probabilistic analysis of the 1 1 evolutionary algorithm
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PROBABILISTIC ANALYSIS OF THE ( 1 + 1 )-EVOLUTIONARY ALGORITHM - PowerPoint PPT Presentation

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 1/59 MIT: EVOLUTONARY COMPUTATION 2/59 WE ARE ALWAYS SEARCHING &


  1. 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 1/59

  2. MIT: EVOLUTONARY COMPUTATION 2/59

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

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

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

  6. EVOLUTIONARY ALGORITHMS 6/59

  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 Pioneers: John Holland, Lawrence J. Fogel, Ingo Rechenberg, . . . 7/59

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

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

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

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

  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 Scalable to higher dimensional problems 12/59

  13. 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 13/59

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

  15. SIMPLEST VERSION: 1 PARENT, 1 CHILD, AND MUTATION ONLY Algorithm ( 1 + 1 ) -EA Choose an initial string x ∈ { 0 , 1 } n 1 uniformly at random Repeat until a terminating condition 2 (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 15/59

  16. ANALYSIS OF (1 + 1)-EA UNDER ONEMAX Known results for O NE M AX f ( x ) = x 1 + · · · + 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 B¨ ack (1992): transition probabilities M¨ uhlenbein (1992): E ( X n ) ≈ n log n Droste et al. (1998, 2002): E ( X n ) ≍ n log n · · · Linear functions � w i x i O NE M AX function Doerr et al. lower bound Jagerskupper upper bound (2010) ( 1 − o ( 1 )) en log ( n ) (2011) 2 . 02 en log ( n ) Sudholt lower bound Doerr et al. upper bound 2010 en log ( n ) − 2 n log log ( n ) (2010) 1 . 39 en log ( n ) Doerr et al. Witt upper bound en log ( n ) − Θ( n ) (2011) (2013) en log ( n ) + O ( n ) Approaches used: Markov chain, martingale, coupon collection, . . . 16/59

  17. KNOWN BOUNDS FOR E ( X n ) UNDER O NE M AX O NE M AX Droste et al. (2002) en ( log n + γ ) Doerr et al. (2011) en log n − 0 . 1369 n = e ( n + 1 2 ) log n − c 1 n + O ( 1 ) Our result c 1 = 1 . 8254 17883 . . . Lehre & Witt (2014) en log n − 7 . 81791 n − O ( log n ) en log n − Θ( n ) Doerr et al. (2011) Sudholt (2010) en log n − 2 n log log n Doerr et al. (2010) ( 1 − o ( 1 )) en log n Droste et al. (2002) 0 . 196 n log n 17/59

  18. 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 1 ≈ − 1 . 9 X n d en − log n − c 1 − → log Exp ( 1 ) (double-exponential) their results had remained obscure in the EA-literature 18/59

  19. OUR RESULTS � log n � E ( X n ) = en log n + c 1 n + 1 2 e log n + c 2 + O n � 1 � �� c 1 = − e log 2 − γ − φ 1 2 ≈ − 1 . 89254 17883 44686 82302 25714 . . . , where γ is Euler’s constant, � z � � S 1 ( t ) − 1 1 φ 1 ( z ) := d t , t 0 z ℓ ( ℓ − j )( 1 − z ) j � � S 1 ( z ) := . ℓ ! j ! ℓ � 1 0 � j <ℓ c ′ k log n + c k � Indeed E ( X n ) ∼ n n k k � 0 19/59

  20. LIMIT GUMBEL DISTRIBUTION � X n � → e − e − x en − log n + log 2 − φ 1 ( 1 2 ) � x P φ 1 ( 1 2 ) ≈ − 0 . 58029 56799 84283 81332 29240 . . . right: e − x − e − x left: n = 10 .. 30 20/59

  21. OUR APPROACH RECURRENCE ⇒ ASYMPTOTICS 21/59

  22. THE RANDOM VARIABLE X n , m � n � � p m q n − m X n , m X n := m 0 � m � n 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 X n , m ) . Then Q n , 0 ( t ) = 1 and � λ n , n − m ,ℓ Q n , m − ℓ ( t ) t 1 � ℓ � m Q n , m ( t ) = � � � 1 − 1 − λ n , n − m ,ℓ t 1 � ℓ � m for 1 � m � n , where ( P ( m 1 ’s �→ m + ℓ 1 ’s ) ) � � n � m �� n − m � 1 − 1 � ( n − 1 ) − ℓ − 2 j λ n , m ,ℓ := n j j + ℓ 0 � j � min { m , n − m − ℓ } 22/59

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

  24. 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 ) 24/59

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

  26. EXPONENTIAL DISTRIBUTION 26/59

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

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

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

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

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

  32. SKETCH OF PROOF: BY INDUCTION r en t � Q n , m ( t ) ∼ � � r 1 − 1 − t en 1 � r � m 1 � � � e s / ( en ) � = ⇒ Q n , m ∼ = ⇒ Exp ( r ) 1 − s r 1 � r � m 1 � r � m Fails when m → ∞ 30/59

  33. SKETCH OF PROOF: BY INDUCTION r en t � Q n , m ( t ) ∼ � � r 1 − 1 − t en 1 � r � m 1 � � � e s / ( en ) � = ⇒ Q n , m ∼ = ⇒ Exp ( r ) 1 − s r 1 � r � m 1 � r � m Fails when m → ∞ Let Y m := � 1 � r � m Exp ( r ) . Then as m → ∞ e − i θ e − i θ � � � e ( Y m − H m ) i θ � r r = e − γ i θ Γ( 1 − i θ ) E = → 1 − i θ 1 − i θ r r 1 � r � m r � 1 30/59

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