我不太了解這個函式中減少 Haskell 的內容：
removeone :: Eq a = > a -> [ a ] -> [ a ]
removeone _ [] = []
removeone x ( y : ys )
| x == y = removeone x ys
| otherwise = y : ( removeone x ys )
remdups :: Eq a = > [ a ] -> [ a ]
remdups [] = []
remdups ( x : xs ) = x : remdups ( removeone x xs )
我如何使用此函式來解釋串列中的歸約步驟（例如remdups [3,7,3,7,5,7]）？
uj5u.com熱心網友回復：
從remdups [3,7,3,7,5,7]盡可能向左開始并簡化運算式。請記住，這[x,y,z]是x:y:z:[]. 因此，remdups(3:7:....) = 3 : remdups (removeone 3 (7:....)) = ...和現在你需要使用的定義，removeone因為沒有等式remdups適用。之后removeone生成:或[]，你將回傳簡化了remdups電話。
一排一減。在注釋中是帶有縮減代碼的行號。我的示例使用最外層（惰性）縮減。
remdups [3,7,3,7,5,7]
3:remdups (removeone 3 [7,3,7,5,7]) -- 9
3:remdups (7:(removeone 3 [3,7,5,7])) -- 5
3:7:remdups (removeone 7 (removeone 3 [3,7,5,7])) -- 9
3:7:remdups (removeone 7 (removeone 3 [7,5,7])) -- 4
3:7:remdups (removeone 7 (7:(removeone 3 [5,7]))) -- 5
3:7:remdups (removeone 7 (removeone 3 [5,7])) -- 4
3:7:remdups (removeone 7 (5:(removeone 3 [7]))) -- 5
3:7:remdups (5:(removeone 7 (removeone 3 [7]))) -- 5
3:7:5:remdups (removeone 5 (removeone 7 (removeone 3 [7]))) -- 9
3:7:5:remdups (removeone 5 (removeone 7 (7:(removeone 3 [])))) -- 5
3:7:5:remdups (removeone 5 (removeone 7 (removeone 3 []))) -- 4
3:7:5:remdups (removeone 5 (removeone 7 [])) -- 2
3:7:5:remdups (removeone 5 []) -- 2
3:7:5:remdups [] -- 2
3:7:5:[] -- 8
為完整起見，還有其他減少程式。例如最左邊最里面（貪婪）的歸約。在這個示例中是相同的結果，但例如head $ remdups [1..]貪婪減少會“掛起”。據我所知，最左邊的最里面僅用于理論上或用于記憶體優化：
remdups [3,7,3,7,5,7]
3:remdups (removeone 3 [7,3,7,5,7])
3:remdups (7:(removeone 3 [3,7,5,7]))
3:remdups (7:(removeone 3 [7,5,7]))
3:remdups (7:(7:(removeone 3 [5,7])))
3:remdups (7:(7:(5:(removeone 3 [7]))))
3:remdups (7:(7:(5:(7:(removeone 3 [])))))
3:remdups (7:(7:(5:(7:([])))))
3:remdups [7,7,5,7] -- shorthand
3:7:remdups (removeone 7 [7,5,7])
3:7:remdups (removeone 7 [5,7])
3:7:remdups (5:(removeone 7 [7]))
3:7:remdups (5:(removeone 7 []))
3:7:remdups (5:([]))
3:7:remdups [5] -- shorthand
3:7:5:remdups (removeone 5 [])
3:7:5:remdups ([])
3:7:5:remdups [] -- shorthand
3:7:5:[]
[3,7,5] -- shorthand
uj5u.com熱心網友回復：
在偽代碼中，我們有
rem1 x (x:t) = rem1 x t
rem1 x (a:t) = a:rem1 x t ; rem1 x [] = []
remdups (a:t) = a:remdups (rem1 a t) ; remdups [] = []
最里面-最左邊（熱切評價）：
remdups [3,7,3,7,5,7]
= 3: remdups (rem1 3 [7,3,7,5,7])
= 3: remdups (7:rem1 3 [3,7,5,7])
= 3: remdups (7: rem1 3 [7,5,7])
= 3: remdups (7:7: rem1 3 [5,7])
= 3: remdups (7:7:5: rem1 3 [7])
= 3: remdups (7:7:5:7: rem1 3 [])
= 3: remdups (7:7:5:7: [] )
= 3: 7: remdups (rem1 7 (7:5:7:[]))
= 3: 7: remdups ( rem1 7 (5:7:[]))
= 3: 7: remdups ( 5:rem1 7 (7:[]))
= 3: 7: remdups ( 5: rem1 7 ([]))
= 3: 7: remdups ( 5: [] )
= 3: 7: 5: remdups (rem1 5 [] )
= 3: 7: 5: remdups []
= 3: 7: 5: [] -- 15 reductions
最外層（Haskell 在做什么）：
remdups [3,7,3,7,5,7]
= 3: remdups (rem1 3 [7,3,7,5,7])
= 3: remdups (7:rem1 3 [3,7,5,7])
= 3: 7: remdups (rem1 7 (rem1 3 [3,7,5,7]))
= 3: 7: remdups (rem1 7 ( rem1 3 [7,5,7]))
= 3: 7: remdups (rem1 7 (7: rem1 3 [5,7]))
= 3: 7: remdups (rem1 7 ( rem1 3 [5,7]))
= 3: 7: remdups (rem1 7 ( 5: rem1 3 [7]))
= 3: 7: remdups (5:rem1 7 ( rem1 3 [7]))
= 3: 7: 5: remdups (rem1 5 (rem1 7 ( rem1 3 [7])))
= 3: 7: 5: remdups (rem1 5 (rem1 7 ( 7: rem1 3 [])))
= 3: 7: 5: remdups (rem1 5 (rem1 7 ( rem1 3 [])))
= 3: 7: 5: remdups (rem1 5 (rem1 7 [] ))
= 3: 7: 5: remdups (rem1 5 [] )
= 3: 7: 5: remdups []
= 3: 7: 5: [] -- 15 reductions
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