Subtract two numbers without using arithmetic operators

01 Nov, 2021
Write a function subtract(x, y) that returns x-y where x and y are integers. The function should not use any of the arithmetic operators (+, ++, –, -, .. etc). The idea is to use bitwise operators. Addition of two numbers has been discussed using Bitwise operators. Like addition, the idea is to use subtractor logic. 
The truth table for the half subtractor is given below. 
X     Y     Diff     Borrow
0     0     0     0
0     1     1     1
1     0     1     0
1     1     0     0
From the above table one can draw the Karnaugh map for “difference” and “borrow”.So, Logic equations are: 
    Diff   = y ⊕ x
    Borrow = x' . y 
Source: Wikipedia page for subtractor
Following is implementation based on above equations.  
C++
C
Java
Python3
C#
PHP
Javascript
// C++ program to Subtract two numbers// without using arithmetic operators#include <iostream>using namespace std; int subtract(int x, int y){    // Iterate till there    // is no carry    while (y != 0)    {        // borrow contains common        // set bits of y and unset        // bits of x        int borrow = (~x) & y;         // Subtraction of bits of x        // and y where at least one        // of the bits is not set        x = x ^ y;         // Borrow is shifted by one        // so that subtracting it from        // x gives the required sum        y = borrow << 1;    }    return x;} // Driver Codeint main(){    int x = 29, y = 13;    cout << "x - y is " << subtract(x, y);    return 0;} // This code is contributed by shivanisinghss2110
// C program to Subtract two numbers// without using arithmetic operators#include<stdio.h> int subtract(int x, int y){    // Iterate till there    // is no carry    while (y != 0)    {        // borrow contains common        // set bits of y and unset        // bits of x        int borrow = (~x) & y;         // Subtraction of bits of x        // and y where at least one        // of the bits is not set        x = x ^ y;         // Borrow is shifted by one        // so that subtracting it from        // x gives the required sum        y = borrow << 1;    }    return x;} // Driver Codeint main(){    int x = 29, y = 13;    printf("x - y is %d", subtract(x, y));    return 0;}
// Java Program to subtract two Number// without using arithmetic operatorimport java.io.*; class GFG{    static int subtract(int x, int y)    {             // Iterate till there    // is no carry    while (y != 0)    {        // borrow contains common        // set bits of y and unset        // bits of x        int borrow = (~x) & y;         // Subtraction of bits of x        // and y where at least one        // of the bits is not set        x = x ^ y;         // Borrow is shifted by one        // so that subtracting it from        // x gives the required sum        y = borrow << 1;    }         return x;}         // Driver Code    public static void main (String[] args)    {        int x = 29, y = 13;                 System.out.println("x - y is " +                        subtract(x, y));    }} // This code is contributed by vt_m
def subtract(x, y):     # Iterate till there    # is no carry    while (y != 0):             # borrow contains common        # set bits of y and unset        # bits of x        borrow = (~x) & y                 # Subtraction of bits of x        # and y where at least one        # of the bits is not set        x = x ^ y         # Borrow is shifted by one        # so that subtracting it from        # x gives the required sum        y = borrow << 1         return x  # Driver Codex = 29y = 13print("x - y is",subtract(x, y)) # This code is contributed by# Smitha Dinesh Semwal
// C# Program to subtract two Number// without using arithmetic operatorusing System; class GFG {         static int subtract(int x, int y)    {                 // Iterate till there        // is no carry        while (y != 0)        {                         // borrow contains common            // set bits of y and unset            // bits of x            int borrow = (~x) & y;                 // Subtraction of bits of x            // and y where at least one            // of the bits is not set            x = x ^ y;                 // Borrow is shifted by one            // so that subtracting it from            // x gives the required sum            y = borrow << 1;        }                 return x;    }         // Driver Code    public static void Main ()    {        int x = 29, y = 13;                 Console.WriteLine("x - y is " +                        subtract(x, y));    }} // This code is contributed by anuj_67.
<?php// PHP Program to subtract two Number// without using arithmetic operator function subtract($x, $y){         // Iterate till there is no carry    while ($y != 0)    {                 // borrow contains common set        // bits of y and unset        // bits of x        $borrow = (~$x) & $y;         // Subtraction of bits of x        // and y where at least        // one of the bits is not set                 $x = $x ^ $y;         // Borrow is shifted by one so        // that subtracting it from        // x gives the required sum                 $y = $borrow << 1;    }    return $x;}         // Driver Code        $x = 29; $y = 13;        echo "x - y is ", subtract($x,$y); // This code is contributed by Ajit?>
<script> // JavaScript program to Subtract two numbers// without using arithmetic operators function subtract(x, y){    // Iterate till there    // is no carry    while (y != 0)    {        // borrow contains common        // set bits of y and unset        // bits of x        let borrow = (~x) & y;         // Subtraction of bits of x        // and y where at least one        // of the bits is not set        x = x ^ y;         // Borrow is shifted by one        // so that subtracting it from        // x gives the required sum        y = borrow << 1;    }    return x;} // Driver Code  let x = 29, y = 13;document.write("x - y is " + subtract(x, y));  // This code is contributed by Surbhi Tyagi. </script>
Output : 
x - y is 16
Time Complexity: O(log y)
Auxiliary Space: O(1)
Following is recursive implementation for the same approach. 
C++
C
Java
Python3
C#
PHP
Javascript
#include <iostream>using namespace std; int subtract(int x, int y){    if (y == 0)        return x;    return subtract(x ^ y, (~x & y) << 1);} // Driver programint main(){    int x = 29, y = 13;    cout << "x - y is "<< subtract(x, y);    return 0;} // this code is contributed by shivanisinghss2110
#include<stdio.h> int subtract(int x, int y){    if (y == 0)        return x;    return subtract(x ^ y, (~x & y) << 1);} // Driver programint main(){    int x = 29, y = 13;    printf("x - y is %d", subtract(x, y));    return 0;}
// Java  Program to subtract two Number// without using arithmetic operator// Recursive implementation.class GFG {     static int subtract(int x, int y)    {         if (y == 0)            return x;         return subtract(x ^ y, (~x & y) << 1);    }     // Driver program    public static void main(String[] args)    {        int x = 29, y = 13;        System.out.printf("x - y is %d",                            subtract(x, y));    }} // This code is contributed by // Smitha Dinesh Semwal.
# Python  Program to# subtract two Number# without using arithmetic operator# Recursive implementation. def subtract(x, y):     if (y == 0):        return x    return subtract(x ^ y, (~x & y) << 1) # Driver programx = 29y = 13print("x - y is", subtract(x, y)) # This code is contributed by# Smitha Dinesh Semwal
// C# Program to subtract two Number// without using arithmetic operator// Recursive implementation.using System; class GFG {     static int subtract(int x, int y)    {        if (y == 0)            return x;         return subtract(x ^ y, (~x & y) << 1);    }     // Driver program    public static void Main()    {        int x = 29, y = 13;        Console.WriteLine("x - y is "+                            subtract(x, y));    }} // This code is contributed by anuj_67.
<?php function subtract($x, $y){    if ($y == 0)        return $x;    return subtract($x ^ $y,                   (~$x & $y) << 1);} // Driver Code$x = 29; $y = 13;echo "x - y is ", subtract($x, $y); # This code is contributed by ajit?>
<script>// javascript  Program to subtract two Number// without using arithmetic operator// Recursive implementation.  function subtract(x , y){     if (y == 0)        return x;     return subtract(x ^ y, (~x & y) << 1);} // Driver programvar x = 29, y = 13;document.write("x - y is "+                    subtract(x, y)); // This code is contributed by Princi Singh</script>
Output : 
x - y is 16
Time Complexity: O(log y)
Auxiliary Space: O(log y)
This article is contributed Dheeraj. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 
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