On a whim today, I decided to take on 4Clojure Problem #178, which gives you as input a collection of 5 strings like "HA" (ace of hearts) or "D2" (2 of diamonds), representing a poker hand, and asks you to write a function that determines the best hand you could play with those cards (straight flush, full house, etc.)
I had already done Problem #128, which asks for a function that will take a string like "HA" or "D2" and return a map with the suit and rank (from 0-12, where 0-8 are 2-10, 9 is Jack, 10 is Queen, 11 is King and 12 is Ace), for example {:suit :heart, :rank 12} for the ace of hearts. So I simply reused my solution for that problem within a let binding.
I should note that this solution is presented here as a single function only because that’s the way you have to write it in order for 4Clojure to test it. Otherwise, I probably would have written each part of the let binding as a def or defn at the top level. Nonetheless, I think this is a good example of how easy it is to write concise, readable code in Clojure. Enjoy!
(fn best-hand [card-strings]
(let [card-parser (fn [[s r]]
(let [suit ({\S :spade, \H :heart,
\D :diamond, \C :club} s)
rank (if (> (Character/digit r 10) -1)
(- (Character/digit r 10) 2)
({\T 8, \J 9,
\Q 10, \K 11, \A 12} r))]
{:suit suit, :rank rank}))
cards (map card-parser card-strings)
suits (map :suit cards)
ranks (map :rank cards)
flush?
(if (= 1 (count (set suits))) :flush nil)
straight?
(let [aces-high (sort ranks)
aces-low (sort (replace {12 -1} ranks))]
(if (or
(= aces-high (take 5 (iterate inc (first aces-high))))
(= aces-low (take 5 (iterate inc (first aces-low)))))
:straight
nil))
straight-flush?
(if (and flush? straight?) :straight-flush nil)
pair?
(if (some (fn [[r num]] (>= num 2)) (frequencies ranks))
:pair
nil)
three-of-a-kind?
(if (some (fn [[r num]] (>= num 3)) (frequencies ranks))
:three-of-a-kind
nil)
four-of-a-kind?
(if (some (fn [[r num]] (= num 4)) (frequencies ranks))
:four-of-a-kind
nil)
two-pair?
(if (or
(some (fn [[r num]] (>= num 4)) (frequencies ranks))
(= 2 (count (filter (fn [[r num]] (>= num 2))
(frequencies ranks)))))
:two-pair
nil)
full-house?
(if (and
(some (fn [[r num]] (= num 3)) (frequencies ranks))
(some (fn [[r num]] (= num 2)) (frequencies ranks)))
:full-house
nil)
possible-hands
(remove nil? [straight-flush? four-of-a-kind? full-house? flush?
straight? three-of-a-kind? two-pair? pair?])]
(if-not (empty? possible-hands)
(first possible-hands)
:high-card)))I’d be curious to see your own solutions to this problem, either in Clojure or any other language!