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Combinatorial Problem

Combinatorial Problem. Graph Partition. Undirected graph G=(V,E) V=V1 V2, V1V2= minimize the number of edges connect V1 and V2. Graph Partition. Binary coding: 0-->V1, 1-->V2 Fitness function: the number of edges connect V1 and V2. Knapsack. max pi xi s.t. sixiC xi={0,1}.

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Combinatorial Problem

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  1. Combinatorial Problem

  2. Graph Partition Undirected graph G=(V,E) V=V1V2, V1V2= minimize the number of edges connect V1 and V2

  3. Graph Partition Binary coding: 0-->V1, 1-->V2 Fitness function: the number of edges connect V1 and V2

  4. Knapsack max pi xi s.t. sixiC xi={0,1}

  5. Binary coding 0-->not in the knapsack 1-->in the knapsack The problem is how to get feasible solution

  6. Penalty function fitness= pi  xi+g(x) If si  xiC, then g(x)=0 If si  xi>C, then g(x)=(C- si  xi)

  7. Repair algorithm If the chromosome is not feasible, choose some genes and convert them to 0 1) Random choose the genes 2) Choose according to the order of pi/si

  8. Crossover c1=p1 AND p2 c2=p1 OR p2

  9. Permutations: Now it Gets Interesting! Use permutations when bit strings fall short: TSP Knapsack Graph coloring N Queens

  10. TSP The TSP was proved to be NP-hard. It arises in numerous applications and the number of cities might be quite significant: 1) Circuit board drilling application with up to 17,000 cities 2) X-ray crystallography instances with up to 14,000 cities 3) VLSI fabrication with as many as 1.2 million cities TSP has an extremely easy fitness function. In a population of tours, we can easily compare any two of them. However the choice of the representation of a tour and the choice of operators to be used are far from clear. There is an agreement in the GA community that the binary representation of tours is not well suited for the TSP.

  11. TSP The expression of the chromosome Ordinal Adjacency Path

  12. Ordinal expression The ith gene belongs to the region [1,n-i+1]. It determines which city will be chosen from the city list. City list (1 2 3 4 5 6 7 8 9) Chromosome ( 1 1 2 1 4 1 3 1 1) Then the path is (1 2 4 3 8 5 9 6 7)

  13. Adjacency expression Why adjacency expression? Schemata analysis

  14. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path:

  15. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path:

  16. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2

  17. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2

  18. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2-4

  19. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2-4

  20. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2-4 3-8

  21. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2-4 3-8

  22. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2-4-3-8

  23. Adjacency expression If and only if go from city i to j, city j is on the position i chromosome (2 4 8 3 9 7 1 5 6) path: 1-2-4-3-8-5-9-6-7 For every path there is only one adjacency expression, some adjacency expression may be illegal. (2 4 8 1 9 3 5 7 6) will leads to (1 2 4 1)

  24. Crossover Alternating edges Subtour chunks Heuristic crossover

  25. Alternating edges Build an offspring by choosing (at random) an edge from the first parent, then selects an appropriate edge from the second parent, etc.----The operator extends the tour by choosing edges from alternating parents. If the new edge (from one of the parents) introduces a cycle into the current (still partial) tour, the operator selects instead a (random) edge from the remaining edges which does not introduce cycles. It can be considered as uniform crossover.

  26. p1=(2 3 8 7 9 4 1 5 6) tour1=(1 2 3 8 5 9 6 4 7) p2=(7 5 1 6 9 2 8 4 3) tour2=(1 7 8 4 6 2 5 9 3) The offspring might be c=(2 5 8 7 9 4 1 6 3) tour=(1 2 5 9 3 8 6 4 7) The process starts from the edge (1,2) from p1. The 6th edge should be (6,4) because of the edge (1,2). The 8th edge could be neither (8 5) nor (8 4). It could only be one of (8 3) and (8 6). Since (8 3) will introduce a premature cycle, the 8th edge has to be (8 6).

  27. Subtour chunks Construct an offspring by choosing a (random length) subtour from one of the parents, then choosing a (random length) subtour from another parent, etc. ---- the operator extends the tour by choosing edges from alternating parents, Again if some edge (from one of the parents) introduces a cycle into the current (still partial) tour, the operator selects instead a (random) edge from the remaining edges which does not introduce cycles.

  28. Heuristic crossover Build an offspring by choosing a random city as the starting point for the offspring’s tour. Then it compares the two edges (from both parents) leaving this city and selects the better (shorter) edge. The city on the other end of the selected edge serves as a starting point in selecting the shorter of the two edges leaving this city, etc. If, at some stage, a new edge would introduce a cycle into the partial tour, then the tour is extended by a random edge from the remaining edges which does not introduce cycles.

  29. Path expression It is perhaps the most natural representation of a tour. A tour(5 1 7 8 9 4 6 2 3) is represented simply as (5 1 7 8 9 4 6 2 3)

  30. How to Make a initial solution for( i = 1; i <= N; i++ ) P[i] = i; for( i = 1; i <= N; i++ ) { k = random_int( N - i ); interchange P[i] with P[i + k]; } This uniformly generates permutations of {1;  ;N}

  31. Crossing Over: the Problem Cross over the two permutations 6 7 | 4 3 1 8 | 5 2 8 1 | 2 3 4 5 | 6 7 by swapping the indicated substrings. The results are not permutations: 6 7 2 3 4 5 52 81 4 3 18 6 7

  32. Repairing the Crossover Damage PMX: "partially matched crossover" OX: "ordered crossover" CX: "cycle crossover"

  33. PMX: Partially Matched Crossover Build an offspring by choosing a subsequence of a tour from one parent and preserving the order and position of as many cities as possible from the other parent. A subsequence of a tour is selected by choosing two random cut points, which serve as boundaries for swapping operations.

  34. p1=(1 2 3 | 4 5 6 7 | 8 9) p2=(4 5 2 | 1 8 7 6 | 9 3) First the segments between cut points are swapped c1=(x x x | 1 8 7 6 | x x) c2=(x x x | 4 5 6 7 | x x) This swap defines also a series of mappings: 1--4 8--5 7--6 6--7

  35. We can fill further cities (from the original parents), for which there is no conflict: p1=(1 2 3 | 4 5 6 7 | 8 9) p2=(4 5 2 | 1 8 7 6 | 9 3) c1=(x 2 3 | 1 8 7 6 | x 9) c2=(x x 2 | 4 5 6 7 | 9 3)

  36. Finally the first x in c1 (which should be 1, but there is a conflict) is replaced by 4 because of the mapping 1--4. Similarly the second x in the offspring c1 is replaced by 5, and the x and x in the c2 are 1 and 8. c1=(4 2 3 | 1 8 7 6 | 5 9) c2=(1 8 2 | 4 5 6 7 | 9 3)

  37. OX: Ordered Crossover Build offspring by choosing a subsequence of a tour from one parent and preserving the relative order of cities from the other parent. p1=(1 2 3 | 4 5 6 7 | 8 9) p2=(4 5 2 | 1 8 7 6 | 9 3)

  38. First the segments between cut points are copied into offspring c1=(x x x | 4 5 6 7 | x x) c2=(x x x | 1 8 7 6 | x x)

  39. Next, starting from the second cut point of one parent, the cities from the other parent are copied in the same order, omitting symbols already present. Reaching the end of the string, we continue from the first place of the string. The sequence of the cities in the second parent (from the second cut point) is (9 3 4 5 2 1 8 7 6) (p2=(4 5 2 | 1 8 7 6 | 9 3))

  40. After removal of cities 4, 5, 6 and 7, which are already in the first offspring, we get (9 3 2 1 8) This sequence is placed in the first offspring (starting from the second cut point) c1=(2 1 8 | 4 5 6 7 | 9 3) Similarly we get the other offspring c2=(3 4 5 | 1 8 7 6 | 9 2)

  41. The OX crossover exploits a property of the path representation, that the order of cities (not their positions) are important, i.e., the two tours (9 3 4 5 2 1 8 7 6) (4 5 2 1 8 7 6 9 3) are in fact identical

  42. CX: Cycle Crossover Build offspring in such a way that each city (and its position) comes from one of the parents. It preserves the absolute position of the elements in the parent sequence.

  43. CX: Cycle Crossover p1=(1 2 3 4 5 6 7 8 9) p2=(4 1 2 8 7 6 9 3 5) would produce the first offspring by taking the first city from the first parent c1=(1 x x x x x x x x) and c2=(4 x x x x x x x x)

  44. p1=(1 2 3 4 5 6 7 8 9) p2=(4 1 2 8 7 6 9 3 5) Since every city in the offspring should be taken from one of its parents (from the same position), we do not have any choice now: the next city to be considered must be city 4, as the city from the p2 just “below” the selected city 1. In p1 this city is at position ‘4’, thus c1=(1 x x 4 x x x x x) and c2=(4 x x 8 x x x x x)

  45. p1=(1 2 3 4 5 6 7 8 9) p2=(4 1 2 8 7 6 9 3 5) This, in turn, implies city 8, as the city from p2 just “below” the selected city 4. Thus c1=(1 x x 4 x x x 8 x) Following this rule, the next cities to be included in the first offspring are 3 and 2. However, that the selection of city 2 requires selection of city 1, which is already on the list----thus we have completed a cycle c1=(1 2 3 4 x x x 8 x)

  46. p1=(1 2 3 4 5 6 7 8 9) p2=(4 1 2 8 7 6 9 3 5) The remaining cities are filled from the other parent c1=(1 2 3 4 769 8 5) Similarly c2=(4 1 2 8 567 3 9)

  47. Comparison p1=(1 2 3 | 4 5 6 7 | 8 9) p2=(4 5 2 | 1 8 7 6 | 9 3) PMX c1=(4 2 3 | 1 8 7 6 | 5 9) c2=(1 8 2 | 4 5 6 7 | 9 3) OX c1=(2 1 8 | 4 5 6 7 | 9 3) c2=(3 4 5 | 1 8 7 6 | 9 2) CX c1=(1 5 2 4 8 6 7 9 3) c2=(4 2 3 1 5 7 6 8 9)

  48. Exercise p1=(9 8 7 | 6 5 4 3 | 2 1) p2=(4 5 2 | 1 8 7 6 | 9 3) PMX c1=( ) c2=( ) OX c1=( ) c2=( ) CX c1=( 9 5 x 6 x x x x x) c2=( )

  49. Answer p1=(9 8 7 | 6 5 4 3 | 2 1) p2=(4 5 2 | 1 8 7 6 | 9 3) PMX c1=(9 5 4 | 1 8 7 6 | 2 3) c2=(7 8 2 | 6 5 4 3 | 9 1) OX c1=(1 8 7 | 6 5 4 3 | 9 2) c2=(5 4 3 | 1 8 7 6 | 2 9) CX c1=(9 5 7 6 8 4 3 2 1) c2=(4 8 2 1 5 7 6 9 3)

  50. Mutation inversion: the substring between the two selected points is reversed 123456789 --> 126543789 insertion: selects a city and inserts it in a random place 123456789 --> 124567839 displacement: selects a subtour and inserts it in a random place 123456789 --> 127834569 reciprocal exchange: swaps two cities 123456789 --> 126453789

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