57
1 Combinatorial Problem

africa
• Category

## Documents

• view

33

0

description

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}. - PowerPoint PPT Presentation

### Transcript of Combinatorial Problem

1

Combinatorial Problem

2

Graph Partition

Undirected graph G=(V,E)

V=V1V2, V1V2=

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 xis.t. sixiC xi={0,1}

5

Binary coding

0-->not in the knapsack1-->in the knapsackThe problem is how to get feasible solution

6

Penalty function

fitness= pi xi+g(x)If si xiC, then g(x)=0If 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 genes2) 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

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

Schemata analysis

14

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

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

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

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

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

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

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

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

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

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 edgesSubtour chunksHeuristic 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 permutations6 7 | 4 3 1 8 | 5 28 1 | 2 3 4 5 | 6 7by swapping the indicated substrings.The results are not permutations:6 7 2 3 4 5 5 28 1 4 3 1 8 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 “belo

w” 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 firs

t offspring are 3 and 2.However, that the selection of city 2 requires selection of c

ity 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 parentc1=(1 2 3 4 7 6 9 8 5)Similarly c2=(4 1 2 8 5 6 7 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

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 --> 126543789insertion: selects a city and inserts it in a random place 123456789 --> 124567839displacement: selects a subtour and inserts it in a random pl

ace 123456789 --> 127834569reciprocal exchange: swaps two cities 123456789 --> 126453789

51

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.

52

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

53

Greedy algorithm

Sort the vertexes according to their weight

Choose the vertex according to the rank, try to color that vertex

Eg: (9 7 8 4 2 6 5 1 3)one colorVertex 9 4 2 are colored.The total weight is 32

54

Eg2: (7 3 4 1 6 5 2)

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.

55

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.

56

N Queens

Place N mutually un-attacking Queens on an N N chess board.

(Queens attack on rows, columns, and diagonals.)

57

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.