XNOR（異或非）閘是一種數位邏輯閘，它接受兩個輸入並給出一個輸出。其功能是異或（XOR）閘的邏輯補。如果兩個輸入相同，則輸出為 TRUE；如果輸入不同，則輸出為 FALSE。下面給出了異或非閘的真值表。
| 一個 | B | 輸出 |
|---|---|---|
| 1 | 1 | 1 |
| 1 | 0 | 0 |
| 0 | 1 | 0 |
| 0 | 0 | 1 |
給定兩個數字 x 和 y。求兩個數的異或。
Input: x = 12, y = 5
Output: 6
(12)<sub>10</sub> = (1100)<sub>2</sub> (5)<sub>10</sub> = (101)<sub>2</sub> XNOR = (110)2 = (6)<sub>10</sub>
Input: x = 16, y = 16
Output: 31
(16)<sub>10</sub> = (10000)<sub>2</sub> (16)<sub>10</sub> = (10000)<sub>2</sub> XNOR = (11111)<sub>2</sub> = (31)<sub>10</sub>
暴力方法是檢查兩個數字的每一位並比較它們是否相同。若相同則加1，否則加0。
procedure xnor (x, y)
if x > y then
swap(x,y)
end if
if x == 0 and y == 0 then
ans = 1
end if
while x
x_rem = x & 1
y_rem = y & 1
if x_rem == y_rem then
ans = ans | (1 << count)
end if
count = count + 1
x = x >> 1
y = y >> 1
end procedure
在下面的程式中，檢查x和y的位元是否相同，然後設定答案中的位元。
#include <bits/stdc++.h>
using namespace std;
// Function to swap values of two variables
void swap(int *a, int *b){
int temp = *a;
*a = *b;
*b = temp;
}
// Function to find the XNOR of two numbers
int xnor(int x, int y){
// Placing the lower number in variable x
if (x > y){
swap(x, y);
}
// Base Condition
if (x == 0 && y == 0){
return 1;
}
// Cnt for counting the bit position Ans stores ans sets the bits of XNOR operation
int cnt = 0, ans = 0;
// executing loop for all the set bits in the lower number
while (x){
// Gets the last bit of x and y
int x_rem = x & 1, y_rem = y & 1;
// If last bits of x and y are same
if (x_rem == y_rem){
ans |= (1 << cnt);
}
// Increase counter for bit position and right shift both x and y
cnt++;
x = x >> 1;
y = y >> 1;
}
return ans;
}
int main(){
int x = 10, y = 11;
cout << "XNOR of " << x << " and " << y << " = " << xnor(x, y);
return 0;
}
XNOR of 10 and 11 = 14
時間複雜度：O(logn)，因為遍歷是對所有 logn 位元進行的。
空間複雜度：O(1)
XNOR是XOR運算的逆運算，它的真值表也是XOR表的逆運算。因此，切換較高數字的位，即將 1 設為 0，將 0 設為 1，然後與較低數字進行異或將產生 XNOR 數字。
Input: x = 12, y = 5
Output: 6
(12)10 = (1100)2 (5)10 = (101)2 Toggled bits of 12 = 0011 0011 ^ 101 = 0110
Input: x = 12, y = 31
Output: 12
(12)10 = (1100)2 (31)10 = (11111)2 Toggled bits of 31 = 00000 00000 ^ 1100 = 01100
procedure toggle (n)
temp = 1
while temp <= n
n = n ^ temp
temp = temp << 1
end procedure
procedure xnor (x, y)
max_num = max(x,y)
min_num = min(x,y)
toggle (max_num)
ans = max_num ^ min_num
end procedure
在下面的程式中，較高數字的所有位元都會被切換，然後與較低數字進行異或。
#include <bits/stdc++.h>
using namespace std;
// Function to toggle all bits of a number
void toggle(int &num){
int temp = 1;
// Execute loop until set bit of temp cross MST of num
while (temp <= num){
// Toggle bit of num corresponding to set bit in temp
num ^= temp;
// Move set bit of temp to left
temp <<= 1;
}
}
// Function to find the XNOR of two numbers
int xnor(int x, int y){
// Finding max and min number
int max_num = max(x, y), min_num = min(x, y);
// Togglinf the max number
toggle(max_num);
// XORing toggled max num and min num
return max_num ^ min_num;
}
int main(){
int x = 5, y = 15;
cout << "XNOR of " << x << " and " << y << " = " << xnor(x, y);
return 0;
}
XNOR of 5 and 15 = 5
時間複雜度：O(logn)，由於在toggle()函數中進行遍歷
空間複雜度：O(1)
從邏輯上講，XNOR是XOR的反向操作，但是在對XOR進行補碼操作時，前導零也會被反轉。為了避免這種情況，可以使用位元遮罩來去除所有不必要的前導位。
Input: x = 12, y = 5
Output: 6
(12)<sub>10</sub> = (1100)<sub>2</sub> (5)<sub>10</sub> = (101)<sub>2</sub> 1100 ^ 101 = 1001 Inverse of 1001 = 0110
Input: x = 12, y = 31
Output: 12
(12)<sub>10</sub> = (1100)<sub>2</sub> (31)<sub>10</sub> = (11111)<sub>2</sub> 1100 ^ 11111 = 10011 Inverse of 10011 = 01100
Procedure xnor (x, y) bit_count = log2 (maximum of a and y) mask = (1 << bit_count) - 1 ans = inverse(x xor y) and mask end procedure
在下面的程式中，位元遮罩用於從 x xor y 的逆中僅取得所需的位元。
#include <bits/stdc++.h>
using namespace std;
// Function to find the XNOR of two numbers
int xnor(int x, int y){
// Maximum number of bits used in both the numbers
int bit_count = log2(max(x, y));
int mask = (1 << bit_count) - 1;
// Inverse of XOR operation
int inv_xor = ~(x ^ y);
// GEtting the required bits and removing the inversion of leading zeroes.
return inv_xor & mask;
}
int main(){
int x = 10, y = 10;
cout << "XNOR of " << x << " and " << y << " = " << xnor(x, y);
return 0;
}
XNOR of 10 and 10 = 7
總之，可以使用各種方法和時間複雜度來找出兩個數字的XNOR，範圍從O(logn)的暴力法到O(1)的最佳化方法。應用位運算更廉價，因此暴力法的複雜度是對數等級的。
以上是兩個數字的XNOR的詳細內容。更多資訊請關注PHP中文網其他相關文章！
