Printing brackets in matrix chain multiplication
Matrix chain multiplication is a common problem in computer science and mathematics that involves finding the most efficient way to multiply a series of matrices. The problem is to determine the optimal order of matrix multiplication, considering that the cost of multiplying two matrices depends on their dimensions. In this article, we will explore the problem statement and provide a solution in Java for printing the brackets in the optimal matrix chain multiplication order.
Problem Statement:
Given a sequence of matrices with their dimensions, the goal is to find the optimal order of matrix multiplication that minimizes the total cost of multiplication. The cost of multiplying two matrices is defined as the number of scalar multiplications required.
For example, given the sequence of matrices A, B, C, and D with dimensions [10x16, 16x12, 12x6, 6x14], we want to find the optimal order of multiplication and print the brackets that represent the order. In this case, the optimal order is ((A(BC))D).
Code Solution
To solve this problem, we can use dynamic programming to find the optimal order of matrix multiplication. We will create two matrices, result and brackets, to store the intermediate results and the optimal brackets respectively.
The result matrix will store the minimum cost of multiplying matrices in the given sequence. The brackets matrix will store the index of the matrix that determines the optimal split at each step.
Here is the step-by-step approach for finding the optimal order and printing the brackets:
Initialize the
resultandbracketsmatrices with appropriate dimensions.Iterate over the matrix chain lengths from 1 to n-1 (n is the number of matrices).
For each chain length, iterate over the matrix chain starting from the first matrix.
Compute the cost of multiplying matrices within the current chain length by considering all possible splits.
Update the
resultmatrix with the minimum cost and record the split index in thebracketsmatrix.Finally, use a recursive function
showBracketsto print the optimal brackets based on thebracketsmatrix.
The showBrackets function recursively prints the brackets by splitting the matrices at the recorded indices in the brackets matrix. If the split index is the same for the start and end matrices, it means there is only one matrix, and we print its name. Otherwise, we enclose the matrices within brackets and recursively call the function for the split sections.
// Java Program
// Printing brackets in matrix chain multiplication
public class Multiplication
{
public char name;
public int count;
public Multiplication()
{
this.name = 'A';
this.count = 0;
}
public void showBrackets(int i, int j, int[][] brackets)
{
if (i == j)
{
if (count != 0)
{
System.out.print(name + count);
}
else
{
System.out.print(name);
}
if (name == 'Z')
{
// Useful when dimension size is exceed A..Z
name = 'A';
count++;
}
else
{
// Change name
name++;
}
}
else
{
System.out.print("(");
showBrackets(i, brackets[i][j], brackets);
showBrackets(brackets[i][j] + 1, j, brackets);
System.out.print(")");
}
}
public void matrixChainMultiplication(int[] dims, int size)
{
int n = size - 1;
int[][] result = new int[n][n];
int[][] brackets = new int[n][n];
int cost = 0;
int j = 0;
for (int len = 1; len < n; len++)
{
for (int i = 0; i < n - len; i++)
{
j = i + len;
result[i][j] = Integer.MAX_VALUE;
for (int k = i; k < j; k++)
{
cost = result[i][k] + result[k + 1][j] +
dims[i] * dims[k + 1] * dims[j + 1];
if (cost < result[i][j])
{
result[i][j] = cost;
brackets[i][j] = k;
}
}
}
}
this.name = 'A';
this.count = 0;
// Show Brackets
showBrackets(0, n - 1, brackets);
System.out.print("\n");
}
public static void main(String args[])
{
Multiplication task = new Multiplication();
int[] dims1 = {
10 , 16 , 12 , 6 , 14
};
int[] dims2 = {
8 , 20 , 16 , 10 , 6
};
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
int n = dims1.length;
task.matrixChainMultiplication(dims1, n);
n = dims2.length;
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n);
}
}Output
((A(BC))D)
(A(B(CD)))// Include header file
#include <iostream>
#include <vector>
#include <limits.h>
using namespace std;
// C++ Program
// Printing brackets in matrix chain multiplication
class Multiplication
{
public: char name;
int count;
Multiplication()
{
this->name = 'A';
this->count = 0;
}
void showBrackets(int i, int j, vector<vector<int>> brackets)
{
if (i == j)
{
if (this->count != 0)
{
cout << this->name + this->count;
}
else
{
cout << this->name;
}
if (this->name == 'Z')
{
// Useful when dimension size is exceed A..Z
this->name = 'A';
this->count++;
}
else
{
// Change name
this->name++;
}
}
else
{
cout << "(";
this->showBrackets(i, brackets[i][j], brackets);
this->showBrackets(brackets[i][j] + 1, j, brackets);
cout << ")";
}
}
void matrixChainMultiplication(int dims[], int size)
{
int n = size - 1;
vector<vector<int>> result (n,vector<int>(n,0));
vector<vector<int>> brackets (n,vector<int>(n,0));
int cost = 0;
int j = 0;
for (int len = 1; len < n; len++)
{
for (int i = 0; i < n - len; i++)
{
j = i + len;
result[i][j] = INT_MAX;
for (int k = i; k < j; k++)
{
cost = result[i][k] + result[k + 1][j] +
dims[i] *dims[k + 1] *dims[j + 1];
if (cost < result[i][j])
{
result[i][j] = cost;
brackets[i][j] = k;
}
}
}
}
this->name = 'A';
this->count = 0;
// Show Brackets
this->showBrackets(0, n - 1, brackets);
cout << "\n";
}
};
int main()
{
Multiplication *task = new Multiplication();
int dims1[] = {
10 , 16 , 12 , 6 , 14
};
int dims2[] = {
8 , 20 , 16 , 10 , 6
};
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
int n = sizeof(dims1) / sizeof(dims1[0]);
task->matrixChainMultiplication(dims1, n);
n = sizeof(dims2) / sizeof(dims2[0]);
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task->matrixChainMultiplication(dims2, n);
return 0;
}Output
((A(BC))D)
(A(B(CD)))// Include namespace system
using System;
// Csharp Program
// Printing brackets in matrix chain multiplication
public class Multiplication
{
public char name;
public int count;
public Multiplication()
{
this.name = 'A';
this.count = 0;
}
public void showBrackets(int i, int j, int[,] brackets)
{
if (i == j)
{
if (this.count != 0)
{
Console.Write(this.name + this.count);
}
else
{
Console.Write(this.name);
}
if (this.name == 'Z')
{
// Useful when dimension size is exceed A..Z
this.name = 'A';
this.count++;
}
else
{
// Change name
this.name++;
}
}
else
{
Console.Write("(");
this.showBrackets(i, brackets[i,j], brackets);
this.showBrackets(brackets[i,j] + 1, j, brackets);
Console.Write(")");
}
}
public void matrixChainMultiplication(int[] dims, int size)
{
int n = size - 1;
int[,] result = new int[n,n];
int[,] brackets = new int[n,n];
int cost = 0;
int j = 0;
for (int len = 1; len < n; len++)
{
for (int i = 0; i < n - len; i++)
{
j = i + len;
result[i,j] = int.MaxValue;
for (int k = i; k < j; k++)
{
cost = result[i,k] + result[k + 1,j] + dims[i] *
dims[k + 1] * dims[j + 1];
if (cost < result[i,j])
{
result[i,j] = cost;
brackets[i,j] = k;
}
}
}
}
this.name = 'A';
this.count = 0;
// Show Brackets
this.showBrackets(0, n - 1, brackets);
Console.Write("\n");
}
public static void Main(String[] args)
{
Multiplication task = new Multiplication();
int[] dims1 = {
10 , 16 , 12 , 6 , 14
};
int[] dims2 = {
8 , 20 , 16 , 10 , 6
};
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
int n = dims1.Length;
task.matrixChainMultiplication(dims1, n);
n = dims2.Length;
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n);
}
}Output
((A(BC))D)
(A(B(CD)))package main
import "math"
import "fmt"
// Go Program
// Printing brackets in matrix chain multiplication
type Multiplication struct {
name byte
count int
}
func getMultiplication() * Multiplication {
var me *Multiplication = &Multiplication {}
me.name = 'A'
me.count = 0
return me
}
func(this *Multiplication) showBrackets(i int,
j int, brackets[][] int) {
if i == j {
if this.count != 0 {
fmt.Print(string(this.name), this.count)
} else {
fmt.Print(string(this.name))
}
if this.name == 'Z' {
// Useful when dimension size is exceed A..Z
this.name = 'A'
this.count++
} else {
// Change name
this.name = this.name + 1
}
} else {
fmt.Print("(")
this.showBrackets(i, brackets[i][j], brackets)
this.showBrackets(brackets[i][j] + 1, j, brackets)
fmt.Print(")")
}
}
func(this *Multiplication) matrixChainMultiplication(dims[] int, size int) {
var n int = size - 1
var result = make([][] int, n)
for i := 0; i < n; i++ {
result[i] = make([]int,n)
}
var brackets = make([][] int, n)
for i := 0; i < n; i++ {
brackets[i] = make([]int,n)
}
var cost int = 0
var j int = 0
for len := 1 ; len < n ; len++ {
for i := 0 ; i < n - len ; i++ {
j = i + len
result[i][j] = math.MaxInt64
for k := i ; k < j ; k++ {
cost = result[i][k] + result[k + 1][j] + dims[i] *
dims[k + 1] * dims[j + 1]
if cost < result[i][j] {
result[i][j] = cost
brackets[i][j] = k
}
}
}
}
this.name = 'A'
this.count = 0
// Show Brackets
this.showBrackets(0, n - 1, brackets)
fmt.Print("\n")
}
func main() {
var task * Multiplication = getMultiplication()
var dims1 = [] int {
10,
16,
12,
6,
14,
}
var dims2 = [] int {
8,
20,
16,
10,
6,
}
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
var n int = len(dims1)
task.matrixChainMultiplication(dims1, n)
n = len(dims2)
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n)
}Output
((A(AB))D)
(A(A(BD)))<?php
// Php Program
// Printing brackets in matrix chain multiplication
class Multiplication
{
public $name;
public $count;
public function __construct()
{
$this->name = 'A';
$this->count = 0;
}
public function showBrackets($i, $j, $brackets)
{
if ($i == $j)
{
if ($this->count != 0)
{
echo(ord($this->name) + $this->count);
}
else
{
echo($this->name);
}
if ($this->name == 'Z')
{
// Useful when dimension size is exceed A..Z
$this->name = 'A';
$this->count++;
}
else
{
// Change name
$this->name++;
}
}
else
{
echo("(");
$this->showBrackets($i, $brackets[$i][$j], $brackets);
$this->showBrackets($brackets[$i][$j] + 1, $j, $brackets);
echo(")");
}
}
public function matrixChainMultiplication($dims, $size)
{
$n = $size - 1;
$result = array_fill(0, $n, array_fill(0, $n, 0));
$brackets = array_fill(0, $n, array_fill(0, $n, 0));
$cost = 0;
$j = 0;
for ($len = 1; $len < $n; $len++)
{
for ($i = 0; $i < $n - $len; $i++)
{
$j = $i + $len;
$result[$i][$j] = PHP_INT_MAX;
for ($k = $i; $k < $j; $k++)
{
$cost = $result[$i][$k] + $result[$k + 1][$j] + $dims[$i] *
$dims[$k + 1] * $dims[$j + 1];
if ($cost < $result[$i][$j])
{
$result[$i][$j] = $cost;
$brackets[$i][$j] = $k;
}
}
}
}
$this->name = 'A';
$this->count = 0;
// Show Brackets
$this->showBrackets(0, $n - 1, $brackets);
echo("\n");
}
}
function main()
{
$task = new Multiplication();
$dims1 = array(10, 16, 12, 6, 14);
$dims2 = array(8, 20, 16, 10, 6);
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
$n = count($dims1);
$task->matrixChainMultiplication($dims1, $n);
$n = count($dims2);
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
$task->matrixChainMultiplication($dims2, $n);
}
main();Output
((A(BC))D)
(A(B(CD)))// Node JS Program
// Printing brackets in matrix chain multiplication
class Multiplication
{
constructor()
{
this.name = 'A';
this.count = 0;
}
showBrackets(i, j, brackets)
{
if (i == j)
{
if (this.count != 0)
{
process.stdout.write("" + this.name.charCodeAt(0) + this.count);
}
else
{
process.stdout.write(this.name);
}
if (this.name == 'Z')
{
// Useful when dimension size is exceed A..Z
this.name = 'A';
this.count++;
}
else
{
// Change name
this.name = String. fromCharCode(this.name.charCodeAt(0) +
1);
}
}
else
{
process.stdout.write("(");
this.showBrackets(i, brackets[i][j], brackets);
this.showBrackets(brackets[i][j] + 1, j, brackets);
process.stdout.write(")");
}
}
matrixChainMultiplication(dims, size)
{
var n = size - 1;
var result = Array(n).fill(0).map(() => new Array(n).fill(0));
var brackets = Array(n).fill(0).map(() => new Array(n).fill(0));
var cost = 0;
var j = 0;
for (var len = 1; len < n; len++)
{
for (var i = 0; i < n - len; i++)
{
j = i + len;
result[i][j] = Number.MAX_VALUE;
for (var k = i; k < j; k++)
{
cost = result[i][k] + result[k + 1][j] + dims[i] *
dims[k + 1] * dims[j + 1];
if (cost < result[i][j])
{
result[i][j] = cost;
brackets[i][j] = k;
}
}
}
}
this.name = 'A';
this.count = 0;
// Show Brackets
this.showBrackets(0, n - 1, brackets);
process.stdout.write("\n");
}
}
function main()
{
var task = new Multiplication();
var dims1 = [10, 16, 12, 6, 14];
var dims2 = [8, 20, 16, 10, 6];
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
var n = dims1.length;
task.matrixChainMultiplication(dims1, n);
n = dims2.length;
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n);
}
main();Output
((A(BC))D)
(A(B(CD)))import sys
# Python 3 Program
# Printing brackets in matrix chain multiplication
class Multiplication :
def __init__(self) :
self.name = 'A'
self.count = 0
def showBrackets(self, i, j, brackets) :
if (i == j) :
if (self.count != 0) :
print(ord(self.name) + self.count, end = "")
else :
print(self.name, end = "")
if (self.name == 'Z') :
# Useful when dimension size is exceed A..Z
self.name = 'A'
self.count += 1
else :
# Change name
self.name = chr(ord(self.name) + 1)
else :
print("(", end = "")
self.showBrackets(i, brackets[i][j], brackets)
self.showBrackets(brackets[i][j] + 1, j, brackets)
print(")", end = "")
def matrixChainMultiplication(self, dims, size) :
n = size - 1
result = [[0] * (n) for _ in range(n) ]
brackets = [[0] * (n) for _ in range(n) ]
cost = 0
j = 0
len = 1
while (len < n) :
i = 0
while (i < n - len) :
j = i + len
result[i][j] = sys.maxsize
k = i
while (k < j) :
cost = result[i][k] + result[k + 1][j] + dims[i] * dims[k + 1] * dims[j + 1]
if (cost < result[i][j]) :
result[i][j] = cost
brackets[i][j] = k
k += 1
i += 1
len += 1
self.name = 'A'
self.count = 0
# Show Brackets
self.showBrackets(0, n - 1, brackets)
print(end = "\n")
def main() :
task = Multiplication()
dims1 = [10, 16, 12, 6, 14]
dims2 = [8, 20, 16, 10, 6]
# dims = [10 , 16 , 12 , 6 , 14]
# matri× A = 10 × 16
# matri× B = 16 × 12
# matri× C = 12 × 6
# matri× D = 6 × 14
# --------------------
# (A(BC))D
# (16×12×6) + (10×16×6) + (10×6×14)
# = 2952
n = len(dims1)
task.matrixChainMultiplication(dims1, n)
n = len(dims2)
# dims = [8 , 20 , 16 , 10 , 6]
# matri× A = 8 × 20
# matri× B = 20 × 16
# matri× C = 16 × 10
# matri× D = 10 × 6
# A(B(CD)) = 3840
# (16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
task.matrixChainMultiplication(dims2, n)
if __name__ == "__main__": main()Output
((A(BC))D)
(A(B(CD)))# Ruby Program
# Printing brackets in matrix chain multiplication
class Multiplication
# Define the accessor and reader of class Multiplication
attr_reader :name, :count
attr_accessor :name, :count
def initialize()
self.name = 'A'
self.count = 0
end
def showBrackets(i, j, brackets)
if (i == j)
if (self.count != 0)
print(self.name.ord + self.count)
else
print(self.name)
end
if (self.name == 'Z')
# Useful when dimension size is exceed A..Z
self.name = 'A'
self.count += 1
else
# Change name
self.name = (self.name.ord+1).chr
end
else
print("(")
self.showBrackets(i, brackets[i][j], brackets)
self.showBrackets(brackets[i][j] + 1, j, brackets)
print(")")
end
end
def matrixChainMultiplication(dims, size)
n = size - 1
result = Array.new(n) {Array.new(n) {0}}
brackets = Array.new(n) {Array.new(n) {0}}
cost = 0
j = 0
len = 1
while (len < n)
i = 0
while (i < n - len)
j = i + len
result[i][j] = (2 ** (0. size * 8 - 2))
k = i
while (k < j)
cost = result[i][k] + result[k + 1][j] + dims[i] *
dims[k + 1] * dims[j + 1]
if (cost < result[i][j])
result[i][j] = cost
brackets[i][j] = k
end
k += 1
end
i += 1
end
len += 1
end
self.name = 'A'
self.count = 0
# Show Brackets
self.showBrackets(0, n - 1, brackets)
print("\n")
end
end
def main()
task = Multiplication.new()
dims1 = [10, 16, 12, 6, 14]
dims2 = [8, 20, 16, 10, 6]
# dims = [10 , 16 , 12 , 6 , 14]
# matri× A = 10 × 16
# matri× B = 16 × 12
# matri× C = 12 × 6
# matri× D = 6 × 14
# --------------------
# (A(BC))D
# (16×12×6) + (10×16×6) + (10×6×14)
# = 2952
n = dims1.length
task.matrixChainMultiplication(dims1, n)
n = dims2.length
# dims = [8 , 20 , 16 , 10 , 6]
# matri× A = 8 × 20
# matri× B = 20 × 16
# matri× C = 16 × 10
# matri× D = 10 × 6
# A(B(CD)) = 3840
# (16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
task.matrixChainMultiplication(dims2, n)
end
main()Output
((A(BC))D)
(A(B(CD)))
// Scala Program
// Printing brackets in matrix chain multiplication
class Multiplication(var name: Char,
var count: Int)
{
def this()
{
this('A', 0)
}
def showBrackets(i: Int, j: Int, brackets: Array[Array[Int]]): Unit = {
if (i == j)
{
if (count != 0)
{
print(name.toInt + count);
}
else
{
print(name);
}
if (name == 'Z')
{
// Useful when dimension size is exceed A..Z
name = 'A';
count += 1;
}
else
{
// Change name
name = (name.toInt + 1).toChar;
}
}
else
{
print("(");
showBrackets(i, brackets(i)(j), brackets);
showBrackets(brackets(i)(j) + 1, j, brackets);
print(")");
}
}
def matrixChainMultiplication(dims: Array[Int], size: Int): Unit = {
var n: Int = size - 1;
var result: Array[Array[Int]] = Array.fill[Int](n, n)(0);
var brackets: Array[Array[Int]] = Array.fill[Int](n, n)(0);
var cost: Int = 0;
var j: Int = 0;
var len: Int = 1;
while (len < n)
{
var i: Int = 0;
while (i < n - len)
{
j = i + len;
result(i)(j) = Int.MaxValue;
var k: Int = i;
while (k < j)
{
cost = result(i)(k) + result(k + 1)(j) + dims(i) *
dims(k + 1) * dims(j + 1);
if (cost < result(i)(j))
{
result(i)(j) = cost;
brackets(i)(j) = k;
}
k += 1;
}
i += 1;
}
len += 1;
}
this.name = 'A';
this.count = 0;
// Show Brackets
showBrackets(0, n - 1, brackets);
print("\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Multiplication = new Multiplication();
var dims1: Array[Int] = Array(10, 16, 12, 6, 14);
var dims2: Array[Int] = Array(8, 20, 16, 10, 6);
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
var n: Int = dims1.length;
task.matrixChainMultiplication(dims1, n);
n = dims2.length;
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n);
}
}Output
((A(BC))D)
(A(B(CD)))import Foundation;
// Swift 4 Program
// Printing brackets in matrix chain multiplication
class Multiplication
{
var name: Character;
var count: Int;
init()
{
self.name = "A";
self.count = 0;
}
func showBrackets(_ i: Int, _ j: Int, _ brackets: [
[Int]
])
{
if (i == j)
{
if (self.count != 0)
{
print(Int(UnicodeScalar(String(self.name))!.value) + self.count, terminator: "");
}
else
{
print(self.name, terminator: "");
}
if (self.name == "Z")
{
// Useful when dimension size is exceed A..Z
self.name = "A";
self.count += 1 ;
}
else
{
// Change name
self.name = Character(UnicodeScalar(
Int(UnicodeScalar(String(self.name))!.value) + i)!)
}
}
else
{
print("(", terminator: "");
self.showBrackets(i, brackets[i][j], brackets);
self.showBrackets(brackets[i][j] + 1, j, brackets);
print(")", terminator: "");
}
}
func matrixChainMultiplication(_ dims: [Int], _ size: Int)
{
let n: Int = size - 1;
var result: [
[Int]
] = Array(repeating: Array(repeating: 0, count: n), count: n);
var brackets: [
[Int]
] = Array(repeating: Array(repeating: 0, count: n), count: n);
var cost: Int = 0;
var j: Int = 0;
var len: Int = 1;
while (len < n)
{
var i: Int = 0;
while (i < n - len)
{
j = i + len;
result[i][j] = Int.max;
var k: Int = i;
while (k < j)
{
cost = result[i][k] + result[k + 1][j] + dims[i] *
dims[k + 1] * dims[j + 1];
if (cost < result[i][j])
{
result[i][j] = cost;
brackets[i][j] = k;
}
k += 1;
}
i += 1;
}
len += 1;
}
self.name = "A";
self.count = 0;
// Show Brackets
self.showBrackets(0, n - 1, brackets);
print(terminator: "\n");
}
}
func main()
{
let task: Multiplication = Multiplication();
let dims1: [Int] = [10, 16, 12, 6, 14];
let dims2: [Int] = [8, 20, 16, 10, 6];
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
var n: Int = dims1.count;
task.matrixChainMultiplication(dims1, n);
n = dims2.count;
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n);
}
main();Output
((A(AB))D)
(A(A(BD)))// Kotlin Program
// Printing brackets in matrix chain multiplication
class Multiplication
{
var name: Char;
var count: Int;
constructor()
{
this.name = 'A';
this.count = 0;
}
fun showBrackets(i: Int, j: Int,
brackets: Array < Array < Int >> ): Unit
{
if (i == j)
{
if (this.count != 0)
{
print(this.name.toInt() + this.count);
}
else
{
print(this.name);
}
if (this.name == 'Z')
{
// Useful when dimension size is exceed A..Z
this.name = 'A';
this.count += 1;
}
else
{
// Change name
this.name += 1;
}
}
else
{
print("(");
this.showBrackets(i, brackets[i][j], brackets);
this.showBrackets(brackets[i][j] + 1, j, brackets);
print(")");
}
}
fun matrixChainMultiplication(dims: Array < Int > , size: Int): Unit
{
var n: Int = size - 1;
val result: Array < Array < Int >> = Array(n)
{
Array(n)
{
0
}
};
val brackets: Array < Array < Int >> = Array(n)
{
Array(n)
{
0
}
};
var cost : Int;
var j : Int;
var len: Int = 1;
while (len < n)
{
var i: Int = 0;
while (i < n - len)
{
j = i + len;
result[i][j] = Int.MAX_VALUE;
var k: Int = i;
while (k < j)
{
cost = result[i][k] + result[k + 1][j] + dims[i] *
dims[k + 1] * dims[j + 1];
if (cost < result[i][j])
{
result[i][j] = cost;
brackets[i][j] = k;
}
k += 1;
}
i += 1;
}
len += 1;
}
this.name = 'A';
this.count = 0;
// Show Brackets
this.showBrackets(0, n - 1, brackets);
print("\n");
}
}
fun main(args: Array < String > ): Unit
{
val task: Multiplication = Multiplication();
val dims1: Array < Int > = arrayOf(10, 16, 12, 6, 14);
val dims2: Array < Int > = arrayOf(8, 20, 16, 10, 6);
/*
dims = [10 , 16 , 12 , 6 , 14]
matri× A = 10 × 16
matri× B = 16 × 12
matri× C = 12 × 6
matri× D = 6 × 14
--------------------
(A(BC))D
(16×12×6) + (10×16×6) + (10×6×14)
= 2952
*/
var n: Int = dims1.count();
task.matrixChainMultiplication(dims1, n);
n = dims2.count();
/*
dims = [8 , 20 , 16 , 10 , 6]
matri× A = 8 × 20
matri× B = 20 × 16
matri× C = 16 × 10
matri× D = 10 × 6
A(B(CD)) = 3840
(16×10×6) + (20×16×6 ) + (8×20×6 ) = 3840
*/
task.matrixChainMultiplication(dims2, n);
}Output
((A(BC))D)
(A(B(CD)))Code Explanation:
Let's understand the provided Java code step by step:
The
Multiplicationclass represents the solution for the problem. It has member variablesnameandcountto keep track of matrix names and counts.The
showBracketsfunction is responsible for recursively printing the brackets based on thebracketsmatrix.The
matrixChainMultiplicationfunction implements the dynamic programming approach to find the optimal order and compute the minimum cost of matrix multiplication.In the
mainfunction, two example sequences of matrices are defined:dims1anddims2. ThematrixChainMultiplicationfunction is called for each sequence to find the optimal order and print the brackets.
Output:
The output of the provided code for the given sequences of matrices is:
((A(BC))D)
(A(B(CD)))These outputs represent the optimal order of matrix multiplication along with the brackets.
The problem of matrix chain multiplication involves finding the optimal order of matrix multiplication to minimize the cost. In this article, we explored a dynamic programming solution in Java for printing the brackets that represent the optimal order. By using the provided code and understanding the underlying approach, you can apply this solution to solve similar matrix chain multiplication problems in other programming languages.
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