In the Theory Lunch of the last week, James Chapman talked about the MU puzzle from Douglas Hofstadter’s book Gödel, Escher, Bach. This puzzle is about a string rewriting system. James presented a Haskell program that computes derivations of strings. Inspired by this, I wrote my own implementation, with the goal of improving efficiency. This blog post presents this implementation. As usual, it is available as a literate Haskell file, which you can load into GHCi.
The puzzle
Let me first describe the MU puzzle shortly. The puzzle deals with strings that may contain the characters , , and . We can derive new strings from old ones using the following rewriting system:
The question is whether it is possible to turn the string into the string using these rules.
You may want to try to solve this puzzle yourself, or you may want to look up the solution on the Wikipedia page.
The code
The code is not only concerned with deriving from , but with derivations as such.
Preliminaries
We import Data.List
:
import Data.List
Basic things
We define the type Sym
of symbols and the type Str
of symbol strings:
data Sym = M | I | U deriving Eq
type Str = [Sym]
instance Show Sym where
show M = "M"
show I = "I"
show U = "U"
showList str = (concatMap show str ++)
Next, we define the type Rule
of rules as well as the list rules
that contains all rules:
data Rule = R1 | R2 | R3 | R4 deriving Show
rules :: [Rule]
rules = [R1,R2,R3,R4]
Rule application
We first introduce a helper function that takes a string and returns the list of all splits of this string. Thereby, a split of a string str
is a pair of strings str1
and str2
such that str1 ++ str2 == str
. A straightforward implementation of splitting is as follows:
splits' :: Str -> [(Str,Str)]
splits' str = zip (inits str) (tails str)
The problem with this implementation is that walking through the result list takes quadratic time, even if the elements of the list are left unevaluated. The following implementation solves this problem:
splits :: Str -> [(Str,Str)]
splits str = zip (map (flip take str) [0 ..]) (tails str)
Next, we define a helper function replace
. An expression replace old new str
yields the list of all strings that can be constructed by replacing the string old
inside str
by new
.
replace :: Str -> Str -> Str -> [Str]
replace old new str = [front ++ new ++ rear |
(front,rest) <- splits str,
old `isPrefixOf` rest,
let rear = drop (length old) rest]
We are now ready to implement the function apply
, which performs rule application. This function takes a rule and a string and produces all strings that can be derived from the given string using the given rule exactly once.
apply :: Rule -> Str -> [Str]
apply R1 str | last str == I = [str ++ [U]]
apply R2 (M : tail) = [M : tail ++ tail]
apply R3 str = replace [I,I,I] [U] str
apply R4 str = replace [U,U] [] str
apply _ _ = []
Derivation trees
Now we want to build derivation trees. A derivation tree for a string str
has the following properties:
- The root is labeled with
str
. - The subtrees of the root are the derivation trees for the strings that can be generated from
str
by a single rule application. - The edges from the root to its subtrees are marked with the respective rules that are applied.
We first define types for representing derivation trees:
data DTree = DTree Str [DSub]
data DSub = DSub Rule DTree
Now we define the function dTree
that turns a string into its derivation tree:
dTree :: Str -> DTree
dTree str = DTree str [DSub rule subtree |
rule <- rules,
subStr <- apply rule str,
let subtree = dTree subStr]
Derivations
A derivation is a sequence of strings with rules between them such that each rule takes the string before it to the string after it. We define types for representing derivations:
data Deriv = Deriv [DStep] Str
data DStep = DStep Str Rule
instance Show Deriv where
show (Deriv steps goal) = " " ++
concatMap show steps ++
show goal ++
"\n"
showList derivs
= (concatMap ((++ "\n") . show) derivs ++)
instance Show DStep where
show (DStep origin rule) = show origin ++
"\n-> (" ++
show rule ++
") "
Now we implement a function derivs
that converts a derivation tree into the list of all derivations that start with the tree’s root label. The function derivs
traverses the tree in breadth-first order.
derivs :: DTree -> [Deriv]
derivs tree = worker [([],tree)] where
worker :: [([DStep],DTree)] -> [Deriv]
worker tasks = rootDerivs tasks ++
worker (subtasks tasks)
rootDerivs :: [([DStep],DTree)] -> [Deriv]
rootDerivs tasks = [Deriv (reverse revSteps) root |
(revSteps,DTree root _) <- tasks]
subtasks :: [([DStep],DTree)] -> [([DStep],DTree)]
subtasks tasks = [(DStep root rule : revSteps,subtree) |
(revSteps,DTree root subs) <- tasks,
DSub rule subtree <- subs]
Finally, we implement the function derivations
which takes two strings and returns the list of those derivations that turn the first string into the second:
derivations :: Str -> Str -> [Deriv]
derivations start end
= [deriv | deriv@(Deriv _ goal) <- derivs (dTree start),
goal == end]
You may want to enter
derivations [M,I] [M,U,I]
at the GHCi prompt to see the derivations
function in action. You can also enter
derivations [M,I] [M,U]
to get an idea about the solution to the MU puzzle.
Pingback: MIU in Haskell, part 2 | Theory Lunch
In the “replace” function, I noticed that you didn’t use the “stripPrefix” function – was that for readability concerns?
The
stripPrefix
function returns aMaybe
value, since it needs to signal whether the given list starts with the given prefix or not. In thereplace
function, I have already ensured that it does; so it is simpler to use the solution withdrop
andlength
, where I do not have to get rid of theMaybe
.By the way, in your Gravatar profile you say, “I like seeing how other languages do things differently.” So I wonder, did you already look at languages that are not indo-european, like Estonian? They are at some points quite different from what we “indo-europeans” are used to.
Why is
split
more efficient thansplit'
?? I was unable to exhibit any difference between the two (Criterion used).Thank you!
Please enter this on the GHCi prompt:
This should give you the result
1000001
quite quickly.Now try this:
I wasn’t able to get a result within a reasonable amount of time.
The problem with
split'
lies in the use ofinits
. Theinits
function is implemented as follows:Note that the recursive application of
inits
is under amap
. So the suffix of an expressioninits [x_1,x_2,…,x_n]
that starts at an index is defined by a nested application ofmap
of depth :map (x_1 : ) (map (x_2 : ) (…(map (x_i : ) […])…))
To detect that the result of an expression
map f xs
is non-empty, we have to detect thatxs
is non-empty. So to detect that the above suffix is non-empty, we need to walk through the layers ofmap
s, which takes time.This means that to fetch the -th element of
inits xs
, we needtime.
The implementation of
split
doesn’t suffer from this problem, since it creates prefixes by a single application ofmap
, and computes every single prefix independently.Pingback: MIU in Curry « Wolfgang Jeltsch