Matrix chain multiplication using dynamic programming
Here given code implementation process.
// C Program
// Matrix chain multiplication
#include <stdio.h>
#include <limits.h>
int matrixChainMultiplication(int dims[], int n)
{
int c[n][n];
int j = 0;
int cost = 0;
for (int i = 1; i < n; ++i)
{
c[i][i] = 0;
}
for (int len = 2; len < n; ++len)
{
for (int i = 1; i < n - len + 1; ++i)
{
j = i + len - 1;
c[i][j] = INT_MAX;
for (int k = i; k <= j - 1 && j < n; ++k)
{
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] * dims[k] * dims[j];
if (cost < c[i][j])
{
c[i][j] = cost;
}
}
}
}
return c[1][n - 1];
}
int main()
{
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]);
printf("\n %d",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
*/
printf("\n %d",matrixChainMultiplication(dims2, n));
return 0;
}
Output
2952
3840
// Java Program
// Matrix chain multiplication using dynamic programming
public class Multiplication
{
public int matrixChainMultiplication(int[] dims, int n)
{
int[][] c = new int[n][n];
int j = 0;
int cost = 0;
for (int i = 1; i < n; ++i)
{
c[i][i] = 0;
}
for (int len = 2; len < n; ++len)
{
for (int i = 1; i < n - len + 1; ++i)
{
j = i + len - 1;
c[i][j] = Integer.MAX_VALUE;
for (int k = i; k <= j - 1 && j < n; ++k)
{
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] * dims[k] * dims[j];
if (cost < c[i][j])
{
c[i][j] = cost;
}
}
}
}
return c[1][n - 1];
}
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;
System.out.print("\n " + 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
*/
System.out.print("\n " + task.matrixChainMultiplication(dims2, n));
}
}
Output
2952
3840
// Include header file
#include <iostream>
#include <limits.h>
using namespace std;
// C++ Program
// Matrix chain multiplication using dynamic programming
class Multiplication
{
public: int matrixChainMultiplication(int dims[], int n)
{
int c[n][n];
int j = 0;
int cost = 0;
for (int i = 1; i < n; ++i)
{
c[i][i] = 0;
}
for (int len = 2; len < n; ++len)
{
for (int i = 1; i < n - len + 1; ++i)
{
j = i + len - 1;
c[i][j] = INT_MAX;
for (int k = i; k <= j - 1 && j < n; ++k)
{
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] *dims[k] *dims[j];
if (cost < c[i][j])
{
c[i][j] = cost;
}
}
}
}
return c[1][n - 1];
}
};
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]);
cout << "\n " << 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
*/
cout << "\n " << task->matrixChainMultiplication(dims2, n);
return 0;
}
Output
2952
3840
// Include namespace system
using System;
// Csharp Program
// Matrix chain multiplication using dynamic programming
public class Multiplication
{
public int matrixChainMultiplication(int[] dims, int n)
{
int[,] c = new int[n,n];
int j = 0;
int cost = 0;
for (int i = 1; i < n; ++i)
{
c[i,i] = 0;
}
for (int len = 2; len < n; ++len)
{
for (int i = 1; i < n - len + 1; ++i)
{
j = i + len - 1;
c[i,j] = int.MaxValue;
for (int k = i; k <= j - 1 && j < n; ++k)
{
cost = c[i,k] + c[k + 1,j] +
dims[i - 1] * dims[k] * dims[j];
if (cost < c[i,j])
{
c[i,j] = cost;
}
}
}
}
return c[1,n - 1];
}
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;
Console.Write("\n " + 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
*/
Console.Write("\n " + task.matrixChainMultiplication(dims2, n));
}
}
Output
2952
3840
package main
import "math"
import "fmt"
// Go Program
// Matrix chain multiplication using dynamic programming
func matrixChainMultiplication(dims[] int, n int) int {
var c = make([][] int, n)
for i :=0; i < n;i++{
c[i] = make([]int, n)
}
var j int = 0
var cost int = 0
for i := 1 ; i < n ; i++ {
c[i][i] = 0
}
for len := 2 ; len < n ; len++ {
for i := 1 ; i < n - len + 1 ; i++ {
j = i + len - 1
c[i][j] = math.MaxInt64
for k := i ; k <= j - 1 && j < n ; k++ {
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] * dims[k] * dims[j]
if cost < c[i][j] {
c[i][j] = cost
}
}
}
}
return c[1][n - 1]
}
func main() {
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)
fmt.Print("\n ", 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
*/
fmt.Print("\n ", matrixChainMultiplication(dims2, n))
}
Output
2952
3840
<?php
// Php Program
// Matrix chain multiplication using dynamic programming
class Multiplication
{
public function matrixChainMultiplication($dims, $n)
{
$c = array_fill(0, $n, array_fill(0, $n, 0));
$j = 0;
$cost = 0;
for ($len = 2; $len < $n; ++$len)
{
for ($i = 1; $i < $n - $len + 1; ++$i)
{
$j = $i + $len - 1;
$c[$i][$j] = PHP_INT_MAX;
for ($k = $i; $k <= $j - 1 && $j < $n; ++$k)
{
$cost = $c[$i][$k] + $c[$k + 1][$j] +
$dims[$i - 1] * $dims[$k] * $dims[$j];
if ($cost < $c[$i][$j])
{
$c[$i][$j] = $cost;
}
}
}
}
return $c[1][$n - 1];
}
}
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);
echo("\n ".$task->matrixChainMultiplication($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
*/
echo("\n ".$task->matrixChainMultiplication($dims2, $n));
}
main();
Output
2952
3840
// Node JS Program
// Matrix chain multiplication using dynamic programming
class Multiplication
{
matrixChainMultiplication(dims, n)
{
var c = Array(n).fill(0).map(() => new Array(n).fill(0));
var j = 0;
var cost = 0;
for (var i = 1; i < n; ++i)
{
c[i][i] = 0;
}
for (var len = 2; len < n; ++len)
{
for (var i = 1; i < n - len + 1; ++i)
{
j = i + len - 1;
c[i][j] = Number.MAX_VALUE;
for (var k = i; k <= j - 1 && j < n; ++k)
{
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] * dims[k] * dims[j];
if (cost < c[i][j])
{
c[i][j] = cost;
}
}
}
}
return c[1][n - 1];
}
}
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;
process.stdout.write("\n " + 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
*/
process.stdout.write("\n " + task.matrixChainMultiplication(dims2, n));
}
main();
Output
2952
3840
import sys
# Python 3 Program
# Matrix chain multiplication using dynamic programming
class Multiplication :
def matrixChainMultiplication(self, dims, n) :
c = [[0] * (n) for _ in range(n) ]
j = 0
cost = 0
i = 1
while (i < n) :
c[i][i] = 0
i += 1
len = 2
while (len < n) :
i = 1
while (i < n - len + 1) :
j = i + len - 1
c[i][j] = sys.maxsize
k = i
while (k <= j - 1 and j < n) :
cost = c[i][k] + c[k + 1][j] + dims[i - 1] * dims[k] * dims[j]
if (cost < c[i][j]) :
c[i][j] = cost
k += 1
i += 1
len += 1
return c[1][n - 1]
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)
print("\n ", task.matrixChainMultiplication(dims1, n), end = "")
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
print("\n ", task.matrixChainMultiplication(dims2, n), end = "")
if __name__ == "__main__": main()
Output
2952
3840
# Ruby Program
# Matrix chain multiplication using dynamic programming
class Multiplication
def matrixChainMultiplication(dims, n)
c = Array.new(n) {Array.new(n) {0}}
j = 0
cost = 0
i = 1
while (i < n)
c[i][i] = 0
i += 1
end
len = 2
while (len < n)
i = 1
while (i < n - len + 1)
j = i + len - 1
c[i][j] = (2 ** (0. size * 8 - 2))
k = i
while (k <= j - 1 && j < n)
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] * dims[k] * dims[j]
if (cost < c[i][j])
c[i][j] = cost
end
k += 1
end
i += 1
end
len += 1
end
return c[1][n - 1]
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
print("\n ", 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
print("\n ", task.matrixChainMultiplication(dims2, n))
end
main()
Output
2952
3840
// Scala Program
// Matrix chain multiplication using dynamic programming
class Multiplication()
{
def matrixChainMultiplication(dims: Array[Int], n: Int): Int = {
var c: Array[Array[Int]] = Array.fill[Int](n, n)(0);
var j: Int = 0;
var cost: Int = 0;
var i: Int = 1;
while (i < n)
{
c(i)(i) = 0;
i += 1;
}
var len: Int = 2;
while (len < n)
{
var i: Int = 1;
while (i < n - len + 1)
{
j = i + len - 1;
c(i)(j) = Int.MaxValue;
var k: Int = i;
while (k <= j - 1 && j < n)
{
cost = c(i)(k) + c(k + 1)(j) +
dims(i - 1) * dims(k) * dims(j);
if (cost < c(i)(j))
{
c(i)(j) = cost;
}
k += 1;
}
i += 1;
}
len += 1;
}
return c(1)(n - 1);
}
}
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;
print("\n " + 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
*/
print("\n " + task.matrixChainMultiplication(dims2, n));
}
}
Output
2952
3840
import Foundation;
// Swift 4 Program
// Matrix chain multiplication using dynamic programming
class Multiplication
{
func matrixChainMultiplication(_ dims: [Int], _ n: Int) -> Int
{
var c: [
[Int]
] = Array(repeating: Array(repeating: 0, count: n), count: n);
var j: Int = 0;
var cost: Int = 0;
var len: Int = 2;
while (len < n)
{
var i: Int = 1;
while (i < n - len + 1)
{
j = i + len - 1;
c[i][j] = Int.max;
var k: Int = i;
while (k <= j - 1 && j < n)
{
cost = c[i][k] + c[k + 1][j] + dims[i - 1] *
dims[k] * dims[j];
if (cost < c[i][j])
{
c[i][j] = cost;
}
k += 1;
}
i += 1;
}
len += 1;
}
return c[1][n - 1];
}
}
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;
print("\n ", task.matrixChainMultiplication(dims1, n), terminator: "");
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
*/
print("\n ", task.matrixChainMultiplication(dims2, n), terminator: "");
}
main();
Output
2952
3840
// Kotlin Program
// Matrix chain multiplication using dynamic programming
class Multiplication
{
fun matrixChainMultiplication(dims: Array < Int > , n: Int): Int
{
var c: Array < Array < Int >> = Array(n)
{
Array(n)
{
0
}
};
var cost: Int;
var len: Int = 2;
while (len < n)
{
var i: Int = 1;
while (i < n - len + 1)
{
var j = i + len - 1;
c[i][j] = Int.MAX_VALUE;
var k: Int = i;
while (k <= j - 1 && j < n)
{
cost = c[i][k] + c[k + 1][j] +
dims[i - 1] * dims[k] * dims[j];
if (cost < c[i][j])
{
c[i][j] = cost;
}
k += 1;
}
i += 1;
}
len += 1;
}
return c[1][n - 1];
}
}
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();
print("\n " + 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
*/
print("\n " + task.matrixChainMultiplication(dims2, n));
}
Output
2952
3840
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment