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

This is a basic logic problem. It is small enough you can do a manual truth table ( or skip ahead to a k-map), minimize and get a solution.

So basically, you need to evaluate first, if it is a draw. Then, you need to evaluate winning relative to other players. Doing this without needing to compare against each user can be a confusing task. Since this only has 5 variables, you can find a minimized solution with a K-map.

You will need to evaluate each user, based on which item they chose with a specific algorithm to determine if they win. Note that with more than 2 players, there can be more than one winner if two people choose the same thing but both beat a 3rd player. Or you can consider that a tie, whatever. I'll assume the former. You should check also that all players didn't choose the same item.

So I've done the first part of the algorithm for you when the user you are evaluating has chosen "rock".

In code, this would look like:

rock=0, paper=0, scissors=0, lizard=0, spock=0, win=0, tie=0
if ( someone chose rock ) rock=1
if ( someone chose paper ) paper=1
if ( someone chose scissors ) scissors=1
if ( someone chose lizard ) lizard=1
if ( someone chose spock ) spock=1

// Check if tie / draw, double check these, but I think I got them all
tie=rock && !paper && spock && lizard || rock && paper && scissors ||  
    rock && paper && lizard || spock && paper && scissors || 
    spock && !rock && paper && lizard || !spock && scissors && lizard && paper

if ( tie ) die()

CheckIfUserWins() {
  if ( user chose rock ) {
    win=rock && !paper && !spock
  if ( user chose paper) {
    // ....  calculate using k-map and fill in

}

return win

Notice that win=rock && !paper && !spock is exactly what would be expected based on the graphic of what beats what at the link you provided. So you can go to that graphic and pretty quickly fill in the rest of the equations.

This solution is not dependent on any number of players other than to say "someone chose X". So it should scale to > 5 players, etc.