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.

`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))`

**Output :**

x - y is 16