So, how should such a proof look? For a given number of labels , there is a single signed set structure, which is just the set of labels itself (with a sign depending on the parity of ). On the other hand, there are lots of ballots on labels; the key is that some are positive and some are negative, since the sign of the ballots depends on the parity of the number of parts, not the number of labels. For example, consider . There is a single (negative) signed set structure:
(I will use a dashed blue line to indicate negative things, and a solid black line for positive things.)
On the other hand, as we saw last time, there are 13 ballot structures on 3 labels, some positive and some negative:
In this example, it is easy to see that most of the positives and negatives cancel, with exactly one negative ballot left over, which corresponds with the one negative set. As another example, when , there is a single positive set, and 75 signed ballots:
This time it is not quite so easy to tell at a glance (at least not the way I have arranged the ballots in the above picture!), but in fact one can verify that there are exactly 37 negative ballots and 38 positive ones, again cancelling to match the one positive set.
What we need to show, then, is that we can pair up the ballots in such a way that positive ballots are matched with negative ballots, with exactly one ballot of the appropriate sign left to be matched with the one signed set. This is known as a signed involution: an involution is a function which is its own inverse, so it matches things up in pairs; a signed involution sends positive things to negative things and vice versa, except for any fixed points.
In order to do this, we will start by assuming the set of labels is linearly ordered. In one sense this is no big deal, since for any finite set of labels we can always just pick an arbitrary ordering, if there isn’t an “obvious” ordering to use already. On the other hand, it means that the correspondence will be specific to the chosen linear ordering. All other things being equal, we would prefer a correspondence that depends solely on the structure of the ballots, and not on any structure inherent to the labels. I will have quite a bit more to say about this in my third and (probably) final post on the topic. But for today, let’s just see how the correspondence works, given the assumption of a linear order on the labels. I came up with this proof independently while contemplating Anders Claesson’s post, though it turns out that the exact same proof is already in a paper by Claesson and Hannah (in any case it is really just a small lemma, the sort of thing you might give as a homework problem in an undergraduate course on combinatorics).
Given some ballot, find the smallest label. For example, if the labels are as in the examples so far, we will find the label .
If the smallest label is contained in some part together with at least one other label, separate it out into its own part by itself, and put it to the right of its former part. Like this:
On the other hand, if the smallest label is in a part by itself, merge it with the part on the left (if one exists). This is clearly the inverse of the above operation.
The only case we haven’t handled is when the smallest label is in a part by itself which is the leftmost part in the ballot. In that case, we leave that part alone, switch to considering the second-smallest label, and recursively carry out the involution on the remainder of the ballot.
For example:
In this case we find the smallest label (1) in a part by itself in the leftmost position, so we leave it where it is and recurse on the remainder of the ballot. Again, we find the smallest remaining label (2) by itself and leftmost, so we recurse again. This time, we find the smallest remaining label (3) in a part with one other label, so we separate it out and place it to the right.
This transformation on ballots is clearly reversible. The only ballots it doesn’t change are ballots with each label in its own singleton part, sorted from smallest to biggest, like this:
In this case the algorithm recurses through the whole ballot and finds each smallest remaining label in the leftmost position, ultimately doing nothing. Notice that a sorted ballot of singletons has the same sign as the signed set on the same labels, namely, . In any other case, we can see that the algorithm matches positive ballots to negative and vice versa, since it always changes the number of parts by 1, either splitting one part into two or merging two parts into one.
Here’s my implementation of the involution in Haskell:
type Ballot = [[Int]]
ballotInv :: Ballot -> Ballot
ballotInv = go 1
where
go _ [] = []
go s ([a]:ps)
| s == a = [a] : go (s+1) ps
go s (p:ps)
| s `elem` p = delete s p : [s] : ps
go s (p:[a]:ps)
| s == a = sort (a:p) : ps
go s (p:ps) = p : go s ps
(The call to sort
is not strictly necessary, but I like to keep each part canonically sorted.)
Here again are the 13 signed ballots for , this time arranged so that the pair of ballots in each row correspond to each other under the involution, with the leftover, sorted ballot by itself at the top.
If you’d like to see an illustration of the correspondence for , you can find it here (I didn’t want to include inline since it’s somewhat large).
This completes the proof that signed sets and signed ballots correspond. But did we really need that linear order on the labels? Tune in next time to find out!
Recall that denotes the species of sets, defined by , that is, the only -structure on a given label set is the set of labels itself. Recall also that the exponential generating function of a species is given by
where counts the number of labelled -structures of size . In the case of , we have for all , so
(This is why is such a good name for the species of sets—though in a fantastic coincidence, it seems to originally come from the French word for set, ensemble, rather than from the fact that (though on the other hand calling it a “coincidence” is probably too strong, since Joyal must surely have picked the notation with the generating function already in mind!).)
Now, from my previous post we know that
Let’s first consider (without the ). This means that we have, for some , a -ary product of structures—in other words, a list of nonempty sets. This is the species of ballots, also known as ordered set partitions, and can also be written . As an example, here is a ballot on the set of labels :
The order of the parts matters, so this is a different ballot:
But the order of labels within each part doesn’t matter (since each part is a set). As another example, here is the complete collection of ballot structures on the labels :
We can see that there are 13 in total: six where the labels are each in their own separate part (corresponding to the six possible permutations of the labels); six where two labels share a part and the other label is a singleton part (corresponding to the three ways to choose the solitary label, times the two ways to order the parts); and one final ballot where all three labels are grouped in the same part. (As an exercise, can you verify that there are 75 different ballot structures on a set of four labels?)
Returning to , we can see that it consists of signed ballots, where the sign of a ballot is the parity of its number of parts, that is, a ballot with parts has sign . The second half of Anders’ post gives a nice combinatorial proof that , via a sign-reversing involution: if we consider -structures, i.e. pairs of sets and signed ballots, there is a natural^{1} way to pair them up, matching positive and negative structures so everything cancels (except in the case of the empty label set, which is why we get instead of ).
However, Anders is trying to do more than that. Note first that since multiplication of EGFs corresponds to multiplication of species, the EGF for is of course . But this ought to also be the EGF for the virtual species , and the rest of his post hinges on identifying and . As Anders and I discovered, however, this is precisely the point where it is worth being more careful.
First of all, what is ? Intuitively, an structure consists of a set of negative atoms; since each set can be thought of as an (unordered) product of atoms, the whole set acquires a sign given by the parity of the number of atoms. In other words, intuitively it seems that should be the species of signed sets, where an even-sized set is considered positive and an odd-sized set negative. That is,
where denotes the species of sets of size exactly . As a sanity check, this makes sense as an EGF equation too, since the EGF of is just and indeed
But hold on a minute, what does really mean, formally? It is the composition of the species with the virtual species , and it turns out that it is not at all a priori obvious how to define composition for virtual species! We can find the definition on p. 127 of Bergeron et al. A special case (which is enough for our present purposes) is
where and are two sorts of atoms, and denotes Cartesian product of species. In our case,
since is the identity for Cartesian product (overlaying an additional structure on a set of labels does not add any structure, since there is only one possible -structure).
All of this is to say, is actually defined as ! So at first glance it may seem we actually have nothing to prove: and are the same by definition, end of story. …but in fact, all we have done is shift the burden of proof elsewhere: now it is our intuitive idea of representing signed sets that requires proof!
To sum up, we know that is the species of signed ballots, with sign given by parity of the number of parts; and intuitively, we also believe that should correspond to parity-signed sets, . So, is there a nice combinatorial proof showing the correspondence between signed sets and signed ballots?
One can use the law of excluded middle to show that the answer must be “yes”: suppose the answer were “no”; but then I would not be writing this blog post, which is a contradiction since I am writing this blog post. But is there a constructive proof? Fear not! This blog post has gotten long enough, so I will stop here for now and let interested readers puzzle over it; in my next post I will explain what I came up with, along with some musings on linear orders and naturality.
I am indeed using the word natural in a technical, categorical sense here! This will play an important role in my second post…↩
Recall that the intuitive idea of a combinatorial species is a family of labelled structures which are invariant under relabelling. If you’ve never seen the formal definition before, don’t worry: just think “data structures” or “algebraic data types” for now.
The basic idea of virtual species is to work with pairs of species where is considered “positive” and “negative”. Formally, we consider equivalence classes of such pairs under the equivalence relation defined by iff .^{1} This parallels the way one typically gives a formal definition of the integers starting from the natural numbers (the “Grothendieck construction”); see my previous post for more details.
How can we build intuition for virtual species, and for additive inverses of species in particular? To be honest I have been struggling with this question for many years.
Multiplicative inverses are much simpler to think about: they are like matter and antimatter. Having both an -structure and an structure is the same as having nothing; they annihilate each other. By “having nothing” we mean “having no information”, that is, having a unit value: .
What about additive inverses? Note first that the species does not correspond to having nothing; the word “nothing” corresponds to the (i.e. unit) species. Instead the (i.e. uninhabited) species corresponds to (logical) impossibility. So to interpret we have to imagine something where having either or is impossible.
…yeah, me neither. This seems deeply strange. If someone says, “I either have an or a ”, you can confidently call them a liar, because it is impossible to have either an or a ; that is, . But surely if you actually have an -structure, it should also be true to say “I have either an or a ”? Well, that works for normal, positive species—in which case we can define a canonical injection . But once we introduce negative species this completely breaks down. As another example, if someone truthfully says, “I have either a tree or a negative non-empty tree”, you should be able to say, “Aha! I know what you have—it must be an empty tree.” In general, it’s strange that expressing a disjunction can cause some possibilities to be ruled out. Normally, we are used to disjunctions only increasing the number of possibilities.
Inspired by James and Sabry’s really cool paper The Two Dualities of Computation: Negative and Fractional Types, I have thought a bit about whether there is some plausible interpretation involving travelling backwards in time, but I haven’t been able to come up with one. I can’t quite seem to make the formalism of the paper match up with my intuition about species (though this may just be a failure of my imagination or intuition).
In any case, let’s see why the ring of virtual species actually has multiplicative inverses—at least, all the ones we could possibly hope for. This is somewhat surprising, since when we build integers from natural numbers by considering equivalence classes of pairs, we certainly don’t get any multiplicative inverses, only additive ones. To get multiplicative inverses we have to do the same process a second time, building the rational numbers as equivalence classes of pairs of integers. But species already have enough extra structure that throwing in additive inverses is all it takes.
First, a caveat: we don’t get multiplicative inverses for all species, but only those species such that : that is, species with only a single structure of size zero, which are of the form . With any constant term other than , we clearly have no hope of finding another species such that , since the constant term of will be a multiple of ’s constant term.
So given such a , write , where denotes “non-empty -structures”. Then we can define the multiplicative inverse of as follows:
That is, a -structure consists of a list of nonempty -structures, except that even-length lists are considered “positive” and odd-length lists considered “negative”.
We can easily check that this indeed defines a multiplicative inverse for :
The infinite sums telescope down to leave only . Notice this really isn’t about species in particular, but really about infinite power series (of which species are the categorification): any infinite power series with integer coefficients and a constant term of has a multiplicative inverse which is also such a power series.
As an example, consider . We know this is “supposed” to be the species of lists (since it results from solving for ), but let’s see what happens. In this case and . So the inverse ought to be
And hey, look at that! Lists!
So what would we need to get a true field, i.e. a multiplicative inverse for every nonzero species? Well, for that we would need to throw in rational coefficients. I forget exactly where I read this—some paper by Baez and Dolan, most likely—but I believe the proper way to interpret this would be as groupoid-valued species, since there is a sense in which the “cardinality” of groupoids can be interpreted as rational numbers. But to be honest I have no idea where this leads.
Note that species sum is cancellative—that is, if then —so this is a coherent definition. This cancellative property is probably worth another post of its own since the reason for it is not entirely trivial.↩
This post is literate Haskell—download it and play along!
> {-# LANGUAGE GADTs #-}
>
> import Control.Arrow ((***))
> import Data.Bits (xor)
> import Data.List (unfoldr)
> import Data.List.Split (chunksOf)
> import Data.Maybe (fromJust)
You can go read the problem description if you like, but it’s rather verbose—I’ll try to give a more concise description here, illustrated with Haskell code.
The problem is concerned with strings of bits:
> type BitString = [Bool]
We’ll start just by defining a few utility functions to view and input bitstrings conveniently.
> readbits :: String -> BitString
> readbits = map (=='1')
>
> showbits :: BitString -> String
> showbits = map (\b -> if b then '1' else '0')
>
> withbits :: (BitString -> BitString) -> String -> String
> withbits f = showbits . f . readbits
Now on to the problem proper. There is a central operation—which I’ll call the “dragon transform”—which makes a longer bitstring from a shorter one. Given a bitstring , append a 0 to the end, and then append a reversed and inverted version of (where “invert” means to flip all the bits). Like so:
> invert :: BitString -> BitString
> invert = map not
>
> dragon :: BitString -> BitString
> dragon s = s ++ [False] ++ invert (reverse s)
For example,
ghci> withbits dragon "1"
"100"
ghci> withbits dragon "1101111"
"110111100000100"
(Incidentally, this operation is called dragon
since it is related to the classic dragon curve. Hint: interpret 0 as a left turn and 1 as a right turn.)
Given a starting bitstring, and a target length, we are supposed to iterate dragon
until we have at least the number of target bits, and then truncate the string to the desired length:
> fill :: Int -> BitString -> BitString
> fill len = take len . head . dropWhile ((< len) . length) . iterate dragon
For example, if we start with 1
, we have to iterate dragon
three times to end up with at least ten bits.
ghci> map showbits . take 4 $ iterate dragon [True]
["1","100","1000110","100011001001110"]
ghci> withbits (fill 10) "1"
"1000110010"
Finally, after extending an initial bitstring to a given length, we perform a checksum operation:
(==)
). This results in a bitstring half as long. Now repeat the process, continuing to halve the length until we have an odd number of bits remaining.In code:
> checksum :: BitString -> BitString
> checksum a
> | odd (length a) = a
> | otherwise = checksum . map xnor . chunksOf 2 $ a
> where
> xnor [x,y] = x == y
So, we now have a simple reference implementation that directly follows the specification. We can use this to solve the first task, which just asks to start with a given short bitstring, extend it to length , and then compute the checksum. I think different logged-in users get different starting strings, but mine was 01000100010010111
:
> input = "01000100010010111"
ghci> withbits (checksum . fill 272) input
"10010010110011010"
Notice that , so after expanding to that length and then repeatedly halving the length, we end up with a checksum of length 17.
That was easy. Bring on the second task! Well… of course, it is much bigger. It asks to use the same starting bitstring, but this time extend it to length before computing the checksum (which will again end up having length 17). Using this naive, unoptimized implementation completely blows up: it turns out that generating a list of 35 million booleans is really not a good idea. Using actual lists with a cons cell for each bit incurs a whole lot of memory and allocation overhead; it just made my computer grind to a halt.
As you may realize, there is a lot of low-hanging fruit here: for example, we can use an unboxed Vector
instead of a list, or even do some deforestation to avoid allocation (the former code is by Eric Mertens aka glguy
, the latter by Daniel Wagner aka dmwit
). Using techniques like that, it’s possible to get the runtime and memory requirements down to something reasonable. But that’s not what I want to talk about. Though more efficient, those solutions are still actually computing every single bit. It seemed to me we shouldn’t have to do that: the computation has a lot of nice structure, and seemingly a lot of opportunity for sharing intermediate results. I went off in search of a way to compute the correct checksum without actually generating the entire intermediate bitstring.
The first order of business was to work out an algebraic understanding of the xnor
operation, which I will denote (the circled plus operator denotes xor
, and the overbar denotes logical negation). One fundamental fact is that
(checking whether and are equal is the same as first negating one and then checking whether they are unequal). From this, and the fact that is associative, we can prove associativity of xnor
:
Associativity, along with the fact that is an identity for the operation, means it forms a monoid. When we repeatedly take the xnor
of adjacent bits, we are therefore basically doing an mconcat
using a strictly balanced combining scheme. But associativity means we can be freer about the order in which we do the combining. If we start with a bitstring of length , the checksumming operation iterates times, and each consecutive sequence of bits gets folded down into a single bit via mconcat
. In other words, the checksum
operation can be reimplemented like this:
> checksum2 :: BitString -> BitString
> checksum2 a = map combine . chunksOf (powTwo (length a)) $ a
> where
> combine = foldr (==) True
>
> -- Find the biggest power of two that divides n
> powTwo n
> | odd n = 1
> | otherwise = 2 * powTwo (n `div` 2)
Let’s check that this works:
ghci> withbits (checksum2 . fill 272) input
"10010010110011010"
ghci> let bits = fill 272 (readbits input) in checksum bits == checksum2 bits
True
Now, this isn’t really any faster yet; but this idea will be important later!
There’s one more thing we can observe about xnor
: if we fold an odd number of bits with xnor
, it’s the same as taking the xor
of all the bits; if we fold an even number of bits, it’s the same as taking the xor
of all the bits and then negating the result. That is,
and so on. The proof is a simple induction argument, making use of the relation we noted before. So when folding xnor
, as a simple optimization, we can avoid doing a lot of negations by just computing the xor
and then negating appropriately based on the parity of the number of bits.
With that under our belts, we can move on to the real meat of the solution. The central idea is that instead of representing bitstrings directly as lists (or vectors, or whatever) of bits, we represent them using a deep embedding of a little bitstring algebra (aka DSL). That is, we represent each bitstring operation as a constructor of an algebraic data type, which allows us to directly manipulate bitstring expressions. The point is that this algebra/DSL has a lot of nice structure that allows us to work at an abstract, algebraic level instead of working directly with bits.
There’s one more twist to note before actually seeing the data type definition. We know that we will need to talk about the length of bitstrings as well as their xnor/xor. Instead of having to recalculate these every time we need them, we can cache them at each node of a bitstring expression. We’ll see how these cached values come in handy later.
> data BitExpr where
So, what does our algebra of bitstrings need? First, it’s useful to have an explicit representation of the empty bitstring, as well as a singleton bit. We don’t need to cache length or xor
values here, since they are obvious and can be computed in constant time.
> Emp :: BitExpr
> Bit :: Bool -> BitExpr
Next, we need to be able to append bitstrings. Notice the Bool
, which represents the cached xor
of the entire bitstring, as well as the Integer
which represents the length.
> App :: !Bool -> !Integer -> BitExpr -> BitExpr -> BitExpr
Finally, we need three unary operations on bitstrings: invert, reverse, and dragon
. Each also carries a cached length and xor
.
> Inv :: !Bool -> !Integer -> BitExpr -> BitExpr
> Rev :: !Bool -> !Integer -> BitExpr -> BitExpr
> Drg :: !Bool -> !Integer -> BitExpr -> BitExpr
>
> deriving Show
Note that Drg
is redundant in some sense, since the dragon transform can be encoded in terms of append, inverse, and reverse. However, it’s critical that we include it explicitly: since the dragon transform uses the input bitstring twice, expanding an iterated application of Drg
in terms of the other constructors would result in an exponential blowup in the size of the expression.
To be concrete, let’s write a straightforward interpreter which formally connects a bitstring expression with its intended semantics as a bitstring. This comes in handy for testing, but other than testing, the whole point is that we will not use this—we want to solve the problem at the level of bitstring expressions, without ever actually generating their corresponding bitstrings.
> toBits :: BitExpr -> BitString
> toBits Emp = []
> toBits (Bit b) = [b]
> toBits (App _ _ s1 s2) = toBits s1 ++ toBits s2
> toBits (Inv _ _ s) = invert (toBits s)
> toBits (Rev _ _ s) = reverse (toBits s)
> toBits (Drg _ _ s) = dragon (toBits s)
Next, let’s write some simple utility functions to extract the cached length or xor
from the root of a bitstring expression:
> bsLen :: BitExpr -> Integer
> bsLen Emp = 0
> bsLen (Bit _) = 1
> bsLen (App _ l _ _) = l
> bsLen (Inv _ l _) = l
> bsLen (Rev _ l _) = l
> bsLen (Drg _ l _) = l
>
> bsXor :: BitExpr -> Bool
> bsXor Emp = False
> bsXor (Bit b) = b
> bsXor (App b _ _ _) = b
> bsXor (Inv b _ _) = b
> bsXor (Rev b _ _) = b
> bsXor (Drg b _ _) = b
Next, we’ll write some smart constructors which automatically take care of properly computing the cached length and xor
.
> bit :: Bool -> BitExpr
> bit = Bit
Appending combines xor
values with xor
and adds lengths. app
also does a bit of optimization when appending with the empty bitstring. For convenience, we can also use app
to create a function bits
to convert a literal bitstring into a BitExpr
.
> app :: BitExpr -> BitExpr -> BitExpr
> app s1 Emp = s1
> app s1 s2 = App (bsXor s1 `xor` bsXor s2) (bsLen s1 + bsLen s2) s1 s2
>
> bits :: String -> BitExpr
> bits = foldr (app . bit . (=='1')) Emp
Inverting a bitstring preserves the xor
when it has even length, and inverts the xor
when it has odd length. Note how we make use of both the cached xor
and length values to compute the new cached xor
.
> inv :: BitExpr -> BitExpr
> inv s = Inv (if even (bsLen s) then bsXor s else not (bsXor s))
> (bsLen s)
> s
Reversing preserves xor
and length.
> rev :: BitExpr -> BitExpr
> rev s = Rev (bsXor s) (bsLen s) s
Finally, the dragon
operation: the xor
of dragon s
is the xor
of s
combined with the xor
of inv s
; the length is one more than twice the length of s
.
> drg :: BitExpr -> BitExpr
> drg s = Drg (bsXor s `xor` bsXor (inv s)) (2*(bsLen s) + 1) s
We can test these:
ghci> let t = drg (bits "11" `app` inv (bits "10000"))
ghci> showbits . toBits $ t
"110111100000100"
ghci> bsLen t
15
Remember that our high-level goal is to take the expanded version of our bitstring, split it into blocks of length , and then separately reduce each block with xnor
. It turns out that we have enough information to split a bitstring expression into two bitstring expressions which correspond to splitting off a block of a given size from the beginning of the corresponding bitstring. That is, we will write a function splitBits :: Integer -> BitExpr -> (BitExpr, BitExpr)
which works like splitAt
, but on bitstring expressions instead of bitstrings. In other words, it will satisfy the property
splitAt n . toBits == (toBits *** toBits) . splitBits n
We’ll go through the implementation case by case. You might like to try implementing splitBits
yourself before peeking at mine; it makes for a nice exercise.
> splitBits :: Integer -> BitExpr -> (BitExpr, BitExpr)
In the base cases, to split zero bits off the front of a bitstring, or if we are asked to split off more bits than there are, just generate the empty bitstring expression.
> splitBits 0 s = (Emp, s)
> splitBits n s | n >= bsLen s = (s, Emp)
To split an App
node, compare the number of bits we want to split off with the length of the first bitstring, and recursively split in either the left or right side appropriately, remembering to subtract the length of the first bitstring from the number of bits to split if we recurse on the right side.
> splitBits n (App _ _ s1 s2)
> | n < bsLen s1
> = let (s1a, s1b) = splitBits n s1 in (s1a, s1b `app` s2)
> | otherwise
> = let (s2a, s2b) = splitBits (n - bsLen s1) s2 in (s1 `app` s2a, s2b)
Inverting commutes with splitting, so to split an Inv
node, we can just split recursively and then rewrap the results with inv
.
> splitBits n (Inv _ _ s) = (inv *** inv) $ splitBits n s
To split Rev
and Drg
nodes, we expand the expressions a bit to get rid of the top-level constructor before re-calling splitBits
.
> splitBits n (Rev _ _ s) = splitBits n (pushRev s)
> splitBits n (Drg _ _ s) = splitBits n (expandDragon s)
In the case of Rev
, we can “push the reverse through” one level, transforming it into an equivalent expression which no longer has a Rev
node at the top. We make use of some nice algebraic properties governing the interaction of reverse with the other operations:
reverse (s1 ++ s2) == reverse s2 ++ reverse s1
reverse . invert = invert . reverse
reverse . reverse = id
reverse . dragon = dragon . invert
, which can be easily proved by expanding dragon
in terms of the other operations and then applying the above algebraic laws.Using these properties, we can implement pushRev
as follows:
> pushRev :: BitExpr -> BitExpr
> pushRev Emp = Emp
> pushRev (Bit b) = Bit b
> pushRev (App _ _ s1 s2) = rev s2 `app` rev s1
> pushRev (Inv _ _ s) = inv (rev s)
> pushRev (Rev _ _ s) = s
> pushRev (Drg _ _ s) = drg (inv s)
Finally, expandDragon
just expands a dragon operation in terms of the other operations. Although this approximately doubles the size of the bitstring expression, we only do this lazily, when we are actually trying to split the result of a dragon transform. It’s only natural that splitting an expression results in somewhat larger expressions.
> expandDragon :: BitExpr -> BitExpr
> expandDragon s = s `app` (bit False `app` inv (rev s))
We’re almost there! We can now implement the fill
and checksum
operations at the level of bitstring expressions.
fill
is straightforward: keep applying the drg
smart constructor until the cached length is sufficient, then use splitBits
to create an expression corresponding to only the first bits.
> fillE :: Integer -> String -> BitExpr
> fillE n str = fst . splitBits n $ go (bits str)
> where
> go s | bsLen s >= n = s
> | otherwise = go (drg s)
Finally, we can implement checksumE
using the same pattern as checksum2
, where we break up the string into chunks of size and then reduce each chunk. The only difference is that now we use splitBits
to split, and the cached xor
to compute the reduction. We know each of the blocks has an even length, so the xnor
is just the negation of the cached xor
.
> checksumE :: BitExpr -> BitString
> checksumE s = map (not . bsXor) . unfoldr doSplit $ s
> where
> doSplit Emp = Nothing
> doSplit s = Just (splitBits blockLen s)
> blockLen = powTwo (bsLen s)
> powTwo n
> | odd n = 1
> | otherwise = 2 * powTwo (n `div` 2)
Let’s check that we get the same answer for the first task:
ghci> showbits $ checksumE (fillE 272 input)
"10010010110011010"
ghci> withbits (checksum . fill 272) input
"10010010110011010"
Great! And now for the second task:
ghci> showbits $ checksumE (fillE (17 * 2^21) input)
"01010100101011100"
On my machine this finishes pretty much instantaneously, taking only 0.02 seconds. In order to generate enough bits, the dragon transform must be applied 21 times, but that just generates a small expression with 21 Drg
constructors. Splitting into chunks of length certainly expands the size of the expressions a bit, but everything stays nice and logarithmic since many of the Drg
constructors can remain unexpanded.
In fact, this can easily handle MUCH larger problem instances. For example:
ghci> showbits $ checksumE (fillE (17 * 2^80) input)
"10000100010001100"
ghci> showbits $ checksumE (fillE (17 * 2^81) input)
"01010100101011100"
Semantically, this corresponds to generating yottabytes worth of bits (I had to look up the proper prefix) and then checksumming them; operationally, though, these are still basically instantaneous. (Interestingly, I also tried , and it instantaneously printed the first 11 bits of the answer and then segfaulted. Perhaps I have found a bug in GHC 8.0.2.)
Notice that the checksum for is actually the same as that for . After playing around with it a bit, the checksums for seem to have a period of 12, but I’m not sure how to prove it!
Last May, I wrote about my ideas for designing a new PL course, and got a lot of great comments and feedback. Well, somehow I survived the semester, and the course is now over. In the end I’m pretty happy with how it went (though of course there are always things that can be improved next time).
I decided to use class time in an unconventional way: for each class meeting I created a “module”, consisting of a literate Haskell file with some example code, explanatory text, and lots of holes where students needed to write answers to exercises or fill in code. I split the students into groups, and they spent class time just working through the module. Instead of standing at the front lecturing, I just wandered around watching them work and answering questions. It took a bit of getting used to—for the first few classes I couldn’t shake the feeling that I wasn’t really doing my job—but it quickly became clear that the students were really learning and engaging with the material in a way that they would not have been able to if I had just lectured.
A happy byproduct of this approach is that the modules are fairly self-contained and can now be used by anyone to learn the material. Reading through all the modules and working through the exercises should be a great option for anyone wishing to learn some basics of programming language design and implementation. For example, I know I will probably reuse it to get summer research students up to speed. Note that the course assumes no knowledge of Haskell (so those familiar with Haskell can safely skip the first few modules), but introduces just enough to get where I want to go.
I don’t plan to release any solutions, so don’t ask. But other than that, questions, comments, bug reports, etc. are welcome!
MonadRandom-0.5
, a package which provides a convenient monadic interface for random number generation in the style of transformers
and mtl
: a Rand
monad (essentially a state monad that threads through a generator), a monad transformer variant called RandT
, and a RandomMonad
class allowing the use of random generation operations in monad stacks containing RandT
.
This release has quite a few small additions as well as a big module reorganization. However, thanks to module re-exports, most existing code using the library should continue to work with no changes; the major version bump reflects the large reorganization and my resultant inability to 100% guarantee that existing user code will not break. If your code does break, please let me know—I would be happy to help you fix it, or simply to know about it so I can help other users.
Here are a few of the biggest changes that may be of interest to users of the library:
A new MonadInterleave
class (see #20), which is a big improvement over MonadSplit
. It provides a method interleave :: m a -> m a
, which works by splitting the generator, running its argument using one half of the generator, and using the other half as the final state of the resulting action (replacing whatever the final generator state otherwise would have been). This can be used, for example, to allow random computations to run in parallel, or to create lazy infinite structures of random values. In the example below, the infinite tree randTree
cannot be evaluated lazily: even though it is cut off at two levels deep by hew 2
, the random value in the right subtree still depends on generation of all the random values in the (infinite) left subtree, even though they are ultimately unneeded. Inserting a call to interleave
, as in randTreeI
, solves the problem: the generator splits at each Node
, so random values in the left and right subtrees are generated independently.
data Tree = Leaf | Node Int Tree Tree deriving Show
hew :: Int -> Tree -> Tree
hew 0 _ = Leaf
hew _ Leaf = Leaf
hew n (Node x l r) = Node x (hew (n-1) l) (hew (n-1) r)
randTree :: Rand StdGen Tree
randTree = Node <$> getRandom <*> randTree <*> randTree
randTreeI :: Rand StdGen Tree
randTreeI = interleave $ Node <$> getRandom <*> randTreeI <*> randTreeI
>>> hew 2 <$> evalRandIO randTree
Node 2168685089479838995 (Node (-1040559818952481847) Leaf Leaf) (Node ^CInterrupted.
>>> hew 2 <$> evalRandIO randTreeI
Node 8243316398511136358 (Node 4139784028141790719 Leaf Leaf) (Node 4473998613878251948 Leaf Leaf)
A new PrimMonad
instance for RandT
(thanks to Koz Ross), allowing it to be used in conjunction with e.g. mutable vectors.
fromList
now raises an error when the total weight of elements is zero.uniform
is generalized to work over any Foldable
.weighted
, weightedMay
, fromListMay
, and uniformMay
have been added. weighted
is like fromList
but generalized to work over any Foldable
. The May
variants, of course, return a Maybe
result instead of raising an error.New lazy vs strict variants of the Rand
monad. If you import Control.Monad.Random
or Control.Monad.Trans.Random
you get the Lazy
variant re-exported by default, but you can explicitly import .Lazy
or .Strict
if you want. They provide the exact same API, but Lazy
is implemented with a lazy state monad and Strict
with a strict one. To be honest it’s not clear what difference this might make, but since the distinction is already there with the underlying state monad for free, why not provide it?
Although there was some discussion of generalizing MonadRandom
to work for a wider range of underlying generators (see the comments on my previous blog post and the discussion on issue #26), I decided to punt on that for now. It seems rather complicated, and that there are already good alternatives like the very nice random-fu
package, so I decided to keep things simple for this release. I’m still open to proposals for generalizing future releases.
For a full rundown of changes in 0.5, see the change log. Comments, questions, and bug reports are always welcome either as a comment on this blog post or on the GitHub issue tracker.
MonadRandom
package, which provides an mtl
-style type class for monads with support for generation of pseudorandom values, along with a concrete random monad transformer RandT
. As of this writing it has 89 reverse dependencies on Hackage—a healthy number, and one that makes me think carefully about any breaking changes to the package.
Recently I got a number of pull requests, and have been working on putting together an 0.5 release which adds a few functions, adds lazy- and strict-state variants of RandT
, and reorganizes things to be closer to standard practice of the transformers
package. Since this release will include some technically breaking changes already, it’s a good time to think about potentially including others.
The one thing I am not sure what to do about is this issue: Allow MonadRandom interface for MWC-random. mwc-random
is a very nice package for psuedorandom number generation, but apparently it does not fit into the MonadRandom
abstraction. First of all, I would like to understand why—I am not very familiar with mwc-random
. Second of all, I’d love to figure out a solution, but ideally one that causes as little breakage to existing code as possible.
Leave a comment (either here or on the github issue) if this is something you know/care about, and let’s see if we can figure out a good solution together!
I recently solved a problem (using Haskell) that ended up being tougher than I thought, but I learned a lot along the way. Rather than just presenting a solution, I’d like to take you through my thought process, crazy detours and all.
Of course, I should preface this with a big spoiler alert: if you want to try solving the problem yourself, you should stop reading now!
> {-# LANGUAGE GADTs #-}
> {-# LANGUAGE DeriveFunctor #-}
>
> module Brackets where
>
> import Data.List (sort, genericLength)
> import Data.MemoTrie (memo, memo2)
> import Prelude hiding ((++))
There’s a lot of extra verbiage at the official problem description, but what it boils down to is this:
Find the th element of the lexicographically ordered sequence of all balanced bracketings of length .
There is a longer description at the problem page, but hopefully a few examples will suffice. A balanced bracketing is a string consisting solely of parentheses, in which opening and closing parens can be matched up in a one-to-one, properly nested way. For example, there are five balanced bracketings of length :
((())), (()()), (())(), ()(()), ()()()
By lexicographically ordered we just mean that the bracketings should be in “dictionary order” where (
comes before )
, that is, bracketing comes before bracketing if and only if in the first position where they differ, has (
and has )
. As you can verify, the list of length- bracketings above is, in fact, lexicographically ordered.
Oh, this is easy, I thought, especially if we consider the well-known isomorphism between balanced bracketings and binary trees. In particular, the empty string corresponds to a leaf, and (L)R
(where L
and R
are themselves balanced bracketings) corresponds to a node with subtrees L
and R
. So the five balanced bracketings of length correspond to the five binary trees with three nodes:
We can easily generate all the binary trees of a given size with a simple recursive algorithm. If , generate a Leaf
; otherwise, decide how many nodes to put on the left and how many on the right, and for each such distribution recursively generate all possible trees on the left and right.
> data Tree where
> Leaf :: Tree
> Node :: Tree -> Tree -> Tree
> deriving (Show, Eq, Ord)
>
> allTrees :: Int -> [Tree]
> allTrees 0 = [Leaf]
> allTrees n =
> [ Node l r
> | k <- [0 .. n-1]
> , l <- allTrees ((n-1) - k)
> , r <- allTrees k
> ]
We generate the trees in “left-biased” order, where we first choose to put all nodes on the left, then on the left and on the right, and so on. Since a subtree on the left will result in another opening paren, but a subtree on the right will result in a closing paren followed by an open paren, it makes intuitive sense that this corresponds to generating bracketings in sorted order. You can see that the size- trees above, generated in left-biased order, indeed have their bracketings sorted.
Writing allTrees
is easy enough, but it’s definitely not going to cut it: the problem states that we could have up to . The number of trees with nodes has 598 digits (!!), so we can’t possibly generate the entire list and then index into it. Instead we need a function that can more efficiently generate the tree with a given index, without having to generate all the other trees before it.
So I immediately launched into writing such a function, but it’s tricky to get right. It involves computing Catalan numbers, and cumulative sums of products of Catalan numbers, and divMod
, and… I never did get that function working properly.
But I never should have written that function in the first place! What I should have done first was to do some simple tests just to confirm my intuition that left-biased tree order corresponds to sorted bracketing order. Because if I had, I would have found this:
> brackets :: Tree -> String
> brackets Leaf = ""
> brackets (Node l r) = mconcat ["(", brackets l, ")", brackets r]
>
> sorted :: Ord a => [a] -> Bool
> sorted xs = xs == sort xs
ghci> sorted (map brackets (allTrees 3))
True
ghci> sorted (map brackets (allTrees 4))
False
As you can see, my intuition actually led me astray! is a small enough case that left-biased order just happens to be the same as sorted bracketing order, but for this breaks down. Let’s see what goes wrong:
In the top row are the size- trees in “left-biased” order, i.e. the order generated by allTrees
. You can see it is nice and symmetric: reflecting the list across a vertical line leaves it unchanged. On the bottom row are the same trees, but sorted lexicographically by their bracketings. You can see that the lists are almost the same except the red tree is in a different place. The issue is the length of the left spine: the red tree has a left spine of three nodes, which means its bracketing will begin with (((
, so it should come before any trees with a left spine of length 2, even if they have all their nodes in the left subtree (whereas the red tree has one of its nodes in the right subtree).
My next idea was to try to somehow enumerate trees in order by the length of their left spine. But since I hadn’t even gotten indexing into the original left-biased order to work, it seemed hopeless to get this to work by implementing it directly. I needed some bigger guns.
At this point I had the good idea to introduce some abstraction. I defined a type of enumerations (a la FEAT or data/enumerate):
> data Enumeration a = Enumeration
> { fromNat :: Integer -> a
> , size :: Integer
> }
> deriving Functor
>
> enumerate :: Enumeration a -> [a]
> enumerate (Enumeration f n) = map f [0..n-1]
An Enumeration
consists of a size
along with a function Integer -> a
, which we think of as being defined on [0 .. size-1]
. That is, an Enumeration
is isomorphic to a finite list of a given length, where instead of explicitly storing the elements, we have a function which can compute the element at a given index on demand. If the enumeration has some nice combinatorial structure, then we expect that this on-demand indexing can be done much more efficiently than simply listing all the elements. The enumerate
function simply turns an Enumeration
into the corresponding finite list, by mapping the indexing function over all possible indices.
Note that Enumeration
has a natural Functor
instance, which GHC can automatically derive for us. Namely, if e
is an Enumeration
, then fmap f e
is the Enumeration
which first computes the element of e
for a given index, and then applies f
to it before returning.
Now, let’s define some combinators for building Enumeration
s. We expect them to have all the nice algebraic flavor of finite lists, aka free monoids.
First, we can create empty or singleton enumerations, or convert any finite list into an enumeration:
> empty :: Enumeration a
> empty = Enumeration (const undefined) 0
>
> singleton :: a -> Enumeration a
> singleton a = Enumeration (\_ -> a) 1
>
> list :: [a] -> Enumeration a
> list as = Enumeration (\n -> as !! fromIntegral n) (genericLength as)
ghci> enumerate (empty :: Enumeration Int)
[]
ghci> enumerate (singleton 3)
[3]
ghci> enumerate (list [4,6,7])
[4,6,7]
We can form the concatenation of two enumerations. The indexing function compares the given index against the size of the first enumeration, and then indexes into the first or second enumeration appropriately. For convenience we can also define union
, which is just an iterated version of (++)
.
> (++) :: Enumeration a -> Enumeration a -> Enumeration a
> e1 ++ e2 = Enumeration
> (\n -> if n < size e1 then fromNat e1 n else fromNat e2 (n - size e1))
> (size e1 + size e2)
>
> union :: [Enumeration a] -> Enumeration a
> union = foldr (++) empty
ghci> enumerate (list [3, 5, 6] ++ empty ++ singleton 8)
[3,5,6,8]
Finally, we can form a Cartesian product: e1 >< e2
is the enumeration of all possible pairs of elements from e1
and e2
, ordered so that all the pairs formed from the first element of e1
come first, followed by all the pairs with the second element of e1
, and so on. The indexing function divides the given index by the size of e2
, and uses the quotient to index into e1
, and the remainder to index into e2
.
> (><) :: Enumeration a -> Enumeration b -> Enumeration (a,b)
> e1 >< e2 = Enumeration
> (\n -> let (l,r) = n `divMod` size e2 in (fromNat e1 l, fromNat e2 r))
> (size e1 * size e2)
ghci> enumerate (list [1,2,3] >< list [10,20])
[(1,10),(1,20),(2,10),(2,20),(3,10),(3,20)]
ghci> let big = list [0..999] >< list [0..999] >< list [0..999] >< list [0..999]
ghci> fromNat big 2973428654
(((2,973),428),654)
Notice in particular how the fourfold product of list [0..999]
has elements, but indexing into it with fromNat
is basically instantaneous.
Since Enumeration
s are isomorphic to finite lists, we expect them to have Applicative
and Monad
instances, too. First, the Applicative
instance is fairly straightforward:
> instance Applicative Enumeration where
> pure = singleton
> f <*> x = uncurry ($) <$> (f >< x)
ghci> enumerate $ (*) <$> list [1,2,3] <*> list [10, 100]
[10,100,20,200,30,300]
pure
creates a singleton enumeration, and applying an enumeration of functions to an enumeration of arguments works by taking a Cartesian product and then applying each pair.
The Monad
instance works by substitution: in e >>= k
, the continuation k
is applied to each element of the enumeration e
, and the resulting enumerations are unioned together in order.
> instance Monad Enumeration where
> return = pure
> e >>= f = union (map f (enumerate e))
ghci> enumerate $ list [1,2,3] >>= \i -> list (replicate i i)
[1,2,2,3,3,3]
Having to actually enumerate the elements of e
is a bit unsatisfying, but there is really no way around it: we otherwise have no way to know how big the resulting enumerations are going to be.
Now, that function I tried (and failed) to write before that generates the tree at a particular index in left-biased order? Using these enumeration combinators, it’s a piece of cake. Basically, since we built up combinators that mirror those available for lists, it’s just as easy to write this indexing version as it is to write the original allTrees
function (which I’ve copied below for comparison):
allTrees :: Int -> [Tree]
allTrees 0 = [Leaf]
allTrees n =
[ Node l r
| k <- [0 .. n-1]
, l <- allTrees ((n-1) - k)
, r <- allTrees k
]
> enumTrees :: Int -> Enumeration Tree
> enumTrees 0 = singleton Leaf
> enumTrees n = union
> [ Node <$> enumTrees (n-k-1) <*> enumTrees k
> | k <- [0 .. n-1]
> ]
(enumTrees
and allTrees
look a bit different, but actually allTrees
can be rewritten in a very similar style:
allTrees :: Int -> [Tree]
allTrees 0 = [Leaf]
allTrees n = concat
[ Node <$> allTrees ((n-1) - k) <*> r <- allTrees k
| k <- [0 .. n-1]
]
Doing as much as possible using the Applicative
interface gives us added “parallelism”, which in this case means the ability to index directly into a product with divMod
, rather than scanning through the results of calling a function on enumerate
until we have accumulated the right size. See the paper on the GHC ApplicativeDo
extension.)
Let’s try it out:
ghci> enumerate (enumTrees 3)
[Node (Node (Node Leaf Leaf) Leaf) Leaf,Node (Node Leaf (Node Leaf Leaf)) Leaf,Node (Node Leaf Leaf) (Node Leaf Leaf),Node Leaf (Node (Node Leaf Leaf) Leaf),Node Leaf (Node Leaf (Node Leaf Leaf))]
ghci> enumerate (enumTrees 3) == allTrees 3
True
ghci> enumerate (enumTrees 7) == allTrees 7
True
ghci> brackets $ fromNat (enumTrees 7) 43
"((((()())))())"
It seems to work! Though actually, if we try larger values of , enumTrees
just seems to hang. The problem is that it ends up making many redundant recursive calls. Well… nothing a bit of memoization can’t fix! (Here I’m using Conal Elliott’s nice MemoTrie package.)
> enumTreesMemo :: Int -> Enumeration Tree
> enumTreesMemo = memo enumTreesMemo'
> where
> enumTreesMemo' 0 = singleton Leaf
> enumTreesMemo' n = union
> [ Node <$> enumTreesMemo (n-k-1) <*> enumTreesMemo k
> | k <- [0 .. n-1]
> ]
ghci> size (enumTreesMemo 10)
16796
ghci> size (enumTreesMemo 100)
896519947090131496687170070074100632420837521538745909320
ghci> size (enumTreesMemo 1000)
2046105521468021692642519982997827217179245642339057975844538099572176010191891863964968026156453752449015750569428595097318163634370154637380666882886375203359653243390929717431080443509007504772912973142253209352126946839844796747697638537600100637918819326569730982083021538057087711176285777909275869648636874856805956580057673173655666887003493944650164153396910927037406301799052584663611016897272893305532116292143271037140718751625839812072682464343153792956281748582435751481498598087586998603921577523657477775758899987954012641033870640665444651660246024318184109046864244732001962029120
ghci> brackets $ fromNat (enumTreesMemo 1000) 8234587623904872309875907638475639485792863458726398487590287348957628934765
"((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((()(((()((((()))())(()()()))()(())(())((()((()))(((())()(((((()(((()()))(((()((((()()(())()())(((()))))(((()()()(()()))))(((()((()))(((()())())))())(()()(())(())()(()())))()))((()()))()))()))()(((()))(()))))))())()()()))((())((()))((((())(())))((())))))()))()(())))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))"
That’s better!
At this point, I thought that I needed to enumerate trees in order by the length of their left spine. Given a tree with a left spine of length , we enumerate all the ways to partition the remaining elements among the right children of the spine nodes, preferring to first put elements as far to the left as possible. As you’ll see, this turns out to be wrong, but it’s fun to see how easy it is to write this using the enumeration framework.
First, we need an enumeration of the partitions of a given into exactly parts, in lexicographic order.
> kPartitions :: Int -> Int -> Enumeration [Int]
There is exactly one way to partition into zero parts.
> kPartitions 0 0 = singleton []
We can’t partition anything other than into zero parts.
> kPartitions _ 0 = empty
Otherwise, pick a number from down to to go in the first spot, and then recursively enumerate partitions of into exactly parts.
> kPartitions n k = do
> i <- list [n, n-1 .. 0]
> (i:) <$> kPartitions (n-i) (k-1)
Let’s try it:
ghci> let p43 = enumerate $ kPartitions 4 3
ghci> p43
[[4,0,0],[3,1,0],[3,0,1],[2,2,0],[2,1,1],[2,0,2],[1,3,0],[1,2,1],[1,1,2],[1,0,3],[0,4,0],[0,3,1],[0,2,2],[0,1,3],[0,0,4]]
ghci> all ((==3) . length) p43
True
ghci> all ((==4) . sum) p43
True
ghci> sorted (reverse p43)
True
Now we can use kPartitions
to build our enumeration of trees:
> spinyTrees :: Int -> Enumeration Tree
> spinyTrees = memo spinyTrees'
> where
> spinyTrees' 0 = singleton Leaf
> spinyTrees' n = do
>
> -- Pick the length of the left spine
> spineLen <- list [n, n-1 .. 1]
>
> -- Partition the remaining elements among the spine nodes
> bushSizes <- kPartitions (n - spineLen) spineLen
> bushes <- traverse spinyTrees bushSizes
> return $ buildSpine (reverse bushes)
>
> buildSpine :: [Tree] -> Tree
> buildSpine [] = Leaf
> buildSpine (b:bs) = Node (buildSpine bs) b
This appears to give us something reasonable:
ghci> size (spinyTrees 7) == size (enumTreesMemo 7)
True
But it’s pretty slow—which is to be expected with all those monadic operations required. And there’s more:
ghci> sorted . map brackets . enumerate $ spinyTrees 3
True
ghci> sorted . map brackets . enumerate $ spinyTrees 4
True
ghci> sorted . map brackets . enumerate $ spinyTrees 5
False
Foiled again! All we did was stave off failure a bit, until . I won’t draw all the trees of size for you, but the failure mode is pretty similar: picking subtrees for the spine based just on how many elements they have doesn’t work, because there are cases where we want to first shift some elements to a later subtree, keeping the left spine of a subtree, before moving the elements back and having a shorter left spine.
It finally occurred to me that there was nothing in the problem statement that said anything about trees. That was just something my overexcited combinatorial brain imposed on it: obviously, since there is a bijection between balanced bracketings and binary trees, we should think about binary trees, right? …well, there is also a bijection between balanced bracketings and permutations avoiding (231), and lattice paths that stay above the main diagonal, and hundreds of other things, so… not necessarily.
In this case, I think trees just end up making things harder. Let’s think instead about enumerating balanced bracket sequences directly. To do it recursively, we need to know how to enumerate possible endings to the start of any balanced bracket sequence. That is, we need to enumerate sequences containing opening brackets and extra closing brackets (so closing brackets in total), which can be appended to a sequence of brackets with more opening brackets than closing brackets.
Given this idea, the code is fairly straightforward:
> enumBrackets :: Int -> Enumeration String
> enumBrackets n = enumBracketsTail n 0
>
> enumBracketsTail :: Int -> Int -> Enumeration String
> enumBracketsTail = memo2 enumBracketsTail'
> where
To enumerate a sequence with no opening brackets, just generate c
closing brackets.
> enumBracketsTail' 0 c = singleton (replicate c ')')
To enumerate balanced sequences with opening brackets and an exactly matching number of closing brackets, start by generating an opening bracket and then continue by generating sequences with opening brackets and one extra closing bracket to match the opening bracket we started with.
> enumBracketsTail' n 0 = ('(':) <$> enumBracketsTail (n-1) 1
In general, a sequence with opening and extra closing brackets is either an opening bracket followed by an (n-1, c+1)
-sequence, or a closing bracket followed by an (n, c-1)
-sequence.
> enumBracketsTail' n c =
> (('(':) <$> enumBracketsTail (n-1) (c+1))
> ++
> ((')':) <$> enumBracketsTail n (c-1))
This is quite fast, and as a quick check, it does indeed seem to give us the same size enumerations as the other tree enumerations:
ghci> fromNat (enumBrackets 40) 16221270422764920820
"((((((((()((())()(()()()())(()))((()()()()(()((()())))((()())))))))()))()())()))"
ghci> size (enumBrackets 100) == size (enumTreesMemo 100)
True
But, are they sorted? It would seem so!
ghci> all sorted (map (enumerate . enumBrackets) [1..10])
True
At this point, you might notice that this can be easily de-abstracted into a fairly simple dynamic programming solution, using a 2D array to keep track of the size of the enumeration for each (n,c)
pair. I’ll leave the details to interested readers.
Although I was a coathor on an ICFP paper in 2011, when it was in Tokyo, I did not go since my son was born the same week. So this was my first time in Japan, or anywhere in Asia, for that matter. (Of course, this time I missed my son’s fifth birthday…)
I’ve been to Europe multiple times, and although it is definitely foreign, the culture is similar enough that I feel like I basically know how to behave. I did not feel that way in Japan. I’m pretty sure I was constantly being offensive without realizing it, but most of the time people were polite and accommodating.
…EXCEPT for that one time I was sitting in a chair chatting with folks during a break between sessions, with my feet up on a (low, plain) table, and an old Japanese guy WHACKED his walking stick on the table and shouted angrily at me in Japanese. That sure got my adrenaline going. Apparently putting your feet on the table is a big no-no, lesson learned.
The food was amazing even though I didn’t know what half of it was. I was grateful that I (a) am not vegetarian, (b) know how to use chopsticks decently well, and (c) am an adventurous eater. If any one of those were otherwise, things might have been more difficult!
On my last day in Japan I had the whole morning before I needed to head to the airport, so Ryan Yates and I wandered around Nara and saw a bunch of temples, climbed the hill, and such. It’s a stunningly beautiful place with a rich history.
As usual, it’s all about the people. I enjoyed meeting some new people, including (but not limited to):
My student Ollie Kwizera came for PLMW—it was fun having him there. I only crossed paths with him three or four times, but I think that was all for the best, since he made his own friends and had his own experiences.
Other people who I enjoyed seeing and remember having interesting conversations with include (but I am probably forgetting someone!) Michael Adams, Daniel Bergey, Jan Bracker, Joachim Breitner, David Christiansen, David Darais, Stephen Dolan, Richard Eisenberg, Kenny Foner, Marco Gaboardi, Jeremy Gibbons, John Hughes, David Janin, Neel Krishnaswami, Dan Licata, Andres Löh, Simon Marlow, Tom Murphy, Peter-Michael Osera, Jennifer Paykin, Simon Peyton Jones, Ryan Scott, Mary Sheeran, Mike Sperber, Luite Stegeman, Wouter Swierstra, David Terei, Ryan Trinkle, Tarmo Uustalu, Stephanie Weirich, Nick Wu, Edward Yang, and Ryan Yates. My apologies if I forgot you, just remind me and I’ll add you to the list! I’m amazed and grateful I get to know all these cool people.
Here are just a few of my favorite talks:
I’m a sucker for anything involving geometry and/or random testing and/or pretty pictures, and Ilya Sergey’s talk Growing and Shrinking Polygons for Random testing of Computational Geometry had them all. In my experience, doing effective random testing in any domain beyond basic functions usually requires some interesting domain-specific insights, and Ilya had some cool insights about ways to generate and shrink polygons in ways that were much more likely to generate small counterexamples for computational geometry algorithms.
Idris gets more impressive by the day, and I always enjoy David Christiansen’s talks.
Sandra Dylus gave a fun talk, All Sorts of Permutations, with the cute observation that a sorting algorithm equipped with a nondeterministic comparison operator generates permutations (though it goes deeper than that). During the question period someone asked whether there is a way to generate all partitions, and someone sitting next to me suggested using the group
function—and indeed, I think this works. I wonder what other sorts of combinatorial objects can be enumerated by this method. In particular I wonder if quicksort with nondeterministic comparisons can be adapted to generate not just all permutations, but all binary trees.
I greatly enjoyed TyDe, especially Jeremy Gibbons’ talk on APLicative Programming with Naperian Functors (I don’t think the video is online yet, if there is one). I’ll be serving as co-chair of the TyDe program committee next year, so start thinking about what you would like to submit!
There were also some fun talks at FARM, for example, Jay McCarthy’s talk on Bithoven. But I don’t think the FARM videos are uploaded yet. Speaking of FARM, the performance evening was incredible. It will be hard to live up to next year.