Question

I m trying to figure out a solution for Problem 27 of 99 Haskell questions.
Here s how I want it to be:

  group :: (Eq a) => [Int] -> [[a]] -> [[[[a]]]]
  group []     _  = []
  group (i:is) xs 
    | sum (i:is) /= length xs = error "invalid arguments"
    | otherwise               = ...

联系的一个例子:

group [2,2,5] ["aldo","beat","carla","david","evi","flip","gary","hugo","ida"]
[[["aldo","beat"],["carla","david"],["evi","flip","gary","hugo","ida"]],...] (altogether 756 solutions)

因此,我想首先检查一下Int名单的总和是否与上述长处清单相同。我认为,无论这两种价值观是否平等,都永远不印刷“无效的论点”。我也试图这样做:

group (i:is) xs 
     | (sum (i:is) == length xs) = ...
     | otherwise                 = error "invalid arguments"

still doesn t work
Any ideas?

<>UPDATES 我的无能为力。职能中可收回的部分:

 group (i:is) xs 
     | (sum (i:is) == length xs) = filter (/= []) $ concatGroups (combinations i xs) (group is xs)
     | otherwise                 = error ("invalid arguments: " ++ show (sum(i:is)) ++ "/=" ++ show(length xs))

如你所知,<代码>组为xs,减少数额,但缩短时间,因此在休补时总是会提出申诉。我认为,我只想去掉这一警卫,希望使用者永远不会这样做。

Answer 1

如果你把你的法典改成黑帮的话,扩大黑帮。

group :: (Eq a) => [Int] -> [[a]] -> [[[[a]]]]
group []     _  = []
group (i:is) xs 
  | sum (i:is) /= length xs = error ("invalid arguments to group: sum "
                                     ++ show (i:is) ++ " /= " ++ show (length xs))
  | otherwise               = ...

这将有助于你追踪<条码>......部分的错误。

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