Why is it that the concatenation of a infinite language and a finite language always finite iff the language is not the empty set? I thought concatenating an infinite language with the empty set would just be the infinite language.
This statement is false. Try concatenating Σ* and Σ. This gives back Σ+, which is infinite.
I think that you meant
The concatenation of an infinite language I and a finite language F is infinite iff F ≠ ∅
This statement is true. If F = ∅ then IF is the empty set because the concatenation of any language and the empty language is the empty language. Specifically, IF = { wx | w in I and x in F }, so if F is empty, there are no x's that satisfy the condition x in F.
On the other hand, if x ≠ ∅ we can prove that IF is infinite. Consider any string w ∈ I whose length is longer than any string in F. Then the set wF = { wx | x in F } is infinite because there's at least one string in wF in-between any two multiples of |w|. Since wF ⊆ IF, this means that IF is infinite.
Hope this helps!
Related
I need to prove whether this language is recognizable or not:
{ ⟨M, w⟩: M is a Turing machine that accepts string w and does not accept string ε }
I figure I could do a reduction on ATM, but how do I handle the empty string?
Here's one way to do a reduction from the complement of ATM to your language.
Given a TM M and a string w, build this TM/string pair:
M' = "On input x:
If x = "iheartquokkas", accept.
Otherwise:
Run M on w.
If M accepts, accept;
If M rejects, reject;
(Implicitly: if M loops, loop.)"
w' = "iheartquokkas"
First, suppose that M does not accept w. Then machine M' does indeed accept the string w', because it's programmed to do so. Additionally, when you run M' on ε, it will run M on w and never accept because M doesn't accept w. Thus M' accepts w' but not ε, so ⟨M', w'⟩ is in your language.
Second, suppose that M accepts w. Then when you run M' on ε, it will accept, so it is not the case that M' accepts w' but not ε. Therefore, ⟨M', w'⟩ is not in your language.
You now have a mapping reduction from ATM's complement to your language, so your language isn't recognizable.
Now, can you prove that it's not co-recognizable?
I want to break a loop in a situation like this:
import Data.Maybe (fromJust, isJust, Maybe(Just))
tryCombination :: Int -> Int -> Maybe String
tryCombination x y
| x * y == 20 = Just "Okay"
| otherwise = Nothing
result :: [String]
result = map (fromJust) $
filter (isJust) [tryCombination x y | x <- [1..5], y <- [1..5]]
main = putStrLn $ unlines $result
Imagine, that "tryCombination" is a lot more complicated like in this example. And it's consuming a lot of cpu power. And it's not a evalutation of 25 possibilities, but 26^3.
So when "tryCombination" finds a solution for a given combination, it returns a Just, otherwise a Nothing. How can I break the loop instantly on the first found solution?
Simple solution: find and join
It looks like you're looking for Data.List.find. find has the type signature
find :: (a -> Bool) -> [a] -> Maybe a
So you'd do something like
result :: Maybe (Maybe String)
result = find isJust [tryCombination x y | x <- [1..5], y <- [1..5]]
Or, if you don't want a Maybe (Maybe String) (why would you?), you can fold them together with Control.Monad.join, which has the signature
join :: Maybe (Maybe a) -> Maybe a
so that you have
result :: Maybe String
result = join $ find isJust [tryCombination x y | x <- [1..5], y <- [1..5]]
More advanced solution: asum
If you wanted a slightly more advanced solution, you could use Data.Foldable.asum, which has the signature
asum :: [Maybe a] -> Maybe a
What it does is pick out the first Just value from a list of many. It does this by using the Alternative instance of Maybe. The Alternative instance of Maybe works like this: (import Control.Applicative to get access to the <|> operator)
λ> Nothing <|> Nothing
Nothing
λ> Nothing <|> Just "world"
Just "world"
λ> Just "hello" <|> Just "world"
Just "hello"
In other words, it picks the first Just value from two alternatives. Imagine putting <|> between every element of your list, so that
[Nothing, Nothing, Just "okay", Nothing, Nothing, Nothing, Just "okay"]
gets turned to
Nothing <|> Nothing <|> Just "okay" <|> Nothing <|> Nothing <|> Nothing <|> Just "okay"
This is exactly what the asum function does! Since <|> is short-circuiting, it will only evaluate up to the first Just value. With that, your function would be as simple as
result :: Maybe String
result = asum [tryCombination x y | x <- [1..5], y <- [1..5]]
Why would you want this more advanced solution? Not only is it shorter; once you know the idiom (i.e. when you are familiar with Alternative and asum) it is much more clear what the function does, just by reading the first few characters of the code.
To answer your question, find function is what you need. After you get Maybe (Maybe String) you can transform it into Maybe String with join
While find is nicer, more readable and surely does only what's needed, I wouldn't be so sure about inefficiency of the code that you have in a question. The lazy evaluation would probably take care of that and compute only what's needed, (extra memory can still be consumed). If you are interested, try to benchmark.
Laziness can actually take care of that in this situation.
By calling unlines you are requesting all of the output of your "loop"1, so obviously it can't stop after the first successful tryCombination. But if you only need one match, just use listToMaybe (from Data.Maybe); it will convert your list to Nothing if there are no matches at all, or Just the first match found.
Laziness means that the results in the list will only be evaluated on demand; if you never demand any more elements of the list, the computations necessary to produce them (or even see whether there are any more elements in the list) will never be run!
This means you often don't have to "break loops" the way you do in imperative languages. You can write the full "loop" as a list generator, and the consumer(s) can decide independently how much of the they want. The extreme case of this idea is that Haskell is perfectly happy to generate and even filter infinite lists; it will only run the generation code just enough to produce exactly as many elements as you later end up examining.
1 Actually even unlines produces a lazy string, so if you e.g. only read the first line of the resulting joined string you could still "break the loop" early! But you print the whole thing here.
The evaluation strategy you are looking for is exactly the purpose of the Maybe instance of MonadPlus. In particular, there is the function msum whose type specializes in this case to
msum :: [Maybe a] -> Maybe a
Intuitively, this version of msum takes a list of potentially failing computations, executes them one after another until the first computations succeeds and returns the according result. So, result would become
result :: Maybe String
result = msum [tryCombination x y | x <- [1..5], y <- [1..5]]
On top of that, you could make your code in some sense agnostic to the exact evaluation strategy by generalizing from Maybe to any instance of MonadPlus:
tryCombination :: MonadPlus m => Int -> Int -> m (Int,Int)
-- For the sake of illustration I changed to a more verbose result than "Okay".
tryCombination x y
| x * y == 20 = return (x,y) -- `return` specializes to `Just`.
| otherwise = mzero -- `mzero` specializes to `Nothing`.
result :: MonadPlus m => m (Int,Int)
result = msum [tryCombination x y | x <- [1..5], y <- [1..5]]
To get your desired behavior, just run the following:
*Main> result :: Maybe (Int,Int)
Just (4,5)
However, if you decide you need not only the first combination but all of them, just use the [] instance of MonadPlus:
*Main> result :: [(Int,Int)]
[(4,5),(5,4)]
I hope this helps more on a conceptual level than just providing a solution.
PS: I just noticed that MonadPlus and msum are indeed a bit too restrictive for this purpose, Alternative and asum would have been enough.
Regular languages are closed under concatenation - this is demonstrable by having the accepting state(s) of one language with an epsilon transition to the start state of the next language.
If we consider the language L = {a^n | n >=0}, this language is regular (it is simply a*). If we concatenate it with another language L = {b^n | n >=0}, which is also regular, we end up with a^nb^n, but we obviously know this isn't regular.
Where am I going wrong with my logic here?
The definition of the concatenation of two languages L1 and L2 is the set of all strings wx where w ∈ L1 and x ∈ L2. This means that L1L2 consists of all possible strings formed by pairing one string from L1 and one string from L2, which isn't necessarily the same as pairing up matching strings from each language.
As a result, as #Oli Charlesworth pointed out, the language you get back here isn't actually { anbn | n in N }. Instead, it's the language { anbm | n in N and m in N }, which is the language a*b*. This language is regular, since it's given by the regular languages.
Hope this helps!
Is the following language context free?
L = {a^i b^k c^r d^s | i+s = k+r, i,k,r,s >= 0}
I've tried to come up with a context free grammar to generate this but I can not, so I'm assuming its not context free. As for my proof through contradiction:
Assume that L is context free,
Let p be the constant given by the pumping lemma,
Choose string S = a^p b^p c^p d^p where S = uvwxy
As |vwx| <= p, then at most vwx can contain two distinct symbols:
case a) vwx contains only a single type of symbol, therefore uv^2wx^2y will result in i+s != k+r
case b) vwx contains two types of symbols:
i) vwx is composed of b's and c's, therefore uv^2wx^2y will result in i+s != k+r
Now my problem is that if vwx is composed of either a's and b's, or c's and d's, then pumping them won't necessary break the language as i and k or s and r could increase in unison resulting in i+s == k+r.
Am I doing something wrong or is this a context free language?
I can't come up with a CFG to generate that particular language at the top of my head either, but we know that a language is context free iff some pushdown automata recognizes it.
Designing such a PDA won't be too difficult. Some ideas to get you started:
we know i+s=k+r. Equivalently, i-k-r+s = 0 (I wrote it in that order since that is the order in they appear). The crux of the problem is deciding what to do with the stack if (k+r)>i.
If you aren't familiar with PDA's or cannot use them to answer the problem, at least you know now that it is Context Free.
Good luck!
Here is a grammar that accepts this language:
A -> aAd
A -> B
A -> C
B -> aBc
B -> D
C -> bCd
C -> D
D -> bDc
D -> ε
I'm trying to program my TI-83 to do a subset sum search. So, given a list of length N, I want to find all lists of given length L, that sum to a given value V.
This is a little bit different than the regular subset sum problem because I am only searching for subsets of given lengths, not all lengths, and recursion is not necessarily the first choice because I can't call the program I'm working in.
I am able to easily accomplish the task with nested loops, but that is becoming cumbersome for values of L greater than 5. I'm trying for dynamic solutions, but am not getting anywhere.
Really, at this point, I am just trying to get the list references correct, so that's what I'm looking at. Let's go with an example:
L1={p,q,r,s,t,u}
so
N=6
let's look for all subsets of length 3 to keep it relatively short, so L = 3 (6c3 = 20 total outputs).
Ideally the list references that would be searched are:
{1,2,3}
{1,2,4}
{1,2,5}
{1,2,6}
{1,3,4}
{1,3,5}
{1,3,6}
{1,4,5}
{1,4,6}
{1,5,6}
{2,3,4}
{2,3,5}
{2,3,6}
{2,4,5}
{2,4,6}
{2,5,6}
{3,4,5}
{3,4,6}
{3,5,6}
{4,5,6}
Obviously accomplished by:
FOR A,1,N-2
FOR B,A+1,N-1
FOR C,B+1,N
display {A,B,C}
END
END
END
I initially sort the data of N descending which allows me to search for criteria that shorten the search, and using FOR loops screws it up a little at different places when I increment the values of A, B and C within the loops.
I am also looking for better dynamic solutions. I've done some research on the web, but I can't seem to adapt what is out there to my particular situation.
Any help would be appreciated. I am trying to keep it brief enough as to not write a novel but explain what I am trying to get at. I can provide more details as needed.
For optimisation, you simply want to skip those sub-trees of the search where you already now they'll exceed the value V. Recursion is the way to go but, since you've already ruled that out, you're going to be best off setting an upper limit on the allowed depths.
I'd go for something like this (for a depth of 3):
N is the total number of array elements.
L is the desired length (3).
V is the desired sum
Y[] is the array
Z is the total
Z = 0
IF Z <= V
FOR A,1,N-L
Z = Z + Y[A]
IF Z <= V
FOR B,A+1,N-L+1
Z = Z + Y[B]
IF Z <= V
FOR C,B+1,N-L+2
Z = Z + Y[C]
IF Z = V
DISPLAY {A,B,C}
END
Z = Z - Y[C]
END
END
Z = Z - Y[B]
END
END
Z = Z - Y[A]
END
END
Now that's pretty convoluted but it basically check at every stage whether you've already exceed the desired value and refuses to check lower sub-trees as an efficiency measure. It also keeps a running total for the current level so that it doesn't have to do a large number of additions when checking at lower levels. That's the adding and subtracting of the array values against Z.
It's going to get even more complicated when you modify it to handle more depth (by using variables from D to K for 11 levels (more if you're willing to move N and L down to W and X or if TI BASIC allows more than one character in a variable name).
The only other non-recursive way I can think of doing that is to use an array of value groups to emulate recursion with iteration, and that will look only slightly less hairy (although the code should be less nested).