n計算串列與自身時間的笛卡爾積的簡單方法是什么？
也就是說，我該如何定義函式cartesianExp :: Int -> [a] -> [[a]]。
例如，[1,2]與自身 3 次 ( n = 3) 的笛卡爾積應該是：
[
[1, 1, 1],
[1, 1, 2],
[1, 2, 1],
[1, 2, 2],
[2, 1, 1],
[2, 1, 2],
[2, 2, 1],
[2, 2, 2]
]
uj5u.com熱心網友回復：
你可以合作replicateM :: Applicative f => Int -> f a -> f [a]來做到這一點。確實：
ghci> import Control.Monad(replicateM)
ghci> replicateM 0 "abc"
[""]
ghci> replicateM 1 "abc"
["a","b","c"]
ghci> replicateM 2 "abc"
["aa","ab","ac","ba","bb","bc","ca","cb","cc"]
ghci> replicateM 3 "abc"
["aaa","aab","aac","aba","abb","abc","aca","acb","acc","baa","bab","bac","bba","bbb","bbc","bca","bcb","bcc","caa","cab","cac","cba","cbb","cbc","cca","ccb","ccc"]
ghci> replicateM 3 [1,2]
[[1,1,1],[1,1,2],[1,2,1],[1,2,2],[2,1,1],[2,1,2],[2,2,1],[2,2,2]]
uj5u.com熱心網友回復：
import Data.List
cartesianExp :: Int -> [a] -> [[a]]
cartesianExp 0 _ = [[]]
cartesianExp n xs = [x:tup | x <- xs, tup <- cartesianExp (n - 1) xs]
按照慣例，集合S與自身 0 次的笛卡爾積S 0是由單個空元組組成的集合。
然后，集合S與其自身n次的笛卡爾積S n可以通過將S n-1中的每個元組擴展為S中的每個元素來歸納定義。
（這里我使用術語集和元組的數學含義，在 Haskell 實作中，它們都對應于這種情況下的串列。）
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標籤：哈斯克尔