# Matrix Chain Multiplication (A O(N^2) Solution)

• Difficulty Level : Hard
• Last Updated : 24 May, 2021

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 matrices. 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```

We have discussed a O(n^3) solution for Matrix Chain Multiplication Problem
Note: Below solution does not work for many cases. For example: for input {2, 40, 2, 40, 5}, the correct answer is 580 but this method returns 720
There is another similar approach to solving this problem. More intuitive and recursive approach.

Assume there are following available method
minCost(M1, M2) -> returns min cost of multiplying matrices M1 and M2
Then, for any chained product of matrices like,
M1.M2.M3.M4…Mn
min cost of chain = min(minCost(M1, M2.M3…Mn), minCost(M1.M2..Mn-1, Mn))
Now we have two subchains (sub problems) :
M2.M3…Mn
M1.M2..Mn-1

## C++

 `// CPP program to implement optimized matrix chain multiplication problem.` `#include` `using` `namespace` `std;`   `// Matrix Mi has dimension p[i-1] x p[i] for i = 1..n` `int` `MatrixChainOrder(``int` `p[], ``int` `n)` `{`   `    ``/* For simplicity of the program, one extra row and one extra column are allocated in ` `    ``dp[][]. 0th row and 0th column of dp[][] are not used */` `    ``int` `dp[n][n];`   `    ``/* dp[i, j] = Minimum number of scalar multiplications needed to compute the matrix M[i]M[i+1]...M[j] ` `                ``= M[i..j] where dimension of M[i] is p[i-1] x p[i] */` `                `  `    ``// cost is zero when multiplying one matrix.` `    ``for` `(``int` `i=1; i

## Java

 `// Java program to implement optimized matrix chain multiplication problem.` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;`     `class` `GFG{` `// Matrix Mi 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 dp[][]. 0th row and 0th column of dp[][] are not used */` `    ``int` `[][]dp=``new` `int``[n][n];`   `    ``/* dp[i, j] = Minimum number of scalar multiplications needed to compute the matrix M[i]M[i+1]...M[j] ` `                ``= M[i..j] where dimension of M[i] is p[i-1] x p[i] */` `                `  `    ``// cost is zero when multiplying one matrix.` `    ``for` `(``int` `i=``1``; i

## Python3

 `# Python3 program to implement optimized ` `# matrix chain multiplication problem. `   `# Matrix Mi 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 dp[][]. 0th row and ` `    ``# 0th column of dp[][] are not used ` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(n)]` `             ``for` `i ``in` `range``(n)] `   `    ``# dp[i, j] = Minimum number of scalar ` `    ``# multiplications needed to compute ` `    ``# the matrix M[i]M[i+1]...M[j] = M[i..j] ` `    ``# where dimension of M[i] is p[i-1] x p[i] ` `                `  `    ``# cost is zero when multiplying one matrix. ` `    ``for` `i ``in` `range``(``1``, n):` `        ``dp[i][i] ``=` `0`   `    ``# Simply following above recursive formula. ` `    ``for` `L ``in` `range``(``1``, n ``-` `1``):` `        ``for` `i ``in` `range``(n ``-` `L):` `            ``dp[i][i ``+` `L] ``=` `min``(dp[i ``+` `1``][i ``+` `L] ``+` `                                ``p[i ``-` `1``] ``*` `p[i] ``*` `p[i ``+` `L], ` `                               ``dp[i][i ``+` `L ``-` `1``] ``+` `                                ``p[i ``-` `1``] ``*` `p[i ``+` `L ``-` `1``] ``*` `p[i ``+` `L]) ` `    `  `    ``return` `dp[``1``][n ``-` `1``]`   `# Driver code ` `arr ``=` `[``10``, ``20``, ``30``, ``40``, ``30``]` `size ``=` `len``(arr)` `print``(``"Minimum number of multiplications is"``, ` `                 ``MatrixChainOrder(arr, size))`   `# This code is contributed ` `# by sahishelangia`

## C#

 `// C# program to implement optimized ` `// matrix chain multiplication problem.` `using` `System;`   `class` `GFG` `{` `// Matrix Mi 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 dp[][]. ` `       ``0th row and 0th column of dp[][] ` `       ``are not used */` `    ``int` `[,]dp = ``new` `int``[n, n];`   `    ``/* dp[i, j] = Minimum number of scalar ` `       ``multiplications needed to compute ` `       ``the matrix M[i]M[i+1]...M[j] = M[i..j]` `       ``where dimension of M[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;`   `    ``// Simply following above ` `    ``// recursive formula.` `    ``for` `(``int` `L = 1; L < n - 1; L++) ` `    ``for` `(``int` `i = 1; i < n - L; i++)     ` `        ``dp[i, i + L] = Math.Min(dp[i + 1, i + L] + ` `                                 ``p[i - 1] * p[i] * p[i + L],` `                                ``dp[i, i + L - 1] + p[i - 1] *` `                                 ``p[i + L - 1] * p[i + L]); ` `    `  `    ``return` `dp[1, n - 1];` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `[]arr = {10, 20, 30, 40, 30} ;` `    ``int` `size = arr.Length;`   `    ``Console.WriteLine(``"Minimum number of multiplications is "` `+` `                                  ``MatrixChainOrder(arr, size));` `}` `}`   `// This code is contributed by anuj_67`

## PHP

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## Javascript

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Output:

`Minimum number of multiplications is 30000`

Thanks to Rishi_Lazy for providing above optimized solution.
Time Complexity : O(n2
Printing Brackets in Matrix Chain Multiplication

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