矩阵链乘法问题中的打印括号

先决条件：动态编程|集合8(矩阵链乘法)
给定一系列矩阵，找到将这些矩阵相乘的最有效方法。问题实际上不是执行乘法，而是仅决定执行乘法的顺序。
因为矩阵乘法是关联的，所以我们有很多选择可以乘法矩阵链。换句话说，无论我们如何对产品加上括号，结果都是一样的。例如，如果我们有四个矩阵A，B，C和D，则将有：

(ABC)D = (AB)(CD) = A(BCD) = ....

但是，括号内乘积的顺序会影响计算乘积所需的简单算术运算的数量或效率。例如，假设A为10×30矩阵，B为30×5矩阵，C为5×60矩阵。然后，

(AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 operations
A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 operations.

显然，第一个括号需要较少的操作。
给定表示矩阵链的数组p []，使第i个矩阵Ai的尺寸为p [i-1] xp [i]。我们需要编写一个函数MatrixChainOrder()，该函数应返回乘法链所需的最小乘法数。

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

这个问题主要是先前帖子的扩展。在上一篇文章中，我们讨论了仅寻找最佳成本的算法。在这里，我们还需要打印括号。

这个想法是在2D数组括号[n] [n]中为每个子表达式(i，j)存储最佳断点。一旦我们构造了括号数组，就可以使用以下代码打印括号。

// 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 ")";
}

以下是上述步骤的C++实现。

CPP

// C++ program to print optimal parenthesization
// in matrix chain multiplication.
#include 
using namespace std;
 
// Function for printing the optimal
// parenthesization of a matrix chain product
void 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/k6EYKj
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[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 code
int main()
{
    int arr[] = { 40, 20, 30, 10, 30 };
    int n = sizeof(arr) / sizeof(arr[0]);
    matrixChainOrder(arr, n);
    return 0;
}

Java

// 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

输出

Optimal Parenthesization is : ((A(BC))D)nOptimal Cost is : 26000

时间复杂度： O(n ³ )
辅助空间： O(n ² )

另一种方法：

——————————————

此解决方案尝试使用基于置换的递归来解决问题。

Let's take example:  {40, 20, 30, 10, 30}
n = 5

让我们将其划分为矩阵

[ [40, 20], [20, 30], [30, 10], [10, 30] ]

[ A , B , C , D ]

it contains 4 matrices i.e. (n - 1)

我们有3个组合相乘，即(n-2)

AB    or    BC    or     CD

算法：

1)给定长度为M的矩阵数组，遍历M – 1次

2)在每个循环中合并连续的矩阵

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)将当前两个矩阵合并为一个，然后从列表中删除合并的矩阵列表。

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)重复：转到STEP – 1，将newMatrices作为输入M —递归

5)当列表中有2个矩阵时，停止递归。

工作流程

矩阵按以下方式减少，

并且在递归过程中必须保留成本并将其与每个父步骤的先前值相加。

[ 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..

返回时，即在每次递归的最后一步，请检查该值是否小于任何其他值。

以下是上述步骤的Java实现。

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

输出

matrices = 
[[40, 20] [20, 30] [30, 10] [10, 30] ]
Final labels: 
((A(BC))D)
Final Cost:
26000