请补充上述evaluate-hand的代码
时间: 2024-03-18 16:39:13 浏览: 16
好的,以下是一个简单的evaluate-hand函数的示例,它可以评估德州扑克中的手牌强度:
```clojure
(defn evaluate-hand [hand board]
(let [cards (concat hand board)
suits (map first cards)
ranks (map second cards)
count-ranks (frequencies ranks)
flush-suit (some #(if (>= % 5) %) (vals (frequencies suits)))]
(cond
(and flush-suit (= (count board) 5))
;; Flush
(+ 500 (apply max ranks))
(some #(= % 4) count-ranks)
;; Four-of-a-kind
(+ 400 (first (->> count-ranks (filter #(= (val %) 4)) first)))
(and (some #(= % 3) count-ranks) (some #(= % 2) count-ranks))
;; Full house
(+ 300 (* 10 (first (->> count-ranks (filter #(= (val %) 3)) first)))
(first (->> count-ranks (filter #(= (val %) 2)) first)))
(and flush-suit (= (count board) 5))
;; Flush
(+ 200 (apply max ranks))
(let [straight-ranks (->> ranks (distinct) sort (partition-by #(apply - %) identity) (filter #(>= (count %) 5)) first)]
(when straight-ranks
(if (= (count straight-ranks) 1)
;; Straight
(+ 100 (last straight-ranks))
;; Straight flush
(+ 800 (last straight-ranks)))))
(some #(= % 3) count-ranks)
;; Three-of-a-kind
(+ 50 (first (->> count-ranks (filter #(= (val %) 3)) first)))
(let [pairs (->> count-ranks (filter #(= (val %) 2)) keys)]
(cond
(= (count pairs) 2)
;; Two pairs
(+ 30 (* 10 (apply max pairs)) (apply min pairs))
(= (count pairs) 1)
;; One pair
(+ 10 (first pairs))
:else
;; High card
(apply max ranks))))))
```
该函数首先将手牌和公共牌组合成一副牌,并将其分成花色和点数两个部分。然后,它计算每个点数的出现次数,并检查是否有同花色的牌。根据规则集,它为每种情况分配一个分数,并返回最终的分数。