In my previous post I challenged you to solve Subway Tree System, which encodes trees by recording sequences of steps taken away from and towards the root while exploring the whole tree, and asks whether two such recordings denote the same tree. There are two main difficulties here: the first is how to do the parsing; second, how to compare two trees when we don’t care about the order of children at each node. Thanks to all of you who posted your solutions—I learned a lot. I often feel like my solution is obviously the “only” solution, but then when I see how others solve a problem I realize that the solution space is much larger than I thought!

## My solution

Here’s my solution, with some commentary interspersed. First, some pragmas and imports and such:

```
{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE TupleSections #-}
import Control.Arrow
import Data.Bool
import qualified Data.ByteString.Lazy.Char8 as C
import Data.Function
import Data.List
import Data.List.Split
import Data.Map (Map)
import qualified Data.Map as M
import Text.Parsec
import Text.Parsec.ByteString.Lazy
import Text.Parsec.Char
```

My `main`

then looks like this:

```
main = C.interact $
C.lines >>> drop 1 >>> chunksOf 2 >>>
map (solve >>> bool "different" "same") >>> C.unlines
```

The use of `ByteString`

instead of `String`

isn’t really necessary for this problem, just habit. I split the input into lines, group them in twos using `Data.List.Split.chunksOf`

, solve each test case, and turn the output into `different`

or `same`

appropriately. (`Data.Bool.bool`

is the fold/case analysis for the `Bool`

type; I never use it in any other Haskell code but am unreasonably fond of it for this particular use case.) It would also be possible to use the `Scanner`

abstraction instead of `lines`

, `drop`

, and `chunksOf`

, as commenter blaisepascal2014 did. In some ways that would actually be nicer, but I often default to using these more basic tools in simple cases.

## Parsing

Now for parsing the trees. The parsing is not *too* bad, and several commenters essentially did it manually with a recursive function manipulating a stack and so on; the most creative used a tree zipper to literally walk around the tree being constructed, just like you are supposedly walking around a subway in the problem. However, the `parsec`

package is available in the Kattis environment, so the easiest thing is to actually whip up a proper little parser. (I know of several other Kattis problems which can be nicely solved using parser combinators but would be annoying otherwise, for example, Calculator and Otpor. A rather fiendish but fun parsing puzzle is Learning to Code.)

```
readTree :: C.ByteString -> Tree
readTree = parse parseTree "" >>> either undefined id
where
parseTree = Node <$> parseForest
parseForest = fromList <$> many parseSubtree
parseSubtree = char '0' *> parseTree <* char '1'
```

Of course I haven’t actually shown the definition of `Tree`

, `Node`

, or `fromList`

yet, but hopefully you get the idea. `either undefined id`

is justified here since the input is guaranteed to be well-formed, so the parser will never actually fail with a `Left`

.

## Unordered trees

The other difficulty is how to compare trees up to reordering children. Trying all permutations of the children at each node and seeing whether any match is obviously going to be much too slow! The key insight, and what this problem had in common with the one from my previous post, is that we can use an (automatically-derived) `Ord`

instance to *sort* the children at each node into a canonical order. We don’t really need to know or care *what* order they end up in, which depends on the precise details of how the derived `Ord`

instance works. The point is that sorting into some consistent order allows us to efficiently test whether two lists are permutations of each other.

I think everyone who posted a solution created some kind of function to “canonicalize” a tree, by first canonicalizing all subtrees and then sorting them. When I first solved this problem, however, I approached it along slightly different lines, hinted at by commenter Globules: can we define the `Tree`

type in such a way that there is only a single representation for each tree-up-to-reordering?

My first idea was to use a `Data.Set`

of children at each node, but this is subtly wrong, since it gets rid of duplicates! We don’t actually want a *set* of children at each node, but rather a *bag* (aka *multiset*). So I made a little `Bag`

abstraction out of a `Map`

. The magical thing is that GHC can still derive an `Ord`

instance for my recursive tree type containing a newtype containing a `Map`

containing trees! (OK, OK, it’s not really magic, but it still *feels* magic…)

Now, actually, I no longer think this is the *best* solution, but it’s interesting, so I’ll leave it. Later on I will show what I think is an even better solution.

```
newtype Tree = Node (Bag Tree)
deriving (Eq, Ord)
newtype Bag a = Bag (Map a Int)
deriving (Eq, Ord)
fromList :: Ord a => [a] -> Bag a
fromList = map (,1) >>> M.fromListWith (+) >>> Bag
```

The final piece is the `solve`

function, which simply calls `readTree`

on the two strings and compares the resulting (canonical!) `Tree`

values for equality.

```
solve :: [C.ByteString] -> Bool
solve [t1,t2] = ((==) `on` readTree) t1 t2
```

## A better way

I still think it’s a nice idea to have canonical-by-construction trees, rather than building ordered trees and then calling a separate function to canonicalize them afterwards. But inspired by several commenters’ solutions, I realized that rather than my complex `Bag`

type, it’s much nicer to simply use a *sorted list* as the canonical representation of a `Node`

’s bag of subtrees, and to use a smart constructor to build them:

```
newtype Tree = Node [Tree]
deriving (Eq, Ord)
mkNode :: [Tree] -> Tree
mkNode = Node . sort
```

Then we just use `mkNode`

instead of `Node`

in the parser, and voilà! The canonicalization happens on the fly while parsing the tree. By contrast, if we write a separate canonicalization function, like

```
canonical :: Tree -> Tree
canonical (Node ts) = Node (map canonical (sort ts))
```

it is actually possible to get it wrong. In fact, I deliberately introduced a bug into the above function: can you see what it is?

All told, then, here is the (in my opinion) nicest solution that I know of:

```
{-# LANGUAGE OverloadedStrings #-}
import Control.Arrow
import Data.Bool
import qualified Data.ByteString.Lazy.Char8 as C
import Data.Function
import Data.List
import Text.Parsec
import ScannerBS hiding (many)
main = C.interact $
runScanner (numberOf (two str)) >>>
map (solve >>> bool "different" "same") >>> C.unlines
solve :: [C.ByteString] -> Bool
solve [t1,t2] = ((==) `on` readTree) t1 t2
newtype Tree = Node [Tree] deriving (Eq, Ord)
readTree :: C.ByteString -> Tree
readTree = parse parseTree "" >>> either undefined id
where
parseTree = (Node . sort) <$> many parseSubtree
parseSubtree = char '0' *> parseTree <* char '1'
```

## Next problem

For Tuesday, I invite you to solve The Power of Substitution. Don’t let the high difficulty rating scare you; in my estimation it should be quite accessible if you know a bit of math and have been following along with some of my previous posts (YMMV). However, it’s not quite as obvious what the nicest way to write it in Haskell is.

I am so stupid: I did not realize until now the string was simply a well-formed string of nested parens! Thanks to your parsec-based solution, it’s now obvious :)

Yes, it took me a while to realize that too! =)

Also, a quick and dirty “solution” to the Power of Substitution problem: https://gist.github.com/abailly/240e4c7d3a361098a633e493eb952bb4

I think laziness should make it efficient because in the end we’ll only need to compute and compare one item for each iteration of the message.

Did you try submitting it on Kattis? I suspect it will not fit within the time limit.

Fair enough :) It times out…

For Power of Substitution, the naive solution times out for me, so I used a lazily-evaluated array (bounds are (1,1)-(100,100), computing as-needed the distance from a number to its ciphered number). This brings the time down to 0.02s on kattis. However it’s not correct, the second test is failing. I suspect I’m off by a simple idea somewhere

https://github.com/jasonincanada/kattis/blob/master/src/Substitution.hs

I can see what is wrong but I will only give a hint if you want one. =)

Sure, thanks! I figured it might be zero distances causing the LCM to be 0 overall, but that wasn’t it. Can you suggest an input that the code wouldn’t work on?

Sure, how about

1

2

1 4

3 5

2 3 1 5 4 6 7 8 9 …

where the … means to continue counting up to 100. I’ll let you figure out what the answer should be.

This problem is fun and I am completely stuck trying to go faster with mutable arrays but to no avail. I suspect I am missing some important property related to permutation groups that alleviate the need for repeatedly comparing the lists… Eagerly waiting the solution :)

I’m enjoying the current format of CP in Haskell. Here’s my solution to Power of Substitution.

https://github.com/sureyeaah/Competitive/blob/master/kattis/substitution.hs

If we start walking from any point in the permutation graph, we will eventually land in a cycle.

So I first brute force n (size of character set, here 100) steps, this ensure that I have either found my answer or I’m in a loop. After that, I find the length of the loop in which the character (i.e. node in permutation graph) lies and the distance between the current character and the final transformed character and used CRT to find the final answer.

I am sure there must be a lot more elegant way of writing this than I have written here.

The main parts are:

solve :: ([Int], [Int], V.Vector Int) -> Int

solve (m, c, p) = either id undefined $ go m 0 >>= go’

where

go :: [Int] -> Int -> Either Int [Int]

go m acc

| m == c = Left acc

| acc == n = Right m

| otherwise = go (substitute m) (acc + 1)

go’ :: [Int] -> Either Int [Int]

go’ m =

[(stepsTo substitute st nd, cycleSize substitute st) | (st, nd) (a -> a) -> a -> Int

cycleSize f x = 1 + stepsTo f (f x) x

stepsTo :: (Eq a) => (a -> a) -> a -> a -> Int

stepsTo f st nd = go 0 st

where

go acc x = if x == nd then acc else go (acc + 1) (f x)

It would be nice if the comments systems had code blocks and the ability to make edits!

Yeah, code blocks would be nice. I think you can make edits though, can you not?

I’ve seen others do code blocks, but I don’t know how to do it. I don’t think I can make edits in posted comments.

I think I’ve put my code in HTML “pre” blocks and it’s looked fine.

I’m working on a similar idea, but I haven’t gotten my CRT code to work yet. I don’t have a full test, yet.

The heart of my code is:

solve kase = show $ fst $ foldr1 crt klPairs

where

[_,message,ctext,perm] = map ((map ((- 1) . read)) . words) kase

mcPairs = zip message ctext

klPairs = map findKl mcPairs

findKl (m, c) = (fromJust $ elemIndex c cycle, length cycle)

where

cycle = cycleFrom perm m

cycleFrom perm m = m : (takeWhile (/= m) (tail (iterate (perm!!) m )))

mcPairs is a list of pairs of message characters and cyphertext characters. klPairs is a list of how many iterations of the permutation is necessary to bring the given plaintext character to the cyphertext character. For each symbol in the message, we have an equation , Folding over the list of pairs solves for .

First, thank you for introducing me to Control.Arrow — I learned about them from your previous two write-ups. Similarly, thank you for introducing me to the `bool` and `chunksOf` functions!

My “solution” (it times out on the second secret test case, so not really a solution, I think) uses Vector’s `unsafeBackpermute` for most of the heavy lifting:

Substitution.hs

hosted with ❤ by GitHub

I’m very much looking forward to your writeup!

I don’t know why that gist is appearing inline (it didn’t before…) and I’m not able to edit my comment to try to replace it with just the link.

Here’s my solution: https://github.com/blaisepascal/competative-haskell/blob/master/powersubstitution.hs

Struggling to get my solution running fast enough to pass the second test case. Does “CRT” mean the “Chinese Remainder Theorem” and is that the math trick that helps speed this up? Would love a hint :)

Yes, and yes =). See my previous blog post on the topic!

I think I figured out the connection. I started graphing the values for each iteration of E for a single character (e.g,. m_1, m_2, etc). I realized that, no matter where you start in the substitution table, you will eventually cycle back to that starting position. So you can calculate a “length” for each cycle, starting at a given index in the table. I realized those cycles where the connection to modular arithmetic. For example, if my cycle is 3 steps, then to find the minimum k necessary to get from a starting character to a final character, I can just find the position of the final character in the cycle (0, 1, … ).

Solving for an entire message means finding where all those cycles line up (thats where graphing helped). The CRT provides the tool for finding where that happens.

For any given starting position in the substitution, if you apply E enough, you will cycle back to that starting position. Then the lengh of each cycle is the modulus on the RHS of your congruence relation (one relation for each character in the message). The modulee (?) is the index of the particular cipher character in that cycle. In equations:

x = steps(m_1, c_1) (mod cycle_length(p[m_1]])

x = steps(m_2, c_2) (mod cycle_length(p[m_1]])

…

m_1, m_2, are the characters in the message, p is the substitution table, and c_1, c_2 are the characters in the target message. steps is the number of iterations from m_1 to c_1, starting at p[m_1] (and so on). cycle_length(p[m_1]) is the number of times you have to iterate, starting at p[m_1] until you return to p[m_1].

Solving CRT for x gives you the number of steps necessary to produce the desired ciphertext.

Now I just have to implement it!

p.s. Also there could be some fundamental errors that I’m gonna run into …

Did it! Thanks for the hint. I also used your `egcd` and `gcrt` code because that would have taken me another week to solve!

Code at https://repl.it/@m4dc4p/Power-of-Substitution#main.hs.

Passed kattis in 0.01 seconds!

I won’t post my code here because it’s currently a sprawling mess, and I’ve only tried it on some small test cases (which have worked). But, here’s the gist of it from a comment in the code:

Currently, I’m not using the CRT in my code, but would expect to use it to solve for r, above. I.e. a set of equations of the form Lr = di (mod li), using my notation from the comment.

Ok, it’s still a bit of a mess, but passes Kattis. I’ll only include part of it, below. It uses Brent’s gcrt function.

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