Dados n dados, cada uno con m caras, numerados del 1 al m, encuentre el número de formas de obtener la suma X. X es la suma de los valores en cada cara cuando se lanzan todos los dados.
El enfoque Naive es encontrar todas las combinaciones posibles de valores de n dados y seguir contando los resultados que suman X.
Este problema se puede resolver de manera eficiente utilizando Programación Dinámica (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)
¿Por qué enfoque DP?
El problema anterior exhibe subproblemas superpuestos. Vea el siguiente diagrama. Además, vea esta implementación recursiva. Sean 3 dados, cada uno con 6 caras y necesitamos encontrar el número de formas de obtener la suma 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)
Por favor, eche un vistazo más de cerca a la recursividad anterior. Los subproblemas en se resuelven la primera vez y los subproblemas en se resuelven nuevamente (exhiba subproblemas superpuestos). Por lo tanto, almacenar los resultados de los subproblemas resueltos ahorra tiempo.
A continuación se muestra la implementación del enfoque de programación dinámica.
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>
Producción :
0 2 21 4 6
Complejidad temporal: O(m * n * x) donde m es el número de caras, n es el número de dados yx es la suma.
Espacio auxiliar: O(n * x)
Podemos agregar las siguientes dos condiciones al comienzo de findWays() para mejorar el rendimiento del programa en casos extremos (x es demasiado alto o x es demasiado bajo)
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
Con las condiciones anteriores añadidas, la complejidad del tiempo se convierte en O(1) cuando x >= m*n o cuando x <= n.
A continuación se muestra la implementación del enfoque de programación dinámica optimizada.
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>
Producción :
0 2 21 4 6
Complejidad de tiempo: O(n * x) donde n es el número de dados yx es la suma.
Complejidad espacial: O(n * x) donde n es el número de dados yx es la suma.
Ejercicio:
Extienda el algoritmo anterior para encontrar la probabilidad de obtener Suma > X.
Este artículo fue compilado por Aashish Barnwal . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA