Python Program for Matrix Chain Multiplication | DP-8

Last Updated : 20 Apr, 2023

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 parenthesization requires less number of operations. 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

Following is a recursive implementation that simply follows the above optimal substructure property.

Python3

   # A naive recursive implementation that
# simply follows the above optimal 
# substructure property 
import sys
 
# Matrix A[i] has dimension p[i-1] x p[i]
# for i = 1..n
def MatrixChainOrder(p, i, j):
 
    if i == j:
        return 0
 
    _min = sys.maxsize
     
    # 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 k in range(i, j):
     
        count = (MatrixChainOrder(p, i, k) 
             + MatrixChainOrder(p, k + 1, j)
                   + p[i-1] * p[k] * p[j])
 
        if count < _min:
            _min = count;
     
 
    # Return minimum count
    return _min;
 
 
# Driver program to test above function
arr = [1, 2, 3, 4, 3];
n = len(arr);
 
print("Minimum number of multiplications is ",
                MatrixChainOrder(arr, 1, n-1));
 
# This code is contributed by Aryan Garg
 
  

Output

Minimum number of multiplications is  30

Time complexity: The time complexity of the algorithm is O(n^3) where n is the number of matrices in the chain. This is because we have three nested loops and each loop runs n times.

Space complexity: The space complexity of the algorithm is O(n^2) because we are using a two-dimensional array to store the intermediate results. The size of the array is n x n.

Dynamic Programming Solution

Python

   # Dynamic Programming Python implementation of Matrix
# Chain Multiplication. See the Cormen book for details
# of the following algorithm
import sys
 
# Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
def MatrixChainOrder(p, n):
    # For simplicity of the program, one extra row and one
    # extra column are allocated in m[][].  0th row and 0th
    # column of m[][] are not used
    m = [[0 for x in range(n)] for x in range(n)]
 
    # m[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 i in range(1, n):
        m[i][i] = 0
 
    # L is chain length.
    for L in range(2, n):
        for i in range(1, n-L + 1):
            j = i + L-1
            m[i][j] = sys.maxsize
            for k in range(i, j):
 
                # q = cost / scalar multiplications
                q = m[i][k] + m[k + 1][j] + p[i-1]*p[k]*p[j]
                if q < m[i][j]:
                    m[i][j] = q
 
    return m[1][n-1]
 
# Driver program to test above function
arr = [1, 2, 3, 4, 3]
size = len(arr)
 
print("Minimum number of multiplications is " +
       str(MatrixChainOrder(arr, size)))
# This Code is contributed by Bhavya Jain
 
  