Imprimiendo todos los subconjuntos de {1,2,3,…n} sin usar array o bucle

Posted on by Rudeus Greyrat

Dado un número natural n , imprime todos los subconjuntos del conjunto  \{1, 2, 3, ..., n\} sin usar ningún arreglo o ciclo (solo se permite el uso de recursividad ).
Ejemplos: 
 

Input : n = 4
Output : { 1 2 3 4 }
         { 1 2 3 }
         { 1 2 4 }
         { 1 2 }
         { 1 3 4 }
         { 1 3 }
         { 1 4 }
         { 1 }
         { 2 3 4 }
         { 2 3 }
         { 2 4 }
         { 2 }
         { 3 4 }
         { 3 }
         { 4 }
         { }

Input : n = 2
Output : { 1 2 }
         { 1 }
         { 2 }
         { }

Acercarse:
 

  • Empezar desde  número = 2^n - 1 hasta 0.
  • Considere la representación binaria de num con n bits.
  • Comience desde el bit más a la izquierda que representa 1, el segundo bit representa 2, y así sucesivamente hasta el bit n-ésimo que representa n .
  • Imprime el número correspondiente al bit si está activado.
  • Realice los pasos anteriores para todos los valores de num hasta que sea igual a 0.

Comprendamos el enfoque anterior a través de un ejemplo:
considerando la entrada n = 4, comience desde  número = 2^n - 1 = 15 .
 

y así sucesivamente… hasta num = 0.
A continuación se muestra la implementación del enfoque anterior: 
 

C++

// C++ code to print all subsets
// of {1, 2, 3, n} without using
// array or loop, just recursion.
#include <bits/stdc++.h>
using namespace std;
 
void subset(int, int, int);
 
// This recursive function calls subset
// function to print the subsets one by one.
// numBits --> number of bits needed to
// represent the number (simply input value n).
// num --> Initially equal to 2 ^ n - 1 and
// decreases by 1 every recursion until 0.
void printSubsets(int numOfBits, int num)
{
    if (num >= 0)
    {
        cout << "{ ";
 
        // Print the subset corresponding to
        // binary representation of num.
        subset(numOfBits - 1, num, numOfBits);
        cout << "}" << endl;
 
        // Call the function recursively to
        // print the next subset.
        printSubsets(numOfBits, num - 1);
    }
    else
        return;
}
 
// This function recursively prints the
// subset corresponding to the binary
// representation of num.
// nthBit --> nth bit from right side
// starting from n and decreases until 0
void subset(int nthBit, int num, int numOfBits)
{
    if (nthBit >= 0)
    {
        // Print number in given subset only
        // if the bit corresponding to it
        // is set in num.
        if (num & (1 << nthBit))
        {
            cout << numOfBits - nthBit << " ";
        }
 
        // Check for the next bit
        subset(nthBit - 1, num, numOfBits);
    }
    else
        return;
}
 
// Driver Code
int main()
{
    int n = 4;
    printSubsets(n, pow(2, n) - 1);
}
 
// This code is contributed by
// sanjeev2552

Java

// Java code to print all subsets
// of {1, 2, 3, n} without using
// array or loop, just recursion.
class GfG
{
 
    // This recursive function calls subset
    // function to print the subsets one by one.
    // numBits --> number of bits needed to
    // represent the number (simply input value n).
    // num --> Initially equal to 2 ^ n - 1 and
    // decreases by 1 every recursion until 0.
    static void printSubSets(int numOfBits, int num)
    {
        if (num >= 0)
        {
            System.out.print("{ ");
             
            // Print the subset corresponding to
            // binary representation of num.
            subset(numOfBits - 1, num, numOfBits);
            System.out.println("}");
             
            // Call the function recursively to
            // print the next subset.
            printSubSets(numOfBits, num - 1);
 
        } else
            return;
    }
 
    // This function recursively prints the
    // subset corresponding to the binary
    // representation of num.
    // nthBit --> nth bit from right side
    // starting from n and decreases until 0.
    static void subset(int nthBit, int num, int numOfBits)
    {
        if (nthBit >= 0)
        {
            // Print number in given subset only
            // if the bit corresponding to it
            // is set in num.
            if ((num & (1 << nthBit)) != 0)
            {
                System.out.print(numOfBits - nthBit + " ");
 
            }
             
            // Check for the next bit
            subset(nthBit - 1, num, numOfBits);
        } else
            return;
    }
     
    // Driver code
    public static void main(String[] args)
    {
        int n = 4;
        printSubSets(n, (int) (Math.pow(2, n)) -1);
    }
}
 
// This code is contributed by laststringx

Python3

# Python3 code to print all subsets
# of {1, 2, 3, …n} without using
# array or loop, just recursion.
 
# This recursive function calls subset
# function to print the subsets one by one.
# numBits --> number of bits needed to
# represent the number (simply input value n).
# num --> Initially equal to 2 ^ n - 1 and
# decreases by 1 every recursion until 0.
def printSubsets(numOfBits, num):
     
    if num >= 0:
        print("{", end = " ")
 
        # Print the subset corresponding to
        # binary representation of num.
        subset(numOfBits-1, num, numOfBits)
        print("}")
 
        # Call the function recursively to
        # print the next subset.
        printSubsets(numOfBits, num-1)
         
    else:
        return
 
# This function recursively prints the
# subset corresponding to the binary
# representation of num.
# nthBit --> nth bit from right side
# starting from n and decreases until 0.
def subset(nthBit, num, numOfBits):
     
    if nthBit >= 0:
         
        # Print number in given subset only
        # if the bit corresponding to it
        # is set in num.
        if num & (1 << nthBit) != 0:
            print(numOfBits - nthBit, end = " ")
         
        # Check for the next bit
        subset(nthBit-1, num, numOfBits)
         
    else:
        return
 
# Driver Code   
n = 4
printSubsets(n, 2**n - 1)

C#

// C# code to print all subsets
// of {1, 2, 3, n} without using
// array or loop, just recursion.
using System;
 
class GfG
{
 
    // This recursive function calls subset
    // function to print the subsets one by one.
    // numBits --> number of bits needed to
    // represent the number (simply input value n).
    // num --> Initially equal to 2 ^ n - 1 and
    // decreases by 1 every recursion until 0.
    static void printSubSets(int numOfBits, int num)
    {
        if (num >= 0)
        {
            Console.Write("{ ");
             
            // Print the subset corresponding to
            // binary representation of num.
            subset(numOfBits - 1, num, numOfBits);
            Console.WriteLine("}");
             
            // Call the function recursively to
            // print the next subset.
            printSubSets(numOfBits, num - 1);
 
        } else
            return;
    }
 
    // This function recursively prints the
    // subset corresponding to the binary
    // representation of num.
    // nthBit --> nth bit from right side
    // starting from n and decreases until 0.
    static void subset(int nthBit, int num, int numOfBits)
    {
        if (nthBit >= 0)
        {
            // Print number in given subset only
            // if the bit corresponding to it
            // is set in num.
            if ((num & (1 << nthBit)) != 0)
            {
                Console.Write(numOfBits - nthBit + " ");
 
            }
             
            // Check for the next bit
            subset(nthBit - 1, num, numOfBits);
        } else
            return;
    }
     
    // Driver codeM
    public static void Main(String[] args)
    {
        int n = 4;
        printSubSets(n, (int) (Math.Pow(2, n)) -1);
    }
}
 
// This code is contributed by Srathore

Javascript

<script>
 
// Javascript code to print all subsets
// of {1, 2, 3, n} without using
// array or loop, just recursion.
 
// This recursive function calls subset
// function to print the subsets one by one.
// numBits --> number of bits needed to
// represent the number (simply input value n).
// num --> Initially equal to 2 ^ n - 1 and
// decreases by 1 every recursion until 0.
function printSubsets(numOfBits, num)
{
    if (num >= 0)
    {
        document.write( "{ ");
 
        // Print the subset corresponding to
        // binary representation of num.
        subset(numOfBits - 1, num, numOfBits);
        document.write( "}<br>" );
 
        // Call the function recursively to
        // print the next subset.
        printSubsets(numOfBits, num - 1);
    }
    else
        return;
}
 
// This function recursively prints the
// subset corresponding to the binary
// representation of num.
// nthBit --> nth bit from right side
// starting from n and decreases until 0
function subset(nthBit, num, numOfBits)
{
    if (nthBit >= 0)
    {
        // Print number in given subset only
        // if the bit corresponding to it
        // is set in num.
        if (num & (1 << nthBit))
        {
            document.write( numOfBits - nthBit + " ");
        }
 
        // Check for the next bit
        subset(nthBit - 1, num, numOfBits);
    }
    else
        return;
}
 
// Driver Code
var n = 4;
printSubsets(n, Math.pow(2, n) - 1);
 
</script>
Producción: 

{ 1 2 3 4 }
{ 1 2 3 }
{ 1 2 4 }
{ 1 2 }
{ 1 3 4 }
{ 1 3 }
{ 1 4 }
{ 1 }
{ 2 3 4 }
{ 2 3 }
{ 2 4 }
{ 2 }
{ 3 4 }
{ 3 }
{ 4 }
{ }

 

Complejidad del tiempo: O(n*2^n)
 

Publicación traducida automáticamente

Artículo escrito por rituraj_jain y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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