XNOR da el reverso de XOR si es un bit binario.
First bit | Second bit | XNOR 0 0 1 0 1 0 1 0 0 1 1 1 It gives 1 if bits are same else if bits are different it gives 0.
Es el reverso de XOR pero no podemos implementarlo directamente así que este es el programa para implementar XNOR
Ejemplos:
Input : 10 20 Output : 1 Binary of 20 is 10100 Binary of 10 is 1010 So the XNOR is 00001 So output is 1 Input : 10 10 Output : 15 Binary of 10 is 1010 Binary of 10 is 1010 So the XNOR is 1111 So output is 15
Primer método: – (O (logn)) En esta solución verificamos un bit a la vez. Si dos bits son iguales, ponemos 1 en el resultado, de lo contrario ponemos 0.
Entendámoslo con el siguiente código
C++
// CPP program to find
// XNOR of two numbers
#include <iostream>
using namespace std;
// log(n) solution
int xnor(int a, int b)
{
// Make sure a is larger
if (a < b)
swap(a, b);
if (a == 0 && b == 0)
return 1;
int a_rem = 0; // for last bit of a
int b_rem = 0; // for last bit of b
// counter for count bit
// and set bit in xnornum
int count = 0;
// to make new xnor number
int xnornum = 0;
// for set bits in new xnor number
while (a)
{
// get last bit of a
a_rem = a & 1;
// get last bit of b
b_rem = b & 1;
// Check if current two
// bits are same
if (a_rem == b_rem)
xnornum |= (1 << count);
// counter for count bit
count++;
a = a >> 1;
b = b >> 1;
}
return xnornum;
}
// Driver code
int main()
{
int a = 10, b = 50;
cout << xnor(a, b);
return 0;
}
Java
// Java program to find
// XNOR of two numbers
import java.util.*;
import java.lang.*;
public class GfG {
public static int xnor(int a, int b)
{
// Make sure a is larger
if (a < b) {
// swapping a and b;
int t = a;
a = b;
b = t;
}
if (a == 0 && b == 0)
return 1;
// for last bit of a
int a_rem = 0;
// for last bit of b
int b_rem = 0;
// counter for count bit
// and set bit in xnornum
int count = 0;
// to make new xnor number
int xnornum = 0;
// for set bits in new xnor number
while (true) {
// get last bit of a
a_rem = a & 1;
// get last bit of b
b_rem = b & 1;
// Check if current two bits are same
if (a_rem == b_rem)
xnornum |= (1 << count);
// counter for count bit
count++;
a = a >> 1;
b = b >> 1;
if (a < 1)
break;
}
return xnornum;
}
// Driver function
public static void main(String argc[])
{
int a = 10, b = 50;
System.out.println(xnor(a, b));
}
}
Python3
# Python3 program to find XNOR # of two numbers import math def swap(a,b): temp=a a=b b=temp # log(n) solution def xnor(a, b): # Make sure a is larger if (a < b): swap(a, b) if (a == 0 and b == 0) : return 1; # for last bit of a a_rem = 0 # for last bit of b b_rem = 0 # counter for count bit and # set bit in xnor num count = 0 # for make new xnor number xnornum = 0 # for set bits in new xnor # number while (a!=0) : # get last bit of a a_rem = a & 1 # get last bit of b b_rem = b & 1 # Check if current two # bits are same if (a_rem == b_rem): xnornum |= (1 << count) # counter for count bit count=count+1 a = a >> 1 b = b >> 1 return xnornum; # Driver method a = 10 b = 50 print(xnor(a, b)) # This code is contributed by Gitanjali
C#
// C# program to find
// XNOR of two numbers
using System;
public class GfG {
public static int xnor(int a, int b)
{
// Make sure a is larger
if (a < b) {
// swapping a and b;
int t = a;
a = b;
b = t;
}
if (a == 0 && b == 0)
return 1;
// for last bit of a
int a_rem = 0;
// for last bit of b
int b_rem = 0;
// counter for count bit
// and set bit in xnornum
int count = 0;
// to make new xnor number
int xnornum = 0;
// for set bits in new xnor number
while (true) {
// get last bit of a
a_rem = a & 1;
// get last bit of b
b_rem = b & 1;
// Check if current two bits are same
if (a_rem == b_rem)
xnornum |= (1 << count);
// counter for count bit
count++;
a = a >> 1;
b = b >> 1;
if (a < 1)
break;
}
return xnornum;
}
// Driver function
public static void Main()
{
int a = 10, b = 50;
Console.WriteLine(xnor(a, b));
}
}
// This code is contributed by vt_m
PHP
<?php
// PHP program to find
// XNOR of two numbers
// log(n) solution
function xnor($a, $b)
{
// Make sure a is larger
if ($a < $b)
list($a, $b) = array($b, $a);
if ($a == 0 && $b == 0)
return 1;
// for last bit of a
$a_rem = 0;
// for last bit of b
$b_rem = 0;
// counter for count bit
// and set bit in xnornum
$count = 0;
// to make new xnor number
$xnornum = 0;
// for set bits in
// new xnor number
while ($a)
{
// get last bit of a
$a_rem = $a & 1;
// get last bit of b
$b_rem = $b & 1;
// Check if current two
// bits are same
if ($a_rem == $b_rem)
$xnornum |= (1 << $count);
// counter for count bit
$count++;
$a = $a >> 1;
$b = $b >> 1;
}
return $xnornum;
}
// Driver code
$a = 10;
$b = 50;
echo xnor($a, $b);
// This code is contributed by mits.
?>
Javascript
<script>
// javascript program to find
// XNOR of two numbers
function xnor(a, b)
{
// Make sure a is larger
if (a < b)
{
// swapping a and b;
let t = a;
a = b;
b = t;
}
if (a == 0 && b == 0)
return 1;
// for last bit of a
let a_rem = 0;
// for last bit of b
let b_rem = 0;
// counter for count bit
// and set bit in xnornum
let count = 0;
// to make new xnor number
let xnornum = 0;
// for set bits in new xnor number
while (true) {
// get last bit of a
a_rem = a & 1;
// get last bit of b
b_rem = b & 1;
// Check if current two bits are same
if (a_rem == b_rem)
xnornum |= (1 << count);
// counter for count bit
count++;
a = a >> 1;
b = b >> 1;
if (a < 1)
break;
}
return xnornum;
}
// Driver Function
let a = 10, b = 50;
document.write(xnor(a, b));
// This code is contributed by susmitakundugoaldanga.
</script>
Producción
7
Segundo método: – O (1)
1) Encuentra el máximo de dos números dados.
2) Alternar todos los bits en mayor de dos números.
3) Devuelve el XOR del número original más pequeño y el número más grande modificado.
C++
// CPP program to find XNOR of two numbers.
#include <iostream>
using namespace std;
// Please refer below post for details of this
// function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
int togglebit(int n)
{
if (n == 0)
return 1;
// Make a copy of n as we are
// going to change it.
int i = n;
// Below steps set bits after
// MSB (including MSB)
// Suppose n is 273 (binary
// is 100010001). It does following
// 100010001 | 010001000 = 110011001
n |= n >> 1;
// This makes sure 4 bits
// (From MSB and including MSB)
// are set. It does following
// 110011001 | 001100110 = 111111111
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return i ^ n;
}
// Returns XNOR of num1 and num2
int XNOR(int num1, int num2)
{
// if num2 is greater then
// we swap this number in num1
if (num1 < num2)
swap(num1, num2);
num1 = togglebit(num1);
return num1 ^ num2;
}
// Driver code
int main()
{
int num1 = 10, num2 = 20;
cout << XNOR(num1, num2);
return 0;
}
Java
// Java program to find XNOR
// of two numbers
import java.io.*;
class GFG {
// Please refer below post for
// details of this function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
static int togglebit(int n)
{
if (n == 0)
return 1;
// Make a copy of n as we are
// going to change it.
int i = n;
// Below steps set bits after
// MSB (including MSB)
// Suppose n is 273 (binary
// is 100010001). It does following
// 100010001 | 010001000 = 110011001
n |= n >> 1;
// This makes sure 4 bits
// (From MSB and including MSB)
// are set. It does following
// 110011001 | 001100110 = 111111111
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return i ^ n;
}
// Returns XNOR of num1 and num2
static int xnor(int num1, int num2)
{
// if num2 is greater then
// we swap this number in num1
if (num1 < num2)
{
int temp = num1;
num1 = num2;
num2 = temp;
}
num1 = togglebit(num1);
return num1 ^ num2;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int a = 10, b = 20;
System.out.println(xnor(a, b));
}
}
// This code is contributed by Gitanjali
Python
# python program to find XNOR of two numbers import math # Please refer below post for details of this function # https://www.geeksforgeeks.org/toggle-bits-significant-bit/ def togglebit( n): if (n == 0): return 1 # Make a copy of n as we are # going to change it. i = n # Below steps set bits after # MSB (including MSB) # Suppose n is 273 (binary # is 100010001). It does following # 100010001 | 010001000 = 110011001 n = n|(n >> 1) # This makes sure 4 bits # (From MSB and including MSB) # are set. It does following # 110011001 | 001100110 = 111111111 n |= n >> 2 n |= n >> 4 n |= n >> 8 n |= n >> 16 return i ^ n # Returns XNOR of num1 and num2 def xnor( num1, num2): # Make sure num1 is larger if (num1 < num2): temp = num1 num1 = num2 num2 = temp num1 = togglebit(num1) return num1 ^ num2 # Driver code a = 10 b = 20 print (xnor(a, b)) # This code is contributed by 'Gitanjali'.
C#
// C# program to find XNOR
// of two numbers
using System;
class GFG {
// Please refer below post for
// details of this function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
static int togglebit(int n)
{
if (n == 0)
return 1;
// Make a copy of n as we are
// going to change it.
int i = n;
// Below steps set bits after
// MSB (including MSB)
// Suppose n is 273 (binary
// is 100010001). It does following
// 100010001 | 010001000 = 110011001
n |= n >> 1;
// This makes sure 4 bits
// (From MSB and including MSB)
// are set. It does following
// 110011001 | 001100110 = 111111111
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return i ^ n;
}
// Returns XNOR of num1 and num2
static int xnor(int num1, int num2)
{
// if num2 is greater then
// we swap this number in num1
if (num1 < num2)
{
int temp = num1;
num1 = num2;
num2 = temp;
}
num1 = togglebit(num1);
return num1 ^ num2;
}
// Driver program
public static void Main()
{
int a = 10, b = 20;
Console.WriteLine(xnor(a, b));
}
}
// This code is contributed by vt_m
PHP
<?php
// PHP program to find XNOR
// of two numbers.
// Please refer below post
// for details of this function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
function togglebit($n)
{
if ($n == 0)
return 1;
// Make a copy of n as we are
// going to change it.
$i = $n;
// Below steps set bits after
// MSB (including MSB)
// Suppose n is 273 (binary
// is 100010001). It does following
// 100010001 | 010001000 = 110011001
$n |= $n >> 1;
// This makes sure 4 bits
// (From MSB and including MSB)
// are set. It does following
// 110011001 | 001100110 = 111111111
$n |= $n >> 2;
$n |= $n >> 4;
$n |= $n >> 8;
$n |= $n >> 16;
return $i ^ $n;
}
// Returns XNOR of num1 and num2
function XNOR($num1, $num2)
{
// if num2 is greater then
// we swap this number in num1
if ($num1 < $num2)
list($num1, $num2)=array($num2, $num1);
$num1 = togglebit($num1);
return $num1 ^ $num2;
}
// Driver code
$num1 = 10;
$num2 = 20;
echo XNOR($num1, $num2);
// This code is contributed by Smitha.
?>
Javascript
<script>
// Javascript program to find XNOR of two numbers.
// Please refer below post for details of this
// function
// https://www.geeksforgeeks.org/toggle-bits-significant-bit/
function togglebit(n)
{
if (n == 0)
return 1;
// Make a copy of n as we are
// going to change it.
let i = n;
// Below steps set bits after
// MSB (including MSB)
// Suppose n is 273 (binary
// is 100010001). It does following
// 100010001 | 010001000 = 110011001
n |= n >> 1;
// This makes sure 4 bits
// (From MSB and including MSB)
// are set. It does following
// 110011001 | 001100110 = 111111111
n |= n >> 2;
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
return i ^ n;
}
// Returns XNOR of num1 and num2
function XNOR(num1, num2)
{
// if num2 is greater then
// we swap this number in num1
if (num1 < num2) {
let temp = num1;
num1 = num2;
num2 = temp;
}
num1 = togglebit(num1);
return num1 ^ num2;
}
// Driver code
let num1 = 10, num2 = 20;
document.write(XNOR(num1, num2));
</script>
Producción
1
Tercer método: usar XOR
XNOR de A y B es simplemente el inverso de A XOR B. El problema de invertir directamente un número binario es que hace que los ceros iniciales también se inviertan, por lo tanto, es necesario construir una máscara de bits para extraer solo los bits reales del numero invertido
Esta implementación se muestra a continuación:
C++
// CPP program to find XNOR of two numbers.
#include <bits/stdc++.h>
using namespace std;
// Returns XNOR of num1 and num2
unsigned int XNOR(int a, int b)
{
//getting the number of bits from
//max of a and b to construct the bit mask
int numOfBits = log2(max(a, b));
//constructing the bit mask
unsigned mask = (1 << numOfBits) - 1;
//xnor = inverted xor
unsigned xnor = ~(a ^ b);
//getting only the required bits
//using the mask
return xnor & mask;
}
// Driver code
int main()
{
int num1 = 7, num2 = 19;
//function call
cout << XNOR(num1, num2);
return 0;
}
//this code is contributed by phasing17
Java
// Java program to find XNOR of two numbers.
class GFG {
// Returns XNOR of num1 and num2
static int XNOR(int a, int b)
{
// getting the number of bits from
// max of a and b to construct the bit mask
int numOfBits = (int)(Math.log(Math.max(a, b)) / Math.log(2));
// constructing the bit mask
int mask = (1 << numOfBits) - 1;
// xnor = inverted xor
int xnor = ~(a ^ b);
// getting only the required bits
// using the mask
return xnor & mask;
}
// Driver code
public static void main(String[] args)
{
int num1 = 7, num2 = 19;
// function call
System.out.print(XNOR(num1, num2));
}
}
// this code is contributed by phasing17
Python3
# Python3 program to find XNOR of two numbers. import math # Returns XNOR of num1 and num2 def XNOR(a, b): # getting the number of bits from # max of a and b to construct the bit mask numOfBits = int(math.log(max(a, b), 2)) # constructing the bit mask mask = (1 << numOfBits) - 1 # xnor = inverted xor xnor = ~(a ^ b) # getting only the required bits # using the mask return xnor & mask # Driver code num1 = 7 num2 = 19 # function call print(XNOR(num1, num2)) # This code is contributed by phasing17
C#
// C# program to find XNOR of two numbers.
using System;
class GFG {
// Returns XNOR of num1 and num2
static int XNOR(int a, int b)
{
// getting the number of bits from
// max of a and b to construct the bit mask
int numOfBits = (int)(Math.Log(Math.Max(a, b)) / Math.Log(2));
// constructing the bit mask
int mask = (1 << numOfBits) - 1;
// xnor = inverted xor
int xnor = ~(a ^ b);
// getting only the required bits
// using the mask
return xnor & mask;
}
// Driver code
public static void Main(string[] args)
{
int num1 = 7, num2 = 19;
// Function call
Console.Write(XNOR(num1, num2));
}
}
// this code is contributed by phasing17
Javascript
// JavaScript program to find XNOR of two numbers.
// Returns XNOR of num1 and num2
function XNOR(a, b)
{
//getting the number of bits from
//max of a and b to construct the bit mask
let numOfBits = Math.floor(Math.log2(Math.max(a, b)));
//constructing the bit mask
let mask = (1 << numOfBits) - 1;
//xnor = inverted xor
let xnor = ~(a ^ b);
//getting only the required bits
//using the mask
return xnor & mask;
}
// Driver code
let num1 = 7, num2 = 19;
//function call
console.log(XNOR(num1, num2));
//this code is contributed by phasing17
Producción
11
Complejidad de tiempo : O(1)
Espacio Auxiliar : O(1)
Publicación traducida automáticamente
Artículo escrito por DevanshuAgarwal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA