importat topic i the theory of evolutioary algorithms (EAs). This paper reports some ew results o the average time complexity of EAs. Based o drift aalysis, some useful drift coditios for derivig the time complexity of EAs are studied, icludig coditios uder which a EA will take o more tha polyomial time (i problem size) to solve a problem ad coditios uder which a EA will take at least expoetial time (i problem size) to solve a problem. The paper first presets the geeral results, ad the uses several problems as examples to illustrate how these geeral results ca be applied to cocrete problems i aalyzig the average time complexity of EAs. While previous work oly cosidered (1 + 1) EAs without ay crossover, the EAs cosidered i this paper are fairly geeral, which use a fiite populatio, crossover, mutatio, ad selectio Elsevier Sciece B.V. All rights reserved. Keywords: Evolutioary algorithms; Time complexity; Radom sequeces; Drift aalysis; Stochastic iequalities 1. Itroductio Evolutioary algorithms (EAs) are a powerful class of adaptive search algorithms [6,11, 14]. They have bee used to solve may combiatorial problems with success i recet years. However, theories o explaiig why ad how EAs work are still relatively few i spite of recet efforts [3]. The computatioal time complexity of EAs is largely ukow, except for a few simple cases [1,2,5,17,18]. Ambati et al. [1] ad Fogel [5] estimated the computatioal time complexity of their EAs o the travelig salesma problem. No * Correspodig author. address: x.yao@cs.bham.ac.uk (X. Yao) /01/$ see frot matter 2001 Elsevier Sciece B.V. All rights reserved. PII: S (01)

2 58 J. He, X. Yao / Artificial Itelligece 127 (2001) theoretical results were give. Rudolph [17] proved that (1 + 1) EAs with mutatio probability p m = 1/, where is the umber of bits i a biary strig (i.e., idividual) ad p m is the mutatio probability, coverge i average time O( log ) for the ONE- MAX problem. Droste et al. [2] carried out a rigorous complexity aalysis of (1 + 1) EAs for liear fuctios with Boolea iputs. However, all of these results were based o EAs with a populatio size of 1 ad without ay crossover operators. Nimwege et al. [22,23] developed a theory which predicts the total umber of fitess fuctio evaluatios eeded to reach a global optimum by epochal dyamics as a fuctio of mutatio rate ad populatio size. However, o relatioship to the problem size was studied. He et al. [8,9] showed that geetic algorithms (GAs) may take expoetial average time to solve some deceptive problems. This paper presets a more geeral theory about the average time complexity of EAs. The motivatio of this study is to establish a geeral theory for a class of EAs, rather tha a particular EA. The theory ca the be used to derive specific complexity results for differet EAs o differet problems. The theory has bee developed usig drift aalysis [7, 21] a very useful techique i aalyzig radom sequeces. It ca be used to estimate the first hittig time by estimatig the drift of a radom sequece. To our best kowledge, this is the first attempt that drift aalysis is itroduced ito the theoretical study of evolutioary computatio. Oe of the major advatages of usig drift aalysis is that it is ofte easier to estimate the drift tha to estimate the first hittig time directly. The techiques of drift aalysis ca also be applied to radom sequeces which are ot Markovia [7]. The basic idea of this paper is as follows. We first model the evolutio of a EA populatio as a radom sequece, e.g., a Markov chai. A populatio of multiple idividuals will be cosidered. Both crossover ad mutatio are icluded i the EA. The we aalyzed the drift of this sequece to ad from the optimal solutio (assumig we are solvig a optimizatio problem). Various bouds o the first hittig time will be derived uder differet drift coditios. Some drift coditios cause the radom sequece to drift away from the optimal solutio, while other drift coditios eable the sequece to drift towards the optimal solutio. We will study the coditios which are used to determie the time complexity of a EA to solve a problem, whether i polyomial time (i problem size) or i expoetial time. To illustrate the applicatio of the above geeral theory, we will apply the theoretical results to several well-kow problems, icludig a classical combiatorial optimizatio problem the subset sum problem. It is show i this paper that a certai family of subset sum problems ca be solved by a EA withi polyomial time, while other families of subset sum problems will eed at least expoetial time to solve. Although the EAs used i our study do ot iclude all possible variatios of EAs, they do represet a fairly large class of EAs which have multiple idividuals ad use both crossover ad mutatio. The rest of this paper is orgaized as follows: Sectio 2 itroduces briefly EAs ad drift aalysis. Sectio 3 studies the coditios uder which EAs ca solve a problem withi polyomial time o average. A geeral theorem is first preseted. The examples, icludig the subset sum problem, are studied to show the applicatio of the theorem. Sectio 4 studies the coditios uder which EAs eed at least expoetial computatio time to solve a problem. Both a geeral theorem ad a applicatio of the theorem are

3 J. He, X. Yao / Artificial Itelligece 127 (2001) give. Sectio 5 discusses some weaker drift coditios for the subset sum problem. Fially, Sectio 6 cocludes with a brief summary of the paper ad some future work. 2. Evolutioary algorithms ad drift aalysis 2.1. Evolutioary algorithms The combiatorial optimizatio problem cosidered i this paper ca be described as follows: Give a fiite state space S ad a fuctio f(x), x S, fid max{f(x); x S}. (1) Assume x is oe state with the maximum fuctio value, ad f max = f(x ). The EA for solvig the combiatorial optimizatio problem ca be described as follows: (1) Iitializatio: geerate, either radomly or heuristically, a iitial populatio of 2N idividuals, deoted by ξ 0 = (x 1,...,x 2N ),adletk 0, where N>0isa iteger. For ay populatio ξ k,defief(ξ k ) = max{f(x i ): x i ξ k }. (2) Geeratio: geerate a ew (itermediate) populatio by crossover ad mutatio (or ay other operators for geeratig offsprig), ad deote it as ξ k+1/2. (3) Selectio: select ad reproduce 2N idividuals from populatios ξ k+1/2 ad ξ k,ad obtai aother (ew itermediate) populatio ξ k+s. (4) If f(ξ k+s ) = f max, the stop; otherwise let ξ k+1 = ξ k+s ad k k + 1,adgoto step (2). Obviously the above descriptio icludes a wide rage of EAs usig crossover, mutatio ad selectio. The descriptio does ot set ay restrictios o the type of crossover, mutatio or selectio schemes used. It icludes EAs which use crossover or mutatio aloe. The EA framework give above is closer to evolutio strategies [19] ad evolutioary programmig [4] tha to GAs [6] i the sese that selectio is applied after crossover ad/or mutatio. However, the mai results give i this paper, i.e., Theorems 1 ad 10 are idepedet of ay such implemetatio details. I fact, they hold for virtually ay stochastic search algorithms Drift aalysis Assume x is a optimal poit, ad let d(x,x ) be the distace betwee a poit x ad x. If there are more tha oe optimal poit (that is, a set S ), we use d(x,s ) = mi{d(x,x ): x S } as the distace betwee idividual x ad the optimal set S.I short we deote the distace by d(x). Usually d(x) satisfies d(x ) = 0add(x) > 0for ay x/ S. However, i some parts of this paper, we will cosider a pseudo-distace d(x) which allows d(x)= 0forsomex/ S. Give a populatio X ={x 1,...,x 2N },let d(x) = mi{d(x): x X}, (2) which is used to measure the distace of the populatio to the optimal solutio.

4 60 J. He, X. Yao / Artificial Itelligece 127 (2001) The sequece {d(ξ k ); k = 0, 1, 2,...} geerated by the EA is a radom sequece. The sequece ca be modeled by a homogeeous Markov chai if o self-adaptatio is used [10]. The drift of the radom sequece {d(ξ k ); k = 0, 1,...} at time k is defied by ( d(ξ k ) ) = d(ξ k+1 ) d(ξ k ). Defie the stoppig time of a EA as τ = mi{k: d(ξ k ) = 0}, which is the first hittig time o the optimal solutio. The task ow is to ivestigate the relatioship betwee the expect first hittig time τ ad the problem size. I this paper, we focus o the followig questio: uder what coditios of the drift (d(ξ k )) ca we estimate the expect first hittig time E[τ]? I particular, we study the coditiosuderwhicha EA is guarateed to fid the optimal solutio i polyomial time o average ad coditios uder which a EA takes at least expoetial time o average to fid the optimal solutio. The idea behid drift aalysis is quite straightforward. It ca be explaied (by sacrificig mathematical rigor) usig a determiistic algorithm as a example. Assume the distace betwee the startig solutio ad the optimal solutio is d, ad a determiistic algorithm is used to solve a optimizatio problem. If the drift towards the optimal solutio is greater tha at each time step (i.e., iteratio), we would eed at most d/ time steps to fid the optimal solutio. Hece the key issue here is to estimate ad d. Sasaki ad Hajek [20] have successfully used this method to estimate the time complexity of simulated aealig for the maximum matchig problem. 3. Coditios for polyomial average computatio time 3.1. Drift coditios I this sectio, we study uder which drift coditios a EA ca solve a optimizatio problem i polyomial average time. Coditio 1. There exists a polyomial of problem size, h 0 () > 0, such that d(x) h 0 () for ay give populatio X. This coditio says that the distace from ay populatio to the optimal solutio is bouded by a polyomial fuctio of the problem size. Coditio 2. At ay time k 0, if populatio ξ k satisfies d(ξ k )>0, the there exists a polyomial of problem size, h 1 () > 0, such that E [ d(ξ k ) d(ξ k+1 ) d(ξ k )>0 ] 1 h 1 (). This coditio idicates that the drift of the radom sequece {d(ξ k ); k = 0, 1, 2,...} toward the optimal solutio is always positive ad bouded by a iverse polyomial.

5 J. He, X. Yao / Artificial Itelligece 127 (2001) Now we give the followig mai result i the sectio. Theorem 1. If {d(ξ k ); k = 0, 1, 2,...} satisfies Coditios 1 ad 2, the startig from ay iitial populatio X with d(x) > 0, E [ τ d(ξ 0 )>0 ] h(), where h() is a polyomial of problem size. Proof. Accordig to Coditio 2, we kow that {d(ξ k ); k = 0, 1, 2,...} i fact is a supermartigale [15]. Sice 0 d(ξ k ) h 0 (), it coverges almost everywhere [15], ad lim E[ d(ξ k ) d(ξ 0 )>0 ] = 0. k Accordig to the defiitio of stoppig time τ,wehaved(ξ τ ) = 0. Hece, E [ d(ξ τ ) d(ξ 0 )>0 ] = 0. For ay time k 1, E [ d(ξ k ) d(ξ 0 )>0 ] = E [ E [ d(ξ k 1 ) + ( d(ξ k 1 ) ) ] ξ k 1 d(ξ0 )>0 ]. Accordig to Coditio 2, we have for k 1 <τ, E [ d(ξ k 1 ) + ( d(ξ k 1 ) ) ] ξ k 1 d(ξk 1 ) 1 h 1 (). Therefore E [ d(ξ k ) d(ξ 0 )>0 ] E [ d(ξ k 1 ) 1 ] d(ξ0 )>0. h 1 () By iductio o k, we ca get E [ d(ξ k ) d(ξ 0 )>0 ] [ E d(ξ 0 ) k ] d(ξ 0 )>0. h 1 () Hece we have 0 = E [ d(ξ τ ) d(ξ 0 )>0 ] [ E d(ξ 0 ) τ ] d(ξ0 )>0 h 1 () E [ d(ξ 0 ) ] 1 h 1 () E[ τ d(ξ 0 )>0 ]. Accordig to the above iequality ad Coditio 1, E [ τ d(ξ 0 )>0 ] E [ d(ξ 0 ) ] h 1 () h 0 ()h 1 (). Let h() = h 0 ()h 1 (). We arrive at E [ τ d(ξ 0 )>0 ] h(). Uder certai stroger coditios, we ca get some stroger results. Coditio 3. Let d max ={d(x): x S}, ad the iterval [0,d max ] be divided ito L + 1 sub-itervals: d 0 0 <d 1 < <d L <d L+1 d max,wherel>0isaiteger.

6 62 J. He, X. Yao / Artificial Itelligece 127 (2001) (a) For ay l(0 l L), if at time k, the populatio ξ k eters the iterval [0,d l ], i.e., d(ξ k ) d l, the after that time, the populatio will ot retur to the iterval (d l,d L+1 ] agai, i.e., for ay t k: d(ξ t ) d l ; (b) At ay time k, if the populatio ξ k is i the iterval (d l,d l+1 ], the the drift satisfies: E [ ] 1 d(ξ k ) d(ξ k+1 ) d l <d(ξ k ) d l+1 h l (), where h l () > 0. Theorem 2. If {d(ξ k ); k = 0, 1,...} satisfies Coditio 3, the startig from ay iitial populatio ξ 0 with d(ξ 0 )>0, E [ τ d(ξ 0 )>0 ] L h l ()(d l+1 d l ). l=0 Proof. Let s cosider the worst case with the iitial populatio d(ξ 0 ) = d L+1. For ay l with 0 l L,defieτ L+1 = 0ad τ l = mi{t: d(ξ t ) d l }. It is easy to see that τ = τ 0 = (τ L τ L+1 ) + (τ L 1 τ L ) + +(τ 0 τ 1 ). Give ay l with 0 l L, accordig to Coditio 3(b), we kow E [ ] 1 d(ξ k ) d(ξ k+1 ) d l <d(ξ k ) d l+1 h l (). The accordig to Coditio 3(a) ad Theorem 1, we have E [ ] τ l τ l+1 d l <d(ξ k ) d l+1 hl ()(d l+1 d l ). Hece L E[τ ξ 0 ] h l ()(d l+1 d l ). l=0 Coditio 4. Let d max ={d(x): x S}, ad the iterval [0,d max ] be divided ito L + 1 sub-itervals: d 0 0 <d 1 < <d L <d L+1 d max,wherel>0isaiteger.atay time k 0, if the populatio ξ k is i the iterval (d l,d max ], the the drift satisfies: E [ ] 1 d(ξ k ) d(ξ k+1 ) d(ξ k ) d l h l (), where h l () > 0. Theorem 3. If {d(ξ k ); k = 0, 1,...} satisfies Coditio 4, the startig from ay iitial populatio ξ 0 with d(ξ 0 )>0, E [ τ d(ξ 0 )>0 ] L h l ()(d l+1 d l ). l=0

7 J. He, X. Yao / Artificial Itelligece 127 (2001) Proof. The proof is similar to that of Theorem 2. Usig the same aalytical techique as those used i Theorem 1, we ca obtai easily the followig results. Coditio 5. For some populatio X, d(x) h 0 (), where h 0 () > 0 is a fuctio of problem size. Coditio 6. There exists a polyomial fuctio, h 1 () > 0, of problem size such that E [ d(ξ k ) d(ξ k+1 ) ξ k = X ] 1 h 1 () for ay time k ad populatio X with d(x) > 0. Theorem 4. If {d(ξ k ); k = 0, 1,...} satisfies Coditios 5 ad 6, the startig from the iitial populatio with d(x) h 0 (), E[τ ξ 0 = X] h(), where h() = h 0 ()h 1 () is a fuctio of problem size. Proof. Similar to the proof of Theorem The subset sum problem EAs have bee applied to the subset sum problem i practice [13]. The problem ca be described as follows: Give a set W ={w 1,...,w } of itegers ad a large iteger C, fid a subset S of W such that the sum of the elemets i S are closest to but ot exceedig C. The subset sum problem is NP-complete. The partitio problem ca be polyomially trasformed to it [12]. A solutio S to the subset sum problem ca be represeted by a strig x = (s 1 s ) where s i {0, 1}. The presece of w i i S meas that s i = 1 while its absece is represeted by s i = 0. A feasible solutio to the subset sum problem is a strig x = (s 1 s ), s i {0, 1}, such that w i s i C, i=1 where F(x)= j=1 w i s i is called the objective fuctio. The optimal solutio is the strig that maximizes the objective fuctio (without exceedigs C). The fitess fuctio ca be defied as f(x)= (C F(x)+ (1 θ)f(x)),whereθ = 1 whe x is feasible ad θ = 0whex otherwise. Notice that (0 0) is a feasible solutio. I this subsectio, we are iterestig i a particular family of subset sum problems {W,= 1, 2,...}, is a iteger, for which a EA ca fid the optimal solutio withi polyomial average time.

8 64 J. He, X. Yao / Artificial Itelligece 127 (2001) The family of problems we focus o is {W 1,W 2,...,W,...}, where W ={w 1,...,w }, (3) w 1,w 2,...,w are positives, (4) C = w i. (5) i=1 It is obvious that x = (1 1) is the uique optimal solutio ad ay subset of W is a feasible solutio. This problem is, i fact, the liear fuctio problem [2]. The EA for solvig the family of subset sum problems follows the structure give i Sectio 2. The crossover, mutatio ad selectio are implemeted as follows. Oe-poit crossover is used. Give two idividuals x = (s (x) 1 s (x) ) ad y = (s (y) 1 s (y) ) from the populatio ξ k, choose a crossover poit m {1,..., 1} at radom ad exchage all bits after the mth bit betwee two idividuals to form two ew idividuals x ad y : x = ( s (x) 1 s (x) m 1 s(y) m s (y) ) m+1 s(y), y = ( s (y) 1 s (y) ) m 1 s(x) m s(x) m+1 s(x). A ew itermediate populatio ξ k+c of 2N idividuals will be formed after crossover. The mutatio operator is the bit mutatio. Give a idividual x = (s 1 s ) i ξ k+c, choose a sigle bit s i at radom from it ad flip the bit. Aother ew itermediate populatio ξ k+m of 2N idividuals is formed after mutatio. Selectio used implemets a kid of probabilistic elitism. 2N idividuals are selected from ξ k ad ξ k+m as follows: the best idividual with the highest fitess is copied with probability at least 1 e to the ew populatio ξ k+s, ad other idividuals are assiged a survival probability accordig to their fitess. Ay selectio scheme ca be used as log as fitter idividuals were assiged higher probabilities. Theorem 5. Give the family of subset sum problems ad the EA to solve them. For ay iitial populatio X with d(x) > 0, E[τ ξ 0 = X] h() where h() is a polyomial of. Proof. Defie the distace fuctio d(x) as: d(x)= s i 1. i=1 Accordig to Theorem 1, we eed to verify that the radom sequece, {d(ξ k ); k = 0, 1,...}, satisfies Coditios 1 ad 2. From the defiitio of the above distace ad Eq. (2), we kow that for ay populatio X: d(x).

9 J. He, X. Yao / Artificial Itelligece 127 (2001) Hece the radom sequece {d(ξ k ); k = 0, 1,...} satisfies Coditio 1. For ay time k 0, ad populatio ξ k with d(ξ k )>0, we ow ivestigate the impact of crossover o the drift. Oe of the three evets may happe after crossover: (1) evet I{d(ξ k+c )<d(ξ k )}, (2) evet I{d(ξ k+c ) = d(ξ k )},or (3) evet I{d(ξ k+c )>d(ξ k )}. We first show that evet I{d(ξ k+c )>d(ξ k )} caot happe. I other words, crossover does ot produce a worse itermediate populatio. Assume x 1 ad x 2 are two idividuals i populatio ξ k,ady 1 ad y 2 are their offsprig. Sice the crossover does ot icrease or decrease the amout of oes i idividuals x 1 ad x 2,wehave d(y 1 ) + d(y 2 ) = d(x 1 ) + d(x 2 ). That is, d(y 1 ) d(x 1 ) = ( d(y 2 ) d(x 2 ) ). This meas that the icrease of oe idividual s drift will be the decrease of aother idividual s drift. So the crossover will ot make the itermediate populatio ξ k+c worse. Evet I{d(ξ k+c )>d(ξ k )} caot happe. Assume that evet I{d(ξ k+c ) = d(ξ k )} has happeed. The oe of the followig three evets may happe subsequetly: (a) evet I{d(ξ k+m )<d(ξ k+c )}, (b) evet I{d(ξ k+m ) = d(ξ k+c )},ad (c) evet I{d(ξ k+m )>d(ξ k+c )}. Evet I{d(ξ k+m ) = d(ξ k+c )} caot happe because the mutatio always happes. The probability of evet I{d(ξ k+m )<d(ξ k+c )} is ot less tha 1/ (if d(ξ k+c )>0), ad the probability of evet I{d(ξ k+m )>d(ξ k+c )} is ot greater tha ( 1)/. If d(ξ k+c ) = 0, the the populatio ξ k+c has oe idividual with the maximum fitess. Assume that evet I{d(ξ k+c )<d(ξ k )} has happeed. The oe of the followig three evets may happe subsequetly: (a )eveti{d(ξ k+m )<d(ξ k+c )}, (b )eveti{d(ξ k+m ) = d(ξ k+c )},ad (c )eveti{d(ξ k+m )>d(ξ k+c )}. The probabilities of the three evets are similar to those aalysed i the case of I{d(ξ k+c ) = d(ξ k )}. Now let s examie the role of selectio: the idividual with the best fitess will appear i the ext populatio ξ k+s with probability 1 e, so the probability P(d(ξ k+s )< d(ξ k ) d(ξ k+m )<d(ξ k )) is ot less tha 1 e ad the probability P(d(ξ k+s )>d(ξ k ) d(ξ k+m )<d(ξ k )) is ot more tha e. The probability P(d(ξ k+s )>d(ξ k ) d(ξ k+m )> d(ξ k )) is ot more tha e. Ad the evet I{d(ξ k+s )<d(ξ k ) d(ξ k+m )>d(ξ k )} caot happe. Cosiderig all the differet cases discussed above, we have E [ d(ξ k+1 ) d(ξ k ) d(ξ k )>0 ] = E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )<d(ξ k ) } d(ξ k )>0 ]

10 66 J. He, X. Yao / Artificial Itelligece 127 (2001) E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )>d(ξ k ) } d(ξ k )>0 ] + E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )<d(ξ k ) } d(ξ k )>0 ] + E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )>d(ξ k ) } d(ξ k )>0 ] + E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )<d(ξ k ) } d(ξ k )>0 ] + E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )>d(ξ k ) } d(ξ k )>0 ] + E [( d(ξ k+1 ) d(ξ k ) ) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )>d(ξ k ) } d(ξ k )>0 ]. I other words, E [ d(ξ k+1 ) d(ξ k ) d(ξ k )>0 ] ( 1)P ( d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ) d(ξ k )>0 )( 1 e ) + ( 1)P ( d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ) d(ξ k )>0 ) e + ( 1)P ( d(ξ k+c )<d(ξ k ), d(ξ k )>d(ξ k+m )>d(ξ k+c ) d(ξ k )>0 )( 1 e ) + ( 1)P ( d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ) d(ξ k )>0 ) e + ( 1)P ( d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ) d(ξ k )>0 )( 1 e ) + ( 1)P ( d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ) d(ξ k )>0 ) e + ( 1)P ( d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ) d(ξ k )>0 ) e. I arrivig at the above iequality, we have used the fact that d(ξ k+1 ) d(ξ k ) 1. Sice ( 1)P ( d(ξ k+c )<d(ξ k ), d(ξ k )>d(ξ k+m )>d(ξ k+c ) d(ξ k )>0 )( 1 e ) < 0 ad P ( d(ξ k+c )<d(ξ k ) d(ξ k > 0)) + P(d(ξ k+c ) = d(ξ k ) d(ξ k > 0) ) = 1, we have E [ d(ξ k+1 ) d(ξ k ) d(ξ k )>0 ] ( 1)P ( d(ξ k+c )<d(ξ k ) d(ξ k )>0 ) 1 ( 1 e ) + ( 1)P ( d(ξ k+c )<d(ξ k ) d(ξ k )>0 ) 1 e + ( 1)P ( d(ξ k+c )<d(ξ k ) d(ξ k )>0 ) 1 e

11 J. He, X. Yao / Artificial Itelligece 127 (2001) ( 1)P ( d(ξ k+c ) = d(ξ k ) d(ξ k )>0 ) 1 ( 1 e ) + ( 1)P ( d(ξ k+c ) = d(ξ k ) d(ξ k )>0 ) 1 e + ( 1)P ( d(ξ k+c ) = d(ξ k ) d(ξ k )>0 ) 1 e ( 1) 1 ( 1 e ) + 2( 1) 1 e 1 e 2( 1) 2 e. Let h 1 () = 1 e 2( 1) 2 e, the whe +, h 1 () = O(). Hece, E [ d(ξ k+1 ) d(ξ k ) d(ξ k )>0 ] 1 h 1 () ad 1 lim h 1 () < 0, where h 1 () = O(). So we have prove that the radom sequece {d(ξ k ); k = 0, 1,...} satisfies Coditio 2. Accordig to Theorem 1, we kow E[τ ξ 0 = X] h() where h() = O( 2 ) Other problems I order to show the power ad geerality of our mai results, we will use some of the problems give i [18] as examples to derive EA s computatio time by verifyig drift coditios give previously. Rudolph s survey [18] is probably the most comprehesive overview of recet results o the fiite time behavior of EAs i fiite space ad discrete time. It is worth otig that the EA used i this sectio is more geeral tha the (1+1) EA used i [18]. A (2N + 2N) EA without crossover is used i this sectio. The results show i this sectio illustrate that drift aalysis ca be used to derive EA s average computatio time for a variety of differet problems. Let S ={(s 1 s ), s i {0, 1}} be the chromosome represetatio. The mutatio ad selectio are implemeted as follows. A kid of uiform bit mutatio is used. For ay idividual x = (s 1 s ) i ξ k, each bit s i will flip with a mutatio rate p m > 0. A ew itermediate populatio ξ k+m of 2N idividuals is formed after mutatio.

12 68 J. He, X. Yao / Artificial Itelligece 127 (2001) (2N + 2N) elitism is implemeted as the selectio scheme. I other words, the 2N idividuals with the highest fitess from populatios ξ k ad ξ k+m are copied to the ew populatio ξ k+s Liear fuctios A fuctio f : S R is liear if f(x)= c 0 + i=1 c i s i where coefficiets c i R [18]. If c i 0forallc i, the it is clear that (1...1) is the oly optimal solutio. Theorem 6. For the liear fuctio with positive coeffiets c 1 >c 2 > >c > 0,theEA with mutatio probability p m = 1/ eeds a average O( l ) steps to reach the optimal solutio. Proof. Defie the distace fuctio d(x) = i=1 (1 s i ).Sice0 d(x), we ca divide [0,d max ] ito itervals d 0 <d 1 <d 2 < <d where d l = l for 0 l. We will use Theorem 3 to prove the result. Assume at time k 0, the populatio ξ k satisfy d(ξ k )>d l 1 where l {1,...,}. Without the loss of geerality, assume d(ξ k ) = d l (other cases ca be prove i the same way). The E [ d(ξ k ) d(ξ k+1 ) ] = E [( d(ξ k ) d(ξ k+1 ) ) I { d(ξ k )>d(ξ k+1 ) }] + E [( d(ξ k ) d(ξ k+1 ) ) I { d(ξ k )<d(ξ k+1 ) }]. First let s cosider E[(d(ξ k ) d(ξ k+1 ))I{d(ξ k )>d(ξ k+1 )}].Letx be the best idividual i populatio ξ k. The probability of its flippig oe of its l 0 bits while keepig its l ( ) l. 1 bits uchaged is Cl Hece E [( d(ξ k ) d(ξ k+1 ) ) I { d(ξ k )>d(ξ k+1 ) }] l ( 1 1 ) l. Secodly let s cosider E[(d(ξ k ) d(ξ k+1 ))I{d(ξ k )<d(ξ k+1 )}]. Letx be the best idividual i ξ k. Assume that evet I{d(ξ k )<d(ξ k+1 )}] happes, the it implies evet I : i.e., oe of its l 0 bits must flip (I fact the leftmost bit amog all flippig bits must flip from 0 to 1 because of c 1 >c 2 > >c ad elitist selectio), ad at least two of its l 1 bits must also flip. So the probability of evet I{d(ξ k )<d(ξ k+1 )} happeig is o more tha that of evet I.EvetI ca be further divided ito the followig cases: (1) Oe of the 0 bits i x becomes 1, ad two of the 1 bits become 0. The probability of this happeig is ( ) l 1 2 ( C2 l 1 1 ) l 2 < l ( l. 2! ) (2) Oe of the 0 bits i x becomes 1, ad three of the 1 bits become 0. The probability of this happeig is l C3 l ( 1 ) 3 ( 1 1 ) l 3 < l 1 3! ( 1 1 ) l.

13 J. He, X. Yao / Artificial Itelligece 127 (2001) (3)... Hece we get E [( d(ξ k ) d(ξ k+1 ) ) I { d(ξ k )<d(ξ k+1 ) }] > l ( 1 2! + 2 3! So E [ d(ξ k ) d(ξ k+1 ) ] l ( )( ) l c l 2! 3!, where c>0 is a costat. I other words, Coditio 4 holds. Accordig to Theorem 3, E[τ] c 1 l= l = O( l ) Pseudo-modular fuctios A fuctio f : S R is pseudo-modular if )( + 1 ) 1 l. mi { f(x),f(y) } max { f(x y),f(x y) }, ad max { f(x),f(y) } mi { f(x y),f(x y) }, for all x,y S [18]. A example of the pseudo-modular fuctio is the fuctio i f(x)= s j. (6) i=1 j=1 Theorem 7. The expected first-hittig time of the EA for the fitess fuctio (6) is E[τ] 2 (e 1) whe mutatio rate p m = 1/. Proof. For fitess fuctio (6), the optimal solutio is (1 1). Defie the distace fuctio d(x) as follows: i d(x)= s j. i=1 j=1 We ca divide [0,] ito subitervals d 0 <d 1 < <d where d l = l for 0 l. Accordig to Theorem 2, we eed to verify that {d(ξ k ); k = 0, 1,...} satisfies Coditio 3. First let s verify Coditio 3(a). Sice the EA adopts a elitist selectio strategy, Coditio 3(a) holds automatically. Secod let s verify Coditio 3(b). At ay time k 0, if populatio ξ k is i the iterval (d l,d l+1 ] where 0 l 1, the there exists at least oe idividual x i ξ k such that d(x)= l + 1. The probability of x becomig better is o less tha ( 1 ) 1 1 l 1. Hece, E [ d(ξ k ) d(ξ k+1 ) ] 1 ( 1 1 l 1. )

14 70 J. He, X. Yao / Artificial Itelligece 127 (2001) That is, Coditio 3(b) holds. Accordig to Theorem 2, we have 1 ( E[τ ξ 0 ] 1 1 ) +l+1 2 (e 1). l= Uimax fuctios A fuctio f : S R is uimax if there is a uique locally maximal poit x S [18]. The log path problem is a well kow uimax problem. A log path P (where the legth of strig is odd) is defied by a recursio with P 1 ={0, 1} as the base path [16]. Give log path P, create a subpath S 00 by prepedig 00 to each poit i P ad aother subpath S 11 by prepedig 11 to each poit i the reverse order i P. The bridge poit is build from the last poit i P prepedig by 01. Fially, cocateate subpath S 00, the bridge poit ad subpath S 11 to obtai log path P +2. The legth of the paths is described by the recurrece equatios P 1 =2, P +2 =2 P +1, whose solutio is P =3 2 ( 1)/2 1 for odd 1. Table 1 shows log path P 5. Give a poit x o a path P,defiePos(x) to be the positio of x o the path which is umbered from 0 to 3 2 ( 1)/2 2. For a poit ot i the path P,defiePos(x) to be 1. The defie the objective fuctio f(x)as: f(x)= ( 3 2 ( 1)/2 2 ) { Pos(x), if Pos(x) 0, + (7) i=1 s i, otherwise. Theorem 8. For the uimax fuctio (7), startig from the bottom of the icreasig path, the expected first hittig time of the EA is E[τ]=O( 3 ) whe mutatio rate p m = 1/. Proof. Decompose space S ito a family of sets {S 1,S 0,...,S ( 1)/2 } as follows [16]: S 0 = { } { } (00 ) P (01 ) P, S 1 = { } { } ( ) P (1101 ) P, S 2 = { } { } ( ) P ( ) P, Table 1 Log path P 5 Pos(x) x Pos(x) x

15 J. He, X. Yao / Artificial Itelligece 127 (2001) S (l 3)/2 = { } { } ( ) P ( ) P, S (l 1)/2 = { } ( ) P, ad S 1 icludes all remaied poits. Defie the distace fuctio d(x) as follows: d(x)= 0, x S ( 1)/2, d(x)= 1, x S ( 3)/2,... d(x)= ( 3)/2, x S 1, d(x)= ( 1)/2, x S 0, { d(x)= ( 1)/2 + mi (x) s i s (y) ; y P }, x S 1. i=1 i I the followig, we will prove that {d(ξ k ), k = 0, 1,...} satisfies Coditio 3. Let d l = l, we ca divide [0,d max ] ito a fiite umber of subitervals d 0 <d 1 < < d max. First, Coditio 3(a) holds because the EA adopts a elitism selectio strategy. Secodly, let s verify Coditio 3(b). At ay time k 0, if populatio ξ k satisfies d(ξ k )>( 1)/2, the o idividual x i the populatio is o path P.Letx is the best idividual i ξ k, the the probability of x havig a drift is at least (1 1/) 1 /. The drift is at least 1. Hece, E [ d(ξ k+1 ) d(ξ k ) d(ξ k )>( 1)/2 ] = ( 1). Now let s estimate the drift alog the path. At ay time k 0, if populatio ξ k is i the iterval (d l,d l+1 ] where 0 l<( 1)/2, the at least oe idividual x satisfies d(x) = d l+1.ifx is a bridge poit, the probability of a drift happeig is at least (1 1/) 1 / ad the drift legth is at least 1. If x is ot a bridge poit, the probability of a drift happeig is at least (1 1/) 2 / 2 ad the drift legth is at least 1. Summarisig both cases, the expected drift o the path is: E [ d(ξ k+1 ) d(ξ k ) d(ξ k ) = l + 1 ] = ( 2). I other words, Coditio 2(b) holds. Accordig to Theorem 2, we come to the coclusio 0 0 E[τ] O() + O ( 2) = O ( 3), l= l=( 1)/2 where the first part is the average time for poits ot o the path to reach the path, ad the secod part is the average time for poits o the path to reach the optimal solutio Almost positive fuctios A fuctio f : S R is almost-positive if the coefficiets of all oliear terms are o-egative [18].

16 72 J. He, X. Yao / Artificial Itelligece 127 (2001) A example of almost-positive fuctio is f(x)= s i + ( + 1) s i. (8) i=1 i=1 We ca defie the distace fuctio as d(x)= i=1 s i 1. Theorem 9. The expected first-hittig time of the EA for the almost positive fuctio (8) is E[τ]= ( ) whe mutatio rate p m = 1/ ad the EA starts from d(ξ 0 ) =. Proof. For the fitess fuctio (8), the optimal solutio is (1 1). Idividual x = (0 0) is the secod best (maximum) poit because f(x)=, but it is farthest from the optimal solutio with d(x) =. Accordig to Theorem 4, we eed to verify that {d(ξ k ); k = 0, 1,...} satisfies Coditios 5 ad 6. First, let s assume that d(ξ 0 ) =, i.e., the iitial populatio is composed of idividuals (0...0) oly. Let h 0 () =, the Coditio 5 holds. Secodly, at ay time k 0, if ξ k satisfies d(ξ k ) =, there are oly two possible evets which may happe after mutatio ad elitist selectio: evet I{d(ξ k+1 ) = } or evet I{d(ξ k+1 ) = 0, because of elitist selectio. Hece, E [ d(ξ k ) d(ξ k+1 ) d(ξ k ) = ] ( ) 1. Let h 1 () = 1, the Coditio 6 holds. Accordig to Theorem 4, we have E[τ]= ( ). 4. Drift coditios for expoetial average computatio time 4.1. Drift coditios I this subsectio, we ivestigate the drift coditios uder which EAs will take the averagetime expoetiali the problemsize to fid the optimal solutio. Our aalysis is based o Hajek s earlier work [7]. I order to cosider the case where a EA might ot be able to fid the exact optimal solutio, but oly a approximate solutio, defie the stoppig time τ of a EA as: τ = mi{k: d(ξ k ) d b } where d b 0. Coditio 7. For ay populatio X with d b <d(x)<d a,whered b 0add a > 0, E [ e (d(ξ k+1) d(ξ k )) ξ k = X, d b <d(ξ k )<d a ] ρ<1, (9) where ρ>0 is a costat.

17 J. He, X. Yao / Artificial Itelligece 127 (2001) This coditio idicates that (d b,d a ) is a very difficult iterval to search. Whe the coditio is satisfied, d(ξ k+1 )>d(ξ k ). I other words, the offsprig populatio is o average driftig away from the optimal solutio, rather tha gettig closer to it. Coditio 8. For ay populatio X with d(x) d a, d a > 0, E [ e (d(ξ k+1) d a ) ] ξ k = X, d(ξ k ) d a D, (10) where D 1 is a costat. The above coditio idicates that a populatio i the iterval [d a, + ) will ot, o average, drift towards the optimal solutio too much because d(ξ k+1 ) d a l D. Give the above two coditios, the followig lemma ad theorem ca be show by followig Hajek s work o drift aalysis [7]. Lemma 1. If {d(ξ k ); k = 0, 1,...} satisfies Coditios 7 ad 8, the for ay iitial populatio ξ 0, ad E [ e d(ξ k) τ>k 1, d(ξ 0 ) ] ρ k e d(ξ 0) + 1 ρk 1 ρ De d a, (11) P [ d(ξ k ) d b τ>k 1, d(ξ 0 ) ] ρ k e (d(ξ 0) d b ) + 1 ρk 1 ρ De (d a d b ). (12) Proof. Iequality (11) is clearly true for k = 0. For k 0adτ>k, E [ e d(ξk+1) τ>k,d(ξ 0 ) ] = E [ E [ e d(ξk+1) τ>k,d(ξ k ) ] d(ξ 0 ) ], where E [ e d(ξk+1) τ>k,d(ξ k ) ] = E [ e d(ξk+1) ] [ τ>k,d(ξ k ) d a + E e d(ξ k+1 ) ] τ>k,d(ξ k )<d a. (13) The first term o the right-had side of iequality (13) is upper-bouded by De d a accordig to Coditio 8, ad the secod term is upper-bouded by ρe d(ξk) accordig to Coditio 7. Usig these bouds we ca arrive at E [ e d(ξk+1) τ>k,d(ξ 0 ) ] ρe [ e d(ξk) τ>k 1, d(ξ 0 ) ] + De d a. By iductio o k, it is easy to show that the above iequality implies iequality (11) for all k 0. Iequality (12) follows from iequality (11): E [ e (d(ξ k) d b ) τ>k 1, d(ξ 0 ) ] = E [ E [ e (d(ξ k) d b ) ] τ>k 1, d(ξ k ) d b d(ξ0 ) ] + E [ E [ e (d(ξ k) d b ) ] τ>k 1, d(ξ k )>d b d(ξ0 ) ] E [ E [ e (d(ξ k) d b ) ] τ>k 1, d(ξ k ) d b d(ξ0 ) ] e 0 P ( d(ξ k ) d b τ>k 1,d(ξ 0 ) ).

18 74 J. He, X. Yao / Artificial Itelligece 127 (2001) That is, P ( d(ξ k ) d b τ>k 1,d(ξ 0 ) ) E [ e (d(ξ k) d b ) τ>k 1, d(ξ 0 ) ]. Accordig to iequality (11) we have P ( d(ξ k ) d b d(ξ 0 ) d a ) ρ k e (d(ξ 0) d b ) + 1 ρk 1 ρ De (d a d b ). The followig theorem is the mai result of this sectio. Theorem 10. Assume Coditios 7 ad 8 hold. If d(ξ 0 ) d a, D 1 ad ρ<1, the there exist some δ 1 > 0 ad δ 2 > 0 such that E [ ] τ d(ξ 0 ) d a δ1 e δ 2(d a d b ) (14) Proof. Because d(ξ 0 ) d a,wehave e (d(ξ 0) d b ) e (d a d b ). Sice D 1adρ<1, we ca obtai ρ k e (d(ξ 0) d b ) ρk 1 ρ De (d a d b ). Accordig to iequality (12) ad the above iequality, P ( d(ξ k ) d b τ>k 1,d(ξ 0 ) ) 1 1 ρ De (d a d b ). By usig the fact that P(τ = k d(ξ 0 )) = P(d(ξ k ) d b,τ >k 1 d(ξ 0 )),wehave P ( τ>k d(ξ 0 ) ) k = 1 P ( τ = j d(ξ 0 ) ) max (0, 1 k De (d a d b ) ). 1 ρ j=1 Therefore E [ τ d(ξ 0 ) ] + = P ( τ>j d(ξ 0 ) ) j=1 Let δ 1 = 1 ρ 2D ad δ 2 = 1, the E [ ] τ d(ξ 0 ) d a δ1 e δ 2(d a d b ) The subset sum problem revisited + k=0 max (0, 1 k De (d a d b ) ) 1 ρ 1 ρ 2D ed a d b. I this subsectio, we cosider aother family of subset sum problems. We will show that some EAs described i this subsectio take at least a expoetial time o average to fid the optimal solutio.

19 J. He, X. Yao / Artificial Itelligece 127 (2001) The family of subset sum problems that we focus o i this subsectio is {W 1,W 2,..., W,...},where ad W ={w 1,...,w }, 1 w 1,w 2,...,w 1 are positives greater tha 2,w = w i 1, C = w. It is easy to see that the subset {w } is the uique optimal solutio ad ay subset of W {w } is a feasible solutio. This is a deceptive problem. The EA used to solve the above family of problems follows the framework give i sectio 2. The crossover, mutatio ad selectio are implemeted as follows. Oe-poit crossover is used i the EA. Give two idividuals x = (s (x) 1 s (x) ) ad y = (s (y) 1 s (y) ) from populatio ξ k, choose a crossover poit m {1,..., 1} at radom ad exchage all bits from the mth bit betwee two idividuals to form two ew idividuals x ad y : x = ( s (x) 1 s (x) m 1 s(y) m s (y) ) m+1 s(y), y = ( s (y) 1 s (y) ) m 1 s(x) m s(x) m+1 s(x). If a offsprig is ifeasible, oe of the parets will be retaied. A ew itermediate populatio ξ k+c of 2N idividuals will be formed after crossover. Bit mutatio is used i the EA. Give a idividual x = (s 1 s ) i ξ k+c, choose a sigle bit s i at radom from it ad flip the bit. If the offsprig is ifeasible, the paret will be retaied. Aother ew itermediate populatio ξ k+m of 2N idividuals is formed after mutatio. Selectio i the EA ca be regarded as a simple form of rakig. 2N idividuals are selected from ξ k ad ξ k+m as follows: the best 2N idividuals are assiged a survival probability of (1 e )/2N each, ad the worst 2N idividuals are assiged a survival probability of e /2N each. Yet aother ew itermediate populatio ξ k+s of 2N idividuals is formed after selectio. It should be oted that the selectio used here is similar to but ot the same as that used i Sectio 3.2. The chromosome represetatio used is the same as that described i Sectio 3.2. Defie d(x)= s i 1 ad d max = max{d(x): x is a feasible solutio}. i=1 Note that the distace is a pseudo-distace. Give two idividuals x 1 ad x 2 with d(x 1 )>d(x 2 )>2, the fitess of x 1 will be higher tha that of x 2,thatisf(x 1 )>f(x 2 ). Let τ = mi{k: d(ξ k ) = 0}, d a = d max = 2, d b = 3/4, ad τ = mi{k: d(ξ k ) d b }. Obviously E[τ] E[τ ]. The followig theorem gives the mai result of this sectio. i=1

20 76 J. He, X. Yao / Artificial Itelligece 127 (2001) Theorem 11. For the radom sequece, {d(ξ k ); k = 0, 1,...}, defied by the family of subset sum problems ad the EA i this sectio, if d(ξ 0 ) d a, the there exist two costats δ 1 > 0 ad δ 2 > 0 such that E [ ] τ d(ξ 0 ) d a δ1 exp(δ 2 ) for sufficietly large. Proof. Accordig to Theorem 10, all we we eed to show is to verify that Coditios 7 ad 8 are satisfied. First we show that Coditio 7 ca be satisfied, i.e., E [ e (d(ξ k+1) d(ξ k )) ] d b <d(ξ k )<d a ρ<1. After the crossover, oe of the followig three evets may happe: evet I{d(ξ k+c )< d(ξ k )},eveti{d(ξ k+c ) = d(ξ k )} or evet I{d(ξ k+c )>d(ξ k )}. Let x 1 ad x 2 be two idividuals i populatio ξ k. Sice crossover does ot icrease or decrease the amout of oes i idividuals x 1 ad x 2,soevetI{d(ξ k+c )>d(ξ k )} caot happe. Because d(ξ k )>d b = 3/4, x 1 ad x 2 caot be the optimal solutio. Both of their th bits will be 0. The bits will still be 0 after crossover. Let y 1 ad y 2 be offsprig of x 1 ad x 2,wehaved(y 1 )>/4add(y 2 )>/4. Therefore, d(ξ k+c )>/4, d(ξk+c ) d(ξ k ) = + 1. (15) 2 Assume that either evet I{d(ξ k+c )<d(ξ k )} or evet I{d(ξ k+c ) = d(ξ k )} has happeed. I either case, oe of the followig three evets may happe after mutatio: evet I{d(ξ k+m )<d(ξ k+c )}, evet I{d(ξ k+m ) = d(ξ k+c )}, or evet I{d(ξ k+m )> d(ξ k+c )}.Siced(ξ k+c )>/4, d(ξ k+m )>/4 1 > 2. As regard to the impact of selectio o the drift, it is easy to see that P ( d(ξ k+s )>d(ξ k ) 2 <d(ξ k )<d(ξ k+m ) ) ( 1 ( 1 e )) 2N 2N ( 1 e 1 ) 2N >, 2N ad P ( d(ξ k+s )<d(ξ k ) d(ξ k )>d(ξ k+m )>2 ) 1 2N e. Summarizig all the above evets, we have E [ e (d(ξ k+1) d(ξ k )) ] = E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )<d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s ) = d(ξ k+m ) } ]

21 J. He, X. Yao / Artificial Itelligece 127 (2001) E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )>d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k+s )<d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k+s ) = d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k+s )>d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )<d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s ) = d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )>d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )<d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s ) = d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k+s )>d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k+s )<d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k+s ) = d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k+s )>d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )<d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s ) = d(ξ k+m ) } ] + E [ e (d(ξ k+1) d(ξ k )) I { d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k+s )>d(ξ k+m ) } ].

22 78 J. He, X. Yao / Artificial Itelligece 127 (2001) Let p 1 = P ( ) d(ξ k+s )>d(ξ k+m ) d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k )>d b, p 2 = P ( ) d(ξ k+s )>d(ξ k+m ) d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k )>d b, p 3 = P ( ) d(ξ k+s )>d(ξ k+m ) d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k )>d b, p 4 = P ( ) d(ξ k+s )>d(ξ k+m ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), d(ξ k )>d b, p 5 = P ( ) d(ξ k+s )>d(ξ k+m ) d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ), d(ξ k )>d b, p 6 = P ( ) d(ξ k+s )>d(ξ k+m ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ), d(ξ k )>d b, where p 1,...,p 6 are all greater tha (1 e 1 /(2N)) 2N. They represet the probabilities of a populatio driftig away from the optimal solutio after selectio. Accordig to Eq. (15) ad the above aalysis, we have E [ e (d(ξ k+1) d(ξ k )) ] e /2+1 P ( ) ( d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ) O e ) + e 0 P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ) (1 p1 ) + e 1 P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ) p1 + e /2+1 P ( ) ( d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ) O e ) + e 0 P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ) (1 p2 ) + e 1 P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ) p2 + e /2+1 P ( ) ( d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ) O e ) + e 0 P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ) (1 p3 ) + e 1 P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ) p3 + e /2+1 P ( ) ( d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ) O e ) + e 0 P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ) (1 p4 ) + e 1 P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ) p4 + e /2+1 P ( ) ( d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ) O e ) + e 0 P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ) (1 p5 ) Let + e 1 P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ) p5 + e /2+1 P ( ) ( d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ) O e ) + e 0 P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ) (1 p6 ) + e 1 P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ) p6. q 1 = P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m )<d(ξ k+c ), q 2 = P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ), q 3 = P ( ) d(ξ k+c )<d(ξ k ), d(ξ k+m )>d(ξ k+c ),

23 J. He, X. Yao / Artificial Itelligece 127 (2001) q 4 = P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ), q 5 = P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m ) = d(ξ k+c ), q 6 = P ( ) d(ξ k+c ) = d(ξ k ), d(ξ k+m )>d(ξ k+c ). Sice evet I{d(ξ k+c )>d(ξ k )} caot happe, we have 6 q i = 1. i=1 Now we arrive at E [ e (d(ξ k+1) d(ξ k )) ] ( ( q1 O e /2+1 ) + 1 p 1 + e 1 ) p 1 ( ( + q 2 O e /2+1 ) + 1 p 2 + e 1 ) p 2 ( ( + q 3 O e /2+1 ) + 1 p 3 + e 1 ) p 3 ( ( + q 4 O e /2+1 ) + 1 p 4 + e 1 ) p 4 ( ( + q 5 O e /2+1 ) + 1 p 5 + e 1 ) p 5 + q 6 ( O ( e /2+1 ) + 1 p 6 + e 1 p 6 ). Sice 1 p i + e 1 p i < 1, 6 i=1 q i = 1, ad lim O(e /2+1 ) = 0, for sufficietly large, there exists a positive costat ρ<1 such that E [ e (d(ξ k+1) d(ξ k )) ] = 6 ( ( q i O e /2+1 ) + 1 p i + e 1 ) p i i=1 6 q i O ( e /2+1) + i=1 6 ( q i 1 pi + e 1 ) p i <ρ<1. i=1 This shows that {d(ξ k ); k = 0, 1,...} satisfies Coditio 7. The followig aalysis shows that Coditio 8 ca also be satisfied. Let populatio ξ k have the property of d(ξ k ) = d a = d max, which implies that all idividuals i the populatio are the same, i.e., x i = (1 10) where x i is a idividual. The the crossover has o ifluece o the drift, i.e., d(ξ k+c ) = d(ξ k ). There are oly two evets which may happe after mutatio: (1) evet I{d(ξ k+m )<d(ξ k+c )},ad (2) evet I{d(ξ k+m ) = d(ξ k+c )}. Evet I{d(ξ k+m )>d(ξ k+c )} caot happe as d(ξ k+c ) = d max. Similarly, there are followig two cases after selectio: (1) evet I{d(ξ k+s )<d(ξ k )},ad (2) evet I{d(ξ k+s ) = d(ξ k )}. Let the probability of the first evet s happeig be p 0 ad that of the secod be 1 p 0. By summarizig all the above evets, we have

24 80 J. He, X. Yao / Artificial Itelligece 127 (2001) E [ e (d(ξ k+1) d a ) ] d(ξ k ) = d a = E [ e 0 I { }] d(ξ k+s )<d(ξ k ), d(ξ k+m ) = d(ξ k+c ) d(ξ k ) = d a + E [ e 0 I { }] d(ξ k+s )<d(ξ k ), d(ξ k+m )<d(ξ k+c ) d(ξ k ) = d a + E [ e 1 I { }] d(ξ k+s ) = d(ξ k ), d(ξ k+m )<d(ξ k+c ) d(ξ k ) = d a e 0 + e 0 p 0 + e 1 (1 p 0 ) 2 + e. Let D = 2 + e, we arrive at Coditio 8. Accordig to Theorem 10 ad the fact that d a d b = 2 3/4 = /4 2, there exist two positive umbers, δ 1 ad δ 2, such that E [ ] [ τ d(ξ 0 ) d a E τ ] d(ξ 0 ) d a δ1 e δ2(/4 2), where δ 1 ad δ 2 are idepedet of. 5. Discussio o weaker drift coditios Coditios 7 ad 8 may ot be easy to uderstad ad verify for some applicatios, e.g., the subset sum problem. The questio ow is whether we could come up with some weaker ad more ituitive drift coditios. This sectio discusses such coditios. Assume that a distace fuctio d(x) has bee defied. Let d max = max{d(x): x X}, ad d b ad d a be positive umbers such that d b <d a d max. Coditio 9. Let X be a populatio such that d b <d(x)<d a.the E [ d(ξ k+1 ) d(ξ k ) ξ k = X ] C 1, where C 1 > 0. This coditio is a simplified versio of Coditio 7. It implies that whe a populatio X is i the area (d a,d b ), its offsprig teds to drift away from, rather tha move closer to, the optimal solutio. Coditio 10. Let X be a populatio such that d(x) d a.the E [ d(ξ k+1 ) d a ξ k = X ] C 2, where C 2 > 0. This coditio is a simplified versio of Coditio 8. It implies that for a populatio X i the area [d a, + ), its offsprig will ot drift too far away from [d a, + ). The drift is bouded by d a C 2. A iterestig questio ow is: Give a radom sequece {d(ξ k ); k = 0, 1...}, geerated by the EA for solvig the subset sum problem, which satisfies Coditios 9 ad 10,

25 J. He, X. Yao / Artificial Itelligece 127 (2001) whether the average time E[τ ξ 0 = X] is still at least expoetial i the problem size startig from a iitial populatio X with d(x) d a? The aswer to this questio is egative. The two coditios are ot sufficiet to derive a positive aswer although Coditio 9 appears to imply a positive aswer. We will show i the followig usig a example that the radom sequece {d(ξ k ); k = 0, 1,...} satisfies both coditios, but EA s average computatio time is polyomial i the problem size. Cosider a family of subset sum problems, {W 1,W 2,...,W,...},where W ={w 1,...,w }, w 1 = w 2 = =w 1 = 2, w = 2C 0 + 1, C = 2C 0 + 1, where C 0 /2 is a iteger greater tha 2. It is obvious that the subset {w } is the uique optimal solutio. Ay subset of W {w } with o more tha C 0 elemets is a feasible solutio. The EA for solvig the above family of subset sum problems follows the framework give i Sectio 2, but its implemetatio is much simpler tha the EAs used i previous sectios. The same chromosome represetatio is used as before. The EA used here is a (1 + 1) EA. It uses bit mutatio. A sigle bit s i is chose at radom from a idividual ad flipped. If the offsprig is ifeasible, the paret will be retaied. A ew itermediate populatio ξ k+m ={y} is formed after mutatio. I terms of selectio, the idividual with better fitess betwee ξ k ad ξ k+m has a higher survival probability of p, ad that with worse fitess has a survival probability of q = 1 p, where p>q. The ew itermediate populatio after selectio will be ξ k+s ={z}. Sice s = (0 01) is the optimal solutio, defie d(x)= s i s ad d max = max{d(x): x is a feasible solutio}. i=1 It is easy to see that if d(x 1 )>d(x 2 )>1, the f(x 1 )>f(x 2 ). Let d b = 2add a = d max.wefirstverifythat{d(ξ k ); k = 0, 1,...} satisfies Coditio 9. Let ξ k ={x} be the curret populatio with d b <d(x)<d a,adξ k+m ={y} be the itermediate populatio after mutatio. The oe of the followig three evets may happe: evet I{d(y) > d(x)},eveti{d(y) = d(x)} or evet I{d(y) < d(x)}. Sice the mutatio is a bit mutatio, it is easy to show that the probability of evet I{d(y)> d(x)} happeig is P ( ) d(y) > d(x) d b <d(x)<d a = ( d(x) 1)/. The probability of evet I{d(y)= d(x)} happeig is P ( ) d(y) = d(x) d b <d(x)<d a = 1/. The probability of evet I{d(y)< d(x)} happeig is P ( ) d(y) < d(x) d b <d(x)<d a = d(x)/. If evet I{d(y) < d(x)} has happeed, oe of the followig two evets may happe: evet I{d(z) < d(x)} or evet I{d(z) = d(x)}. Accordig to our selectio scheme, the

26 82 J. He, X. Yao / Artificial Itelligece 127 (2001) probability of evet I{d(z) = d(x)} happeig is p ad the probability of evet I{d(z) < d(x)} happeig is q. If evet I{d(y) > d(x)} has happeed, oe of the followig two evets may happe: evet I{d(z) > d(x)} or evet I{d(z) = d(x)}. It is easy to see that the probability of evet I{d(z)> d(x)} happeig is p ad the probability of evet I{d(z) = d(x)} happeig is q. Summarizig all the above evets, we have E [ ] d(ξ k+1 ) d(ξ k ) d b <d(ξ k )<d a = E [ (d(z) d(x))i { d(y) > d(x), d(z) = d(y) } ] d b <d(ξ k )<d a + E [( d(z) d(x) ) I { d(y) > d(y), d(z) = d(x) } ] d b <d(ξ k )<d a + E [( d(z) d(x) ) I { d(y) < d(x), d(z) = d(y) } ] d b <d(ξ k )<d a + E [( d(z) d(x) ) I { d(y) < d(x), d(z) = d(x) } ] d b <d(ξ k )<d a = p d(x) 1 p d a 1 q d(x) q d a. Sice d a = d max <C 0 /2 adp>q, E [ ] /2 1 d(ξ k+1 ) d(ξ k ) d b <d(ξ k )<d a p q /2 1 (p q) 4 whe is sufficietly large. Let C 1 = (p q)/4 > 0. The Coditio 9 is satisfied. Now we verify that {d(ξ k ); k = 0, 1,...} satisfies Coditio 10. From the defiitio of feasible solutios, we kow that ay subset with a cardiality greater tha C 0 is a ifeasible solutio. For ay give idividual x with d(x)= d a = d max, let y be its offsprig after mutatio. The oe of the followig evets may happe: evet I{d(y)= d a } or evet I{d(y)< d a }. The probability of evet I{d(y)= d a } happeig is P ( ) d(y) = d(x) d(x) = d a = d max = ( da )/. The probability of evet I{d(y)< d a } happeig is P ( ) d(y) < d(x) d(x) = d a = d max = da /. If evet I{d(y) = d a } happes, the evet I{d(z) = d a } happes with probability 1. If evet I{d(y) < d a } happes, oe of the followig two may happe: evet I{d(z) = d a } or evet I{d(z) < d a }. It is clear that the probability of evet I{d(z) = d a } happeig is p ad the probability of evet I{d(z)< d a } happeig is q. Summarizig all the above evets, we obtai E [ ] d(ξ k+1 ) d a d(ξ k ) d a = E [( ) { } ] d(z) d a I d(z) = d(y),d(y)< da d(ξk ) d a + E [( ) { ] d(z) d a I d(z) = da } d(ξ k ) d a = q d a. Let C 2 = qd a /. The Coditio 10 is satisfied.