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: 26000There 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*30Input: p[] = {10, 20, 30, 40, 30}Output: 30000There 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*30Input: p[] = {10, 20, 30}Output: 6000There 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

## Recommended: Please solve it on “__PRACTICE__ ” first, before moving on to the solution.

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

## Java

`/* A naive recursive implementation that simply follows` ` ` `the above optimal substructure property */` `class` `MatrixChainMultiplication {` ` ` `// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n` ` ` `static` `int` `MatrixChainOrder(` `int` `p[], ` `int` `i, ` `int` `j)` ` ` `{` ` ` `if` `(i == j)` ` ` `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 = i; k < j; k++) {` ` ` `int` `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` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `arr[] = ` `new` `int` `[] { ` `1` `, ` `2` `, ` `3` `, ` `4` `, ` `3` `};` ` ` `int` `n = arr.length;` ` ` ` ` `System.out.println(` `"Minimum number of multiplications is "` ` ` `+ MatrixChainOrder(arr, ` `1` `, n - ` `1` `));` ` ` `}` `}` `/* This code is contributed by Rajat Mishra*/` |

**Output:**

Minimum number of multiplications is 30

**Dynamic Programming Solution**

## Java

`// Dynamic Programming Python implementation of Matrix` `// Chain Multiplication.` `// See the Cormen book for details of the following algorithm` `class` `MatrixChainMultiplication {` ` ` `// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n` ` ` `static` `int` `MatrixChainOrder(` `int` `p[], ` `int` `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 */` ` ` `int` `m[][] = ` `new` `int` `[n][n];` ` ` ` ` `int` `i, j, k, L, q;` ` ` ` ` `/* 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 = ` `1` `; i < n; i++)` ` ` `m[i][i] = ` `0` `;` ` ` ` ` `// L is chain length.` ` ` `for` `(L = ` `2` `; L < n; L++) {` ` ` `for` `(i = ` `1` `; i < n - L + ` `1` `; i++) {` ` ` `j = i + L - ` `1` `;` ` ` `if` `(j == n)` ` ` `continue` `;` ` ` `m[i][j] = Integer.MAX_VALUE;` ` ` `for` `(k = i; k <= j - ` `1` `; k++) {` ` ` `// 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` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `arr[] = ` `new` `int` `[] { ` `1` `, ` `2` `, ` `3` `, ` `4` `};` ` ` `int` `size = arr.length;` ` ` ` ` `System.out.println(` `"Minimum number of multiplications is "` ` ` `+ MatrixChainOrder(arr, size));` ` ` `}` `}` `/* This code is contributed by Rajat Mishra*/` |

**Output:**

Minimum number of multiplications is 18

Please refer complete article on Matrix Chain Multiplication | DP-8 for more details!

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