Rock, Paper, Scissors. Determine win/loss/tie using math?

Question

So I was writing a rock paper scissors game when I came to writing this function:

a is player one's move, b is player two's move. All I need to figure out is if player one won, lost, or tied.

//rock=0, paper=1, scissors=2
processMove(a, b) {
    if(a == b) ties++;
    else {
             if(a==0 && b==2) wins++;
        else if(a==0 && b==1) losses++;
        else if(a==1 && b==2) losses++;
        else if(a==1 && b==0) wins++;
        else if(a==2 && b==1) wins++;
        else if(a==2 && b==0) losses++;
    }
}

My question is: What's the most elegant way this function can be written?

Edit: I'm looking for a one-liner.

Distillery

Looking for readable correct code is much better than looking for one-liners. – Obligee

Have a look at my answer to a similar question: https://mcmap.net/q/1116288/-scalable-solution-for-rock-paper-scissor – Capsulate

Answer 1

if (a == b) ties++;
else if ((a - b) % 3 == 1) wins++;
else losses++;

I need to know exactly which language you are using to turn it into a strictly one-liner...

For JavaScript (or other languages with strange Modulus) use:

if (a == b) ties++;
else if ((a - b + 3) % 3 == 1) wins++;
else losses++;

Comedic

rock vs scissors doesn't work: (0 - 2 % 3) = -2. Using javascript – Distillery

Looks like it's missing a pair of parenthesis: ((a-b) % 3 == 1). In C operators modulo is higher in order of operations than subtraction. – Trigeminal

It's not a strange modulus its: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… > While in most languages, % is a remainder operator, in some (e.g. Python, Perl) it is a modulo operator. For two values of the same sign, the two are equivalent, but when the operands are of different signs, the modulo result always has the same sign as the divisor, while the remainder has the same sign as the dividend, which can make them differ by one unit of the divisor. – Turpin

For languages where % is a remainder operator, use ((n % d) + d) % d or in this case (((a - b) % 3) + 3) % 3 – Turpin

Answer 2

A 3x3 matrix would be "more elegant", I suppose.

char result = "TWLLTWWLT".charAt(a * 3 + b);

(Edited: Forgot that a and b were already zero-origin.)

Amply

Answer 3

I suppose you could use the terniary operator like this -

if (b==0) a==1? wins++ : loss++;

if (b==1) a==1? loss++ : wins++;

if (b==2) a==1? loss++ : wins++;

Crookes

Answer 4

You can do it with a simple mathematical formula to get the result and then compare with if like this:

var moves = {
  'rock': 0, 
  'paper': 1,
  'scissors': 2
};
var result = {
  'wins': 0,
  'losses': 0,
  'ties': 0
};
var processMove = function (a, b) {
  var processResult = (3 + b - a) % 3;
  if (!processResult) {
    ++result['ties'];
  } else if(1 == processResult) {
    ++result['losses'];
  } else {
    ++result['wins'];
  }
  return result;
};

jsFiddle Demo

---

One line processMove function without return:

var processMove = function (a, b) {
  ((3 + b - a) % 3) ? 1 == ((3 + b - a) % 3) ? ++result.losses : ++result.wins : ++result.ties;
};

Hallway

Answer 5

how do you do it in java?

result = (comp - x ) % 3 ;

System.out.println (result);
 if (result == 0 )// if the game is tie
 {
     System.out.println ("A Tie!") ;
 }

 else if (result == 1 || result == 2 )
 {
    //System.out.println (user + " " +   "beats" + " " + computer_choice + " you win" );
     System.out.println ("comp win");
 }

 else
 {
     System.out.println ("you win");
    //System.out.println (computer_choice  + " " +  "beats" + " " + user + "you lose");
 }

Dacy

Answer 6

Maybe this will come in handy someone (JavaScript):

if (a == b) ties++;
else if ((a - b) == 1 || (b - a) == 2) losses++;
else wins++;

PHP:

function check($a, $b) {
    if ($a == $b) {
        return "Tie";
    }
    if (($a - $b) == 1 || ($b - $a) == 2) {
        return "You Lose";
    } else 
        return "You Win";
}

Superposition