Recall that for scans and reductions, the operator given must be associative. This means that we can move around the parantheses in an application:
(x `f` y) `f` z == x `f` (y `f` z)
While it is undecidable for any compiler to determine statically whether an arbitrary function
f is associative, it’s not so hard to write a test program that looks for a counterexample to associativity by brute force.
let testassoc 'a [n] (eq: a -> a -> bool) (op: a -> a -> a) (ne: a) (arr: [n]a) =let hom xs = reduce op ne xs let golden = hom arr in (.1) <| loop (ok, i) = (true, 0) while ok && i < n do let (bef, aft) = split i arr in if hom bef `op` hom aft `eq` golden then (ok, i + 1) else (false, i)
It works by splitting the input
arr at every possible point, reducing the two parts separately, combining these partial results with
f, and then comparing them to reducing
arr as a whole. If there is a difference (as determined with
eq), then we have a counterexample.
For example, we can use it to show that subtraction is not associative:
> testassoc (==) (-) 0 [1,0,0,-1] 1i32
This says that reducing
 (the first 1 elements) and
[0,0,-1] (the remaining 3 elements) of the input array, then combining them with
(-), differs from reducing the whole test array in one go. Ergo, something is wrong with the operator. If the operator is indeed associative, then
testassoc returns the size of the array:
> testassoc (==) (+) 0 [0,0,0,2,2,2,2] 7i32
While capable of emitting false positives, in practice this technique works well, in particular when combined with large randomly generated input arrays.