Scalable solution for Rock-Paper-Scissor

Question

Just went through a variant of the game : Rock-Paper-Scissor-Lizard-Spock

I have written a Java code for traditional R-P-S problem, but when I tried extending my code

Answer 1

In, Rock-Paper-Scissor games, it is easy to decide if move a wins against move b using their index at a cycle. So you don't have to manually decide in your code the result of every combination as other answers here suggest.

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For the Rock-Paper-Scissor-Spock-Lizard version:

Let's assign a number to each move (0, 1, 2, 3, 4).

Notice that every move beats two moves:

1. The move previous to it in the cycle (or four cases ahead)
2. The move two cases ahead in the cycle

So let d = (5 + a - b) % 5. Then:

1. d = 1 or d = 3 => a wins
2. d = 2 or d = 4 => b wins
3. d = 0 => tie

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For the Rock-Paper-Scissor version:

let d = (3 + a - b) % 3. Then:

1. d = 1 => a wins
2. d = 2 => b wins
3. d = 0 => tie

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Generalization For n >= 3 and n odd:

Let d = (n + a - b) % n. Then:

1. If d = 0 => tie
2. If d % 2 = 1 => a wins
3. If d % 2 = 0 => b wins