我想对集合进行一些改组,每次我的程序运行时它们都是一样的:
这是一种实现方法:
(def colours ["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"])
(defn colour-shuffle [n]
(let [cs (nth (clojure.math.combinatorics/permutations colours) n)]
[(first cs) (drop 1 cs)]))
; use (rand-int 40320) to make up numbers, then hard code:
(def colour-shuffle-39038 (colour-shuffle 39038))
(def colour-shuffle-28193 (colour-shuffle 28193))
(def colour-shuffle-5667 (colour-shuffle 5667))
(def colour-shuffle-8194 (colour-shuffle 8194))
(def colour-shuffle-13895 (colour-shuffle 13895))
(def colour-shuffle-2345 (colour-shuffle 2345))
colour-shuffle-39038 ; ["white" ("magenta" "blue" "green" "cyan" "yellow" "red" "black")]
但是评估需要一些时间,而且看起来很浪费而且很微不足道。
是否有某种方法可以直接生成洗牌39038,而无需生成和使用所有序列?
(我已经意识到我可以对它们进行硬编码,或者将精力花在使用宏进行编译时。这似乎也有些垃圾。)
最佳答案
听起来像您想number permutations:
(def factorial (reductions * 1 (drop 1 (range))))
(defn factoradic [n] {:pre [(>= n 0)]}
(loop [a (list 0) n n p 2]
(if (zero? n) a (recur (conj a (mod n p)) (quot n p) (inc p)))))
(defn nth-permutation [s n] {:pre [(< n (nth factorial (count s)))]}
(let [d (factoradic n)
choices (concat (repeat (- (count s) (count d)) 0) d)]
((reduce
(fn [m i]
(let [[left [item & right]] (split-at i (m :rem))]
(assoc m :rem (concat left right)
:acc (conj (m :acc) item))))
{:rem s :acc []} choices) :acc)))
让我们尝试一下:
(def colours ["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"])
(nth-permutation colours 39038)
=> ["white" "magenta" "blue" "green" "cyan" "yellow" "red" "black"]
...就像问题中一样,但没有生成任何其他排列。
很好,但是我们能全部得到吗?
(def x (map (partial nth-permutation colours) (range (nth factorial (count colours)))))
(count x)
=> 40320
(count (distinct x))
=> 40320
(nth factorial (count colours))
=> 40320
请注意,排列是按(按索引的字典顺序)顺序生成的:
user=> (pprint (take 24 x))
(["red" "blue" "green" "yellow" "cyan" "magenta" "black" "white"]
["red" "blue" "green" "yellow" "cyan" "magenta" "white" "black"]
["red" "blue" "green" "yellow" "cyan" "black" "magenta" "white"]
["red" "blue" "green" "yellow" "cyan" "black" "white" "magenta"]
["red" "blue" "green" "yellow" "cyan" "white" "magenta" "black"]
["red" "blue" "green" "yellow" "cyan" "white" "black" "magenta"]
["red" "blue" "green" "yellow" "magenta" "cyan" "black" "white"]
["red" "blue" "green" "yellow" "magenta" "cyan" "white" "black"]
["red" "blue" "green" "yellow" "magenta" "black" "cyan" "white"]
["red" "blue" "green" "yellow" "magenta" "black" "white" "cyan"]
["red" "blue" "green" "yellow" "magenta" "white" "cyan" "black"]
["red" "blue" "green" "yellow" "magenta" "white" "black" "cyan"]
["red" "blue" "green" "yellow" "black" "cyan" "magenta" "white"]
["red" "blue" "green" "yellow" "black" "cyan" "white" "magenta"]
["red" "blue" "green" "yellow" "black" "magenta" "cyan" "white"]
["red" "blue" "green" "yellow" "black" "magenta" "white" "cyan"]
["red" "blue" "green" "yellow" "black" "white" "cyan" "magenta"]
["red" "blue" "green" "yellow" "black" "white" "magenta" "cyan"]
["red" "blue" "green" "yellow" "white" "cyan" "magenta" "black"]
["red" "blue" "green" "yellow" "white" "cyan" "black" "magenta"]
["red" "blue" "green" "yellow" "white" "magenta" "cyan" "black"]
["red" "blue" "green" "yellow" "white" "magenta" "black" "cyan"]
["red" "blue" "green" "yellow" "white" "black" "cyan" "magenta"]
["red" "blue" "green" "yellow" "white" "black" "magenta" "cyan"])
关于random - 我可以在Clojure中进行确定性的随机播放吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14836414/