Printing brackets in Matrix Chain Multiplication Problem - GeeksforGeeks

23 Dec, 2021
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 parenthesis in 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 parenthesis in 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, cost of which is 10*20*30
This problem is mainly an extension of previous post. In the previous post, we have discussed algorithm for finding optimal cost only. Here we need print parenthesization also.
The idea is to store optimal break point for every subexpression (i, j) in a 2D array bracket[n][n]. Once we have bracket array us constructed, we can print parenthesization using below code. 
// Prints parenthesization in subexpression (i, j)
printParenthesis(i, j, bracket[n][n], name)
{
    // If only one matrix left in current segment
    if (i == j)
    {
        print name;
        name++;
        return;
    }

    print "(";

    // Recursively put brackets around subexpression
    // from i to bracket[i][j].
    printParenthesis(i, bracket[i][j], bracket, name);

    // Recursively put brackets around subexpression
    // from bracket[i][j] + 1 to j.
    printParenthesis(bracket[i][j]+1, j, bracket, name);

    print ")";
}
Below is the implementation of the above steps.
C++
Java
C#
Javascript
// 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[i][j].    // Note that "*((bracket+i*n)+j)" is similar to    // bracket[i][j]    printParenthesis(i, *((bracket + i * n) + j), n,                     bracket, name);     // Recursively put brackets around subexpression    // from bracket[i][j] + 1 to j.    printParenthesis(*((bracket + i * n) + j) + 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// function// https://goo.gl/k6EYKjvoid 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];     // bracket[i][j] stores optimal break point in    // subexpression from i to j.    int bracket[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 bracket[i,j]=k shows                    // where to split the product arr                    // i,i+1....j for the minimum cost.                    bracket[i][j] = 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*)bracket, 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 program to print optimal parenthesization// in matrix chain multiplication.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[i][j].    // Note that "*((bracket+i*n)+j)" is similar to    // bracket[i][j]    printParenthesis(i, bracket[i][j], n, bracket);     // Recursively put brackets around subexpression    // from bracket[i][j] + 1 to j.    printParenthesis(bracket[i][j] + 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  // https://goo.gl/k6EYKj  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];     // bracket[i][j] stores optimal break point in    // subexpression from i to j.    int[][] bracket = 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 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] = 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 bracket[i,j]=k shows            // where to split the product arr            // i,i+1....j for the minimum cost.            bracket[i][j] = 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, bracket);    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
// 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[i,j].    // Note that "*((bracket+i*n)+j)" is similar to    // bracket[i,j]    printParenthesis(i, bracket[i, j], n, bracket);         // Recursively put brackets around subexpression    // from bracket[i,j] + 1 to j.    printParenthesis(bracket[i, j] + 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// function// https://goo.gl/k6EYKjstatic 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];         // bracket[i,j] stores optimal break point in    // subexpression from i to j.    int[,] bracket = 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 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.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 bracket[i,j]=k shows                    // where to split the product arr                    // i,i+1....j for the minimum cost.                    bracket[i, j] = 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, bracket);    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 29AjayKumar
<script>// javascript program to print optimal parenthesization// in matrix chain multiplication.   var name=0;   // Function for printing the optimal  // parenthesization of a matrix chain product  function printParenthesis(i , j, n, bracket)  {         // If only one matrix left in current segment    if (i == j)    {      document.write(name++);      return;    }    document.write("(");     // Recursively put brackets around subexpression    // from i to bracket[i][j].    // Note that "*((bracket+i*n)+j)" is similar to    // bracket[i][j]    printParenthesis(i, bracket[i][j], n, bracket);     // Recursively put brackets around subexpression    // from bracket[i][j] + 1 to j.    printParenthesis(bracket[i][j] + 1, j, n, bracket);    document.write(")");  }   // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n  // Please refer below article for details of this  // function  // https://goo.gl/k6EYKj  function matrixChainOrder( p , 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         */    var m = Array(n).fill(0).map(x => Array(n).fill(0));     // bracket[i][j] stores optimal break point in    // subexpression from i to j.    var bracket = Array(n).fill(0).map(x => Array(n).fill(0));     /*         * 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 (var i = 1; i < n; i++)      m[i][i] = 0;     // L is chain length.    for (var L = 2; L < n; L++)    {      for (var i = 1; i < n - L + 1; i++)      {        var j = i + L - 1;        m[i][j] = Number.MAX_VALUE;        for (var k = i; k <= j - 1; k++)        {           // q = cost/scalar multiplications          var 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 bracket[i,j]=k shows            // where to split the product arr            // i,i+1....j for the minimum cost.            bracket[i][j] = k;          }        }      }    }     // The first matrix is printed as 'A', next as 'B',    // and so on    name = 'A';    document.write("Optimal Parenthesization is : ");    printParenthesis(1, n - 1, n, bracket);    document.write("\nOptimal Cost is : " + m[1][n - 1]);  }   // Driver code  var arr = [ 40, 20, 30, 10, 30 ];  var n = arr.length;  matrixChainOrder(arr, n); // This code is contributed by 29AjayKumar</script>
Optimal Parenthesization is : ((A(BC))D)nOptimal Cost is : 26000
Time Complexity: O(n3) Auxiliary Space: O(n2)
Another Approach:
——————————————
This solution try to solve the problem using Recursion using permutations.
Let's take example:  {40, 20, 30, 10, 30}
n = 5
Let’s divide that into a Matrix
[ [40, 20], [20, 30], [30, 10], [10, 30] ]

[ A , B , C , D ]

it contains 4 matrices i.e. (n - 1)
We have 3 combinations to multiply  i.e.  (n-2)
AB    or    BC    or     CD
1) Given array of matrices with length M, Loop through  M – 1 times
2) Merge consecutive matrices in each loop
for (int i = 0; i < M - 1; i++) {
   int cost =  (matrices[i][0] * 
                 matrices[i][1] * matrices[i+1][1]);
   
   // STEP - 3
   // STEP - 4
}
3) Merge the current two matrices into one, and remove merged matrices list from list.
If  A, B merged, then A, B must be removed from the List

and NEW matrix list will be like
newMatrices = [  AB,  C ,  D ]

We have now 3 matrices, in any loop
Loop#1:  [ AB,  C,   D ]
Loop#2:  [ A,   BC,  D ]
Loop#3   [ A,   B,   CD ]
4) Repeat: Go to STEP – 1  with  newMatrices as input M — recursion
5) Stop recursion, when we get 2 matrices in the list.
Matrices are reduced in following way, 
and cost’s must be retained and summed-up during recursion with previous values of each parent step.
[ A, B , C, D ]

[(AB), C, D ]
 [ ((AB)C), D ]--> [ (((AB)C)D) ] 
 - return & sum-up total cost of this step.
 [ (AB),  (CD)] --> [ ((AB)(CD)) ] 
 - return .. ditto..

 [ A, (BC), D ]
 [ (A(BC)), D ]--> [ ((A(BC))D) ] 
  - return
 [ A, ((BC)D) ]--> [ (A((BC)D)) ] 
  - return
    
 [ A, B, (CD) ]
 [ A, (B(CD)) ]--> [ (A(B(CD))) ] 
  - return
 [ (AB), (CD) ]--> [ ((AB)(CD)) ] 
  - return .. ditto..
on return i.e. at final step of each recursion, check if  this value smaller than of any other.
Below is JAVA implementation of above steps.
Java
import java.util.Arrays; public class MatrixMultiplyCost {     static class FinalCost    {        public String label = "";        public int cost = Integer.MAX_VALUE;    }     private void optimalCost(int[][] matrices,                             String[] labels, int prevCost,                             FinalCost finalCost)    {        int len = matrices.length;         if (len < 2)        {            finalCost.cost = 0;            return;        }        else if (len == 2)        {            int cost = prevCost                       + (matrices[0][0] *                          matrices[0][1] *                          matrices[1][1]);             // This is where minimal cost has been caught            // for whole program            if (cost < finalCost.cost)            {                finalCost.cost = cost;                finalCost.label                    = "(" + labels[0]                    + labels[1] + ")";            }            return;        }         // recursive Reduce        for (int i = 0; i < len - 1; i++)        {            int j;            int[][] newMatrix = new int[len - 1][2];            String[] newLabels = new String[len - 1];            int subIndex = 0;             // STEP-1:            //   - Merge two matrices's into one - in each            //   loop, you move merge position            //        - if i = 0 THEN  (AB) C D ...            //        - if i = 1 THEN  A (BC) D ...            //        - if i = 2 THEN  A B (CD) ...            //   - and find the cost of this two matrices            //   multiplication            int cost = (matrices[i][0] * matrices[i][1]                        * matrices[i + 1][1]);             // STEP - 2:            //    - Build new matrices after merge            //    - Keep track of the merged labels too            for (j = 0; j < i; j++) {                newMatrix[subIndex] = matrices[j];                newLabels[subIndex++] = labels[j];            }             newMatrix[subIndex][0] = matrices[i][0];            newMatrix[subIndex][1] = matrices[i + 1][1];            newLabels[subIndex++]                = "(" + labels[i] + labels[i + 1] + ")";             for (j = i + 2; j < len; j++) {                newMatrix[subIndex] = matrices[j];                newLabels[subIndex++] = labels[j];            }             optimalCost(newMatrix, newLabels,                        prevCost + cost, finalCost);        }    }     public FinalCost findOptionalCost(int[] arr)    {        // STEP -1 : Prepare and convert inout as Matrix        int[][] matrices = new int[arr.length - 1][2];        String[] labels = new String[arr.length - 1];         for (int i = 0; i < arr.length - 1; i++) {            matrices[i][0] = arr[i];            matrices[i][1] = arr[i + 1];            labels[i] = Character.toString((char)(65 + i));        }        printMatrix(matrices);         FinalCost finalCost = new FinalCost();        optimalCost(matrices, labels, 0, finalCost);         return finalCost;    }     /**     * Driver Code     */    public static void main(String[] args)    {        MatrixMultiplyCost calc = new MatrixMultiplyCost();         // ======= *** TEST CASES **** ============         int[] arr = { 40, 20, 30, 10, 30 };        FinalCost cost = calc.findOptionalCost(arr);        System.out.println("Final labels: \n" + cost.label);        System.out.println("Final Cost:\n" + cost.cost                           + "\n");    }     /**     * Ignore this method     * - THIS IS for DISPLAY purpose only     */    private static void printMatrix(int[][] matrices)    {        System.out.print("matrices = \n[");        for (int[] row : matrices) {            System.out.print(Arrays.toString(row) + " ");        }        System.out.println("]");    }} // This code is contributed by suvera
matrices = 
[[40, 20] [20, 30] [30, 10] [10, 30] ]
Final labels: 
((A(BC))D)
Final Cost:
26000
This article is contributed by Yasin Zafar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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