Time for me to reveal the example I had in mind that led to the generalization in my previous post. Thanks for all the interesting comments: it seems like there are some interesting connections to be explored (e.g. to the algebra of graphs, formal group laws, …?)!
This is a literate Haskell post; download it and play along in ghci! (Note it requires GHC 8.6 since I couldn’t resist making use of DerivingVia
…)
> {# LANGUAGE DefaultSignatures #}
> {# LANGUAGE FlexibleInstances #}
> {# LANGUAGE MultiParamTypeClasses #}
> {# LANGUAGE DerivingStrategies #}
> {# LANGUAGE GeneralizedNewtypeDeriving #}
> {# LANGUAGE DerivingVia #}
> {# LANGUAGE TypeApplications #}
> {# LANGUAGE ScopedTypeVariables #}
>
> import Data.Semigroup
> import Data.Coerce
Consider a sequence of integers . (Actually, integers is too specific; any linearly ordered domain will do.) An inversion is a pair of positions in which are “out of order”: that is, such that but . So, for example, has six inversions, namely . (Here I’ve written the elements that are out of order rather than their positions, which doesn’t matter much when the elements are all distinct; but it’s important to keep in mind that e.g. has two inversions, not one, because each copy of makes an inversion with the .) The total number of inversions of is denoted .
One way to think about the inversion count is as a measure of how far away the sequence is from being sorted. In particular, bubble sort will make precisely adjacent swaps while sorting . The highest possible value of is , when is sorted in reverse order.
The obvious bruteforce algorithm to count inversions is to use two nested loops to enumerate all possible pairs of elements, and increment a counter each time we discover a pair which is out of order. This clearly takes time. Can it be done any faster?
It turns out the (generally wellknown) answer is yes, using a variant of mergesort. The trick is to generalize to counting inversions and sorting the sequence at the same time. First split the sequence in half, and recursively sort and count the inversions in each half. Any inversion in the original sequence must either be entirely contained in one of the two halves (these will be counted by the recursive calls), or have one endpoint in the left half and one in the right. One key observation at this point is that any inversion with one endpoint in each half will still be an inversion even after independently sorting the two halves. The other key observation is that we can merge the two sorted subsequences and count inversions between them in linear time. Use the usual twofinger algorithm for merging two sorted sequences; each time we take an element from the right subsequence, it’s because it is less than all the remaining elements in the left subsequence, but it was to the right of all of them, so we can add the length of the remaining left subsequence to the inversion count. Intuitively, it’s this ability to count a bunch of inversions in one step which allows this algorithm to be more efficient, since any algorithm which only ever increments an inversion counter is doomed to be no matter how cleverly it splits up the counting. In the end, the number of total inversions is the sum of the inversions counted recursively in the two sublists, plus any inversions between the two sublists.
Here’s some Haskell code implementing this sortedmergeandinversioncount. We have to be a bit careful because we don’t want to call length
on the remaining sublist at every step (that would ruin the asymptotic performance!), so we precompute the length and pass along the length of the left subsequence as an extra parameter which we keep uptodate as we recurse.
>  Precondition: the input lists are sorted.
>  Output the sorted merge of the two lists, and the number of pairs
>  (a,b) such that a \in xs, b \in ys with a > b.
> mergeAndCount :: Ord a => [a] > [a] > ([a], Int)
> mergeAndCount xs ys = go xs (length xs) ys
>  precondition/invariant for go xs n ys: n == length xs
> where
> go [] _ ys = (ys, 0)
> go xs _ [] = (xs, 0)
> go (x:xs) n (y:ys)
>  x <= y = let (m, i) = go xs (n1) (y:ys) in (x:m, i)
>  otherwise = let (m, i) = go (x:xs) n ys in (y:m, i + n)
>
> merge :: Ord a => [a] > [a] > [a]
> merge xs ys = fst (mergeAndCount xs ys)
>
> inversionsBetween :: Ord a => [a] > [a] > Int
> inversionsBetween xs ys = snd (mergeAndCount xs ys)
Do you see how this is an instance of the sparky monoid construction in my previous post? is the set of sorted lists with merge as the monoid operation; is the natural numbers under addition. The spark operation takes two sorted lists and counts the number of inversions between them. So the monoid on pairs merges the lists, and adds the inversion counts together with the number of inversions between the two lsits.
We have to verify that this satisfies the laws: let be any sorted list, then we need

, that is, a `inversionsBetween` [] = 0
. This is true since there are never any inversions between and an empty list. Likewise for .

a `inversionsBetween` (a1 `merge` a2) == (a `inversionsBetween` a1) + (a `inversionsBetween` a2)
. This is also true since a1 `merge` a2
contains the same elements as a1
and a2
: any inversion between a
and a1 `merge` a2
will be an inversion between a
and a1
, or between a
and a2
, and vice versa. The same reasoning shows that (a1 `merge` a2) `inversionsBetween` a == (a1 `inversionsBetween` a) + (a2 `inversionsBetween` a)
.
Note that and are commutative monoids, but the spark operation isn’t commutative; in fact, any given pair of elements is an inversion between a1
and a2
precisely iff they are not an inversion between a2
and a1
. Note also that and aren’t idempotent; for example merging a sorted list with itself produces not the same list, but a new list with two copies of each element.
So let’s see some more Haskell code to implement the entire algorithm in a nicely modular way. First, let’s encode sparky monoids in general. The Sparky
class is for pairs of types with a spark operation. As we saw in the example above, sometimes it may be more efficient to compute and the spark at the same time, so we bake that possibility into the class.
> class Sparky a b where
The basic spark operation, with a default implementation that projects the result out of the prodSpark
method.
> (<.>) :: a > a > b
> a1 <.> a2 = snd (prodSpark a1 a2)
prodSpark
does the monoidal product and spark at the same time, with a default implementation that just does them separately.
> prodSpark :: a > a > (a,b)
> default prodSpark :: Semigroup a => a > a > (a,b)
> prodSpark a1 a2 = (a1 <> a2, a1 <.> a2)
Finally we can specify that we have to implement one or the other of these methods.
> {# MINIMAL (<.>)  prodSpark #}
Sparked a b
is just a pair type, but with Semigroup
and Monoid
instances that implement the sparky product.
> data Sparked a b = S { getA :: a, getSpark :: b }
> deriving Show
>
> class Semigroup a => CommutativeSemigroup a
> class (Monoid a, CommutativeSemigroup a) => CommutativeMonoid a
>
> instance (Semigroup a, CommutativeSemigroup b, Sparky a b) => Semigroup (Sparked a b) where
> S a1 b1 <> S a2 b2 = S a' (b1 <> b2 <> b')
> where (a', b') = prodSpark a1 a2
>
> instance (Monoid a, CommutativeMonoid b, Sparky a b) => Monoid (Sparked a b) where
> mempty = S mempty mempty
Now we can make instances for sorted lists under merge…
> newtype Sorted a = Sorted [a]
> deriving Show
>
> instance Ord a => Semigroup (Sorted a) where
> Sorted xs <> Sorted ys = Sorted (merge xs ys)
> instance Ord a => Monoid (Sorted a) where
> mempty = Sorted []
>
> instance Ord a => CommutativeSemigroup (Sorted a)
> instance Ord a => CommutativeMonoid (Sorted a)
…and for inversion counts.
> newtype InvCount = InvCount Int
> deriving newtype Num
> deriving (Semigroup, Monoid) via Sum Int
>
> instance CommutativeSemigroup InvCount
> instance CommutativeMonoid InvCount
Finally we make the Sparky (Sorted a) InvCount
instance, which is just mergeAndCount
(some conversion between newtype
s is required, but we can get the compiler to do it automagically via coerce
and a bit of explicit type application).
> instance Ord a => Sparky (Sorted a) InvCount where
> prodSpark = coerce (mergeAndCount @a)
And here’s a function to turn a single a
value into a sorted singleton list paired with an inversion count of zero, which will come in handy later.
> single :: a > Sparked (Sorted a) InvCount
> single a = S (Sorted [a]) 0
Finally, we can make some generic infrastructure for doing monoidal folds. First, Parens a
encodes lists of a
which have been explicitly associated, i.e. fully parenthesized:
> data Parens a = Leaf a  Node (Parens a) (Parens a)
> deriving Show
We can make a generic fold for Parens a
values, which maps each Leaf
into the result type b
, and replaces each Node
with a binary operation:
> foldParens :: (a > b) > (b > b > b) > Parens a > b
> foldParens lf _ (Leaf a) = lf a
> foldParens lf nd (Node l r) = nd (foldParens lf nd l) (foldParens lf nd r)
Now for a function which splits a list in half recursively to produce a balanced parenthesization.
> balanced :: [a] > Parens a
> balanced [] = error "List must be nonempty"
> balanced [a] = Leaf a
> balanced as = Node (balanced as1) (balanced as2)
> where (as1, as2) = splitAt (length as `div` 2) as
Finally, we can make a balanced variant of foldMap
: instead of just mapping a function over a list and then reducing with mconcat
, as foldMap
does, it first creates a balanced parenthesization for the list and then reduces via the given monoid. This will always give the same result as foldMap
due to associativity, but in some cases it may be more efficient.
> foldMapB :: Monoid m => (e > m) > [e] > m
> foldMapB leaf = foldParens leaf (<>) . balanced
Let’s try it out!
λ> :set +s
λ> getSpark $ foldMap single [3000, 2999 .. 1 :: Int]
Sum {getSum = 4498500}
(34.94 secs, 3,469,354,896 bytes)
λ> getSpark $ foldMapB single [3000, 2999 .. 1 :: Int]
Sum {getSum = 4498500}
(0.09 secs, 20,016,936 bytes)
Empirically, it does seem that we are getting quadratic performance with normal foldMap
, but with foldMapB
. We can verify that we are getting the correct inversion count in either case, since we know there should be when the list is reversed, and sure enough, .