Breaking News: Grepper is joining You.com. Read the official announcement!

matrix chain multiplication

Obnoxious Osprey answered on October 30, 2020 Popularity 10/10 Helpfulness 1/10

answer matrix chain multiplication

related matrix chain multiplication

matrix chain multiplication

Comment

    minimum_multiplication(n: integer; d: array(0..n))
        M: array(1..n, 1..n) := (others => 0);

        for diagonal in 1 .. n-1 loop 
            for i in 1 .. n - diagonal loop 
                j := i + diagonal;

                min := integer'last;
                for k in i .. j-1 loop
                    current := M[i, k] + M[k+1, j] + d(i-1)*d(k)*d(j) 
                    if current < min then
                        min := current
                    end if
                end loop
                M[i, j] := min
            end loop
        end loop

Popularity 10/10 Helpfulness 1/10 Language whatever

Source: www.radford.edu

Tags: chain matrix whatever

Link to this answer
Share Copy Link

Contributed on Oct 30 2020

Obnoxious Osprey

0 Answers Avg Quality 2/10

Closely Related Answers

matrix chain multiplication

Comment

// C++ program using memoization
#include <bits/stdc++.h>
using namespace std;
int dp[100][100];
 
// Function for matrix chain multiplication
int matrixChainMemoised(int* p, int i, int j)
{
    if (i == j)
    {
        return 0;
    }
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }
    dp[i][j] = INT_MAX;
    for (int k = i; k < j; k++)
    {
        dp[i][j] = min(
            dp[i][j], matrixChainMemoised(p, i, k)
                     + matrixChainMemoised(p, k + 1, j)
                       + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}
int MatrixChainOrder(int* p, int n)
{
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    memset(dp, -1, sizeof dp);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, n);
}
 
// This code is contributed by Sumit_Yadav

Popularity 10/10 Helpfulness 2/10 Language cpp

Source: www.geeksforgeeks.org

Tags: c++

Link to this answer
Share Copy Link

Contributed on Nov 23 2022

Aditya N Panchal

0 Answers Avg Quality 2/10

matrix chain multiplication

Comment

Input: N = 5
arr = {40, 20, 30, 10, 30}
Code(Java):
static int matrixMultiplication(int N, int arr[])
    {
        // code here
        int c[][] = new int[N][N]; //optimal sol array
        int min, tmp, j;
        //d = 0 and i = 0 is not needed
        for(int d = 1; d < N-1; d++) { //diagonal iteration ex: for 4: 1,2,3
            for(int i = 1; i < N-d; i++) { //rows for the particular diagonal
                j = i+d;
                min = Integer.MAX_VALUE;
                for(int k = i; k <= (j-1); k++) { //bracket at k
                    tmp = c[i][k] + c[k+1][j] + (arr[i-1]*arr[k]*arr[j]);
                    
                    if(tmp < min)
                        min = tmp;
                }
                c[i][j] = min; // choose brackets with min cost
            }
        }
        
        return c[1][N-1]; // return min cost for multiplying
    }

Popularity 10/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: chain chai

Link to this answer
Share Copy Link

Contributed on May 22 2022

Animesh Singh

0 Answers Avg Quality 2/10

matrix chain multiplication

Contents

More Related Answers

matrix chain multiplication

Closely Related Answers

matrix chain multiplication

matrix chain multiplication

Grepper

Documentation

Social

Legal

Contact

Oops, You will need to install Grepper and log-in to perform this action.