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Chakrit Likitkhajorn
Chakrit Likitkhajorn

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Design a program in Functional #2 Poker Kata

#functional #scala #architecture

From last article, I demonstrated how to think about program design in functional way.

In this article, I will demonstrate this again using Poker kata. The question is very simple, given two poker hand, determine who is winning.

Start modeling

So at the end, what we need is a function

compare(hand1, hand2) -> Win, Lose, Draw
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And we can model a data shape of hand, card as followed:

object Suit extends Enumeration { type Suit = Value val Heart = Value("Heart") val Spade = Value("Spade") val Club = Value("Club") val Diamond = Value("Diamond") } case class Card(value: Int, suit: Suit.Suit) type Hand = (Card, Card, Card, Card, Card)
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Basically, we just told the compiler that a card is data consists of value as an integer, and a suit which can be either Heart, Spade, Club or Diamond. Then, a hand is a set of 5 cards.

I will represent Jack, Queen, King and Ace with 11, 12, 13 and 14 respectively.

First design

At the end, we want to implement

def comparePokerHand(hand1: Hand, hand2: Hand): CompareResult.CompareResult
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Ok. I will be the first to admit that implement comparePokerHand from the get go will be hard and complicated.

So I need to break it down.

Here comes the magic question: What is a data A in such that?

  1. It is simpler to calculate a hand comparison result from data A.
  2. It is simpler to calculate data A from each hand.

Creative thinking time.....
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Ok, if I have the rank of each hand (Straight, Flush, Two Pairs, etc.) and highs of each hand, then I can simply compare those. I will compare hand power first, and if hand power is the same, then I will compare each highs until the end.

Here is the first breakdown pseudo code.

case class HandPower(rank: HandRank, highs: List[Int]) def comparePokerHand(hand1: Hand, hand2: Hand): CompareResult = { val handPower1 = handPower(hand1) val handPower2 = handPower(hand2) compareHandPower(handPower1, handPower2) }
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Please note that in order to make above statement to be true, the highs must be ordered from the most significant card value to least significant card value.

For example:

Flush(1 3 10 5 4) -> Highs(10 5 4 3 1) // Order from highest value to lowest value TwoPairs(3 3 5 5 10) -> Highs(5 3 10) // The pairs come first, sorted by higher value, then the non-pair FullHouse(5 5 5 10 10) -> Highs(5 10) // Three of kind first, then pair
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If we have this, then we can easily compare highs by looking at each element one-by-one.

For example:

FullHouse(5 5 5 10 10) lose FullHouse(9 9 9 2 2) Highs (5 10) lose to Highs(9 2) 5 < 9 Two Pairs(10 10 14 14 4) win TwoPairs(9 9 14 14 5) Highs(14 10 4) win Highs(14 9 5) 14 == 14 10 > 9
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This is how having a data highs with above properties make it easier to determine a winner.

Having this all said, next I need to implement two functions: HandPower and compareHandPower.

I will go through the easier one first: compareHandPower. This function will be implemented according to what I thought

I will compare hand power first, and if hand power is the same, then I will compare each highs until the end.

def compareRank( handRank1: HandRank.HandRank, handRank2: HandRank.HandRank ): CompareResult.CompareResult = { if (handRank1.id < handRank2.id) return CompareResult.Win if (handRank1.id > handRank2.id) return CompareResult.Lost CompareResult.Draw } def compareHighs( high1: List[Int], high2: List[Int] ): CompareResult.CompareResult = { high1 .zip(high2) .foldLeft(CompareResult.Draw)((result, pair) => { if (result != CompareResult.Draw) { return result } else if (pair._1 > pair._2) { return CompareResult.Win } else if (pair._1 < pair._2) { return CompareResult.Lost } else { return CompareResult.Draw } }) } def compareHandPower( handPower1: HandPower, handPower2: HandPower ): CompareResult.CompareResult = { val rankComparison = compareRank(handPower1.rank, handPower2.rank) if (rankComparison == CompareResult.Draw) { return compareHighs( handPower1.highs, handPower2.highs ) } return rankComparison }
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Easy peasy!! (Hope you find this simple as well). First function done 😄

Then next, the handPower function.

The hand power function must consist of two ability

  • Ability to generate rank from hand
  • Ability to generate highs from hand

I have to admit that at first, I cannot figure out how to break it down.

So I start with a very simple and naive question: How can I determine if this hand is a four card hand?

The answer is simple: I need to check if any card value occurs four times within a hand. That means I need to determine the frequencies of each card value.

So I start implemented a freqMap like this:

val freqMap = hand.foldLeft(Map[Int, Int]())((acc, card: Card) => { acc.get(card.value) match { case None => acc + (card.value -> 1) case Some(currentFreq) => acc + (card.value -> (currentFreq + 1)) } })
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Basically, freqMap is a map between card value and card frequency.

With this, I can write the logic for four cards to be

if (freqMap.exists(keyValuePair => { val cardFrequency = keyValuePair._2 cardFrequency == 4 })) { return HandPower(HandRank.FourCard, highs) }
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Ok. Now I can check four cards. Let's check for the full house.

if ( freqMap.exists(keyValuePair => { val cardFrequency = keyValuePair._2 cardFrequency == 3 }) && freqMap.exists(keyValuePair => { val cardFrequency = keyValuePair._2 cardFrequency == 2 }) ) { return HandPower(HandRank.Fullhouse, highs) }
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Now I realize that things start to get messy and duplicates.

Back to the core question: Is there any data A that will make our life easier?

Answer: Yes. If we have data A that is a list of frequencies of each card value sorted, we can easily determine many card ranks.

How? I think it is easier to show you by this code:

cardFrequenciesPattern match { case 1 :: 4 :: Nil => return HandPower(HandRank.FourCard, highs) case 2 :: 3 :: Nil => return HandPower(HandRank.Fullhouse, highs) case 1 :: 1 :: 3 :: Nil => return HandPower(HandRank.ThreeOfKind, highs) case 1 :: 2 :: 2 :: Nil => return HandPower(HandRank.TwoPairs, highs) case 1 :: 1 :: 1 :: 2 :: Nil => return HandPower(HandRank.OnePair, highs) case _ => { // Something } }
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You can see that if a cardFrequenciesPattern exists, and it is a frequencies of each card value sorted, we can easily match it to a corresponding pattern.

So the data A for this case is cardFrequenciesPatter.

I wrote a transformation from current freqMap to easier cardFrequenciesPattern as followed:

val cardFrequenciesPattern = freqMap.values.toList.sorted
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What's left for me is to write some if-else for straight, flush and nothing in the last case.

Since we were able to determine a rank, let us determine highs.

Definition: highs is a list of card values in hand, ordered from the most significant card value to least significant card value.

I notice that the most significant card values in any hand are also the most frequent card in hand. For example: In full house hand, three kinds of cards are considered to be more significant than a pair of cards.

So the logic could be order card values based on frequency first, then order by the values itself.

And since we already have a map of card frequency and card value, then here it is:

val highs = freqMap.toSeq .sortWith((pair1, pair2) => { val cardVal1 = pair1._1 val cardVal2 = pair2._1 val cardFreq1 = pair1._2 val cardFreq2 = pair2._2 if (cardFreq1 == cardFreq2) { cardVal1 > cardVal2 } else { cardFreq1 > cardFreq2 } }) .map(_._1) .toList
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Surprisingly, the intermediate freqMap data make it very easy to determine both highs and rank.

Now we got everything: We have a function to create a hand power from a hand. We have a function to compare two hand power. That's mean we solved Poker Kata.

Done deal! 🎉🎉

Full implementation in Scala: https://github.com/chrisza4/scala-poker-kata

Full implementation in Clojure: https://github.com/chrisza4/clojure-poker-kata

Recap and Lesson

In this article, I demonstrate the thought process and program design that repeatedly ask a simple question: Is there data A?

And by asking ourselves this question, we found that:

  • Intermediate data HandPower make it way easier to compare poker hand easier.
  • Intermediate data freqMap make it way easier to both create a rank and highs.
  • Intermediate data cardFrequenciesPattern make it way easier to determine a card rank

This is the functional thinking. That is how we breakdown a problem in a functional paradigm. This is what it means to "write more functions".

Again, as I wrote in the previous article, this thinking and thought process line is useful even if you write a program in an imperative or object-oriented paradigm.

So I hope this article is useful for every programmer.

Thanks for reading!!

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