Question

Deceptively simple algorithm question I came across. I’m trying to pull it off in 3 or less operations, and I’m reasonably convinced it can be solved with math, but I can’t figure it out (and the source for the question didn’t have an answer).

EDIT:

(a[0] == a[1] + a[0] == a[2] + a[1] == a[2]) == 1

is what I originally thought, but I’d like to see if it can be done in less operations (1 comparison being an operation).

Answer 1

Assuming the 3 numbers are a, b and c,

(b == c) ? (a != c) : (a == b || a == c)

• If (a, b, c) = (1, 1, 1), then we will call b == c (true) and then a != c (false) and done.
• If (a, b, c) = (1, 1, 2), then we will call b == c (false) and then a == b (true) and done.
• If (a, b, c) = (1, 2, 1), then we will call b == c (false) and then a == b (false) and a == c (true) and done.
• If (a, b, c) = (2, 1, 1), then we will call b == c (true) and then a != c (true) and done.
• If (a, b, c) = (1, 2, 3), then we will call b == c (false) and then a == b (false) and a == c (false) and done.

So at most 3 comparison are performed.

There are also 2 conditional branching points at ?: and || but OP does not count it.