Dice Throw | DP-30

Given n dice each with m faces, numbered from 1 to m, find the number of ways to get sum X. X is the summation of values on each face when all the dice are thrown.

The Naive approach is to find all the possible combinations of values from n dice and keep on counting the results that sum to X.

This problem can be efficiently solved using Dynamic Programming (DP).  

Let the function to find X from n dice is: Sum(m, n, X)
The function can be represented as:
Sum(m, n, X) = Finding Sum (X - 1) from (n - 1) dice plus 1 from nth dice
               + Finding Sum (X - 2) from (n - 1) dice plus 2 from nth dice
               + Finding Sum (X - 3) from (n - 1) dice plus 3 from nth dice
                  ...................................................
                  ...................................................
                  ...................................................
              + Finding Sum (X - m) from (n - 1) dice plus m from nth dice

So we can recursively write Sum(m, n, x) as following
Sum(m, n, X) = Sum(m, n - 1, X - 1) + 
               Sum(m, n - 1, X - 2) +
               .................... + 
               Sum(m, n - 1, X - m)

Why DP approach?
The above problem exhibits overlapping subproblems. See the below diagram. Also, see this recursive implementation. Let there be 3 dice, each with 6 faces and we need to find the number of ways to get sum 8:

Sum(6, 3, 8) = Sum(6, 2, 7) + Sum(6, 2, 6) + Sum(6, 2, 5) + 
               Sum(6, 2, 4) + Sum(6, 2, 3) + Sum(6, 2, 2)

To evaluate Sum(6, 3, 8), we need to evaluate Sum(6, 2, 7) which can 
recursively written as following:
Sum(6, 2, 7) = Sum(6, 1, 6) + Sum(6, 1, 5) + Sum(6, 1, 4) + 
               Sum(6, 1, 3) + Sum(6, 1, 2) + Sum(6, 1, 1)

We also need to evaluate Sum(6, 2, 6) which can recursively written
as following:
Sum(6, 2, 6) = Sum(6, 1, 5) + Sum(6, 1, 4) + Sum(6, 1, 3) +
               Sum(6, 1, 2) + Sum(6, 1, 1)
..............................................
..............................................
Sum(6, 2, 2) = Sum(6, 1, 1)

Please take a closer look at the above recursion. The sub-problems in RED are solved first time and sub-problems in BLUE are solved again (exhibit overlapping sub-problems). Hence, storing the results of the solved sub-problems saves time.

Following is implementation of Dynamic Programming approach.  

C++

// C++ program to find number of ways to get sum 'x' with 'n'
// dice where every dice has 'm' faces
#include <iostream>
#include <string.h>
using namespace std;

//  The main function that returns number of ways to get sum 'x'
//  with 'n' dice and 'm' with m faces.
int findWays(int m, int n, int x)
{
    // Create a table to store results of subproblems.  One extra
    // row and column are used for simplicity (Number of dice
    // is directly used as row index and sum is directly used
    // as column index).  The entries in 0th row and 0th column
    // are never used.
    int table[n + 1][x + 1];
    memset(table, 0, sizeof(table)); // Initialize all entries as 0

    // Table entries for only one dice
    for (int j = 1; j <= m && j <= x; j++)
        table[1][j] = 1;

    // Fill rest of the entries in table using recursive relation
    // i: number of dice, j: sum
    for (int i = 2; i <= n; i++)
        for (int j = 1; j <= x; j++)
            for (int k = 1; k <= m && k < j; k++)
                table[i][j] += table[i-1][j-k];

    /* Uncomment these lines to see content of table
    for (int i = 0; i <= n; i++)
    {
      for (int j = 0; j <= x; j++)
        cout << table[i][j] << " ";
      cout << endl;
    } */
    return table[n][x];
}

// Driver program to test above functions
int main()
{
    cout << findWays(4, 2, 1) << endl;
    cout << findWays(2, 2, 3) << endl;
    cout << findWays(6, 3, 8) << endl;
    cout << findWays(4, 2, 5) << endl;
    cout << findWays(4, 3, 5) << endl;

    return 0;
}

Java

// Java program to find number of ways to get sum 'x' with 'n'
// dice where every dice has 'm' faces
import java.util.*;
import java.lang.*;
import java.io.*;

class GFG {
    /* The main function that returns the number of ways to get sum 'x' with 'n' dice and 'm' with m faces. */
    public static long findWays(int m, int n, int x){

    /* Create a table to store the results of subproblems.
    One extra row and column are used for simplicity
    (Number of dice is directly used as row index and sum is directly used as column index).
    The entries in 0th row and 0th column are never used. */
    long[][] table = new long[n+1][x+1];

    /* Table entries for only one dice */
    for(int j = 1; j <= m && j <= x; j++)
                table[1][j] = 1;

    /* Fill rest of the entries in table using recursive relation
    i: number of dice, j: sum */
    for(int i = 2; i <= n;i ++){
                for(int j = 1; j <= x; j++){
                    for(int k = 1; k < j && k <= m; k++)
                        table[i][j] += table[i-1][j-k];
                }
        }

        /* Uncomment these lines to see content of table
        for(int i = 0; i< n+1; i++){
            for(int j = 0; j< x+1; j++)
                System.out.print(dt[i][j] + " ");
            System.out.println();
        } */

        return table[n][x];
    }

    // Driver Code
    public static void main (String[] args) {
        System.out.println(findWays(4, 2, 1));
        System.out.println(findWays(2, 2, 3));
        System.out.println(findWays(6, 3, 8));
        System.out.println(findWays(4, 2, 5));
        System.out.println(findWays(4, 3, 5));
    }
}

// This code is contributed by MaheshwariPiyush

Python3

# Python3 program to find the number of ways to get sum 'x' with 'n' dice
# where every dice has 'm' faces

# The main function that returns number of ways to get sum 'x'
# with 'n' dice and 'm' with m faces.
def findWays(m,n,x):
    # Create a table to store results of subproblems. One extra
    # row and column are used for simplicity (Number of dice
    # is directly used as row index and sum is directly used
    # as column index). The entries in 0th row and 0th column
    # are never used.
    table=[[0]*(x+1) for i in range(n+1)] #Initialize all entries as 0

    for j in range(1,min(m+1,x+1)): #Table entries for only one dice
        table[1][j]=1

    # Fill rest of the entries in table using recursive relation
    # i: number of dice, j: sum
    for i in range(2,n+1):
        for j in range(1,x+1):
            for k in range(1,min(m+1,j)):
                table[i][j]+=table[i-1][j-k]

    #print(dt)
    # Uncomment above line to see content of table

    return table[-1][-1]

# Driver code
print(findWays(4,2,1))
print(findWays(2,2,3))
print(findWays(6,3,8))
print(findWays(4,2,5))
print(findWays(4,3,5))

# This code is contributed by MaheshwariPiyush

C#

// C# program to find number
// of ways to get sum 'x'
// with 'n' dice where every
// dice has 'm' faces
using System;

class GFG
{
// The main function that returns
// number of ways to get sum 'x'
// with 'n' dice and 'm' with m faces.
static int findWays(int m,
                    int n, int x)
{
    // Create a table to store
    // results of subproblems.
    // row and column are used
    // for simplicity (Number
    // of dice is directly used
    // as row index and sum is
    // directly used as column
    // index). The entries in 0th
    // row and 0th column are
    // never used.
    int[,] table = new int[n + 1,
                           x + 1];

    // Initialize all
    // entries as 0
    for (int i = 0; i <= n; i++)
    for (int j = 0; j <= x; j++)
    table[i, j] = 0;

    // Table entries for
    // only one dice
    for (int j = 1;
             j <= m && j <= x; j++)
        table[1, j] = 1;

    // Fill rest of the entries
    // in table using recursive
    // relation i: number of
    // dice, j: sum
    for (int i = 2; i <= n; i++)
        for (int j = 1; j <= x; j++)
            for (int k = 1;
                     k <= m && k < j; k++)
                table[i, j] += table[i - 1,
                                     j - k];

    /* Uncomment these lines to
    see content of table
    for (int i = 0; i <= n; i++)
    {
    for (int j = 0; j <= x; j++)
        cout << table[i][j] << " ";
    cout << endl;
    } */
    return table[n, x];
}

// Driver Code
public static void Main()
{
    Console.WriteLine(findWays(4, 2, 1));
    Console.WriteLine(findWays(2, 2, 3));
    Console.WriteLine(findWays(6, 3, 8));
    Console.WriteLine(findWays(4, 2, 5));
    Console.WriteLine(findWays(4, 3, 5));
}
}

// This code is contributed by mits.

PHP

<?php
// PHP program to find number
// of ways to get sum 'x' with
// 'n' dice where every dice
// has 'm' faces

// The main function that returns
// number of ways to get sum 'x'
// with 'n' dice and 'm' with m faces.
function findWays($m, $n, $x)
{
    // Create a table to store results
    // of subproblems. One extra row
    // and column are used for
    // simplicity (Number of dice is
    // directly used as row index and
    // sum is directly used as column
    // index). The entries in 0th row
    // and 0th column are never used.
    $table;

    // Initialize all entries as 0
        for ($i = 1; $i < $n + 1; $i++)
        for ($j = 1; $j < $x + 1; $j++)
        $table[$i][$j] = 0;

    // Table entries for
    // only one dice
    for ($j = 1; $j <= $m &&
                 $j <= $x; $j++)
        $table[1][$j] = 1;

    // Fill rest of the entries
    // in table using recursive
    // relation i: number of dice,
    // j: sum
    for ($i = 2; $i <= $n; $i++)
        for ($j = 1; $j <= $x; $j++)
            for ($k = 1; $k <= $m &&
                         $k < $j; $k++)
                $table[$i][$j] +=
                        $table[$i - 1][$j - $k];

    return $table[$n][$x];
}

// Driver Code
echo findWays(4, 2, 1). "\n";
echo findWays(2, 2, 3). "\n";
echo findWays(6, 3, 8). "\n";
echo findWays(4, 2, 5). "\n";
echo findWays(4, 3, 5). "\n";

// This code is contributed by mits.
?>

Javascript

<script>
// Javascript program to find number of ways to get sum 'x' with 'n'
// dice where every dice has 'm' faces


 /* The main function that returns the number of ways to get sum 'x' with 'n' dice and 'm' with m faces. */
function findWays(m,n,x)
{
    /* Create a table to store the results of subproblems.
    One extra row and column are used for simplicity
    (Number of dice is directly used as row index and sum is directly used as column index).
    The entries in 0th row and 0th column are never used. */
    let table = new Array(n+1);
    for(let i=0;i<(n+1);i++)
    {
        table[i]=new Array(x+1);
        for(let j=0;j<(x+1);j++)
        {
            table[i][j]=0;
        }
    }

    /* Table entries for only one dice */
    for(let j = 1; j <= m && j <= x; j++)
                table[1][j] = 1;

    /* Fill rest of the entries in table using recursive relation
    i: number of dice, j: sum */
    for(let i = 2; i <= n;i ++){
                for(let j = 1; j <= x; j++){
                    for(let k = 1; k < j && k <= m; k++)
                        table[i][j] += table[i-1][j-k];
                }
        }

        /* Uncomment these lines to see content of table
        for(int i = 0; i< n+1; i++){
            for(int j = 0; j< x+1; j++)
                System.out.print(dt[i][j] + " ");
            System.out.println();
        } */

        return table[n][x];
}

// Driver Code
document.write(findWays(4, 2, 1)+"<br>");
document.write(findWays(2, 2, 3)+"<br>");
document.write(findWays(6, 3, 8)+"<br>");
document.write(findWays(4, 2, 5)+"<br>");
document.write(findWays(4, 3, 5)+"<br>");


// This code is contributed by rag2127
</script>

Output : 

0
2
21
4
6

Time Complexity: O(m * n * x) where m is number of faces, n is number of dice and x is given sum.

Auxiliary Space: O(n * x)
We can add the following two conditions at the beginning of findWays() to improve performance of the program for extreme cases (x is too high or x is too low) 

C++

// When x is so high that sum can not go beyond x even when we
// get maximum value in every dice throw.
if (m*n <= x)
    return (m*n == x);

// When x is too low
if (n >= x)
    return (n == x);

Java

    // When x is so high that sum can not go beyond x even when we
    // get maximum value in every dice throw.
    if (m*n <= x)
        return (m*n == x);

    // When x is too low
    if (n >= x)
        return (n == x);

// This code is contributed by umadevi9616

Python3

# Python3 implementation
def condition(m, n, x):

    # When x is so high that sum can not go beyond x even when we
    # get maximum value in every dice throw.
    if (m*n <= x):
        return (m*n == x)

    # When x is too low
    if (n >= x):
        return (n == x)

# This code is contributed by phasing17

C#

// C# code to implement the approach

    // When x is so high that sum can not go beyond x even when we
    // get maximum value in every dice throw.
    if (m*n <= x)
        return (m*n == x);

    // When x is too low
    if (n >= x)
        return (n == x);

// This code is contributed by phasing17

Javascript

// JavaScript implementation

    // When x is so high that sum can not go beyond x even when we
    // get maximum value in every dice throw.
    if (m*n <= x)
        return (m*n == x);

    // When x is too low
    if (n >= x)
        return (n == x);


// This code is contributed by phasing17

With above conditions added, time complexity becomes O(1) when x >= m*n or when x <= n.

Following is the implementation of the Optimized Dynamic Programming approach. 

C++

//  C++ program
//  The main function that returns number of ways to get sum 'x'
//  with 'n' dice and 'm' with m faces.
#include<bits/stdc++.h>
using namespace std;

long findWays(int f, int d, int s)
{
    // Create a table to store results of subproblems. One extra
    // row and column are used for simplicity (Number of dice
    // is directly used as row index and sum is directly used
    // as column index). The entries in 0th row and 0th column
    // are never used.
    long mem[d + 1][s + 1];
    memset(mem,0,sizeof mem);
    // Table entries for no dices
    // If you do not have any data, then the value must be 0, so the result is 1
    mem[0][0] = 1;
    // Iterate over dices
    for (int i = 1; i <= d; i++)
    {
        // Iterate over sum
        for (int j = i; j <= s; j++)
        {
            // The result is obtained in two ways, pin the current dice and spending 1 of the value,
            // so we have mem[i-1][j-1] remaining combinations, to find the remaining combinations we
            // would have to pin the values ??above 1 then we use mem[i][j-1] to sum all combinations
            // that pin the remaining j-1's. But there is a way, when "j-f-1> = 0" we would be adding
            // extra combinations, so we remove the combinations that only pin the extrapolated dice face and
            // subtract the extrapolated combinations.
            mem[i][j] = mem[i][j - 1] + mem[i - 1][j - 1];
            if (j - f - 1 >= 0)
                mem[i][j] -= mem[i - 1][j - f - 1];
        }
    }
    return mem[d][s];
}

// Driver code
int main(void)
{
    cout << findWays(4, 2, 1) << endl;
    cout << findWays(2, 2, 3) << endl;
    cout << findWays(6, 3, 8) << endl;
    cout << findWays(4, 2, 5) << endl;
    cout << findWays(4, 3, 5) << endl;
    return 0;
}

// This code is contributed by ankush_953

Java

/**
 * The main function that returns number of ways to get sum 'x'
 * with 'n' dice and 'm' with m faces.
 *
 * @author Pedro H. Chaves <pedrohcd@hotmail.com> <https://github.com/pedrohcdo>
 */
public class GFG {

    /**
     * Count ways
     *
     * @param f
     * @param d
     * @param s
     * @return
     */
    public static long findWays(int f, int d, int s) {
        // Create a table to store results of subproblems.  One extra
        // row and column are used for simplicity (Number of dice
        // is directly used as row index and sum is directly used
        // as column index).  The entries in 0th row and 0th column
        // are never used.
        long mem[][] = new long[d + 1][s + 1];
        // Table entries for no dices
        // If you do not have any data, then the value must be 0, so the result is 1
        mem[0][0] = 1;
        // Iterate over dices
        for(int i=1; i<=d; i++) {
            // Iterate over sum
            for(int j=i; j<=s; j++) {
                // The result is obtained in two ways, pin the current dice and spending 1 of the value,
                // so we have mem[i-1][j-1] remaining combinations, to find the remaining combinations we
                // would have to pin the values ??above 1 then we use mem[i][j-1] to sum all combinations
                // that pin the remaining j-1's. But there is a way, when "j-f-1> = 0" we would be adding
                // extra combinations, so we remove the combinations that only pin the extrapolated dice face and
                // subtract the extrapolated combinations.
                mem[i][j] = mem[i][j-1] + mem[i-1][j-1];
                if(j-f-1 >= 0)
                    mem[i][j] -= mem[i-1][j-f-1];
            }
        }
        return mem[d][s];
    }


    /**
     * Main
     *
     * @param args
     */
    public static void main(String[] args) {
        System.out.println(findWays(4, 2, 1));
        System.out.println(findWays(2, 2, 3));
        System.out.println(findWays(6, 3, 8));
        System.out.println(findWays(4, 2, 5));
        System.out.println(findWays(4, 3, 5));
    }
}

Python3

#  Python program
#  The main function that returns number of ways to get sum 'x'
#  with 'n' dice and 'm' with m faces.


def findWays(f, d, s):
    # Create a table to store results of subproblems. One extra
    # row and column are used for simplicity (Number of dice
    # is directly used as row index and sum is directly used
    # as column index). The entries in 0th row and 0th column
    # are never used.
    mem = [[0 for i in range(s+1)] for j in range(d+1)]
    # Table entries for no dices
    # If you do not have any data, then the value must be 0, so the result is 1
    mem[0][0] = 1
    # Iterate over dices
    for i in range(1, d+1):

        # Iterate over sum
        for j in range(1, s+1):
            # The result is obtained in two ways, pin the current dice and spending 1 of the value,
            # so we have mem[i-1][j-1] remaining combinations, to find the remaining combinations we
            # would have to pin the values ??above 1 then we use mem[i][j-1] to sum all combinations
            # that pin the remaining j-1's. But there is a way, when "j-f-1> = 0" we would be adding
            # extra combinations, so we remove the combinations that only pin the extrapolated dice face and
            # subtract the extrapolated combinations.
            mem[i][j] = mem[i][j - 1] + mem[i - 1][j - 1]
            if j - f - 1 >= 0:
                mem[i][j] -= mem[i - 1][j - f - 1]
    return mem[d][s]

# Driver code

print(findWays(4, 2, 1))
print(findWays(2, 2, 3))
print(findWays(6, 3, 8))
print(findWays(4, 2, 5))
print(findWays(4, 3, 5))

# This code is contributed by ankush_953

C#

// C# program
// The main function that returns number of ways to get sum 'x'
// with 'n' dice and 'm' with m faces.
using System;

class GFG
{

    /**
    * Count ways
    *
    * @param f
    * @param d
    * @param s
    * @return
    */
    public static long findWays(int f, int d, int s)
    {
        // Create a table to store results of subproblems. One extra
        // row and column are used for simplicity (Number of dice
        // is directly used as row index and sum is directly used
        // as column index). The entries in 0th row and 0th column
        // are never used.
        long [,]mem = new long[d + 1,s + 1];

        // Table entries for no dices
        // If you do not have any data, then the value must be 0, so the result is 1
        mem[0,0] = 1;
        // Iterate over dices
        for(int i = 1; i <= d; i++)
        {
            // Iterate over sum
            for(int j = i; j <= s; j++)
            {
                // The result is obtained in two ways, pin the current dice and spending 1 of the value,
                // so we have mem[i-1,j-1] remaining combinations, to find the remaining combinations we
                // would have to pin the values ??above 1 then we use mem[i,j-1] to sum all combinations
                // that pin the remaining j-1's. But there is a way, when "j-f-1> = 0" we would be adding
                // extra combinations, so we remove the combinations that only pin the extrapolated dice face and
                // subtract the extrapolated combinations.
                mem[i,j] = mem[i,j-1] + mem[i-1,j-1];
                if(j-f-1 >= 0)
                    mem[i,j] -= mem[i-1,j-f-1];
            }
        }
        return mem[d,s];
    }

    // Driver code
    public static void Main(String[] args)
    {
        Console.WriteLine(findWays(4, 2, 1));
        Console.WriteLine(findWays(2, 2, 3));
        Console.WriteLine(findWays(6, 3, 8));
        Console.WriteLine(findWays(4, 2, 5));
        Console.WriteLine(findWays(4, 3, 5));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program
// The main function that returns number
// of ways to get sum 'x' with 'n' dice
// and 'm' with m faces.

/**
     * Count ways
     *
     * @param f
     * @param d
     * @param s
     * @return
     */
function findWays(f, d, s)
{

    // Create a table to store results of subproblems.
    // One extra row and column are used for simplicity
    // (Number of dice is directly used as row index and
    // sum is directly used as column index). The entries
    // in 0th row and 0th column are never used.
    let mem = new Array(d + 1);
    for(let i = 0; i < (d + 1); i++)
    {
        mem[i] = new Array(s + 1);
        for(let j = 0; j < s + 1; j++)
        {
            mem[i][j] = 0;
        }
    }

    // Table entries for no dices
    // If you do not have any data,
    // then the value must be 0,
    // so the result is 1
    mem[0][0] = 1;

    // Iterate over dices
    for(let i = 1; i <= d; i++)
    {

        // Iterate over sum
        for(let j = i; j <= s; j++)
        {

            // The result is obtained in two ways,
            // pin the current dice and spending 1
            // of the value, so we have mem[i-1][j-1]
            // remaining combinations, to find the
            // remaining combinations we would have
            // to pin the values ??above 1 then we
            // use mem[i][j-1] to sum all combinations
            // that pin the remaining j-1's. But there
            // is a way, when "j-f-1> = 0" we would be
            // adding extra combinations, so we remove
            // the combinations that only pin the
            // extrapolated dice face and subtract the
            // extrapolated combinations.
            mem[i][j] = mem[i][j - 1] + mem[i - 1][j - 1];

            if (j - f - 1 >= 0)
                mem[i][j] -= mem[i - 1][j - f - 1];
        }
    }
    return mem[d][s];
}

// Driver code
document.write(findWays(4, 2, 1) + "<br>");
document.write(findWays(2, 2, 3) + "<br>");
document.write(findWays(6, 3, 8) + "<br>");
document.write(findWays(4, 2, 5) + "<br>");
document.write(findWays(4, 3, 5) + "<br>");

// This code is contributed by avanitrachhadiya2155

</script>

Output : 

0
2
21
4
6

Time Complexity: O(n * x) where n is number of dice and x is given sum.

Space Complexity: O(n * x) where n is number of dice and x is given sum.

Exercise:
Extend the above algorithm to find the probability to get Sum > X.
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