Haskell type checker complaint 184 of 698
I want to build an adjacency matrix for the vertices of a cube; this
is a matrix that has m[a][b] = 1 exactly when vertices a and b
share an edge. We can enumerate the vertices arbitrarily but a
convenient way to do it is to assign them the numbers 0 through 7 and
then say that vertices !!a!! and !!b!! are adjacent if, regarded as
binary numerals, they differ in exactly one bit, so:
import Data.Bits
a `adj` b = if (elem (xor a b) [1, 2, 4]) then 1 else 0
This compiles and GHC infers the type
adj :: (Bits a, Num a, Num t) => a > a > t
Fine.
Now I want to build the adjacency matrix, which is completely
straightforward:
cube = [ [a `adj` b  b < [0 .. 7] ]  a < [0 .. 7] ] where
a `adj` b = if (elem (xor a b) [1, 2, 4]) then 1 else 0
Ha ha, no it isn't; in Haskell nothing is straightforward. This
produces 106 lines of type whining, followed by a failed compilation.
Apparently this is because because 0 and 7 are overloaded, and
could mean some weird values in some freakish instance of Num , and
then 0 .. 7 might generate an infinite list of 1graded torsion
rings or something.
To fix this I have to say explicitly what I mean by 0 . “Oh, yeah,
by the way, that there zero is intended to denote the integer zero,
and not the 1graded torsion ring with no elements.”
cube = [ [a `adj` b  b < [0 :: Integer .. 7] ]  a < [0 .. 7] ] where
a `adj` b = if (elem (xor a b) [1, 2, 4]) then 1 else 0
Here's another way I could accomplish this:
zero_i_really_mean_it = 0 :: Integer
cube = [ [a `adj` b  b < [zero_i_really_mean_it .. 7] ]  a < [0 .. 7] ] where
a `adj` b = if (elem (xor a b) [1, 2, 4]) then 1 else 0
Or how about this?
cube = [ [a `adj` b  b < numbers_dammit [0 .. 7] ]  a < [0 .. 7] ] where
p `adj` q = if (elem (xor p q) [1, 2, 4]) then 1 else 0
numbers_dammit = id :: [Integer] > [Integer]
I think there must be something really wrong with the language design
here. I don't know exactly what it is, but I think someone must have
made the wrong tradeoff at some point.
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