# Smallest of three integers without comparison operators

Write a C program to find the smallest of three integers, without using any of the comparison operators.

Let 3 input numbers be x, y and z.

Method 1 (Repeated Subtraction)
Take a counter variable c and initialize it with 0. In a loop, repeatedly subtract x, y and z by 1 and increment c. The number which becomes 0 first is the smallest. After the loop terminates, c will hold the minimum of 3.

```#include<stdio.h>

int smallest(int x, int y, int z)
{
int c = 0;
while ( x && y && z )
{
x--;  y--; z--; c++;
}
return c;
}

int main()
{
int x = 12, y = 15, z = 5;
printf("Minimum of 3 numbers is %d", smallest(x, y, z));
return 0;
}
```

This methid doesn’t work for negative numbers. Method 2 works for negative nnumbers also.

Method 2 (Use Bit Operations)
Use method 2 of this post to find minimum of two numbers (We can’t use Method 1 as Method 1 uses comparison operator). Once we have functionality to find minimum of 2 numbers, we can use this to find minimum of 3 numbers.

```// See mthod 2 of http://www.geeksforgeeks.org/archives/2643
#include<stdio.h>
#define CHAR_BIT 8

/*Function to find minimum of x and y*/
int min(int x, int y)
{
return  y + ((x - y) & ((x - y) >>
(sizeof(int) * CHAR_BIT - 1)));
}

/* Function to find minimum of 3 numbers x, y and z*/
int smallest(int x, int y, int z)
{
return min(x, min(y, z));
}

int main()
{
int x = 12, y = 15, z = 5;
printf("Minimum of 3 numbers is %d", smallest(x, y, z));
return 0;
}
```

Method 3 (Use Division operator)
We can also use division operator to find minimum of two numbers. If value of (a/b) is zero, then b is greater than a, else a is greater. Thanks to gopinath and Vignesh for suggesting this method.

```#include <stdio.h>

// Using division operator to find minimum of three numbers
int smallest(int x, int y, int z)
{
if (!(y/x))  // Same as "if (y < x)"
return (!(y/z))? y : z;
return (!(x/z))? x : z;
}

int main()
{
int x = 78, y = 88, z = 68;
printf("Minimum of 3 numbers is %d", smallest(x, y, z));
return 0;
}

```

Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.

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