Haskell中的高阶函数谜题求解函数

我正在尝试用金字塔重建一个谜语：

金字塔的最后一层是数字从1到n的排列，其中n是字段数。然后，不在最低层中的所有其他字段都是该数字对角线下方数字的总和。

所以我想做一个函数，当给定左边的谜语时，返回右边的解。我打算通过枚举类似这样的层来做到这一点：

对于各层，我做了一个自定义数据类型：

data Layer = One | Two | Three | Four | Five | Six
deriving (Eq,Ord,Enum,Show)

和其他类型：

type LayerDepth = Layer
type FieldNumber = Int
type FieldContent = Int
type FieldAssignment = (FieldNumber -> FieldContent)
type PyramidRiddle = FieldAssignment
type PyramidSolution = FieldAssignment
type PyramidSolutions = [PyramidSolution]

和功能：

solveRiddle :: LayerDepth -> PyramidRiddle -> Pyramidsolutions

对于所示的示例，我将创建类型为（FieldNumber-> FieldContent）的匿名函数：

fieldAssignment1 = \n -> if (n==6) || n==8) then 3 else 0

此功能将用数字3标记第6和第8字段

然后调用： solveRiddle Four fieldAssignment1 ->> [pyramidSolution1, pyramidSolution2]

Four means four layers, and PyramidSolutions is the list of FieldAssignments that has a solutions of the riddle

My problem:

I would somehow need to return a function that will given the field assignment calculate the permutations in the last layer and based on that assign the numbers to the rest of fields.

Somehow like this:

pyramidSolution1 = \n -> case n of 1 -> 18
                                   2 -> 11 
                                   3 -> 7
                                   4 -> 7 
                                   5 -> 4 
                                   6 -> 3 
                                   7 -> 4 
                                   8 -> 3 
                                   9 -> 1 
                                  10 -> 2
                                   _ -> 0 

and

pyramidSolution2 = \n -> case n of 1 -> 20
                                   2 -> 12 
                                   3 -> 8
                                   4 -> 7 
                                   5 -> 5 
                                   6 -> 3 
                                   7 -> 4 
                                   8 -> 3 
                                   9 -> 2 
                                  10 -> 1
                                   _ -> 0 

But what is the best way to approach this?

How do I assign a permutation of the numbers and know how to place them that the number up is the sum of the numbers bellow?

What is the best way to implement the anonymous function pyramidSolution1 and pyramidSolution2 into the code above?