How many equivalence classes in the RL relation for
{w in {a, b}* | (#a(w) mod m) = ((#b(w)+1) mod m)}
I am looking at a past test question which gives me the options
m(m+1)
2m
m^2
m^2+1
infinite
However, i claim that its m, and I came up with an automaton that I believe accepts this language which contains 3 states (for m=3).
Am I right?
Actually you're right. To see this, observe that the difference of #a(w) and #b(w), #a(w) - #b(w) modulo m, is all that matters; and there are only m possible values of this difference modulo m. So, m states are always sufficient to accept a language of this form: simply make the state corresponding to the appropriate difference the accepting state.
In your DFA, a2 corresponds to a difference of zero, a1 to a difference of one and a3 to a difference of two.
I've seen this post about how to convert context free grammar to a DFA:
Automata theory : Conversion of a Context free grammar to a DFA
However, just wondering can all context free grammars be converted to DFA/NFA? What about context free grammars that cannot be expressed as a regular expression? Ex. S->(S) | ()
Thanks!
Only regular languages can be converted to a DFA, and not all CFGs represent regular languages, including the one in the question.
So the answer is "no".
NFAs are not more expressive than DFAs, so the above statement would still be true if you replaced DFA with NFA
A CFG represents a regular language if it is right- or left-linear. But the mere fact that a CFG is not left- or right-linear proves nothing. For example, S→a | a S a happens to generate the same language as S→a | S a a.
Yes ... if the F in "DFA" is replaced by I to get "DIA", but no ... for DFA, itself; and I will show how this works for your example at the end. In fact, all languages have DIA's whose state diagrams reside on a single Universal State Diagram as sub-diagrams thereof.
Consider your example, but rewrite it as S → u S v, S → w. This grammar, like all grammars, is algebraically a system of inequations over a certain partially ordered algebra. In particular, it can be rewritten as
S ⊇ {u}S{v}, S ⊇ {w},
or equivalently as
S ⊇ {u}S{v} ∪ {w}.
The object identified by the grammar is the least solution to the system. Since the system is a fixed point system S ⊇ f(S) = {u}S{v} ∪ {w}, then the least solution may also be described as the least fixed point solution and it is denoted μx f(x) = μx({u}x{v} ∪ {w}).
The ordering relation, for this algebra here, is subset ordering y ⊆ x ⇔ x ⊇ y. The operations include a product AB ≡ { ab: a ∈ A, b ∈ B }, defined element-wise (where, component-wise, the product is word concatenation, with ab being the concatenation of a and b). The product has {1} as an identity, where 1 denotes the empty word. Both word concatenation and product satisfy the fundamental properties
(xy)z = x(yz) [Associativity]
and
xe = x = ex [Identity property]
with the respective identities e = 1 (for concatenation) or e = {1} (for set product). The algebra is called a Monoid.
The simplest and most direct monoid formed from the elements X = {u,v,w} is the Free Monoid X* = {u,v,w}*, which is equivalently described as the set of all words of finite length (including the empty word, 1, of length 0) formed from u, v and w. It is possible to frame the question in terms of more general monoids, but (as the literature usually does) I will restrict it to free monoids.
The family of languages over X is one and the same as the family 𝔓M of subsets A ⊆ M of the monoid M = X*; the defining condition being A ∈ 𝔓M ⇔ A ⊆ M. Other distinguished subfamilies exist, such as the families ℜM ⊆ ℭM ⊆ 𝔗M ⊆ 𝔓M, respectively, of rational, context-free and Turing (or recursively enumerable) languages. The second of these ℭM, which is what your question is concerned with, are given by context-free grammars and are identified as the least fixed point solutions to the corresponding fixed point system of inequations.
Over 𝔓M, one can define the left-quotient operation v\A = { w ∈ M: vw ∈ A }, for each word v ∈ M and subset A ∈ 𝔓M. Because M = X* is a free monoid, it can be decomposed uniquely into left-quotients on the individual elements of X, by the properties 1\A = A, and (vw)\A = w\(v\A).
Correspondingly, one can define a state transition on each x ∈ X by x: A → x\A, treating each subset A ∈ 𝔓M as a state. Together, 𝔓M comprises the state set of the Universal State Diagram over M. Because M = X* is a free monoid, every element of M is either of the form xw for some x ∈ X and w ∈ X*, or is the empty word 1. The decomposition is unique: xw ≠ 1 for any x ∈ X or w ∈ X* and xw = x'w' for x, x' ∈ X and w, w' ∈ X*, only if x = x' and w = w'. Therefore, every A ∈ 𝔓M decomposes uniquely into a partition in a manner analogous to Taylor's Theorem as
A = A₀ ∪ ⋃_{x∈X} {x} x\A.
where A₀ ≡ A ∩ {1} is either {1} if 1 ∈ A or is ∅ if 1 ∉ A. The states for which A₀ = {1} may be regarded as the Final States in the Universal State Diagram.
The analogy to Taylor's Theorem is not too far-removed, since the left-quotient satisfies an analogue of the Product Rule
x\(AB) = (x\A) B ∪ A₀ (x\B)
so it is also denoted as a partial derivative x\A = ∂A/∂x: the Brzozowski Derivative, so that the decomposition rule could just as well be written as:
A = A₀ ∪ ⋃_{x∈X} {x} ∂A/∂x.
What you actually have is an infinite fixed-point system of inequations
A ⊇ A₀ ∪ ⋃_{x∈X} {x} ∂A/∂x for all A ∈ 𝔓M,
with variables A ∈ 𝔓M ranging over all of 𝔓M, whose right-hand sides are all right-linear in the variables. The sets, themselves, are the least fixed point solution to their own system (and to all closed subsystems of the universal system that contain that set as a variable).
Choosing different states as start states yields the different DIA's contained within it. Every minimal DIA (and every minimal DFA) of every language over X is contained in it.
In particular, in this diagram, you can consider the largest subdiagram accessible from a specific state A ∈ 𝔓M. All the states that can be accessed from A are left-quotients by words in M. So, together they comprise a family δA ≡ { v\A: v ∈ M }. The subdiagram consisting only of these states gives you the minimal DIA for the language A, where A, itself, is treated as the start state of the DIA.
If δA is finite, then the I is an F and it's actually a DFA - and that's what you're looking for. Which states in 𝔓M have DIA that are actually DFA's? The regular ones - the ones in the subfamily ℜM ⊆ 𝔓M. This is the case when M = X* is a free monoid. I'm not totally sure if this can also be proven for non-free monoids (like X* × Y*, whose rational subsets ℜ(X* × Y*) are one and the same as what are known as rational transductions) ... because of the reliance on the Taylor's Formula decomposition. There is still something like a Taylor's Theorem, but the decompositions are not necessarily partitions or unique, any longer.
For larger subfamilies of 𝔓M, the DIA are necessarily infinite; but their transitions may possess a sufficient degree of symmetry to allow both the states and transition rules to be wrapped up more succinctly. Correspondingly, one can distinguish different families of DIA by what symmetry properties they possess.
For your example, X = {u,v,w} and M = {u,v,w}*. The subset identified by your grammar is S = {uⁿ w vⁿ: n = 0, 1, 2, ...}. We can define the following sets
S(n) = S {vⁿ}, T(n) = {vⁿ}, for n = 0, 1, 2, ...
The sub-diagram of states accessible from S consists of all the states
δS = { S(n): n = 0, 1, 2, ... } ∪ { T(n): n = 0, 1, 2, ... } ∪ { ∅ }
The state transitions are the following
u: S(n) → S(n+1)
v: T(n+1) → T(n)
w: S(n) → T(n)
with x: A → ∅ in all other cases for x ∈ {u,v,w} and A ∈ δS. The sole final state is T(0).
As you can see, the DIA is infinite and is not a DFA at all. If you were to draw out the diagram, you would see an infinite ladder with S = S(0) being the start state T(0) = {1} the final state, with all the u transitions climbing up a rung, all the v transitions climbing down a rung, and the w transitions crossing over on a rung.
The symmetry is captured by factoring the state set into
δS = {S,T}×{0,1,2,3,⋯} ∪ {∅}
with S(n) rewritten as (S,n) and T(n) as (T,n). This includes a finite set of states Q = {S,T} for a finite state "control" and a set of states D = {0,1,2,3,⋯} for a "device"; as well as the empty set ∅ for the fail state. That device is none other than a counter, and this DIA is just a one-counter automaton in disguise.
All of the classical automata models posed in the literature have a similar form, when expressed as DIA. They contain a state set Q×D ∪ {∅} that includes a finite set Q for the "finite state control" and a (generally infinite) state set D for the device, along with the fail state ∅. The restrictions or constraints on the device correspond to what types of symmetries are contained in the underlying DIA. A deterministic PDA, with two stack symbols {a,b} for instance, has a device state set D = {a,b}* (consisting of all stack words), and an underlying DIA that has the form of an infinite binary tree with copies of Q residing at each node.
You can best see this by writing out and graphing the DIA for the Dyck language, which is given by the grammar D₂ → b D₂ d D₂, D₂ → p D₂ q D₂, D₂ → 1 as a language over X = {b,d,p,q} and subset of M = X* = {b,d,p,q}*; i.e. as the least-fixed point D₂ = μx ({b}x{d}x ∪ {p}x{q}x ∪ {1}).
Every subset in A ⊆ ℭM can be expressed in terms of a subset in A' ⊆ ℜM[b,d,p,q] of the free extension of the monoid M by indeterminates {b,d,p,q}, by carrying out insertions of {b,d,p,q} in suitable places in A, such that the result upon applying the identities {bd} = {1} = {pq}, {bq} = ∅ = {pd}, and xy = yx for x ∈ M and y ∈ {b,d,p,q} will yield A, itself, from A'.
This result (known, but unpublished since the 1990's and published only in 2022) is the algebraic form of the Chomsky-Schützenberger Theorem and is true for all monoids M. For instance, it holds for the non-free monoid M = X* × Y*, where the corresponding family ℭ(X* × Y*) comprise the push-down transductions from X to Y (or "simple syntax directed translations"; aka yacc-like grammars).
So, there is also something like a DFA even for these classes of DIA; provided you include transition arrows for {b,d,p,q}. For your example, A = μx({u}x{v} ∪ {w}), you have A' = {b}{up,qv,w}*{d} and you can easily write down the corresponding DFA. That automaton is just the one-counter machine, itself, with "b" interpreted as "start up at count 0", "d" as "check for count 0 and finish", "p" as "add one to the count" and "q" as "check for count greater than 0 and subtract 1". With respect to the algebraic rules given for {b,d,p,q}, A' is not just a representation of A, it is actually is A: A' = A.
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.
I am quite confused about formal description of PDA (push down automata)
If we write down L(M), this means that PDA M recognizes language L, correct ?
then , L(M)* means that PDA M recognizes that L* right ?
but what is the L* means ? how PDA can recognize infinite combination of L ?
L(M)* means that PDA M recognizes that L* right ?
No! It means a language which is constituted from the concatenation of any number of sentences (including zero) that are valid in the language recognized by the PDA M.
Given that L(M) is context-free, it is easy to prove that L(M)* is context-free too.
To construct L(M)* , just take all the grammar of L(M), take the starting symbol S from L(M) and add two new productions R -> S R and R -> empty where R is the starting symbol of L(M)*.
So, given that L(M)* is context-free, then there is some PDA that recognize it. If you could construct a PDA for L(M), then a PDA for L(M)* should be trivial to construct, given that it is almost the same as L(M) just with two extra productions.