Minimum bitwise operations to convert given a into b.

Given two positive integer a and b you have to change a to b by applying any of the three operations on binary form of a. You can select ai and aj (any two bits where i!=j) from binary form of a and then perform operation as:

  • AND operation as : temp = ai & aj, ai = temp & ai, aj = temp & aj
  • OR operation as : temp = ai | aj, ai = temp | ai, aj = temp | aj
  • XOR operation as : temp = ai ^ aj, ai = temp ^ ai, aj = temp ^ aj

where & = bitwise AND, | = bitwiese OR and ^ = bitwise XOR.
Find the minimum operation required for conversion of a to b. Also, if conversion of a to b is not possible then print -1.

Examples:



Input : a = 12 (1100), b = 10 (1010)
Output : 1
Explanation : select a2 and a3 and perform XOR 

Input : a = 15 (1111), b = 10 (1010)
Output : -1
Explanation : Conversion from a to b is not possible

Explanation : First of all let’s understand the working of all three operation.

  1. AND operation as : temp = ai & aj, ai = temp & ai, aj = temp & aj
  2. If any of ai or aj is 0 then it makes both as 0 otherwise no effect on ai and aj.

  3. OR operation as : temp = ai | aj, ai = temp | ai, aj = temp | aj
  4. If any of ai or aj is 1 then it makes both as 1 otherwise no effect on ai and aj.

  5. XOR operation as : temp = ai ^ aj, ai = temp ^ ai, aj = temp ^ aj
  6. Simply interchange value of ai and aj.

Some conclusion on basis of working of operations :

  1. If all bits of a are 1 or 0 then we can not change value of a.
  2. If a equals to b then no operation required.
  3. Let n be number of indices i, where ai = 0 and bi = 1.
    Let m be number of indices i, where ai = 1 and bi = 0.

    Let us think about the n elements, where ai = 0 and bi = 1. We have to change all of these zeros into ones. Note that this will require at least n operations.
    Similarly for all the m elements, where ai = 1 and bi = 0. We have to change all of these ones into zeros. Note that this will require at least m operations.

    Let res = max(n, m). We can make the a an b equal in res operations as follows.

    Let n >= m. Take m 1’s and n 0’s in A and apply the XOR operation to swap 0’s with 1’s. After that you will be left with total n-m zeros elements to change to one. That you can do by taking each of these zeros with some single one and applying the OR operation.
    Let m >= n. Take m 1’s and n 0’s in A and apply the XOR operation to swap 0’s with 1’s. After that you will be left with total m-n ones elements to change to zero. That you can do by taking each of these ones with some single zero and applying the OR operation.

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// Cpp program to find min operation to convert a to b
#include <bits/stdc++.h>
using namespace std;
  
// function to return min operation
int minOp(bitset<32> a1, bitset<32> b1)
{
    // if a1 == b1 return 0
    if (a1 == b1)
        return 0;
  
    // if all bits of a = 0
    if (a1 == 0)
        return -1;
  
    // if all bits of a =1
    // first convert a1 to int and then cal a1 & a1+1
    if (((int)a1.to_ulong() & ((int)a1.to_ulong() + 1)) 
                                               == 0)
        return -1;
  
    // convert a and b to binary string
    string a = a1.to_string();
    string b = b1.to_string();
  
    // check where ai and bi are different
    // and count n where ai = 1 and m where ai = 0
    int n = 0, m = 0;
    for (int i = 0; i < b.size(); i++) {
        if (b[i] != a[i]) {
            if (a[i] == '1')
                n++;
            else
                m++;
        }
    }
  
    // return result
    return max(n, m);
}
  
// driver program
int main()
{
    bitset<32> a = 14, b = 1;
    cout << minOp(a, b);
    return 0;
}

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Output:

3

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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