Summary: given an injective function $A \times N \hookrightarrow B \times N$, it is possible to constructively “divide by $N$” to obtain an injection $A \hookrightarrow B$, as shown recently by Peter Doyle and Cecil Qiu and expounded by Richard Schwartz. Their algorithm is nontrivial to come up with—this had been a longstanding open question—but it’s not too difficult to explain. I exhibit some Haskell code implementing the algorithm, and show some examples.

# Introduction: division by two

Suppose someone hands you the following:

• A Haskell function f :: (A, Bool) -> (B, Bool), where A and B are abstract types (i.e. their constructors are not exported, and you have no other functions whose types mention A or B).

• A promise that the function f is injective, that is, no two values of (A, Bool) map to the same (B, Bool) value. (Thus (B, Bool) must contain at least as many inhabitants as (A, Bool).)

• A list as :: [A], with a promise that it contains every value of type A exactly once, at a finite position.

Can you explicitly produce an injective function f' :: A -> B? Moreover, your answer should not depend on the order of elements in as.

It really seems like this ought to be possible. After all, if (B, Bool) has at least as many inhabitants as (A, Bool), then surely B must have at least as many inhabitants as A. But it is not enough to reason merely that some injection must exist; we have to actually construct one. This, it turns out, is tricky. As a first attempt, we might try f' a = fst (f (a, True)). That is certainly a function of type A -> B, but there is no guarantee that it is injective. There could be a1, a2 :: A which both map to the same b, that is, one maps to (b, False) and the other to (b, True). The picture below illustrates such a situation: (a1, True) and (a2, True) both map to b2. So the function f may be injective overall, but we can’t say much about f restricted to a particular Bool value.

The requirement that the answer not depend on the order of as also makes things difficult. (Over in math-land, depending on a particular ordering of the elements in as would amount to the well-ordering principle, which is equivalent to the axiom of choice, which in turn implies the law of excluded middle—and as we all know, every time someone uses the law of excluded middle, a puppy dies. …I feel like I’m in one of those DirecTV commercials. “Don’t let a puppy die. Ignore the order of elements in as.”) Anyway, making use of the order of values in as, we could do something like the following:

• For each a :: A:
• Look at the B values generated by f (a,True) and f (a,False). (Note that there might only be one distinct such B value).
• If neither B value has been used so far, pick the one that corresponds to (a,True), and add the other one to a queue of available B values.
• If one is used and one unused, pick the unused one.
• If both are used, pick the next available B value from the queue.

It is not too hard I couldn’t be bothered to show that this will always successfully result in a total function A -> B, which is injective by construction. (One has to show that there will always be an available B value in the queue when you need it.) The only problem is that the particular function we get depends on the order in which we iterate through the A values. The above example illustrates this as well: if the A values are listed in the order $[a_1, a_2]$, then we first choose $a_1 \mapsto b_2$, and then $a_2 \mapsto b_3$. If they are listed in the other order, we end up with $a_2 \mapsto b_2$ and $a_1 \mapsto b_1$. Whichever value comes first “steals” $b_2$, and then the other one takes whatever is left. We’d like to avoid this sort of dependence on order. That is, we want a well-defined algorithm which will yield a total, injective function A -> B, which is canonical in the sense that the algorithm yields the same function given any permutation of as.

It is possible—you might enjoy puzzling over this a bit before reading on!

# Division by N

The above example is a somewhat special case. More generally, let $N = \{0, \dots, n-1\}$ denote a canonical finite set of size $n$, and let $A$ and $B$ be arbitrary sets. Then, given an injection $f : A \times N \hookrightarrow B \times N$, is it possible to effectively (that is, without excluded middle or the axiom of choice) compute an injection $A \hookrightarrow B$?

Translating down to the world of numbers representing set cardinalities—natural numbers if $A$ and $B$ are finite, or cardinal numbers in general—this just says that if $an \leq bn$ then $a \leq b$. This statement about numbers is obviously true, so it would be nice if we could say something similar about sets, so that this fact about numbers and inequalities can be seen as just a “shadow” of a more general theorem about sets and injections.

As hinted in the introduction, the interesting part of this problem is really the word “effectively”. Using the Axiom of Choice/Law of Excluded Middle makes the problem a lot easier, but either fails to yield an actual function that we can compute with, instead merely guaranteeing the existence of such a function, or gives us a function that depends on a particular ordering of $A$.

Apparently this has been a longstanding open question, recently answered in the affirmative by Peter Doyle and Cecil Qiu in their paper Division By Four. It’s a really great paper: they give some fascinating historical context for the problem, and explain their algorithm (which is conceptually not all that difficult) using an intuitive analogy to a card game with certain rules. (It is not a “game” in the usual sense of having winners and losers, but really just an algorithm implemented with “players” and “cards”. In fact, you could get some friends together and actually perform this algorithm in parallel (if you have sufficiently nerdy friends).) Richard Schwartz’s companion article is also great fun and easy to follow (you should read it first).

# A Game of Thrones Cards

Here’s a quick introduction to the way Doyle, Qiu, and Schwartz use a card game to formulate their algorithm. (Porting this framework to use “thrones” and “claimants” instead of “spots” and “cards” is left as an exercise to the reader.)

The finite set $N$ is to be thought of as a set of suits. The set $A$ will correspond to a set of players, and $B$ to a set of ranks or values (for example, Ace, 2, 3, …) In that case $B \times N$ corresponds to a deck of cards, each card having a rank and a suit; and we can think of $A \times N$ in terms of each player having in front of them a number of “spots” or “slots”, each labelled by a suit. An injection $A \times N \hookrightarrow B \times N$ is then a particular “deal” where one card has been dealt into each of the spots in front of the players. (There may be some cards left over in the deck, but the fact that the function is total means every spot has a card, and the fact that it is injective is encoded in the common-sense idea that a given card cannot be in two spots at once.) For example, the example function from before:

corresponds to the following deal:

Here each column corresponds to one player’s hand, and the rows correspond to suit spots (with the spade spots on top and the heart spots beneath). We have mapped $\{b_1, b_2, b_3\}$ to the ranks A, 2, 3, and mapped T and F to Spades and Hearts respectively. The spades are also highlighted in green, since later we will want to pay particular attention to what is happening with them. You might want to take a moment to convince yourself that the deal above really does correspond to the example function from before.

Of course, doing everything effectively means we are really talking about computation. Doyle and Qiu do talk a bit about computation, but it’s still pretty abstract, in the sort of way that mathematicians talk about computation, so I thought it would be interesting to actually implement the algorithm in Haskell.

The algorithm “works” for infinite sets, but only (as far as I understand) if you consider some notion of transfinite recursion. It still counts as “effective” in math-land, but over here in programming-land I’d like to stick to (finitely) terminating computations, so we will stick to finite sets $A$ and $B$.

First, some extensions and imports. Nothing too controversial.

> {-# LANGUAGE DataKinds                  #-}
> {-# LANGUAGE GeneralizedNewtypeDeriving #-}
> {-# LANGUAGE KindSignatures             #-}
> {-# LANGUAGE RankNTypes                 #-}
> {-# LANGUAGE ScopedTypeVariables        #-}
> {-# LANGUAGE StandaloneDeriving         #-}
> {-# LANGUAGE TypeOperators              #-}
>
> module PanGalacticDivision where
>
> import           Control.Arrow (second, (&&&), (***))
> import           Data.Char
> import           Data.List     (find, findIndex, transpose)
> import           Data.Maybe
>
> import           Diagrams.Prelude hiding (universe, value)
> import           Diagrams.Backend.Rasterific.CmdLine
> import           Graphics.SVGFonts


We’ll need some standard machinery for type-level natural numbers. Probably all this stuff is in a library somewhere but I couldn’t be bothered to find out. Pointers welcome.

> -- Standard unary natural number type
> data Nat :: * where
>   Z :: Nat
>   Suc :: Nat -> Nat
>
> type One = Suc Z
> type Two = Suc One
> type Three = Suc Two
> type Four = Suc Three
> type Six = Suc (Suc Four)
> type Eight = Suc (Suc Six)
> type Ten = Suc (Suc Eight)
> type Thirteen = Suc (Suc (Suc Ten))
>
> -- Singleton Nat-indexed natural numbers, to connect value-level and
> -- type-level Nats
> data SNat :: Nat -> * where
>   SZ :: SNat Z
>   SS :: Natural n => SNat n -> SNat (Suc n)
>
> -- A class for converting type-level nats to value-level ones
> class Natural n where
>   toSNat :: SNat n
>
> instance Natural Z where
>   toSNat = SZ
>
> instance Natural n => Natural (Suc n) where
>   toSNat = SS toSNat
>
> -- A function for turning explicit nat evidence into implicit
> natty :: SNat n -> (Natural n => r) -> r
> natty SZ r     = r
> natty (SS n) r = natty n r
>
> -- The usual canonical finite type.  Fin n has exactly n
> -- (non-bottom) values.
> data Fin :: Nat -> * where
>   FZ :: Fin (Suc n)
>   FS :: Fin n -> Fin (Suc n)
>
> finToInt :: Fin n -> Int
> finToInt FZ     = 0
> finToInt (FS n) = 1 + finToInt n
>
> deriving instance Eq (Fin n)


# Finiteness

Next, a type class to represent finiteness. For our purposes, a type a is finite if we can explicitly list its elements. For convenience we throw in decidable equality as well, since we will usually need that in conjunction. Of course, we have to be careful: although we can get a list of elements for a finite type, we don’t want to depend on the ordering. We must ensure that the output of the algorithm is independent of the order of elements.1 This is in fact true, although somewhat nontrivial to prove formally; I mention some of the intuitive ideas behind the proof below.

While we are at it, we give Finite instances for Fin n and for products of finite types.

> class Eq a => Finite a where
>   universe :: [a]
>
> instance Natural n => Finite (Fin n) where
>   universe = fins toSNat
>
> fins :: SNat n -> [Fin n]
> fins SZ     = []
> fins (SS n) = FZ : map FS (fins n)
>
> -- The product of two finite types is finite.
> instance (Finite a, Finite b) => Finite (a,b) where
>   universe = [(a,b) | a <- universe, b <- universe]


# Division, inductively

Now we come to the division algorithm proper. The idea is that panGalacticPred turns an injection $A \times N \hookrightarrow B \times N$ into an injection $A \times (N-1) \hookrightarrow B \times (N-1)$, and then we use induction on $N$ to repeatedly apply panGalacticPred until we get an injection $A \times 1 \hookrightarrow B \times 1$.

> panGalacticDivision
>   :: forall a b n. (Finite a, Eq b)
>   => SNat n -> ((a, Fin (Suc n)) -> (b, Fin (Suc n))) -> (a -> b)


In the base case, we are given an injection $A \times 1 \hookrightarrow B \times 1$, so we just pass a unit value in along with the $A$ and project out the $B$.

> panGalacticDivision SZ f = \a -> fst (f (a, FZ))


In the inductive case, we call panGalacticPred and recurse.

> panGalacticDivision (SS n') f = panGalacticDivision n' (panGalacticPred n' f)


# Pan-Galactic Predecessor

And now for the real meat of the algorithm, the panGalacticPred function. The idea is that we swap outputs around until the function has the property that every output of the form $(b,0)$ corresponds to an input also of the form $(a,0)$. That is, using the card game analogy, every spade in play should be in the leftmost spot (the spades spot) of some player’s hand (some spades can also be in the deck). Then simply dropping the leftmost card in everyone’s hand (and all the spades in the deck) yields a game with no spades. That is, we will have an injection $A \times \{1, \dots, n-1\} \hookrightarrow B \times \{1, \dots, n-1\}$. Taking predecessors everywhere (i.e. “hearts are the new spades”) yields the desired injection $A \times (N-1) \hookrightarrow B \times (N-1)$.

We need a Finite constraint on a so that we can enumerate all possible inputs to the function, and an Eq constraint on b so that we can compare functions for extensional equality (we iterate until reaching a fixed point). Note that whether two functions are extensionally equal does not depend on the order in which we enumerate their inputs, so far validating my claim that nothing depends on the order of elements returned by universe.

> panGalacticPred
>   :: (Finite a, Eq b, Natural n)
>   => SNat n
>   -> ((a, Fin (Suc (Suc n))) -> (b, Fin (Suc (Suc n))))
>   -> ((a, Fin (Suc n)) -> (b, Fin (Suc n)))


We construct a function f' which is related to f by a series of swaps, and has the property that it only outputs FZ when given FZ as an input. So given (a,i) we can call f' on (a, FS i) which is guaranteed to give us something of the form (b, FS j). Thus it is safe to strip off the FS and return (b, j) (though the Haskell type checker most certainly does not know this, so we just have to tell it to trust us).

> panGalacticPred n f = \(a,i) -> second unFS (f' (a, FS i))
>   where
>     unFS :: Fin (Suc n) -> Fin n
>     unFS FZ = error "impossible!"
>     unFS (FS i) = i


To construct f' we iterate a certain transformation until reaching a fixed point. For finite sets $A$ and $B$ this is guaranteed to terminate, though it is certainly not obvious from the Haskell code. (Encoding this in Agda so that it is accepted by the termination checker would be a fun (?) exercise.)

One round of the algorithm consists of two phases called “shape up” and “ship out” (to be described shortly).

>     oneRound = natty n $shipOut . shapeUp > > -- iterate 'oneRound' beginning with the original function... > fs = iterate oneRound f > -- ... and stop when we reach a fixed point. > f' = fst . head . dropWhile (uncurry (=/=))$ zip fs (tail fs)
>     f1 =/= f2 = all (\x -> f1 x == f2 x) universe


# Encoding Card Games

Recall that a “card” is a pair of a value and a suit; we think of $B$ as the set of values and $N$ as the set of suits.

> type Card v s = (v, s)
>
> value :: Card v s -> v
> value = fst
>
> suit :: Card v s -> s
> suit = snd


Again, there are a number of players (one for each element of $A$), each of which has a “hand” of cards. A hand has a number of “spots” for cards, each one labelled by a different suit (which may not have any relation to the actual suit of the card in that position).

> type PlayerSpot p s = (p, s)
> type Hand v s = s -> Card v s


A “game” is an injective function from player spots to cards. Of course, the type system is not enforcing injectivity here.

> type Game p v s = PlayerSpot p s -> Card v s


Some utility functions. First, a function to project out the hand of a given player.

> hand :: p -> Game p v s -> Hand v s
> hand p g = \s -> g (p, s)


A function to swap two cards, yielding a bijection on cards.

> swap :: (Eq s, Eq v) => Card v s -> Card v s -> (Card v s -> Card v s)
> swap c1 c2 = f
>   where
>     f c
>       | c == c1   = c2
>       | c == c2   = c1
>       | otherwise = c


leftmost finds the leftmost card in a player’s hand which has a given suit.

> leftmost :: Finite s => s -> Hand v s -> Maybe s
> leftmost targetSuit h = find (\s -> suit (h s) == targetSuit) universe


# Playing Rounds

playRound abstracts out a pattern that is used by both shapeUp and shipOut. The first argument is a function which, given a hand, produces a function on cards; that is, based on looking at a single hand, it decides how to swap some cards around.2 playRound then applies that function to every hand, and composes together all the resulting permutations.

Note that playRound has both Finite s and Finite p constraints, so we should think about whether the result depends on the order of elements returned by any call to universe—I claimed it does not. Finite s corresponds to suits/spots, which corresponds to $N$ in the original problem formulation. $N$ explicitly has a canonical ordering, so this is not a problem. The Finite p constraint, on the face of it, is more problematic. We will have to think carefully about each of the rounds implemented in terms of playRound and make sure they do not depend on the order of players. Put another way, it should be possible for all the players to take their turn simultaneously.

> playRound :: (Finite s, Finite p, Eq v) => (Hand v s -> Card v s -> Card v s) -> Game p v s -> Game p v s
> playRound withHand g = foldr (.) id swaps . g
>   where
>     swaps = map (withHand . flip hand g) players
>     players = universe


# Shape Up and Ship Out

Finally, we can describe the “shape up” and “ship out” phases, beginning with “shape up”. A “bad” card is defined as one having the lowest suit; make sure every hand with any bad cards has one in the leftmost spot (by swapping the leftmost bad card with the card in the leftmost spot, if necessary).

> shapeUp :: (Finite s, Finite p, Eq v) => Game p v s -> Game p v s
> shapeUp = playRound shapeUp1
>   where
>     shapeUp1 theHand =
>       case leftmost badSuit theHand of
>         Nothing      -> id


And now for the “ship out” phase. Send any “bad” cards not in the leftmost spot somewhere else, by swapping with a replacement, namely, the card whose suit is the same as the suit of the spot, and whose value is the same as the value of the bad card in the leftmost spot. The point is that bad cards in the leftmost spot are OK, since we will eventually just ignore the leftmost spot. So we have to keep shipping out bad cards not in the leftmost spot until they all end up in the leftmost spot. For some intuition as to why this is guaranteed to terminate, consult Schwartz; note that columns tend to acquire more and more cards that have the same rank as a spade in the top spot (which never moves).

> shipOut :: (Finite s, Finite p, Eq v) => Game p v s -> Game p v s
> shipOut = playRound shipOutHand
>   where
>     spots = universe
>     shipOutHand theHand = foldr (.) id swaps
>       where
>         swaps = map (shipOut1 . (theHand &&& id)) (drop 1 spots)
>         shipOut1 ((_,s), spot)
>           | s == badSuit = swap (theHand spot) (value (theHand badSuit), spot)
>           | otherwise    = id


And that’s it! Note that both shapeUp and shipOut are implemented by composing a bunch of swaps; in fact, in both cases, all the swaps commute, so the order in which they are composed does not matter. (For proof, see Schwartz.) Thus, the result is independent of the order of the players (i.e. the set A).

Enough code, let’s see an example! This example is taken directly from Doyle and Qiu’s paper, and the diagrams are being generated literally (literately?) by running the code in this blog post. Here’s the starting configuration:

Again, the spades are all highlighted in green. Recall that our goal is to get them all to be in the first row, but we have to do it in a completely deterministic, canonical way. After shaping up, we have:

Notice how the 6, K, 5, A, and 8 of spades have all been swapped to the top of their column. However, there are still spades which are not at the top of their column (in particular the 10, 9, and J) so we are not done yet.

Now, we ship out. For example, the 10 of spades is in the diamonds position in the column with the Ace of spades, so we swap it with the Ace of diamonds. Similarly, we swap the 9 of spades with the Queen of diamonds, and the Jack of spades with the 4 of hearts.

Shaping up does nothing at this point so we ship out again, and then continue to alternate rounds.

In the final deal above, all the spades are at the top of a column, so there is an injection from the set of all non-spade spots to the deck of cards with all spades removed. This example was, I suspect, carefully constructed so that none of the spades get swapped out into the undealt portion of the deck, and so that we end up with only spades in the top row. In general, we might end up with some non-spades also in the top row, but that’s not a problem. The point is that ignoring the top row gets rid of all the spades.

Anyway, I hope to write more about some “practical” examples and about what this has to do with combinatorial species, but this post is long enough already. Doyle and Qiu also describe a “short division” algorithm (the above is “long division”) that I hope to explore as well.

# The rest of the code

For completeness, here’s the code I used to represent the example game above, and to render all the card diagrams (using diagrams 1.3).

> type Suit = Fin
> type Rank = Fin
> type Player = Fin
>
> readRank :: SNat n -> Char -> Rank n
> readRank n c = fins n !! (fromJust $findIndex (==c) "A23456789TJQK") > > readSuit :: SNat n -> Char -> Suit n > readSuit (SS _) 'S' = FZ > readSuit (SS (SS _)) 'H' = FS FZ > readSuit (SS (SS (SS _))) 'D' = FS (FS FZ) > readSuit (SS (SS (SS (SS _)))) 'C' = FS (FS (FS FZ)) > > readGame :: SNat a -> SNat b -> SNat n -> String -> Game (Player a) (Rank b) (Suit n) > readGame a b n str = \(p, s) -> table !! finToInt p !! finToInt s > where > table = transpose . map (map readCard . words) . lines$ str
>
> -- Example game from Doyle & Qiu
> exampleGameStr :: String
> exampleGameStr = unlines
>   [ "4D 6H QD 8D 9H QS 4C AD 6C 4S"
>   , "JH AH 9C 8H AS TC TD 5H QC JS"
>   , "KC 6S 4H 6D TS 9S JC KD 8S 8C"
>   , "5C 5D KS 5S TH JD AC QH 9D KH"
>   ]
>
> exampleGame :: Game (Player Ten) (Rank Thirteen) (Suit Four)
> exampleGame = readGame toSNat toSNat toSNat exampleGameStr
>
> suitSymbol :: Suit n -> String
> suitSymbol = (:[]) . ("♠♥♦♣"!!) . finToInt  -- Huzzah for Unicode
>
> suitDia :: Suit n -> Diagram B
> suitDia = (suitDias!!) . finToInt
>
> suitDias = map mkSuitDia (fins (toSNat :: SNat Four))
> mkSuitDia s = text' (suitSymbol s) # fc (suitColor s) # lw none
>
> suitColor :: Suit n -> Colour Double
> suitColor n
>   | finToInt n elem [0,3] = black
>   | otherwise               = red
>
> rankStr :: Rank n -> String
> rankStr n = rankStr' (finToInt n + 1)
>   where
>     rankStr' 1 = "A"
>     rankStr' i | i <= 10    = show i
>                | otherwise = ["JQK" !! (i - 11)]
>
> text' t = stroke (textSVG' (TextOpts lin INSIDE_H KERN False 1 1) t)
>
> renderCard :: (Rank b, Suit n) -> Diagram B
> renderCard (r, s) = mconcat
>   [ mirror label
>   , cardContent (finToInt r + 1)
>   , back
>   ]
>   where
>     cardWidth  = 2.25
>     cardHeight = 3.5
>     cardCorners = 0.1
>     mirror d = d  d # rotateBy (1/2)
>     back  = roundedRect cardWidth cardHeight cardCorners # fc white
>           # lc (case s of { FZ -> green; _ -> black })
>     label = vsep 0.1 [text' (rankStr r), text' (suitSymbol s)]
>           # scale 0.6 # fc (suitColor s) # lw none
>           # translate ((-0.9) ^& 1.5)
>     cardContent n
>       | n <= 10   = pips n
>       | otherwise = face n # fc (suitColor s) # lw none
>                            # sized (mkWidth (cardWidth * 0.6))
>     pip = suitDia s # scale 1.1
>     pips 1 = pip # scale 2
>     pips 2 = mirror (pip # up 2)
>     pips 3 = pips 2  pip
>     pips 4 = mirror (pair pip # up 2)
>     pips 5 = pips 4  pip
>     pips 6 = mirror (pair pip # up 2)  pair pip
>     pips 7 = pips 6  pip # up 1
>     pips 8 = pips 6  mirror (pip # up 1)
>     pips 9 = mirror (pair (pip # up (2/3)  pip # up 2))  pip # up (case finToInt s of {1 -> -0.1; 3 -> 0; _ -> 0.1})
>     pips 10 = mirror (pair (pip # up (2/3)  pip # up 2)  pip # up (4/3))
>     pips _ = mempty
>     up n = translateY (0.5*n)
>     pair d = hsep 0.4 [d, d] # centerX
>     face 11 = squares # frame 0.1
>     face 12 = loopyStar
>     face 13 = burst # centerXY
>     squares
>       = strokeP (mirror (square 1 # translate (0.2 ^& 0.2)))
>       # fillRule EvenOdd
>     loopyStar
>       = regPoly 7 1
>       # star (StarSkip 3)
>       # pathVertices
>       # map (cubicSpline True)
>       # mconcat
>       # fillRule EvenOdd
>     burst
>       = [(1,5), (1,-5)] # map r2 # fromOffsets
>       # iterateN 13 (rotateBy (-1/13))
>       # mconcat # glueLine
>       # strokeLoop
>
> renderGame :: (Natural n, Natural a) => Game (Player a) (Rank b) (Suit n) -> Diagram B
> renderGame g = hsep 0.5 $map (\p -> renderHand p$ hand p g) universe
>
> renderHand :: Natural n => Player a -> Hand (Rank b) (Suit n) -> Diagram B
> renderHand p h = vsep 0.2 \$ map (renderCard . h) universe


1. If we could program in Homotopy Type Theory, we could make this very formal by using the notion of cardinal-finiteness developed in my dissertation (see section 2.4).

2. In practice this function on cards will always be a permutation, though the Haskell type system is not enforcing that at all. An early version of this code used the Iso type from lens, but it wasn’t really paying its way.

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in haskell, math, species and tagged , , , , , , , , , . Bookmark the permalink.

### 6 Responses to Pan-Galactic Division in Haskell

1. > Thus (B, Bool) must contain more inhabitants than (A, Bool), or else both are infinite.

And that’s where you lost me. Isn’t the following function injective:

data A = A | B
data B = C | D

f (A, True) = (C,True)
f (A, False) = (C, False)
f (B, True) = (D, True)
f (B, False) = (D, False)

?

I guess that the fact that A and B are abstract somehow comes into play here. Or is it the bottoms?

• Brent says:

Ah, sorry, that is sloppy wording. I should have said something like “(B,Bool) must contain at least as many inhabitants as (A,Bool)”. That is, greater-than-or-equal-to, not strictly greater-than.

2. By assuming decidable equality, aren’t you assuming LEM?

• Brent says:

No, LEM implies decidable equality for everything, but having decidable equality for a particular type does not imply LEM. The only reason I’m using decidable equality is so I can tell when the iteration has converged.

3. s says:

Should rows and columns be switched in the paragraph beginning, “Here each column…”?

• Brent says:

Hmm, you’re right, it does seem like I got something backwards there.