Rock Paper Scissors Recursion

Question

var rockPaperScissors = function(rounds) {
  var outcomes = [];
  var plays = ['rock', 'paper', 'scissors'];
  var playedSoFar = [];

  var combos = function(roundsToGo) {
    // base case
    if (roundsToGo === 0) {
      outcomes.push(playedSoFar.slice());
      return;
    }

    for (var i = 0; i < plays.length; i++) {
      playedSoFar.push(plays[i]);
      combos(roundsToGo - 1);
      playedSoFar.pop();
    }
  };
  combos(rounds);
  return outcomes;
};

console.log(rockPaperScissors(2));

If I take out the slice() from the playedSoFar, outcomes just gets returned as nine empty arrays, instead of all the combinations from a 2 round rock paper scissors game:

[ [ 'rock', 'rock' ],
  [ 'rock', 'paper' ],
  [ 'rock', 'scissors' ],
  [ 'paper', 'rock' ],
  [ 'paper', 'paper' ],
  [ 'paper', 'scissors' ],
  [ 'scissors', 'rock' ],
  [ 'scissors', 'paper' ],
  [ 'scissors', 'scissors' ] ]

Why doesn't this work when I take out the slice() from the playedSoFar?

if I change this line of code to outcomes.push(playedSoFar); it doesn't work for some reason, what is the slice doing — WinchenzoMagnifico, Aug 01 '15 at 20:05
yeah it returns a copy of the array, but i'm not sure why on the first run through, outcomes doesn't have ['rock', 'rock] in it — WinchenzoMagnifico, Aug 01 '15 at 20:09
Why don't you go through the recursion step by step (not a long way to go) and see where the problem lies... — EugenSunic, Aug 01 '15 at 20:13
Yeah I know it works when the slice is there, why doesn't it work when there is no slice? did you even read the bottom part of my question — WinchenzoMagnifico, Aug 01 '15 at 20:16
you can use this also, it will be clearer to you: outcomes=outcomes.concat(playedSoFar); — EugenSunic, Aug 01 '15 at 20:25
@AaditMShah this isn't a duplicate. The question is unclear, but I believe OP is asking how to generate all the possible combinations from two rounds of rock, paper, & scissors. — royhowie, Aug 01 '15 at 22:34

score 1 · Answer 1 · edited May 23 '17 at 10:26

1

.slice() returns a copy of your array, if you don't use .slice() and just push the playedSoFar array then its updated contents will be reflected throughout the scope, therefore they will be reflected in the combos array. More details here.

edited May 23 '17 at 10:26

Community

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1

answered Aug 01 '15 at 20:24

Lucian

644
3
13

thanks this was a lot more helpful – WinchenzoMagnifico Aug 01 '15 at 20:27
No problem. At first, I didn't quite understand what the question was. There are more questions like 'JavaScript pass by reference vs. pass by value' throughout SO if you really want to understand how JavaScript works under the hood. – Lucian Aug 01 '15 at 20:29

score 1 · Answer 2 · answered Aug 01 '15 at 20:25

1

If you want to retain playedSoFar, then this is the way to go:

  outcomes=outcomes.concat(playedSoFar);

answered Aug 01 '15 at 20:25

EugenSunic

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Rock Paper Scissors Recursion

2 Answers2