Finding the middle number out of three (without if-else and such)

Question

I've been asked to make a short program that gets 3 numbers, and print the middle one i.e. 3 2 0 -> "middle number is 2" another example : 4 4 5 -> middle number is 4 (the user may enter the same number twice

it would be pretty easy with the if function and so but

• I'm not allowed to use conditionals or loops

edit: cant use arrays I've tried to divide numbers by each other to get a 0 which indicates one is bigger or smaller

Answer 1

As mentioned in the almost duplicate question and in previous answer, one way is to sum the three values and then to substract the min and max values.

However, using min and max function of the library is cheating a little bit, as it is likely that internally conditionals are used.

A way forward is to use the relations:

min(a, b) = (a + b - abs(a-b))/2;
max(a, b) = (a + b + abs(a-b))/2;

etc.

It was mentioned in a comment that using abs is cheating also. In practice, I don't know how it is implemented internally.

One possibility is to used the tricks mentioned in this post: get absolute values....

For example, abs(a) = sqrt (a*a);

Answer 2

Use these two functions:

int maximum(int a, int b, int c)
{
    // initialize max with a
    int max = a;
    
    // set max to b if and only if max is less than b
    (max < b) && (max = b); // these are not conditional statements.
    
    // set max to c if and only if max is less than c
    (max < c) && (max = c); // these are just Boolean expressions.

    return max;
}

int minimum(int a, int b, int c)
{
    // initialize min with a
    int min = a;
    
    // set min to b if and only if min is more than b
    (min > b) && (min = b);
    
    // set min to c if and only if min is more than c
    (min > c) && (min = c);
    
    return min;
}

Then the mid-value can be calculated easily using:

(a+b+c)-minimum(a, b, c)-maximum(a, b, c)

Edit:

To run the code without defining any function, we can use the following. It is exactly same as above, but does all the works inside main function.

int main() {
    int a=3, b=2, c=0, min, max;

    // initialize min with a
    min = a;

    // set min to b if and only if min is more than b
    (min > b) && (min = b);

    // set min to c if and only if min is more than c
    (min > c) && (min = c);

    // similarly find max
    max = a;
    (max < b) && (max = b);
    (max < c) && (max = c);

    printf("%d", a+b+c-min-max);
    return 0;
}

Answer 3

Use wider than int math to handle range limitations.

Find absolute value by division trick.¹

((3 * d) / (3 * d + 1)) -->  0 when d >= 0
((3 * d) / (3 * d + 1)) --> -1 when d < 0

No conditions used.

long long my_abs_diff(long long a, long long b) {
  long long d = 0LL + a - b;
  return d * (1 - 2 * ((3 * d) / (3 * d + 1)));
}

long long my_min(long long a, long long b) {
  return ((a + b) - my_abs_diff(a, b)) / 2;
}

long long my_max(long long a, long long b) {
  return ((a + b) + my_abs_diff(a, b)) / 2;
}

int median3(int a, int b, int c) {
  long long mn = my_min(my_min(a, b), c);
  long long mx = my_max(my_max(a, b), c);
  return (int) (-mn - mx + a + b + c);
}

Simplification may exist.

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¹In C, since C99, integer division is defined as truncating towards zero. It is this directed-ness towards zero code is exploiting to effect an "if". A cost to this approach is that we need a few more bits of integer range. For pre-C99, code can use div() to insure the desired quotient truncation.

Tested successfully with many edge cases and 10,000,00,00 random combinations over entire int range: [INT_MIN ... INT_MAX].

Answer 4

Hint : first find the min and max values by using std::min and std::max
Then sum all the numbers and subtract the min and max values. What you are left with is the middle value.