Is the language given below a CFL? If not prove via Pumping Lemma.
L = {www| w belongs to {0,1}*}
That much I know that it's not a cfl! But I'm facing difficulty solving it via pumping lemma. Kindly help me
Related
The Z3 input format is an extension of the one defined by SMT-LIB 2.0 standard. The input expressions need to write in prefix form. As for example rise4fun,
x + (y * 2) = 20 needs to be given input in the form of " (= (+ x (* 2 y)) 20)) ".
Z3 supports JAVA API. As for example, let us consider the below code which evaluates and checks satisfiability expressions: x+y = 500 and x + (y * 2) = 20.
final Context ctx = new Context();
final Solver solver = ctx.mkSimpleSolver();
IntExpr x = ctx.mkIntConst("x");
IntExpr y = ctx.mkIntConst("y");
IntExpr th = ctx.mkInt(500);
IntExpr th1 = ctx.mkInt(2);
IntExpr th2 = ctx.mkInt(20);
BoolExpr t1 = ctx.mkEq(ctx.mkAdd(x,y), th);
BoolExpr t2 = ctx.mkEq(ctx.mkAdd(x,ctx.mkMul(th1, y)), th2);
solver.add(t1);
solver.add(t2);
solver.check()
The problem is if an external user wants to give input to the solver, he cannot give it in the form of the general formula as " x+y = 500, x + (y * 2) = 20 ".
The input needs to be parsed and then should be written manually using JAVA API in prefix form (Note BoolExpr t2 in above code) to give the final expressions to Solver.
Is there any parser/library/API (Preferably JAVA or in any other language) which parses the general expressions with arithmetic operators(+, -, <, >, =), propositional logic connectors (And, OR), Quantifiers(ForAll, Exists) and then gives input to the Z3 Solvers ?
Please suggest and help.
This is precisely why people build high-level interfaces to SMT solvers. There are many choices here for z3:
Official Python interface to z3: https://ericpony.github.io/z3py-tutorial/guide-examples.htm This is supported by Microsoft directly, and is probably the easiest to use.
PySMT, which aims to cover as many SMT solvers as they can: https://github.com/pysmt/pysmt
Scala bindings: https://github.com/epfl-lara/ScalaZ3
Haskell SBV: http://hackage.haskell.org/package/sbv
Many others at various maturity levels for Racket, Go, Lisp.. You name it. You can find many of them on github.
The goal of these "wrappers" is to precisely save the end-user from all the details of intricate bindings, and provide something much easier and less-error prone to use. On the flip side, they require you to learn yet another library. In my experience, if the host language of your choice has such an implementation, it'd pay off nicely to use it. If not, you should build one!
I recently had an assignment where I was asked to use pumping lemma to show that a language was not regular, and unfortunately got the wrong answer.
The language to prove is non-regular is as follows:
L = {ai bj ck: i = j or j = k}
The definition of a pumping lemma that I was given is as follows:
opponent picks m
I want to pick w to contradict the pumping lemma. Use m to pick a string w ∈ L where |w| ≥ m
opponent picks a decomposition of w subject to constraints.
I try to pick an i so that the pumped string wi ∉ L. If found, L is not regular
This subject has proven to be very difficult for me to understand and I feel like a complete airhead because of it, so a detailed explanation as to how I would properly apply a pumping lemma would be appreciated.
Intuitively, the pumping lemma says that long enough words (the length depends only on the language L) in a regular language L must contain a section (of length > 0) which can be repeated as often as desired. Repeating that section ('pumping' the original word) any number of time results in some longer words which are also in the language L.
The minimal length for the word is the m in the first step of the above definition; the number of times the section is repeated is the i in the 4th step of the above definition.
The pumping lemma is usually used to show that a language L is not regular. It is a proof by contradiction: assume that L is regular and thus the pumping lemma for regular languages is true for L. Then pick a word w which is in L of sufficient length* and show that regardless of how it is decomposed* some pumped word is not in the language. This contradicts the pumping lemma - which we know to be true. Thus our assumption that the language is regular was wrong and thus the language is not regular. The parts marked with an * cannot be chosen to make things easy - that's why in steps 1 and 3 the 'opponent' selects them.
The word w is rewritten as w = x y z, where | y | > 0 and |x y| <= m. Both x and z may be of length 0.
The usual approach is to pick xy to be a string consisting of the same letter such that having more of that same letter (after the pumping) leads to a word not in L.
No restrictions are specified for the i or the k in the language L in the post, so assuming that i = 0 is allowed, a suitable word might be b^m c^m (that is m bs followed by m cs). Now whatever decomposition the opponent might select, the y will always consist of some bs. Repeating those bs leads to a word with more bs than cs and no as, and thus i != j and j!= k and the word is not in the language.
This is the simple finite automata I tried, what am I doing it wrong?
This matches , not . Aka ababab, not aaabbb.
The b transition leads to a final state which halts the machine. Your machine will only halt if given a sequence of 'ab' of length 1 or more.
The language a^n b^n where n>=1 is not regular, and it can be proved using the pumping lemma. Assume there is a finite state automaton that can accept the language. This finite automaton has a finite number of states k, and there is string x in the language such that n > k. According to the pumping lemma, x can be decomposed such that x=uvw, and any finite automaton that accepts x must also accept uv*w. v is non-empty and can be made to consist of only a's or only b's since n > k. Let v consist of only a's. If the finite automaton accepts x = uvw, it must also accept x = uvvw, which has more a's than b's and is not of the form a^n b^n. This is a contradiction, so a^n b^n cannot be a regular language.
This matches (ab)^n, not a^nb^n.
What you're looking for is Pumping lemma for regular languages.
Examples:
Let L = {a^mb^m | m ≥ 1}.
Then L is not regular.
Proof: Let n be as in Pumping Lemma.
Let w = a^nb^n.
Let w = xyz be as in Pumping Lemma.
Thus, xy^2z ∈ L, however, xy^2z contains more a’s than b’s.
I'm trying to understand what is this 'magical' number 'n' that is used in every application of the Pumping lemma. After hours of research on the subject, I came to the following website: http://elvis.rowan.edu/~nlt/TheoryNotes/PumpingLemma.pdf
It states
n is
the longest a string can be without having a loop. The biggest n can
be is s, though it might be smaller for some particular language.
From what I understand if there is a Language L then the pumping length of L is the amount of states in the Finite State Automata that recognizes L. Is this true?
If it is then what exactly does the last line from above say "though it might be smaller for some particular language"? Complete mess in my head. Could somebody clear it up, please?
A state doesn't recognise a language. A DFA recognises a language by accepting exactly the set of words in the languages and no others. A DFA has many states.
If there is a regular language L, which can be modelled by the Pumping Lemma, it will have a property n.
For a DFA with s states, in order for it to accept L, s must be >= n.
The last line merely states that there are some languages in which s is greater than n, rather than equal.
This is probably more suited for https://cstheory.stackexchange.com/ or https://cs.stackexchange.com/ (not quite sure of the value of both myself).
Note: Trivially, not all DFA's with sufficient states will accept the language. Also, the fact that a language passes the pumping lemma doesn't mean it's regular (but failing it means definitely isn't).
Note also, the language changes between FA and DFA - this is a bit lax, but because NDFAs have the same power as DFAs and DFAs are easier to write and understand, DFAs are used for the proof.
Edit I'll give an example of a regular language, so you can see an idea of u,v,w,z and n. Then we'll discuss s.
L = xy^nz, n > 2 (i.e. xyyz, xyyyz, xyyyyz)
u = xy
v = y
w = z
n = 4
Hence:
|z| = 3: xyz (i = 0) Not in L, but |z| < n
|z| = 4: xyyz (i = 1)
|z| = 5: xyyyz (i = 2)
etc
Hence, it's modelled by the Pumping Lemma. A DFA would need at least 4 states. So let's think of one.
-> State 1: x
State 1:
-> State 2: y
State 2:
-> State 3: y
State 3:
-> State 3: y
-> State 4: z
State 4:
Accepting state
Terminating state
4 states, as expected. Not possible to do it in less, as expected by n = 4. In this case, the example is quite simple so I don't think you can build one with more than 4 states (but that would be okay if it were needed).
I think a possibility is when you have a FA with an unreachable state. The FA has s states, but 1 is unreachable, so all strings recognizing L will be comprised of (s-1)=n states, so n<s.
What is "monadic reflection"?
How can I use it in F#-program?
Is the meaning of term "reflection" there same as .NET-reflection?
Monadic reflection is essentially a grammar for describing layered monads or monad layering. In Haskell describing also means constructing monads. This is a higher level system so the code looks like functional but the result is monad composition - meaning that without actual monads (which are non-functional) there's nothing real / runnable at the end of the day. Filinski did it originally to try to bring a kind of monad emulation to Scheme but much more to explore theoretical aspects of monads.
Correction from the comment - F# has a Monad equivalent named "Computation Expressions"
Filinski's paper at POPL 2010 - no code but a lot of theory, and of course his original paper from 1994 - Representing Monads. Plus one that has some code: Monad Transformers and Modular Interpreters (1995)
Oh and for people who like code - Filinski's code is on-line. I'll list just one - go one step up and see another 7 and readme. Also just a bit of F# code which claims to be inspired by Filinski
I read through the first Google hit, some slides:
http://www.cs.ioc.ee/mpc-amast06/msfp/filinski-slides.pdf
From this, it looks like
This is not the same as .NET reflection. The name seems to refer to turning data into code (and vice-versa, with reification).
The code uses standard pure-functional operations, so implementation should be easy in F#. (once you understand it)
I have no idea if this would be useful for implementing an immutable cache for a recursive function. It look like you can define mutable operations and convert them to equivalent immutable operations automatically? I don't really understand the slides.
Oleg Kiselyov also has an article, but I didn't even try to read it. There's also a paper from Jonathan Sobel (et al). Hit number 5 is this question, so I stopped looking after that.
As previous answers links describes, Monadic reflection is a concept to bridge call/cc style and Church style programming. To describe these two concepts some more:
F# Computation expressions (=monads) are created with custom Builder type.
Don Syme has a good blog post about this. If I write code to use a builder and use syntax like:
attempt { let! n1 = f inp1
let! n2 = failIfBig inp2
let sum = n1 + n2
return sum }
the syntax is translated to call/cc "call-with-current-continuation" style program:
attempt.Delay(fun () ->
attempt.Bind(f inp1,(fun n1 ->
attempt.Bind(f inp2,(fun n2 ->
attempt.Let(n1 + n2,(fun sum ->
attempt.Return(sum))))))))
The last parameter is the next-command-to-be-executed until the end.
(Scheme-style programming.)
F# is based on OCaml.
F# has partial function application, but it also is strongly typed and has value restriction.
But OCaml don't have value restriction.
OCaml can be used in Church kind of programming, where combinator-functions are used to construct any other functions (or programs):
// S K I combinators:
let I x = x
let K x y = x
let S x y z = x z (y z)
//examples:
let seven = S (K) (K) 7
let doubleI = I I //Won't work in F#
// y-combinator to make recursion
let Y = S (K (S I I)) (S (S (K S) K) (K (S I I)))
Church numerals is a way to represent numbers with pure functions.
let zero f x = x
//same as: let zero = fun f -> fun x -> x
let succ n f x = f (n f x)
let one = succ zero
let two = succ (succ zero)
let add n1 n2 f x = n1 f (n2 f x)
let multiply n1 n2 f = n2(n1(f))
let exp n1 n2 = n2(n1)
Here, zero is a function that takes two functions as parameters: f is applied zero times so this represent the number zero, and x is used to function combination in other calculations (like add). succ function is like plusOne so one = zero |> plusOne.
To execute the functions, the last function will call the other functions with last parameter (x) as null.
(Haskell-style programming.)
In F# value restriction makes this hard. Church numerals can be made with C# 4.0 dynamic keyword (which uses .NET reflection inside). I think there are workarounds to do that also in F#.