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 expression of the chromosome Ordinal Adjacency Path

11 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 ( ) Chromosome ( ) Then the path is ( )

12 Adjacency expression If and only if go from city i to j, city j is on the position i chromosome ( ) path ( ) For every path there is only one adjacency expression, some adjacency expression may be illegal. ( ) will leads to ( )

13 Crossover Alternating edges Subtour chunks Heuristic crossover

14 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.

15 p1=( ) tour1=( ) p2=( ) tour2=( ) The offspring might be c=( ) tour=( ) 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).

16 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.

17 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.

18 Path expression It is perhaps the most natural representation of a tour. A tour( ) is represented simply as ( )

19 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}

20 Crossing Over: the Problem Cross over the two permutations 6 7 | | | | 6 7 by swapping the indicated substrings. The results are not permutations:

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

22 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.

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

24 We can fill further cities (from the original parents), for which there is no conflict: c1=(x 2 3 | | x 9) c2=(x x 2 | | 9 3)

25 Finally the first x in c1 (which should be 1, but there is a conflict) is replaced by 4 because of the mapping 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 | | 5 9) c2=(1 8 2 | | 9 3)

26 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 | | 8 9) p2=(4 5 2 | | 9 3)

27 First the segments between cut points are copied into offspring c1=(x x x | | x x) c2=(x x x | | x x)

28 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 ( ) (p2=(4 5 2 | | 9 3))

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

30 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 ( ) ( ) are in fact identical

31 CX: Cycle Crossover Build offspring in such a way that each city (and its position) comes from one of the parents. p1=( ) p2=( ) 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)

32 Since every city in the offspring should be taken from one of its parents (from the same position), we dot 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=(1 x x 8 x x x x x)

33 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=( x x x 8 x)

34 The remaining cities are filled from the other parent c1=( ) Similarly c2=( ) The CX preserves the absolute position of the elements in the parent sequence.

35 Comparison p1=(1 2 3 | | 8 9) p2=(4 5 2 | | 9 3) PMX c1=(4 2 3 | | 5 9) c2=(1 8 2 | | 9 3) OX c1=(2 1 8 | | 9 3) c2=(3 4 5 | | 9 2) CX c1=( ) c2=( )

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

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

38 Mutation inversion: the substring between the two selected points is reversed insertion: selects a city and inserts it in a random place displacement: selects a subtour and inserts it in a random place reciprocal exchange: swaps two cities

39 Knapsack again: Permutation expression Let the GA creatures be permutations of (1;  ;n): Interpret this as a object placement: If the placement of one object cause the solution infeasible, delete this object and consider the next one.

40 Graph coloring Undirected graph G=(V,E) and n colors Every two vertexes connected can not be given the same color Maximize the total weight of the colored vertexes

41 Greedy algorithm Sort the vertexes according to their weight Choose the vertex according to the rank, try to color that vertex Eg: ( ) one color Vertex are colored. The total weight is 32

42 Eg2: ( ) one color Vertex 7 will be chosen and the total weight is 15. The optimal solution is (1 5 3) and the total weight is 35. Only the weight of vertex 7 is larger than 35, greedy algorithm will get the global optimum.

43 Permutation expression Let the GA creatures be permutations of (1;  ;n) Individual = (c1;  ; cN). Interpret this as the order of coloring process. First try to give a color to c1, then consider c2, and so on.

44 N Queens Place N mutually un-attacking Queens on an N  N chess board. (Queens attack on rows, columns, and diagonals.)

45 Permutation Is Placement Let the GA creatures be permutations of (1; 2;  ;N): Individual = (c1;  ; cN). Interpret this as a Queens placement: The Queen in row k is in column ck. Fitness: number of Queens unattacked by other Queens.