Smallest of three integers without comparison operators

Write a 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.

C++

 // C++ program to find Smallest // of three integers without // comparison operators #include using namespace std; int smallest(int x, int y, int z) {     int c = 0;     while (x && y && z) {         x--;         y--;         z--;         c++;     }     return c; }    // Driver Code int main() {     int x = 12, y = 15, z = 5;     cout << "Minimum of 3 numbers is "          << smallest(x, y, z);     return 0; }    // This code is contributed // by Akanksha Rai

C

 // C program to find Smallest // of three integers without // comparison operators #include    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; }

Java

 // Java program to find Smallest // of three integers without // comparison operators class GFG {        static int smallest(int x, int y, int z)     {         int c = 0;            while (x != 0 && y != 0 && z != 0) {             x--;             y--;             z--;             c++;         }            return c;     }        public static void main(String[] args)     {         int x = 12, y = 15, z = 5;            System.out.printf("Minimum of 3"                               + " numbers is %d",                           smallest(x, y, z));     } }    // This code is contributed by  Smitha Dinesh Semwal.

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Python3

 # Python3 program to find Smallest # of three integers without  # comparison operators    def smallest(x, y, z):     c = 0            while ( x and y and z ):         x = x-1         y = y-1         z = z-1         c = c + 1        return c    # Driver Code x = 12 y = 15 z = 5 print("Minimum of 3 numbers is",        smallest(x, y, z))    # This code is contributed by Anshika Goyal

C#

 // C# program to find Smallest of three // integers without comparison operators using System;    class GFG {     static int smallest(int x, int y, int z)     {         int c = 0;            while (x != 0 && y != 0 && z != 0) {             x--;             y--;             z--;             c++;         }            return c;     }        // Driver Code     public static void Main()     {         int x = 12, y = 15, z = 5;            Console.Write("Minimum of 3"                       + " numbers is " + smallest(x, y, z));     } }    // This code is contributed by Sam007

PHP



Output:

Minimum of 3 numbers is 5

This method doesn’t work for negative numbers. Method 2 works for negative numbers 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.

 #include #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; }

Output:

Minimum of 3 numbers is 5

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.

C++

 #include    // 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; }

Java

 // Java program of above approach class GfG {        // Using division operator to     // find minimum of three numbers     static int smallest(int x, int y, int z)     {         if ((y / x) != 1) // Same as "if (y < x)"             return ((y / z) != 1) ? y : z;         return ((x / z) != 1) ? x : z;     }        // Driver code     public static void main(String[] args)     {         int x = 78, y = 88, z = 68;         System.out.printf("Minimum of 3 numbers"                               + " is %d",                           smallest(x, y, z));     } }    // This code has been contributed by 29AjayKumar

python3

 # Using division operator to find # minimum of three numbers def smallest(x, y, z):        if (not (y / x)): # Same as "if (y < x)"         return y if (not (y / z)) else z     return x if (not (x / z)) else z    # Driver Code if __name__== "__main__":        x = 78     y = 88     z = 68     print("Minimum of 3 numbers is",                   smallest(x, y, z))    # This code is contributed  # by ChitraNayal

C#

 // C# program of above approach using System; public class GfG {        // Using division operator to     // find minimum of three numbers     static int smallest(int x, int y, int z)     {         if ((y / x) != 1) // Same as "if (y < x)"             return ((y / z) != 1) ? y : z;         return ((x / z) != 1) ? x : z;     }        // Driver code     public static void Main()     {         int x = 78, y = 88, z = 68;         Console.Write("Minimum of 3 numbers"                           + " is {0}",                       smallest(x, y, z));     } } /* This code contributed by PrinciRaj1992 */

Output:

Minimum of 3 numbers is 68

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

tags:

Bit Magic Bit Magic