Question

Find an optimal parenthesization for matrix-chain multiplications using any PYTHON/java/c++/c for the number {26, 9, 41, 18, 13, 22, 28, 32, 25, 26, 30, 37, 19, 47, 11, 24, 20} using a top-down memorized solution. The output must be three lines:

1) the first line contains the optimal number of multiplication

2) the second line contains the optimal parenthesization and

3) the third line is the time required to compute the optimal parenthesization

Answer 1

//C++ program

#include<iostream>
#include <chrono>
#include <ctime>
#include<limits.h>
using namespace std;

void Parenthesis(int , int , int ,int *, char &) ;
void matrixChainMultiplication(int *, int );

int main()
{
int arr[] = {26, 9, 41, 18, 13, 22, 28, 32, 25, 26, 30, 37, 19, 47, 11, 24, 20};
int n = sizeof(arr)/sizeof(arr[0]);
matrixChainMultiplication(arr, n);
return 0;
}

void matrixChainMultiplication(int arr[], int n)
{
int DP[n][n];
int bracket[n][n];
for (int i=1; i<n; i++)
DP[i][i] = 0;
  

for (int L=2; L<n; L++)
{
for (int i=1; i<n-L+1; i++)
{
int j = i+L-1;
DP[i][j] = INT_MAX;
for (int k=i; k<=j-1; k++)
{
int q = DP[i][k] + DP[k+1][j] + arr[i-1]*arr[k]*arr[j];
if (q < DP[i][j])
{
DP[i][j] = q;
bracket[i][j] = k;
}
}
}
}
  
cout<<"Optimal number of multiplication : "<<DP[1][n-1]<<"\n";
char name = 'A';
  
std::chrono::time_point<std::chrono::system_clock> start, end;
   double time;
   std::chrono::duration<double> elapsed_seconds;
  
cout << "Optimal Parenthesization is : ";
start = std::chrono::system_clock::now();
Parenthesis(1, n-1, n, (int *)bracket, name);
end = std::chrono::system_clock::now();
elapsed_seconds = end - start;
time= elapsed_seconds.count();
  
cout<<"\nTime required to compute the optimal parenthesization "<<time<<"\n\n";
  
  
}
  

void Parenthesis(int i, int j, int n,int *bracket, char &name)
{
  
if (i == j)
{
cout << name++;
return;
}
  
cout << "(";

Parenthesis(i, *((bracket+i*n)+j), n,bracket, name);
  
Parenthesis(*((bracket+i*n)+j) + 1, j,n, bracket, name);
cout << ")";
}
//sample output