Given n dices each with m faces, numbered from 1 to m, the task is to find the number of ways to get sum x. x is the summation of values on each face when all the dice are thrown.
Examples:
Input: m = 6, n = 3, x = 12
Output: 25
Explanation: There are 25 total ways to get the Sum 12 using 3 dices with faces from 1 to 6.
Input: m = 2, n = 3, x = 6
Output: 1
Explanation: There is only 1 way to get the Sum 6 using 3 dices with faces from 1 to 2. All the dices will have to land on 2.
Using Recursion - O(m^n) Time and O(m) Space
The idea is to explore all possible ways to achieve the target sum. For each die, the approach tries every possible face value (from 1 to m), and recursively checks the number of ways to achieve the remaining sum with the remaining dice. At each recursive step, the function reduces the number of dice by one and subtracts the current die's value from the target sum. Base cases handle scenarios where all dice have been used or the target sum cannot be achieved, returning 1 for a valid combination and 0 for invalid scenarios.
Mathematically the recurrence relation will look like the following:
- noOfWays(m, n, x) = sum (noOfWays(m, n-1, x-j)) for j in range[1, m].
Base Cases:
- noOfWays(m, n, x) = 1, if n = 0 and x = 0.
- noOfWays(m, n, x) = 0, if n = 0 or x < 0.
C++
// C++ program to implement
// dice throw problem using recursion
#include <bits/stdc++.h>
using namespace std;
int noOfWays(int m, int n, int x) {
// Base case: Valid combination
if (n == 0 && x == 0)
return 1;
// Base case: Invalid combination
if (n == 0 || x <= 0)
return 0;
int ans = 0;
// Check for all values of m.
for (int i = 1; i <= m; i++) {
ans += noOfWays(m, n - 1, x - i);
}
return ans;
}
int main() {
int m = 6, n = 3, x = 8;
cout << noOfWays(m, n, x);
}
// Java program to implement
// dice throw problem using recursion
import java.util.*;
class GfG {
static int noOfWays(int m, int n, int x) {
// Base case: Valid combination
if (n == 0 && x == 0) return 1;
// Base case: Invalid combination
if (n == 0 || x <= 0) return 0;
int ans = 0;
// Check for all values of m.
for (int i = 1; i <= m; i++) {
ans += noOfWays(m, n - 1, x - i);
}
return ans;
}
public static void main(String[] args) {
int m = 6, n = 3, x = 8;
System.out.println(noOfWays(m, n, x));
}
}
# Python program to implement
# dice throw problem using recursion
def noOfWays(m, n, x):
# Base case: Valid combination
if n == 0 and x == 0:
return 1
# Base case: Invalid combination
if n == 0 or x <= 0:
return 0
ans = 0
# Check for all values of m.
for i in range(1, m + 1):
ans += noOfWays(m, n - 1, x - i)
return ans
if __name__ == "__main__":
m = 6
n = 3
x = 8
print(noOfWays(m, n, x))
// C# program to implement
// dice throw problem using recursion
using System;
class GfG {
static int noOfWays(int m, int n, int x) {
// Base case: Valid combination
if (n == 0 && x == 0) return 1;
// Base case: Invalid combination
if (n == 0 || x <= 0) return 0;
int ans = 0;
// Check for all values of m.
for (int i = 1; i <= m; i++) {
ans += noOfWays(m, n - 1, x - i);
}
return ans;
}
static void Main(string[] args) {
int m = 6, n = 3, x = 8;
Console.WriteLine(noOfWays(m, n, x));
}
}
// JavaScript program to implement
// dice throw problem using recursion
function noOfWays(m, n, x) {
// Base case: Valid combination
if (n === 0 && x === 0) return 1;
// Base case: Invalid combination
if (n === 0 || x <= 0) return 0;
let ans = 0;
// Check for all values of m.
for (let i = 1; i <= m; i++) {
ans += noOfWays(m, n - 1, x - i);
}
return ans;
}
let m = 6, n = 3, x = 8;
console.log(noOfWays(m, n, x));
Using Top-Down DP (Memoization) - O(n*x*m) Time and O(n*x) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: Number of ways to make sum at dice n, i.e., noOfWays(m, n, x), depends on the solutions of the subproblems noOfWays(m, n-1, x-j) for j in range [1, m]. By combining these optimal substructures, we can efficiently calculate the number of ways to make target sum at dice n.
2. Overlapping Subproblems: While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times.
- There are only are two parameters: n and x that changes in the recursive solution. So we create a 2D matrix of size (n+1)*(x+1) for memoization.
- We initialize this matrix as -1 to indicate nothing is computed initially.
- Now we modify our recursive solution to first check if the value is -1, then only make recursive calls. This way, we avoid re-computations of the same subproblems.
C++
// C++ program to implement
// dice throw problem using memoization
#include <bits/stdc++.h>
using namespace std;
int countRecur(int m, int n, int x, vector<vector<int>> &memo) {
// Base case: Valid combination
if (n==0 && x==0) return 1;
// Base case: Invalid combination
if (n==0 || x<=0) return 0;
// If value is memoized
if (memo[n][x] != -1) return memo[n][x];
int ans = 0;
// Check for all values of m.
for (int i=1; i<=m; i++) {
ans += countRecur(m, n-1, x-i, memo);
}
return memo[n][x] = ans;
}
int noOfWays(int m, int n, int x) {
vector<vector<int>> memo(n+1, vector<int>(x+1, -1));
return countRecur(m, n, x, memo);
}
int main() {
int m = 6, n = 3, x = 8;
cout << noOfWays(m, n, x);
}
// Java program to implement
// dice throw problem using memoization
import java.util.Arrays;
class GfG {
static int countRecur(int m, int n, int x, int[][] memo) {
// Base case: Valid combination
if (n == 0 && x == 0) return 1;
// Base case: Invalid combination
if (n == 0 || x <= 0) return 0;
// If value is memoized
if (memo[n][x] != -1) return memo[n][x];
int ans = 0;
// Check for all values of m.
for (int i = 1; i <= m; i++) {
ans += countRecur(m, n - 1, x - i, memo);
}
return memo[n][x] = ans;
}
static int noOfWays(int m, int n, int x) {
int[][] memo = new int[n + 1][x + 1];
for (int[] row : memo) Arrays.fill(row, -1);
return countRecur(m, n, x, memo);
}
public static void main(String[] args) {
int m = 6, n = 3, x = 8;
System.out.println(noOfWays(m, n, x));
}
}
# Python program to implement
# dice throw problem using memoization
def countRecur(m, n, x, memo):
# Base case: Valid combination
if n == 0 and x == 0:
return 1
# Base case: Invalid combination
if n == 0 or x <= 0:
return 0
# If value is memoized
if memo[n][x] != -1:
return memo[n][x]
ans = 0
# Check for all values of m.
for i in range(1, m + 1):
ans += countRecur(m, n - 1, x - i, memo)
memo[n][x] = ans
return ans
def noOfWays(m, n, x):
memo = [[-1 for _ in range(x + 1)] for _ in range(n + 1)]
return countRecur(m, n, x, memo)
if __name__ == "__main__":
m = 6
n = 3
x = 8
print(noOfWays(m, n, x))
// C# program to implement
// dice throw problem using memoization
using System;
class GfG {
static int countRecur(int m, int n, int x, int[,] memo) {
// Base case: Valid combination
if (n == 0 && x == 0) return 1;
// Base case: Invalid combination
if (n == 0 || x <= 0) return 0;
// If value is memoized
if (memo[n, x] != -1) return memo[n, x];
int ans = 0;
// Check for all values of m.
for (int i = 1; i <= m; i++) {
ans += countRecur(m, n - 1, x - i, memo);
}
return memo[n, x] = ans;
}
static int noOfWays(int m, int n, int x) {
int[,] memo = new int[n + 1, x + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= x; j++) {
memo[i, j] = -1;
}
}
return countRecur(m, n, x, memo);
}
static void Main(string[] args) {
int m = 6, n = 3, x = 8;
Console.WriteLine(noOfWays(m, n, x));
}
}
// JavaScript program to implement
// dice throw problem using memoization
function countRecur(m, n, x, memo) {
// Base case: Valid combination
if (n === 0 && x === 0) return 1;
// Base case: Invalid combination
if (n === 0 || x <= 0) return 0;
// If value is memoized
if (memo[n][x] !== -1) return memo[n][x];
let ans = 0;
// Check for all values of m.
for (let i = 1; i <= m; i++) {
ans += countRecur(m, n - 1, x - i, memo);
}
memo[n][x] = ans;
return ans;
}
function noOfWays(m, n, x) {
const memo = Array.from({ length: n + 1 }, () => Array(x + 1).fill(-1));
return countRecur(m, n, x, memo);
}
const m = 6, n = 3, x = 8;
console.log(noOfWays(m, n, x));
Using Bottom-Up DP (Tabulation) - O(n*x*m) Time and O(n*x) Space
The idea is to fill the DP table based on previous values. For each dice i and sum j, we try all the values k from 1 to m. The table is filled in an iterative manner from i = 1 to i = n and for each sum j from 1 to x.
The dynamic programming relation is as follows:
- dp[i][j] = sum(dp[i-1][j-k]) where k is in range [1, m] and j-k >= 0.
C++
// C++ program to implement
// dice throw problem using tabulation
#include <bits/stdc++.h>
using namespace std;
int noOfWays(int m, int n, int x) {
// Create a 2D dp array with (n+1) rows and (x+1) columns
// dp[i][j] will store the number of ways to get a sum
// of 'j' using 'i' dice
vector<vector<int>> dp(n + 1, vector<int>(x + 1, 0));
// Base case: There is 1 way to get
// a sum of 0 with 0 dice
dp[0][0] = 1;
// Loop through each dice (i) from 1 to n
for (int i = 1; i <= n; i++) {
// Loop through each sum (j) from 1 to x
for (int j = 1; j <= x; j++) {
// Loop through all possible dice values (k) from 1 to m
// and if the sum j - k is valid (non-negative),
// add the number of ways from dp[i-1][j-k]
for (int k = 1; k <= m && j - k >= 0; k++) {
dp[i][j] += dp[i - 1][j - k];
}
}
}
// The result will be in dp[n][x], which contains
// the number of ways to get sum 'x' using 'n' dice
return dp[n][x];
}
int main() {
int m = 6, n = 3, x = 8;
cout << noOfWays(m, n, x);
}
// Java program to implement
// dice throw problem using tabulation
class GfG {
static int noOfWays(int m, int n, int x) {
// Create a 2D dp array with (n+1) rows and (x+1)
// columns dp[i][j] will store the number of ways to
// get a sum of 'j' using 'i' dice
int[][] dp = new int[n + 1][x + 1];
// Base case: There is 1 way to get a sum of 0 with
// 0 dice
dp[0][0] = 1;
// Loop through each dice (i) from 1 to n
for (int i = 1; i <= n; i++) {
// Loop through each sum (j) from 1 to x
for (int j = 1; j <= x; j++) {
// Loop through all possible dice values (k)
// from 1 to m and if the sum j - k is valid
// (non-negative), add the number of ways
// from dp[i-1][j-k]
for (int k = 1; k <= m && j - k >= 0; k++) {
dp[i][j] += dp[i - 1][j - k];
}
}
}
// The result will be in dp[n][x], which contains
// the number of ways to get sum 'x' using 'n' dice
return dp[n][x];
}
public static void main(String[] args) {
int m = 6, n = 3, x = 8;
System.out.println(noOfWays(m, n, x));
}
}
# Python program to implement
# dice throw problem using tabulation
def noOfWays(m, n, x):
# Create a 2D dp array with (n+1) rows and (x+1) columns
# dp[i][j] will store the number of ways to get a
# sum of 'j' using 'i' dice
dp = [[0 for _ in range(x + 1)] for _ in range(n + 1)]
# Base case: There is 1 way to get a sum
# of 0 with 0 dice
dp[0][0] = 1
# Loop through each dice (i) from 1 to n
for i in range(1, n + 1):
# Loop through each sum (j) from 1 to x
for j in range(1, x + 1):
# Loop through all possible dice values (k) from 1 to m
# and if the sum j - k is valid (non-negative),
# add the number of ways from dp[i-1][j-k]
for k in range(1, m + 1):
if j - k >= 0:
dp[i][j] += dp[i - 1][j - k]
# The result will be in dp[n][x], which contains
# the number of ways to get sum 'x' using 'n' dice
return dp[n][x]
if __name__ == "__main__":
m = 6
n = 3
x = 8
print(noOfWays(m, n, x))
// C# program to implement
// dice throw problem using tabulation
using System;
class GfG {
static int noOfWays(int m, int n, int x) {
// Create a 2D dp array with (n+1) rows and (x+1)
// columns dp[i, j] will store the number of ways to
// get a sum of 'j' using 'i' dice
int[, ] dp = new int[n + 1, x + 1];
// Base case: There is 1 way to get a sum of 0 with
// 0 dice
dp[0, 0] = 1;
// Loop through each dice (i) from 1 to n
for (int i = 1; i <= n; i++) {
// Loop through each sum (j) from 1 to x
for (int j = 1; j <= x; j++) {
// Loop through all possible dice values (k)
// from 1 to m and if the sum j - k is valid
// (non-negative), add the number of ways
// from dp[i-1, j-k]
for (int k = 1; k <= m && j - k >= 0; k++) {
dp[i, j] += dp[i - 1, j - k];
}
}
}
// The result will be in dp[n, x], which contains
// the number of ways to get sum 'x' using 'n' dice
return dp[n, x];
}
static void Main(string[] args) {
int m = 6, n = 3,x = 8;
Console.WriteLine(noOfWays(m, n, x));
}
}
// JavaScript program to implement
// dice throw problem using tabulation
function noOfWays(m, n, x) {
const dp = Array.from({length : n + 1},
() => Array(x + 1).fill(0));
// Base case: There's 1 way to achieve a sum of 0 using
// 0 dice
dp[0][0] = 1;
// Loop through each dice (i) from 1 to n (number of
// dice)
for (let i = 1; i <= n; i++) {
// Loop through each sum (j) from 1 to x (target
// sum)
for (let j = 1; j <= x; j++) {
// Loop through all possible outcomes of the
// dice (k) from 1 to m (faces of the dice) If
// the sum j - k is valid (non-negative),
// increment the number of ways to achieve sum j
for (let k = 1; k <= m && j - k >= 0; k++) {
dp[i][j] += dp[i - 1][j - k];
}
}
}
// Return the number of ways to achieve the sum x using
// n dice
return dp[n][x];
}
const m = 6, n = 3, x = 8;
console.log(noOfWays(m, n, x));
Using Space Optimized DP - O(n*x*m) Time and O(x) Space
In previous approach of dynamic programming we have derive the relation between states as given below:
- dp[i][j] = sum(dp[i-1][j-k]) where k is in range [1, m] and j-k >= 0.
If we observe that for calculating current dp[i][j] state we only need previous row. There is no need to store all the previous states just one previous state is used to compute result.
C++
// C++ program to implement
// dice throw problem using space optimised
#include <bits/stdc++.h>
using namespace std;
int noOfWays(int m, int n, int x) {
// Initialize a 1D dp array with size (x + 1), all values set to 0.
// dp[j] will store the number of ways to get a sum of 'j' using 'i' dice.
vector<int> dp(x + 1, 0);
// Base case: There is 1 way to get
// sum 0 (using no dice)
dp[0] = 1;
// Iterate through each dice (i) from 1 to n (number of dice)
for (int i = 1; i <= n; i++) {
// Iterate backwards through all possible sums (j) from x to 1
// to ensure that the results from previous dice
// counts are not overwritten
for (int j = x; j >= 1; j--) {
// Reset dp[j] before calculating its
// new value for the current dice
dp[j] = 0;
// Loop through all possible outcomes of the dice (k)
// from 1 to m (faces of the dice)
// If j - k is a valid sum (non-negative), update dp[j]
for (int k = 1; k <= m && j - k >= 0; k++) {
dp[j] += dp[j - k];
}
}
// After each dice iteration, set dp[0] to 0
// since there are no ways to achieve sum 0
dp[0] = 0;
}
// Return the number of ways to get
// sum x using n dice
return dp[x];
}
int main() {
int m = 6, n = 3, x = 8;
cout << noOfWays(m, n, x);
}
// Java program to implement
// dice throw problem using space optimised
class GfG {
// Function to find the number of ways to get a sum 'x'
// with 'n' dice
static int noOfWays(int m, int n, int x) {
// Initialize a 1D dp array of size (x + 1), all
// values initially 0 dp[j] will store the number of
// ways to get a sum of 'j' using 'i' dice
int[] dp = new int[x + 1];
// Base case: There is 1 way to get sum 0 (using no
// dice)
dp[0] = 1;
// Iterate through each dice (i) from 1 to n (total
// number of dice)
for (int i = 1; i <= n; i++) {
// Iterate backwards through all possible sums
// (j) from x to 1 to avoid overwriting the
// results from the previous dice count
for (int j = x; j >= 1; j--) {
// Reset dp[j] before calculating its new
// value for the current dice
dp[j] = 0;
// Loop through all possible outcomes of the
// dice (k) from 1 to m (faces of the dice)
// If j - k is a valid sum (non-negative),
// update dp[j]
for (int k = 1; k <= m && j - k >= 0; k++) {
dp[j] += dp[j - k];
}
}
// After each dice iteration, set dp[0] to 0
// since there are no ways to achieve sum 0
dp[0] = 0;
}
// Return the number of ways to get sum 'x' using
// 'n' dice
return dp[x];
}
public static void main(String[] args) {
int m = 6, n = 3, x = 8;
System.out.println(noOfWays(m, n, x));
}
}
# Python program to implement
# dice throw problem using space optimised
def noOfWays(m, n, x):
# Initialize a 1D dp array where dp[j] will store
# the number of ways to get a sum of 'j' using 'i' dice
dp = [0] * (x + 1)
# Base case: There is 1 way to get a sum
# of 0 (using no dice)
dp[0] = 1
# Iterate through each dice (i) from 1 to n (total number of dice)
for i in range(1, n + 1):
# Iterate backwards through all possible sums (j) from x to 1
# to avoid overwriting the results from the previous dice count
for j in range(x, 0, -1):
# Reset dp[j] before calculating its new value
# for the current dice
dp[j] = 0
# Loop through all possible outcomes of the dice
# (k) from 1 to m (faces of the dice)
# If j - k is a valid sum (non-negative), update dp[j]
for k in range(1, m + 1):
if j - k >= 0:
dp[j] += dp[j - k]
# After each dice iteration, set dp[0] to 0
# since there are no ways to achieve sum 0
dp[0] = 0
# Return the number of ways to get
# sum 'x' using 'n' dice
return dp[x]
if __name__ == "__main__":
m = 6
n = 3
x = 8
print(noOfWays(m, n, x))
// C# program to implement
// dice throw problem using space optimized
using System;
class GfG {
// Function to calculate the number of ways to get a sum
// 'x' using 'n' dice
static int noOfWays(int m, int n, int x) {
// Create a 1D dp array to store the number of ways
// to achieve each sum from 0 to x
int[] dp = new int[x + 1];
// Base case: there is exactly 1 way to get a sum of
// 0 (using no dice)
dp[0] = 1;
// Iterate over the number of dice
for (int i = 1; i <= n; i++) {
// Iterate backwards through all possible sums
// (from x down to 1) to prevent overwriting the
// dp values
for (int j = x; j >= 1; j--) {
// Reset dp[j] before calculating its new
// value for the current dice
dp[j] = 0;
// Loop through all possible dice outcomes
// (1 to m faces of the dice) and add the
// number of ways to achieve sum (j - k) to
// dp[j]
for (int k = 1; k <= m && j - k >= 0; k++) {
dp[j] += dp[j - k];
}
}
// After processing a dice, we set dp[0] to 0
// because there are no ways to achieve sum 0
// after using any dice
dp[0] = 0;
}
// Return the number of ways to achieve the sum 'x'
// using 'n' dice
return dp[x];
}
static void Main(string[] args) {
int m = 6, n = 3, x = 8;
Console.WriteLine(noOfWays(m, n, x));
}
}
// JavaScript program to implement
// dice throw problem using space optimized
function noOfWays(m, n, x) {
// Initialize dp array to store the number of ways to
// achieve each sum from 0 to x All values initially set
// to 0
const dp = Array(x + 1).fill(0);
// Base case: there is exactly 1 way to achieve sum 0
// (using no dice)
dp[0] = 1;
// Iterate over the number of dice
for (let i = 1; i <= n; i++) {
// Iterate backwards through all possible sums from
// x to 1 to avoid overwriting dp values
for (let j = x; j >= 1; j--) {
// Reset dp[j] to 0 before calculating its new
// value
dp[j] = 0;
// Loop through all possible dice outcomes (1 to
// m faces of the dice) Add the number of ways
// to achieve sum (j - k) to dp[j]
for (let k = 1; k <= m && j - k >= 0; k++) {
dp[j] += dp[j - k];
}
}
// After processing each dice, set dp[0] to 0 since
// it's not valid to have sum 0 with any dice
dp[0] = 0;
}
// Return the number of ways to achieve the sum x using
// n dice
return dp[x];
}
const m = 6, n = 3, x = 8;
console.log(noOfWays(m, n, x));
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Your task is to count the number of ways to construct sum N by throwing a dice one or more times. Each throw produces an outcome between 1 and 6. Examples: Input: N = 3Output: 4Explanation: There are 4 ways to make sum = 3. Using 1 die {3}, sum = 3.Using 2 dice {1, 2}, sum = 1 + 2 = 3.Using 2 dice {
8 min read
Probability of getting more value in third dice throw
Given that the three players playing a game of rolling dice. Player1 rolled a die got A and player2 rolled a die got B. The task is to find the probability of player3 to win the match and Player3 wins if he gets more than both of them.Examples: Input: A = 2, B = 3Output: 1/2Player3 wins if he gets 4
5 min read
JavaScript Math random() Method
The JavaScript Math.random() function gives you a random number between 0 and just under 1. You can use this number as a base to get random numbers within any range you want. Now, let's see the syntax to use the JavaScript random method along with multiple examples to easily understand how we can ge
3 min read
Probability of getting a sum on throwing 2 Dices N times
Given the sum. The task is to find out the probability of occurring that sum on the thrown of the two dice N times. Probability is defined as the favorable numbers of outcomes upon total numbers of the outcome. Probability always lies between 0 and 1.Examples: Input: sum = 11, times = 1 Output: 2 /
7 min read