I am trying to prove that the following language is not regular using the pumping lemma
L= { a^i b^j | i^2 > j}
Any tips on this? I am completely stuck.
Thanks.
The pumping lemma says:
If a language A is regular => there is a number p (pumping length) where, if s is any string in L such that |s| >= p, then s may be divided into three pieces s=xyz, satisfying the following condition:
xyiz is in L for each i>=0
|y|>=0
p>=|xy|
The right way to show that a certain language L is not regular is to suppose L regular and try to reach a contradiction.
Lets try to demonstrate that L = {0n1n}|n>=0} is not regular.
We start assuming to the contrary that L is regular.
You can think about this kind of demonstration as a game:
Challenger: He choose the pumping length p. You cannot do any presumption on it.
You: Now it is your turn: choose the "kind" of string that represents the irregularity of the language.
Lets say that the string is in the form 0p1p.
A good tip in this step is to try to limit the adversary next move.
Challenger: He presents to you a string s in the form 0p1p.
You: It's time to pump! If you chose correctly the form of the string in your previous move, you can do some assumption. In our case, for example, we know that the substring y consists only of 0s (at least one 0 because |y|>0), because |xy|<=p and first p-elements are 0s.
Now we show that it exists i>=0 such that xyiz is not in L. For example, for i=2 the string xyyz has more 0s than 1s and so is not a member of L. This case is a contradiction. => L is not regular.
Never forget to demonstrate why the pumped string cannot be a member of L.
If you have any doubt, feel free to ask :)
Cheers.
To the above answer, "The pumping lemma says: If a language A is regular => there is a number p (pumping length) where, if s is any string in L such that |s| >= p, then s may be divided into three pieces s=xyz, satisfying the following condition:"
You mean "If a language L is regular"
Also, the three conditions
1. xy^iz is in L for each i>=0
2. |y|>=0
3. p>=|xy|
The second should be just |y| > 0 not >=
Say you choose the string:
a^2b^5
aabbbbb. Which is in the language.
Now your opponent can choose XYZ.
Their options:
1.) X(empty)Y(some a's)
2.) X(some a's)Y(some a's and some b's)
3.) X(some a's)Y(some a's)
Based on their possible choices, you pump up Y using Y^i where i is an arbitrary number of your choice.
Say they choose 1.)
X(-)Y(a)Z(abbbbb)
If you "pump" up Y^i choosing i = 0. The new string becomes abbbbb. Which is not in the language.
Repeat this for each possible choice of the opponent, if you can pump up Y in a way that produces a string that is not in the language L, then you've succeeded in proving that the language is not regular.
Related
Where can I find the proof for the linear context free languages pumping lemma?
I am looking for the proof that is specific for the linear context free language
I also looked for the formal prof and could not find one.Not sure if the below is a formal prof but it may give you some idea.
The lemma : For every linear context free languages L there is an n>0 so that for every w in L with |w| > n we can write w as uvxyz such that |vy|> 0,|uvyz| <= n and uv^ixy^iz for every i>= 0 is in L.
"Proof":
Imagine a parse tree for some long string w in L with a start symbol S. Also lets assume that the tree does not contains non useful nodes. If w is long enough, there will be at least one non terminal repeating more than once. Lets call the first repeating non terminal going down the tree X, its first occurrence (from the top) as X[1] and its second occurrence as X[2].Let x be the string in w generated by X[2], vxy the string generated by X[1]and uvxyz the full string w generated by S. Since the movement from X[1] to X[2] generates v,y we could theoretically generate a new tree where we replicate this move multiple times before moving from X[1] down.This proves that uv^ixy^iz for every i>= 0 is in L. Since our tree contains no useless nodes, moving from X[1] to X[2] must generate some terminals and this proves that |vy|> 0.L is linear which means that on every level of the tree we have a single non terminal symbol. Each node in the tree covers some substring in w that its length is bounded by a linear function of the node height. Moving from S to X[2] covers uv and yz from w and the number of tree levels traveled is bounded by (2 * the number of non-terminals symbols + 1). Since the number of levels traveled is bounded and the tree is linear it also puts a bound on the yield of the movement from S to X[2] which means ,|uvyz| <= n for some n >= 0.
Note: Keep in mind that we construct X[1] , X[2] top down , in contradiction to how we prove the “regular” pumping lemma for context free grammar in general. In the "regular” pumping lemma there is a bound on the height of X[1] and therefore a bound on |vxy|. In our case there is no bound on the height of X[1]and it can be as high as required by the length of w. There is a bound,however,on the number of tree levels from S to X[2].This does not means much if the grammar is not linear as the output going from S to X[2] is still bounded only by the high of S (that is unbounded). But in the linear case,this output is bounded and therefore |uvyz| <= n
If L is a regular language, then there exists a constant n (which depends on L) such that for every string w in the language L, such that the length of w is greater than or equal to n, we can divide w into three strings, w = xyz.
w = length of string. n = Number of States.
Why should we pick w greater than or equal to n?
and what is Pumping length?
If you look at the complete statement of the lemma (http://en.wikipedia.org/wiki/Pumping_lemma_for_regular_languages), you can see that it is actually stating that every string is formed by a prefix x, a part that can be repeated any number of times y and a suffix z. Now it is obvious that, in the shortest case (when the repeating part is taken only once), the length of w equals the number of states needed for the language. This Wikipedia image is very useful:
http://en.wikipedia.org/wiki/File:Pumping-Lemma_xyz_svg.svg
You seem to be misunderstanding the lemma (which you also have not stated completely), and mixing aspects of a proof with what you did state. The lemma says that for every regular language L, there is a constant p such that every string of at least p symbols that belongs to L has a non-empty substring of length no greater than p that can be "pumped", always yielding another element of L. The constant p is the (a) "pumping length".
This can be proved by observing that if a language is regular then there is a finite state automaton that accepts it, and taking p to be the number of states in that automaton (details omitted).
That does not imply, however, that the number of states in the smallest FSA the recognizes a given regular language is the smallest possible pumping length for that language. For instance, consider the language consisting of the union of { an } and { bn } for all n. You need a four-state FSA to recognize this language, but its minimum pumping length is 1.
Proving that a^n b^n, n >= 0, is non-regular.
Using the string a^p b^p.
Every example I've seen claims that y can either contain a's, b's, or both. But I don't see how y can contain anything other than a's, because if y contains any b's, then the length of xy must be greater than p, which makes it invalid.
Conversely, for examples such as:
www, w is {a, b}*, the string used is a^p b a^p b a^p b. In the proofs I've seen, it claims that y cannot contain anything other than a's, for the reason I stated above. Why is this different?
Also throwing in another question:
Describe the error in the following "proof" that 0* 1* is not a regular language. (An
error must exist because 0* 1* is regular.) The proof is by contradiction. Assume
that 0* 1* is regular. Let p be the pumping length for 0* 1* given by the pumping
lemma. Choose s to be the string OP P. You know that s is a member of 0* 1*, but
a^p b^p cannot be pumped. Thus you have a contradiction. So 0* 1* is not regular.
I can't find any problem with this proof. I only know that 0*1* is a regular language because I can construct a DFA.
The pumping lemma states that for a regular language L:
for all strings s greater than p there exists a subdivision s=xyz such that:
For all i, xyiz is in L;
|y|>0; and
|xy|<p.
Now the claim that y can only contain a's or b's originates from the first item. Since if it contained both a's and b's, with i=2, this would result in a string of the form aa...abb...baa...b, etc. That's what the statement wants to say.
The third part indeed, makes it obvious that y can only contain a's. In other words, what the textbooks say is a conclusion derived from the first item.
Finally if you combine 1., 2. and 3., one reaches contradiction, because we know y must contain at least one character (2.), the string can only contain a's. Say y contains k a's. If we would "pump" this with i=2, the result is that we generate a string:
s'=xy2z=ap+kbp
We know however that s' is not part of L, which it should be by 1., so we reach inconsistency.
You can thus only make the proof work by combining the three items. It's not enough to know that y consist only out of a's: that doesn't result in contradiction. It's because there is no subdivision available that satisfies all three constraints simultaneously.
About your second question. In that case, L looks different. You can't reuse the proof of a^nb^n because L is perfectly happy if the string contains more a's. In other words, you can't find a contradiction. In other words, the last item of the proof fails. As long as y contains only one type of characters - regardless of its length - it can satisfy all three constraints.
When I prove some theorem, my goal evolves as I apply more and more tactics. Generally speaking the goal tends to split into sub goals, where the subgoals are more simple. At some final point Coq decides that the goal is proven. How this "proven" goal may look like? These goals seems to be fine:
a = a. (* Any object is identical to itself (?) *)
myFunc x y = myFunc x y. (* Result of the same function with the same params
is always the same (?) *)
What else can be here or can it be that examples are fundamentally wrong?
In other words, when I finally apply reflexivity, Coq just says ** Got it ** without any explanation. Is there any way to get more details on what it actually did or why it decided that the goal is proven?
You're actually facing a very general notion that seems not so general because Coq has some user-friendly facility for reasoning with equality in particular.
In general, Coq accepts a goal as solved as soon as it receives a term whose type is the type of the goal: it has been convinced the proposition is true because it has been convinced the type that this proposition describes is inhabited, and what convinced it is the actual witness you helped build along your proof.
For the particular case of inductive datatypes, the two ways you are going to be able to proved the proposition P a b c are:
by constructing a term of type P a b c, using the constructors of the inductive type P, and providing all the necessary arguments.
or by reusing an existing proof or an axiom in the environment whose type you can get to match P a b c.
For the even more particular case of equality proofs (equality is just an inductive datatype in Coq), the same two ways I list above degenerate to this:
the only constructor of equality is eq_refl, and to apply it you need to show that the two sides are judgementally equal. For most purposes, this corresponds to goals that look like T a b c = T a b c, but it is actually a slightly more broad notion of equality (see below). For these, all you have to do is apply the eq_refl constructor. In a nutshell, that is what reflexivity does!
the second case consists in proving that the equality holds because you have other equalities in your context, nothing special here.
Now one part of your question was: when does Coq accept that two sides of an equality are equal by reflexivity?
If I am not mistaken, the answer is when the two sides of the equality are αβδιζ-convertible.
What this grossly means is that there is a way to make them syntactically equal by repeated applications of:
α : sane renaming of non-free variables
β : computing reducible expressions
δ : unfolding definitions
ι : simplifying matches
ζ : expanding let-bound expressions
[someone please correct me if more rules apply or if I got one wrong]
For instance some of the things that are not captured by these rules are:
equality of functions that do more or less the same thing in different ways:
(fun x => 0 + x) = (fun x => x + 0)
quicksort = mergesort
equality of terms that are stuck reducing but would be equal:
forall n, 0 + n = n + 0
I know that {a^i b^j | i = j } is not regular and I can prove with pumping lemma; similarly I can use pumping lemma to prove this one not regular too. But I think I see some similar problem that says such language is actually regular. And because I'm not confident about my knowledge in pumping lemma I'm asking this bad question. Sorry.
This is how I prove it: let w be a^p b^(19k+p) , clearly this is in the language. Then if I pump a, making it a^(p+1) b^(19k+p) . It fails. Therefore, it's not regular.
Is my proof right?
Take a look at this answer. In short, you're not pumping the string correctly. The pumping lemma states that your string w can be divided as w = xyz where |xy| ≥ p, and y is not empty. You can then pump the string as xyiz for all i ≥ 0.
The key here, is that the pumping lemma states that there exists a division of the string w satisfying these properties, you do not get to choose the division, and that you can only pump the string as xyiz.
However, this language is regular, so the pumping lemma can't be used to prove if a language is regular, it can only prove if a language is irregular (it's a necessary but not sufficient condition). To show that a language is regular you can construct a DFA, NFA or regular expression that describes your language exactly. One such regular expression is:
(a^19)*(e|ab|aabb|aaabbb|...|a^18b^18)(b^19)*
where e is the empty string.
I suspect that your language is an example from an introductory course in automata or computation. If you're interested, the Myhill-Nerode theorem is not often covered in introductory material, but in this case offers a very easy proof of regularity: consider the distinguising extensions b, bb, bbb, ..., b^19 and the proof follows relatively easily from that.