DP8 矩阵链相乘 Matrix Chain Multiplication @geeksforgeeks

思路是在矩阵链的每一个地方分别分开，然后找最小

Given a sequence of matrices, find the most efficient way to multiply these matrices together. The problem is not actually to perform the multiplications, but merely to decide in which order to perform the multiplications.

We have many options to multiply a chain of matrices because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result will be the same. For example, if we had four matrices A, B, C, and D, we would have:

    (ABC)D = (AB)(CD) = A(BCD) = ....

However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. For example, suppose A is a 10 × 30 matrix, B is a 30 × 5 matrix, and C is a 5 × 60 matrix. Then,

    (AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 operations
    A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 operations.

Clearly the first method is the more efficient.

Given an array p[] which represents the chain of matrices such that the ith matrix Ai is of dimension p[i-1] x p[i]. We need to write a function MatrixChainOrder() that should return the minimum number of multiplications needed to multiply the chain.

  Input: p[] = {40, 20, 30, 10, 30}   
  Output: 26000  
  There are 4 matrices of dimensions 40x20, 20x30, 30x10 and 10x30.
  Let the input 4 matrices be A, B, C and D.  The minimum number of 
  multiplications are obtained by putting parenthesis in following way
  (A(BC))D --> 20*30*10 + 40*20*10 + 40*10*30

  Input: p[] = {10, 20, 30, 40, 30} 
  Output: 30000 
  There are 4 matrices of dimensions 10x20, 20x30, 30x40 and 40x30. 
  Let the input 4 matrices be A, B, C and D.  The minimum number of 
  multiplications are obtained by putting parenthesis in following way
  ((AB)C)D --> 10*20*30 + 10*30*40 + 10*40*30

  Input: p[] = {10, 20, 30}  
  Output: 6000  
  There are only two matrices of dimensions 10x20 and 20x30. So there 
  is only one way to multiply the matrices, cost of which is 10*20*30

1) Optimal Substructure:
A simple solution is to place parenthesis at all possible places, calculate the cost for each placement and return the minimum value. In a chain of matrices of size n, we can place the first set of parenthesis in n-1 ways. For example, if the given chain is of 4 matrices. let the chain be ABCD, then there are 3 way to place first set of parenthesis: A(BCD), (AB)CD and (ABC)D. So when we place a set of parenthesis, we divide the problem into subproblems of smaller size. Therefore, the problem has optimal substructure property and can be easily solved using recursion.

Minimum number of multiplication needed to multiply a chain of size n = Minimum of all n-1 placements (these placements create subproblems of smaller size)

2) Overlapping Subproblems
Following is a recursive implementation that simply follows the above optimal substructure property.

package DP;

public class MatrixChainMultiplication {

	public static void main(String[] args) {
		int[] dm = {1, 2, 3, 4};
		int begin = 1;
		int end = dm.length-1;
		System.out.println(matrixChainOrderRec(dm, begin, end));
		System.out.println(matrixChainOrderDP(dm, dm.length));
	}

	// Matrix Ai has dimension dm[i-1] x dm[i] for i = 1..n
	public static int matrixChainOrderRec(int[] dm, int begin, int end){
		if(begin == end){
			return 0;
		}
		int min = Integer.MAX_VALUE;
		
		// place parenthesis at different places between first and last matrix,
	    // recursively calculate count of multiplications for each parenthesis 
	    // placement and return the minimum count
		for(int k=begin; k<end; k++){
			int count = matrixChainOrderRec(dm, begin, k) +
							 matrixChainOrderRec(dm, k+1, end) +
							 dm[begin-1]*dm[k]*dm[end];
			min = Math.min(min, count);
		}
		return min;
	}
	
	// Matrix Ai has dimension dm[i-1] x dm[i] for i = 1..n
	// Time Complexity: O(n^3), Auxiliary Space: O(n^2)
	public static int matrixChainOrderDP(int[] dm, int n){
		/* For simplicity of the program, one extra row and one extra column are
	       allocated in dp[][].  0th row and 0th column of m[][] are not used */
		int[][] dp = new int[n+1][n+1];
		/* dp[i,j] = Minimum number of scalar multiplications needed to compute
	       the matrix A[i]A[i+1]...A[j] = A[i..j] where dimension of A[i] is p[i-1] x p[i] */
		
		// cost is zero when multiplying one matrix.
		for(int i=1; i<n; i++){
			dp[i][i] = 0;
		}
		
		// L is chain length. L starts from 2
		for(int L=2; L<n; L++){
			for(int begin=1; begin<n-L+1; begin++){
				int end = begin+L-1;
				dp[begin][end] = Integer.MAX_VALUE;
				for(int k=begin; k<=end-1; k++){
					// q = cost/scalar multiplications
					int q = dp[begin][k] + dp[k+1][end] + dm[begin-1]*dm[k]*dm[end];
					dp[begin][end] = Math.min(dp[begin][end], q);
				}
			}
		}
		return dp[1][n-1];
	}
	
}

http://www.geeksforgeeks.org/dynamic-programming-set-8-matrix-chain-multiplication/