I am trying to recreate a riddle with pyramid:
The last layer of the pyramid is the permutation of the numbers from 1 to n, where n is the number of fields. Then every other field that is not in the lowest layer is the sum of the numbers diagonally under that number.
So I want to make a function that when given the riddle on the left returns solutions on the right. I was planning to do that by enumerating layers like this:
for the layers I made a custom data type:
data Layer = One | Two | Three | Four | Five | Six
deriving (Eq,Ord,Enum,Show)
and other types:
type LayerDepth = Layer
type FieldNumber = Int
type FieldContent = Int
type FieldAssignment = (FieldNumber -> FieldContent)
type PyramidRiddle = FieldAssignment
type PyramidSolution = FieldAssignment
type PyramidSolutions = [PyramidSolution]
and the function:
solveRiddle :: LayerDepth -> PyramidRiddle -> Pyramidsolutions
for the illustrated example I would make anonymous function of type (FieldNumber -> FieldContent):
fieldAssignment1 = \n -> if (n==6) || n==8) then 3 else 0
This function will mark 6th and 8th field with number 3
Then calling:
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?
