Matrix-chain multiplication

问题描述

Input:

• {p₀, p₁ ,... , p_n}. a chain {A₁, A₂,..., A_n} of n matrices for i = 1, 2,...,n , matrix A_i has dimension p_i-1 * p_i.

值得注意的该问题的输入定义及其巧妙，不仅定义了矩阵的行以及列，还说明了它们是可乘的。

Output:

• fully parenthesize the product A₁, A₂,..., A_n in a way that minimizes the number of scalar multiplications.

定义1, A product of matrices is fully parenthesized, if it is either a single matrix or the product of two fully parenthesized matrix products, surrounded by parentheses.

当一个矩阵链乘问题是 fully parenthesized 时，就可以使用两个矩阵相乘的简单算法进行处理整个矩阵链。对于p * q的矩阵乘以q * r的矩阵，其所进行的 scalar multiplications 的次数为p * q * r(参见下方伪代码中第8行的执行次数)。而问题中的 the number of scalar multiplications 为计算以该算法计算整个矩阵链乘时所进行的 scalar multiplications 次数。

两个矩阵相乘的简单算法的算法伪代码如下：

MATRIX-MULTIPLY(A,B)
1. if A.columns != B.rows 
2.     error “incompatible dimensions”
3. else let C be a new A.rows * B.columns matrix 
4.     for i = 1 to A.rows
5.        for j = 1 to B.columns 
6.            c[i,j] = 0
7.            for k = 1 to A.columns 
8.               c[i,j] = c[i,j] + A[i,k] * B[k,j]
9.      return C

暴力算法分析

Denote the number of alternative parenthesizations of a sequence of n matrices by P(n).

• When n = 1, P(n) = 1。
• When n > 1, a fully parenthesized matrix product is the product of two fully parenthesized matrix subproducts, and the split between the two subproducts may occur between the kth and (k + 1)st matrices for any k = 1,2,3,...,n-1.

so

The number of solutions is thus exponential in n, and the brute-force method of exhaustive search makes for a poor strategy when determining how to optimally parenthesize a matrix chain.

Applying dynamic programming

Step 1: The structure of an optimal parenthesization

定理1，若A_i~j 的最优解中最后一次乘法在k处，即在A_k与A_k+1之间，那么，A_i~j的最优解的乘法序列的可以在在k处断开矩阵乘法链。则，A_i~j的最优解乘法序列对应子问题A_i~k的解必须是子问题计算A_i *...* A_k的优化解，A_i~j的最优解乘法序列对应子问题A_k+1~j的解必须是子问题计算A_k+1 *...* A_j的优化解。

证明,如果A_i~j的优化顺序对应子问题A_i~k的解不是子问题A_i~k的优化解，那么，完全可以使用子问题A_i~k的优化解来取代A_i~j的优化顺序对应子问题A_i~k的解，最终可以得到比A_i~j此时的优化解代价更小解，矛盾，故而假设错误，A1_{n的优化顺序对应子问题A1}k的解必是子问题A1~k的优化解。

同理可以证明Ak+1 ~ n部分。

Step 2: overlapping subproblems

Step 3: A recursive solution

定义：m[ i, j ] = 计算A_i~j的最小乘法数。

• m[ i,j ] = 0 , if i =j
• m[ i,j ] = min_i≤k<j { m[ i,k ] + m[ k+1,j ] + pi-1pkpj } , if i < j, 其中，Pm表示第m个矩阵的列数。

Step 4: Computing the optimal costs

Matrix-Chain-Order(p) 
1. n=length(p)-1;
2. let m[1..n , 1..n] and s[1..n-1 , 2..n] be new chain length
3. FOR  i=1  TO  n  DO 
4.        m[i, i]=0; 
5. FOR  l=2  TO  n  DO  /* 计算从对角线开始，第l条对角线 */ 
6.        FOR  i=1  TO  n-l+1  DO  /*n-l+1是第l条对角线上的元素*/
7.                j = i+l-1;   /*计算纵坐标*/            
8.                m[i, j] = ∞;  
9.                FOR  k =i  To  j-1  DO /* 计算m[i,j] */ 
10.                      q=m[i, k]+m[k+1, j]+Pi-1PkPj                    
11.                      IF  q<m[i, j]  
12.                      THEN  
13.                               m[i,j]=q;  
14.                               s[i.j]=k;
15. Return m 

显然其算法时间复杂度：Θ(n³)，空间复杂度O(n²)。

Step 4: Constructing an optimal solution

PRINT-OPTIMAL-PARENS(s,i,j)
1. if i == j 
2.    print “A”
3. else 
4.    print “(” 
5.    PRINT-OPTIMAL-PARENS(s,i,s[i,j]) 
6.    PRINT-OPTIMAL-PARENS(s,s[i,j]+1,j)
7.    print “)”