8 -1 Dynamic Programming Khujaev Otabek Fibonacci sequence Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, … F i = i if i 1 F i = F i-1 + F i-2. - презентация

1 8 -1 Dynamic Programming Khujaev Otabek

2 8 -2 Fibonacci sequence Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, … F i = i if i 1 F i = F i-1 + F i-2 if i 2 Solved by a recursive program: Much replicated computation is done. It should be solved by a simple loop.

3 8 -3 Dynamic Programming Dynamic Programming is an algorithm design method that can be used when the solution to a problem may be viewed as the result of a sequence of decisions

4 8 -4 The shortest path To find a shortest path in a multi-stage graph Apply the greedy method : the shortest path from S to T : = 8

5 8 -5 The shortest path in multistage graphs e.g. The greedy method can not be applied to this case: (S, A, D, T) = 23. The real shortest path is: (S, C, F, T) = 9.

6 8 -6 Dynamic programming approach Dynamic programming approach (forward approach): d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} d(A,T) = min{4+d(D,T), 11+d(E,T)} = min{4+18, 11+13} = 22.

7 8 -7 Dynamic programming d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F, T)} = min{9+18, 5+13, 16+2} = 18. d(C, T) = min{ 2+d(F, T) } = 2+2 = 4 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)} = min{1+22, 2+18, 5+4} = 9. The above way of reasoning is called backward reasoning.

8 8 -8 Backward approach (forward reasoning) d(S, A) = 1 d(S, B) = 2 d(S, C) = 5 d(S,D)=min{d(S, A)+d(A, D),d(S, B)+d(B, D)} = min{ 1+4, 2+9 } = 5 d(S,E)=min{d(S, A)+d(A, E),d(S, B)+d(B, E)} = min{ 1+11, 2+5 } = 7 d(S,F)=min{d(S, A)+d(A, F),d(S, B)+d(B, F)} = min{ 2+16, 5+2 } = 7

9 8 -9 d(S,T) = min{d(S, D)+d(D, T),d(S,E)+ d(E,T), d(S, F)+d(F, T)} = min{ 5+18, 7+13, 7+2 } = 9

10 8 -10 Principle of optimality Principle of optimality: Suppose that in solving a problem, we have to make a sequence of decisions D 1, D 2, …, D n. If this sequence is optimal, then the last k decisions, 1 k n must be optimal. e.g. the shortest path problem If i, i 1, i 2, …, j is a shortest path from i to j, then i 1, i 2, …, j must be a shortest path from i 1 to j In summary, if a problem can be described by a multistage graph, then it can be solved by dynamic programming.

11 8 -11 Forward approach and backward approach: Note that if the recurrence relations are formulated using the forward approach then the relations are solved backwards. i.e., beginning with the last decision On the other hand if the relations are formulated using the backward approach, they are solved forwards. To solve a problem by using dynamic programming: Find out the recurrence relations. Represent the problem by a multistage graph. Dynamic programming

12 8 -12 The resource allocation problem m resources, n projects profit p(i, j) : j resources are allocated to project i. maximize the total profit.

13 8 -13 The multistage graph solution The resource allocation problem can be described as a multistage graph. (i, j) : i resources allocated to projects 1, 2, …, j e.g. node H=(3, 2) : 3 resources allocated to projects 1, 2.

14 8 -14 Find the longest path from S to T : (S, C, H, L, T), =13 2 resources allocated to project 1. 1 resource allocated to project 2. 0 resource allocated to projects 3, 4.

15 8 -15 The traveling salesperson (TSP) problem e.g. a directed graph : Cost matrix:

16 8 -16 A multistage graph can describe all possible tours of a directed graph. Find the shortest path: (1, 4, 3, 2, 1) =17 The multistage graph solution

17 8 -17 The representation of a node Suppose that we have 6 vertices in the graph. We can combine {1, 2, 3, 4} and {1, 3, 2, 4} into one node. (3),(4,5,6) means that the last vertex visited is 3 and the remaining vertices to be visited are (4, 5, 6).

18 8 -18 The dynamic programming approach Let g(i, S) be the length of a shortest path starting at vertex i, going through all vertices in S and terminating at vertex 1. The length of an optimal tour : The general form: Time complexity: ( ),( ) (n-k) (n-1)