# Subtract two numbers without using arithmetic operators

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. Like addition, Write a function Add() that returns sum of two integers. The function should not use any of the arithmetic operators (+, ++, –, -, … etc). Sum of two bits can be obtained by performing XOR (^) of the two bits. Carry bit can be obtained by performing AND (&) of two bits. We can extend this logic for integers. If x and y don’t have set bits at same position(s), then bitwise XOR (^) of x and y gives the sum of x and y. To incorporate common set bits also, bitwise AND (&) is used. Bitwise AND of x and y gives all carry bits. We calculate (x & y) << 1 and add it to x ^ y to get the required result. 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
``````

Following is implementation based on above equations.

`// 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));`

` ` `}`

`}`

Output :

x - y is 16