.each not iterating over nested arrays

Question

For some reason, this is not iterating over the collection. Throwing a pry in there shows that only the first index is getting looked at. [0,1,2] I'm sure I'm just tired and it's something small, but I'm drawing a black here,

WIN_COMBINATIONS = [
  [0, 1, 2],
  [3, 4, 5],
  [6, 7, 8],
  [0, 3, 6],
  [1, 4, 7],
  [2, 5, 8],
  [0, 4, 8],
  [2, 4, 6]
]

def won?(board)
  WIN_COMBINATIONS.each do |combo|
    if board[combo[0]] != " " && board[combo[0]] == board[combo[1]] && board[combo[0]] == board[combo[2]]
      return combo
    else
      return false
    end
  end
end

@MarekLipka it's an array. board = ["O", "O", " ", "X", "X", "X", " ", " ", " "] in the test is where it's getting passed in. — Int'l Man Of Coding Mystery
– Int'l Man Of Coding Mystery, Commented Jan 2, 2018 at 15:09

Aleksei Matiushkin · Accepted Answer · 2018-01-03 08:19:12Z

5

def won?(board)
  WIN_COMBINATIONS.any? do |combo|
    board[combo[0]] != " " &&
      board[combo[0]] == board[combo[1]] &&
      board[combo[0]] == board[combo[2]]
  end
end

or, even more succinct with Array#values_at:

def won?(board)
  WIN_COMBINATIONS.any? do |combo|
    [%w|X| * 3, %w|O| * 3].include? board.values_at(*combo)
  end
end

edited Jan 3, 2018 at 8:19

answered Jan 2, 2018 at 15:17

Aleksei Matiushkin

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7 Comments

Shimu Over a year ago

@mike-heft In your actual code you should use this. This is more idiomatic ruby

Cary Swoveland Over a year ago

...or arr = board.values_at(*combo).uniq; arr.size == 1 && arr.first != ' '.

Aleksei Matiushkin Over a year ago

@CarySwoveland you might want to check the previous version of this answer :)

Cary Swoveland Over a year ago

Ah, yes. I prefer your previous revision. Adding a variable and a second statement to remove the select clause is imo a wash, but mainly, I think the earlier one reads better.

Stefan Pochmann Over a year ago

You can make your second version even more succinct: def won?(board) false end :-P

|

Shimu · Accepted Answer · 2018-01-02 15:13:28Z

4

The reason is that you return from your method on the first iteration.
Inside your iteration you always return after the first step. Just put the return false statement after your iteration and it will work.

answered Jan 2, 2018 at 15:13

Shimu

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1 Comment

Int'l Man Of Coding Mystery Over a year ago

Yup that was it. Was just coming back to answer Thank you :D

Stefan Pochmann · Accepted Answer · 2018-01-03 01:17:44Z

1

Just three more solutions along with test code checking them against each other:

def won1?(board)
  WIN_COMBINATIONS.any? do |i, j, k|
    board[i] != " " && board[i] == board[j] && board[i] == board[k]
  end
end

def won2?(board)
  WIN_COMBINATIONS.any? do |combo|
    a, b = board.values_at(*combo).minmax; a != " " && a == b
  end
end

def won3?(board)
  WIN_COMBINATIONS.any? do |combo|
    [["X"], ["O"]].member?(board.values_at(*combo).uniq)
  end
end

WIN_COMBINATIONS = [
  [0, 1, 2],
  [3, 4, 5],
  [6, 7, 8],
  [0, 3, 6],
  [1, 4, 7],
  [2, 5, 8],
  [0, 4, 8],
  [2, 4, 6]
]

a = ["O", "X", " "]
a.product(*([a] * 8)).each { |board|
  w1 = won1?(board)
  w2 = won2?(board)
  w3 = won3?(board)
  p [w1, w2, w3, board] if w2 != w1 || w3 != w1
}

answered Jan 3, 2018 at 1:17

Stefan Pochmann

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1 Comment

Mark Thomas Over a year ago

Another: WIN_COMBINATIONS.any? {|combo| ["X", "O"].one? {|token| combo.all? {|i| board[i] == token}}

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