Dice Throw | DP-30 - Tutorialspoint.dev

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 simpilicity (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 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 results of subproblems.  
    One extra row and column are used for simpilicity  
    (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 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 simpilicity (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 simpilicity (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  
    // simpilicity (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). "
"; 
echo findWays(2, 2, 3). "
"; 
echo findWays(6, 3, 8). "
"; 
echo findWays(4, 2, 5). "
"; 
echo findWays(4, 3, 5). "
"; 
  
// This code is contributed by mits. 
?> 

Output :

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

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

// 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); 

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

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

This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Dice Throw | DP-30

C++

Java

Python3

C#

PHP

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