Prerequisite : Dynamic Programming | Set 8 (Matrix Chain Multiplication)
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: Optimal parenthesization is ((A(BC))D)
Optimal cost of parenthesization is 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 the parenthesis in the following way
(A(BC))D --> 20*30*10 + 40*20*10 + 40*10*30
Input: p[] = {10, 20, 30, 40, 30}
Output: Optimal parenthesization is (((AB)C)D)
Optimal cost of parenthesization is 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 the parenthesis in the following way
((AB)C)D --> 10*20*30 + 10*30*40 + 10*40*30
Input: p[] = {10, 20, 30}
Output: Optimal parenthesization is (AB)
Optimal cost of parenthesization is 6000
There are only two matrices of dimensions 10x20 and 20x30. So there
is only one way to multiply the matrices, the cost of which is 10*20*30
This problem is mainly an extension of Finding Optimal cost of Matrix Chain Multiplication . Here we also need to print brackets.
We have discussed a solution in a post that uses two matrices. In this post, a space-optimized solution is discussed that uses a single matrix.
1) To find the optimal cost, we create a matrix whose only upper triangle is filled and the rest of the cells are not used.
2) The idea is to use the lower triangular part of the same matrix (that is not used) for storing brackets.
The idea is to store optimal break point for every subexpression (i, j) at m [ j ][ i ] and optimal cost at m [ i ] [ j ].
Below is the implementation of above steps.
C++
// A space optimized C++ program to print optimal// parenthesization in matrix chain multiplication.#include<bits/stdc++.h>using namespace std;// Function for printing the optimal// parenthesization of a matrix chain productvoid printParenthesis(int i, int j, int n, int *bracket, char &name){ // If only one matrix left in current segment if (i == j) { cout << name++; return; } cout << "("; // Recursively put brackets around subexpression // from i to bracket[j][i]. // Note that "*((bracket+j*n)+i)" is similar to // bracket[j][i] printParenthesis(i, *((bracket+j*n)+i), n, bracket, name); // Recursively put brackets around subexpression // from bracket[j][i] + 1 to i. printParenthesis(*((bracket+j*n)+i) + 1, j, n, bracket, name); cout << ")";}// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n// Please refer below article for details of this// functionvoid 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[n][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 (int i=1; i<n; i++) m[i][i] = 0; // L is chain length. for (int L=2; L<n; L++) { for (int i=1; i<n-L+1; i++) { int j = i+L-1; m[i][j] = INT_MAX; for (int k=i; k<=j-1; k++) { // q = cost/scalar multiplications int q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j]; if (q < m[i][j]) { m[i][j] = q; // Each entry m[j,ji=k shows // where to split the product arr // i,i+1....j for the minimum cost. m[j][i] = k; } } } } // The first matrix is printed as 'A', next as 'B', // and so on char name = 'A'; cout << "Optimal Parenthesization is: "; printParenthesis(1, n-1, n, (int *)m, name); cout << "\nOptimal Cost is : " << m[1][n-1];}// Driver codeint main(){ int arr[] = {40, 20, 30, 10, 30}; int n = sizeof(arr)/sizeof(arr[0]); matrixChainOrder(arr, n); return 0;} |
Java
// A space optimized Java program to print optimal // parenthesization in matrix chain multiplication.import java.util.*;class GFG { static char name; // Function for printing the optimal // parenthesization of a matrix chain product static void printParenthesis(int i, int j, int n, int[][] bracket) { // If only one matrix left in current segment if (i == j) { System.out.print(name++); return; } System.out.print('('); // Recursively put brackets around subexpression // from i to bracket[j][i]. // Note that "*((bracket+j*n)+i)" is similar to // bracket[i][j] printParenthesis(i, bracket[j][i], n, bracket); // Recursively put brackets around subexpression // from bracket[j][i] + 1 to i. printParenthesis(bracket[j][i] + 1, j, n, bracket); System.out.print(')'); } // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n // Please refer below article for details of this // function static void 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]; /* * 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 (int L = 2; L < n; L++) { for (int i = 1; i < n - L + 1; i++) { int j = i + L - 1; m[i][j] = Integer.MAX_VALUE; for (int k = i; k <= j - 1; k++) { // q = cost/scalar multiplications int q = m[i][k] + m[k + 1][j] + p[i - 1] * p[k] * p[j]; if (q < m[i][j]) { m[i][j] = q; // Each entry m[j,ji=k shows // where to split the product arr // i,i+1....j for the minimum cost. m[j][i] = k; } } } } // The first matrix is printed as 'A', next as 'B', // and so on name = 'A'; System.out.print("Optimal Parenthesization is: "); printParenthesis(1, n - 1, n, m); System.out.print("\nOptimal Cost is :" + m[1][n - 1]); } // Driver Code public static void main(String[] args) { int[] arr = { 40, 20, 30, 10, 30 }; int n = arr.length; matrixChainOrder(arr, n); }}// This code is contributed by// sanjeev2552 |
Python3
# A space optimized python3 program to# print optimal parenthesization in # matrix chain multiplication.def printParenthesis(m, j, i ): # Displaying the parenthesis. if j == i: # The first matrix is printed as # 'A', next as 'B', and so on print(chr(65 + j), end = "") return; else: print("(", end = "") # Passing (m, k, i) instead of (s, i, k) printParenthesis(m, m[j][i] - 1, i) # (m, j, k+1) instead of (s, k+1, j) printParenthesis(m, j, m[j][i]) print (")", end = "" )def matrixChainOrder(p, n): # Creating a matrix of order # n*n in the memory. m = [[0 for i in range(n)] for i in range (n)] for l in range (2, n + 1): for i in range (n - l + 1): j = i + l - 1 # Initializing infinity value. m[i][j] = float('Inf') for k in range (i, j): q = (m[i][k] + m[k + 1][j] + (p[i] * p[k + 1] * p[j + 1])); if (q < m[i][j]): m[i][j] = q # Storing k value in opposite index. m[j][i] = k + 1 return m;# Driver Codearr = [40, 20, 30, 10, 30]n = len(arr) - 1m = matrixChainOrder(arr, n) # Forming the matrix mprint("Optimal Parenthesization is: ", end = "")# Passing the index of the bottom left# corner of the 'm' matrix instead of# passing the index of the top right# corner of the 's' matrix as we used# to do earlier. Everything is just opposite# as we are using the bottom half of the# matrix so assume everything opposite even# the index, take m[j][i].printParenthesis(m, n - 1, 0)print("\nOptimal Cost is :", m[0][n - 1])# This code is contributed by Akash Gupta. |
C#
// A space optimized C# program to print optimal // parenthesization in matrix chain multiplication.using System;class GFG{static char name;// Function for printing the optimal// parenthesization of a matrix chain productstatic void printParenthesis(int i, int j, int n, int[,] bracket) { // If only one matrix left in current segment if (i == j) { Console.Write(name++); return; } Console.Write('('); // Recursively put brackets around subexpression // from i to bracket[j,i]. // Note that "*((bracket+j*n)+i)" is similar to // bracket[i,j] printParenthesis(i, bracket[j, i], n, bracket); // Recursively put brackets around subexpression // from bracket[j,i] + 1 to i. printParenthesis(bracket[j, i] + 1, j, n, bracket); Console.Write(')');}// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n// Please refer below article for details of this// functionstatic void 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]; /* * 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(int L = 2; L < n; L++) { for(int i = 1; i < n - L + 1; i++) { int j = i + L - 1; m[i, j] = int.MaxValue; for(int k = i; k <= j - 1; k++) { // q = cost/scalar multiplications int q = m[i, k] + m[k + 1, j] + p[i - 1] * p[k] * p[j]; if (q < m[i, j]) { m[i, j] = q; // Each entry m[j,ji=k shows // where to split the product arr // i,i+1....j for the minimum cost. m[j, i] = k; } } } } // The first matrix is printed as 'A', next as 'B', // and so on name = 'A'; Console.Write("Optimal Parenthesization is: "); printParenthesis(1, n - 1, n, m); Console.Write("\nOptimal Cost is :" + m[1, n - 1]);}// Driver Codepublic static void Main(String[] args) { int[] arr = { 40, 20, 30, 10, 30 }; int n = arr.Length; matrixChainOrder(arr, n);}}// This code is contributed by Princi Singh |
Javascript
<script>// A space optimized python3 program to// print optimal parenthesization in// matrix chain multiplication.function printParenthesis(m, j, i){ // Displaying the parenthesis. if(j == i){ // The first matrix is printed as // 'A', next as 'B', and so on document.write(String.fromCharCode(65 + j), end = "") return; } else{ document.write("(", end = "") // Passing (m, k, i) instead of (s, i, k) printParenthesis(m, m[j][i] - 1, i) // (m, j, k+1) instead of (s, k+1, j) printParenthesis(m, j, m[j][i]) document.write(")") }}function matrixChainOrder(p, n){ // Creating a matrix of order // n*n in the memory. let m = new Array(n).fill(0).map(()=>new Array(n).fill(0)) for(let l=2;l<n + 1;l++){ for(let i=0;i<(n - l + 1);i++){ let j = i + l - 1 // Initializing a very value m[i][j] = Number.MAX_VALUE for(let k=i;k<j;k++){ let q = (m[i][k] + m[k + 1][j] + (p[i] * p[k + 1] * p[j + 1])); if (q < m[i][j]){ m[i][j] = q // Storing k value in opposite index. m[j][i] = k + 1 } } } } return m;}// Driver Codelet arr = [40, 20, 30, 10, 30]let n = arr.length - 1let m = matrixChainOrder(arr, n) // Forming the matrix mdocument.write("Optimal Parenthesization is: ")// Passing the index of the bottom left// corner of the 'm' matrix instead of// passing the index of the top right// corner of the 's' matrix as we used// to do earlier. Everything is just opposite// as we are using the bottom half of the// matrix so assume everything opposite even// the index, take m[j][i].printParenthesis(m, n - 1, 0)document.write("</br>","Optimal Cost is :", m[0][n - 1])// This code is contributed by shinjanpatra</script> |
Output:
Optimal Parenthesization is : ((A(BC))D) Optimal Cost is : 26000
Time Complexity: O(n^3)
Auxiliary Space: O(n^2): The asymptotic value remains same, but we reduce the auxiliary space to half.
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