2

I'm supposed to create a method in ruby that will take in a structured,multi-dimensioned array, such as:

my_arr = [
[ [['Armando', 'P'], ['Dave', 'S']], [['Richard', 'R'], ['Michael', 'S']] ],
[ [['Allen', 'S'], ['Omer', 'P']], [['David E.', 'R'], ['Richard X.', 'P']] ]
]

This array is supposed to represent a tournament of Rock, paper & scissors, the number of players will always be 2^n and no repetitions (of players) are made.

The code I wrote is as follows:

class WrongNumberOfPlayersError < StandardError ; end
class NoSuchStrategyError < StandardError ; end

def rps_game_winner(game)
  raise WrongNumberOfPlayersError unless game.length == 2
  valid = ["r","p","s"]
  a1=[(game[0][1]).downcase]
  a2=[(game[1][1]).downcase]
  raise NoSuchStrategyError unless (valid & a1) && (valid & a2)

  return (game[0]) if a1 === a2
  case (a1[0])
  when "r"
    case (a2[0])
    when "p"
      return (game[1])
    else 
      return (game[0])
    end
  when "p"
    case (a2[0])
    when "s"
      return (game[1])
    else
      return (game[0])
    end
  when "s"
    case (a2[0])
    when "r"
      return (game[1])
    else 
      return (game[0])
    end
  end
end

def rps_tournament_winner(tournament)
  if tournament[0][0].is_a?(Array)
    rps_tournament_winner(tournament[0])
  elsif tournament[1][0].is_a?(Array)
    rps_tournament_winner(tournament[1])
  else
    rps_game_winner(tournament)
  end
end

So my problem is that given the use of array i mentioned earlier being passed to rps_tournament_winner Dave always wins instead of Richard and i haven't been able to figure out where i went wrong.

Ty for reading the wall of text/code :)

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13
  • 1
    I'd recommend using the :paper,:rock,:scissor instead of strings. It's easier to read the code and faster for the interpreter. Commented Feb 29, 2012 at 17:44
  • the array i'm provided uses the strings like "R" "p", etc. i suppose i could have internally used symbols? but then would have to convert it back to one character strings for the required result format (which is like ['Richard', 'R'] ? Commented Feb 29, 2012 at 17:51
  • Where do you initialize the game array? Commented Feb 29, 2012 at 17:53
  • Could you explain better how the tournament is translated into the array? Why should Richard be the winner? Why is this a recursion problem? Commented Feb 29, 2012 at 17:59
  • 1
    To clean up your case statement you could also do: case [a1[0], a2[0]] when ['r','s'], ['s','p'], ['p','r'], ['r', 'r'], ['p', 'p'], ['s', 's'] game[0] else game[1] end Commented Mar 2, 2012 at 7:27

3 Answers 3

Reset to default
3

One thing I noticed is that your use of 'valid' doesn't do anything to check if your input is actually valid. If you're trying to check that a1 and a2 are either "r" "p" or "s" you should use a regular expression:

valid = /[rps]/    # matches either "r" "p" or "s"
raise NoSuchStrategyError unless (a1 =~ valid) && (a2 =~ valid)

Your array of players is nested very deep. You'll make your life simpler by thinning it out:

my_arr = [['Armando', 'P'], ['Dave', 'S'], ['Richard', 'R'], ['Michael', 'S'],
        ['Allen', 'S'], ['Omer', 'P'], ['David E.', 'R'], ['Richard X.', 'P']]

You can make it easier to read and maintain by breaking your program into parts. For example, create a method for determining a win:

# This is incomplete as it doesn't deal with ties. I'll let you do that part
def win(first, second)
  if (first == "p" && second == "r") || (first == "r" && second == "s") || (first == "s" && second == "p")
    return true
  else
    return false
  end
end

Now it's easier to write and understand the game itself, using the above method:

def rps_game_winner(player1, player2)
  first = player1[1].downcase
  second = player2[1].downcase
  win(first, second) ? player1 : player2  # ternary operator returns the winner
end

You now have a method to put all this into play (sort of the main logic), and we'll use recursion here:

def rps_tournament_winner(player_list)
  round_winners = []    # place to hold the winners for each round

  if player_list.size == 1  # final condition to stop the recursion
    puts "The winner is #{player_list[0][0]}."
  else
    player_list.each_slice(2) do |l1, l2|  # take pairs from your list to play each other
      round_winners << rps_game_winner(l1, l2)
    end

    rps_tournament_winner(round_winners)  # use recursion to play the winners against each other
  end
end

# put it into play
puts test_array(my_arr)

That's it. The winner is Richard and it'll always be Richard because the play is deterministic.

While this will run, you should know that I left out some important things like dealing with ties and odd numbers of players. The program will not work right under these conditions. But I'll leave it up to you to solve those parts.

EDIT: Modifying the original array:

new_array = []
my_arr.flatten.each_slice(2) do |name, pick|
  new_array << [name, pick]
end
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3 Comments

There was a typo that I fixed so it should work now if it didn't before... unless there's another typo I missed. :o
Thank you soo much, i'm new to very new to ruby and coding in general, each of your point is a gem indeed, I will take all of them to heart. As for the number of players, it will always be even (thats how my method is supposed to receive the tournament array) getting the "Tie" logic will be easy. Once again thank you for dissecting such hideous code and for the priceless tips. One last question, how am I to "thin out" my array? maybe Array#flatten
Hi, I'll edit my post to show a way of thinning out your original array. The comment field doesn't make it easy to show code. See above.
1

Your recursion only goes down tournament[0] before ending never tournament[1]. You need to play both tournament[0] and tournament[1] then play them against each other.

Something like. ( I leave it to you write the ruby code)

rps_game_winner( [rps_tournament_winner(tournament[0], rps_tournament_winner(tournament[1]] )

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1 Comment

Yep - you are right the above example produces the wrong result since the winner in the "bracketed array" originally described by Osama Husain - would lead to Richard X. winning. The important point to note is that the league is bracketed.
0

It's very hard to debug your code.

But I think that it is fundamentally flawed.

When you call rps_tournament_winner you pass in

[
  [ [['Armando', 'P'], ['Dave', 'S']], [['Richard', 'R'], ['Michael', 'S']] ],
  [ [['Allen', 'S'], ['Omer', 'P']], [['David E.', 'R'], ['Richard X.', 'P']] ]
]

It asks tournament[0][0].is_a?(Array)

[['Armando', 'P'], ['Dave', 'S']], [['Richard', 'R'], ['Michael', 'S']]

Yes it is.

It calls rps_tournament_winner(tournament[0])

tournament = [ [['Armando', 'P'], ['Dave', 'S']], [['Richard', 'R'], ['Michael', 'S']] ]

It asks tournament[0][0].is_a?(Array)

['Armando', 'P']

Yes it is.

it calls rps_tournament_winner(tournament[0])

tournament = [['Armando', 'P'], ['Dave', 'S']]

It asks tournament[0][0].is_a?(Array)

'Armando'

No its not.

It calls rps_game_winner(tournament)

And in the game Dave beats Armando!

I don't think this is the functionality you intended.

I suggest you rewrite this, trying to keep things simple.

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1 Comment

TY, i figured that after 'Armando' wins, then 'Richard' and 'Michael' battle it out in which 'Richard' wins and gets to beat 'Armando' in the following recursion. Off-topic: can you suggest ways for me to make my code simpler, easier to read and to debug?

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