From: Ruby Quiz Date: 2005-11-25T00:24:39+09:00 Subject: [SUMMARY] Euchre Hands (#55) My wife can't stand how Euchre reorders the suits. "Everyone know Jacks come after Tens," she would say. Funny how most of us except the odd Ace concept where the low card is frequently the high card, or even both. Once you start moving Jacks though, a lot of people have trouble following the game. Of course, my wife would also say, "How can you call that Jack a Club (assuming Clubs are Trump), when it has Spades printed on it?" She's got me there. My wife's analysis of Euchre's card order is really right on: It's unusual. Because of that, the traditional sort() tool won't get you all the way to a solution. That makes us think about other options. Don't get me wrong, many solutions bent sort() to their will. However, as I said in the quiz, I was looking as much at how to know the answer is correct as I was at just having an answer. To me, the weighted sorts are hard to prove and no one did it to my satisfaction. Given that, I want to look at the other major approach people hit on. The following is the beginning of Bob Showalter's solution: #!/usr/local/bin/ruby -w # Euchre hand sorter CARDS = %w(A K Q J T 9) suits = %w(Spades Hearts Clubs Diamonds) # read and check input data trump = gets.chomp raise "Invalid Trump Suit" unless suits.include? trump hand = [] 5.times do card = gets.chomp raise "Invalid card #{card}" unless card.length == 2 && CARDS.include?(card[0,1]) && suits.find { |suit| suit[0,1].downcase == card[1,1] } raise "Duplicate card #{card}" if hand.include? card hand << card end # ... Bob begins by defining the traditional card order, and the four suits. The next chunk of code reads the trump suit and hand into variables. This code is careful to ensure proper input. Suits must be one of the known four and cards must be two characters, have a known face and suit, and not be duplicated. # ... # rotate trump suit to front suits.push(suits.shift) while suits.first != trump # if hand is void in second suit, swap second and fourth # (this keeps output in alternating colors) unless hand.find { |card| card[1,1] == suits[1][0,1].downcase } suits[1], suits[3] = suits[3], suits[1] end # ... This chunk of code does two things. First, the suits are "rotated" until trump is the first suit. Note that this is why "CARDS" was declared as a constant but "suits" is just a variable. Good hints for the reader in this code. The second thing the code does is the trick many hit on to defeat my extra red/black requirement. If we have four suits ordered red, black, red, black (or black, red, black, red), the only edge case we can do anything about is when the player doesn't have the second suit. Here's the mapping of possible orderings (assuming Spades is trump) to show what I mean: all four suits (black, red, black, red ordering works fine) Spades, Hearts, and Clubs (already correct) Hearts, Clubs, and Diamonds (already correct) Spades, Hearts, and Diamonds (can't move trump to fix) Spades, Clubs, and Diamonds (the fixable case) just two suits (moves won't change anything) a single suit (already correct) In the fixable case, swapping the (missing) second and (present) fourth suits resolves the issue. That's what you're seeing in the code above. # ... # generate a sort order deck = [] suits.each do |suit| CARDS.each do |card| deck << card + suit[0,1].downcase end end # ... Given that the correct suit order has been established, the entire Euchre deck is built, in traditional card order. # ... # move bowers to front deck.insert(0, deck.delete_at(3)) deck.insert(1, deck.delete_at(15)) # ... With the full deck in proper suit order, the only thing that remains is to shift the bowers to the head of the pack. Their location is assured, thus the use of magic numbers above, though index() is an alternative. # ... # output sorted hand (n.b. Array#& (intersection) seemed to work, but # is the order guaranteed?) puts trump puts deck.select { |card| hand.include? card } With the work out of the way, showing the results is trivial. Print trump and then just walk the deck printing any card in the hand. I think that solution is very straight-forward and easy to accept as correct, even without a large group of tests. Let's look at one more variation of the same theme, by Dominik Bathon: order = { "Spades" => "JsJcAsKsQsTs9sAdKdQdJdTd9dAcKcQcTc9cAhKhQhJhTh9h", "Hearts" => "JhJdAhKhQhTh9hAsKsQsJsTs9sAdKdQdTd9dAcKcQcJcTc9c", "Clubs" => "JcJsAcKcQcTc9cAhKhQhJhTh9hAsKsQsTs9sAdKdQdJdTd9d", "Diamonds" => "JdJhAdKdQdTd9dAcKcQcJcTc9cAhKhQhTh9hAsKsQsJsTs9s" } trump = gets.strip cards = readlines.map { |l| l.strip } o = order[trump].dup # do we have a card of the 2nd suit unless cards.any? { |card| card[1] == o[15] } # if not replace second suit by the last o[14, 12] = o[36, 12] end puts trump, cards.sort_by { |card| o.index(card) } This is basically the same thing, with prebuilt decks. The code here just selects which of the four possible decks to use base on the trump indication in the input. Again we see the second and fourth suits swapped, if needed. This time it is done with a simple String copy. Finally, cards are sorted, by their position in the complete deck. My thanks to all the card sharks who solved the quiz and even taught me alternate hand ordering schemes. Tomorrow we're off to the Pinewood Derby, so everyone pick up a car kit sometime this evening...