Mathematically Find Max Value without Conditional Comparison

Question

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codymanix and moonshadow have been a big help thus far. I was able to solve my problem using the equations and instead of using right

Answer 1

It depends which language you're using, but the Ternary Operator might be useful.

But then, if you can't perform conditional checks in your 'scripting application', you probably don't have the ternary operator.

Answer 2

If A is always greater than B .. [ we can use] .. MAX = (A - B) + B;

No need. Just use: int maxA(int A, int B){ return A;}

(1) If conditionals are allowed you do max = a>b ? a : b.

(2) Any other method either use a defined set of numbers or rely on the implicit conditional checks.

(2a) max = a-((a-b)&((a-b)>>31)) this is neat, but it only works if you use 32 bit numbers. You can expand it arbitrary large number N, but the method will fail if you try to find max(N-1, N+1). This algorithm works for finite state automata, but not a Turing machine.

(2b) Magnitude |a-b| is a condition |a-b| = a-b>0 a-b : b-a

What about:

Square root is also a condition. Whenever c>0 and c^2 = d we have second solution -c, because (-c)^2 = (-1)^2*c^2 = 1*c^2 = d. Square root returns the greatest in the pair. I comes with a build in int max(int c1, int c2){return max(c1, c2);}

Without comparison operator math is very symmetric as well as limited in power. Positive and negative numbers cannot be distinguished without if of some sort.

Answer 3

Using logical operations only, short circuit evaluation and assuming the C convention of rounding towards zero, it is possible to express this as:

int lt0(int x) {
    return x && (!!((x-1)/x));
}

int mymax(int a, int b) {
    return lt0(a-b)*b+lt0(b-a)*a;
}

The basic idea is to implement a comparison operator that will return 0 or 1. It's possible to do a similar trick if your scripting language follows the convention of rounding toward the floor value like python does.

Answer 4

function Min(x,y:integer):integer;
  Var
   d:integer;
   abs:integer;
 begin
  d:=x-y;
  abs:=d*(1-2*((3*d) div (3*d+1)));
  Result:=(x+y-abs) div 2;
 end;

Answer 5

finding the maximum of 2 variables:

max = a-((a-b)&((a-b)>>31))

where >> is bitwise right-shift (also called SHR or ASR depeding on signedness).

Instead of 31 you use the number of bits your numbers have minus one.

Answer 6

try this, (but be aware for overflows) (Code in C#)

    public static Int32 Maximum(params Int32[] values)
    {
        Int32 retVal = Int32.MinValue;
        foreach (Int32 i in values)
            retVal += (((i - retVal) >> 31) & (i - retVal));
        return retVal;        
    }