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 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 Code int main() { int x = 29, y = 13; printf ( "x - y is %d" , subtract(x, y)); return 0; } |
Java
// Java Program to subtract two Number // without using arithetic operater import 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 |
Python3
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 Code x = 29 y = 13 print ( "x - y is" ,subtract(x, y)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# Program to subtract two Number // without using arithetic operater using 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 // PHP Program to subtract two Number // without using arithetic operater 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 ?> |
Javascript
<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
Following is recursive implementation for the same approach.
C
#include<stdio.h> int subtract( int x, int y) { if (y == 0) return x; return subtract(x ^ y, (~x & y) << 1); } // Driver program int main() { int x = 29, y = 13; printf ( "x - y is %d" , subtract(x, y)); return 0; } |
Java
// Java Program to subtract two Number // without using arithetic operater // 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. |
Python3
# 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 program x = 29 y = 13 print ( "x - y is" , subtract(x, y)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# Program to subtract two Number // without using arithetic operater // 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
<?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 ?> |
Output :
x - y is 16
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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