What is the Pumping Lemma in Layman's terms?

Question

I saw this question, and was curious as to what the pumping lemma was (Wikipedia didn\'t help much).

I understand that it\'s basically a theoretical p

Answer 1

Its a difficult thing to get in layman's terms, but basically regular expressions should have a non-empty substring within it that can be repeated as many times as you wish while the entire new word remains valid for the language.

In practice, pumping lemmas are not sufficient to PROVE a language correct, but rather as a way to do a proof by contradiction and show a language does not fit in the class of languages (Regular or Context-Free) by showing the pumping lemma does not work for it.

Answer 2

The pumping lemma is a simple proof to show that a language is not regular, meaning that a Finite State Machine cannot be built for it. The canonical example is the language (a^n)(b^n). This is the simple language which is just any number of as, followed by the same number of bs. So the strings

ab
aabb
aaabbb
aaaabbbb

etc. are in the language, but

aab
bab
aaabbbbbb

etc. are not.

It's simple enough to build a FSM for these examples:

This one will work all the way up to n=4. The problem is that our language didn't put any constraint on n, and Finite State Machines have to be, well, finite. No matter how many states I add to this machine, someone can give me an input where n equals the number of states plus one and my machine will fail. So if there can be a machine built to read this language, there must be a loop somewhere in there to keep the number of states finite. With these loops added:

all of the strings in our language will be accepted, but there is a problem. After the first four as, the machine loses count of how many as have been input because it stays in the same state. That means that after four, I can add as many as as I want to the string, without adding any bs, and still get the same return value. This means that the strings:

aaaa(a*)bbbb

with (a*) representing any number of as, will all be accepted by the machine even though they obviously aren't all in the language. In this context, we would say that the part of the string (a*) can be pumped. The fact that the Finite State Machine is finite and n is not bounded, guarantees that any machine which accepts all strings in the language MUST have this property. The machine must loop at some point, and at the point that it loops the language can be pumped. Therefore no Finite State Machine can be built for this language, and the language is not regular.

Remember that Regular Expressions and Finite State Machines are equivalent, then replace a and b with opening and closing Html tags which can be embedded within each other, and you can see why it is not possible to use regular expressions to parse Html

Answer 3

This is not an explanation as such but it is simple. For a^n b^n our FSM should be built in such a way that b must know the number of a's already parsed and will accept the same n number of b's. A FSM can not simply do stuff like that.

Answer 4

The simple pumping lemma is the one for regular languages, which are the sets of strings described by finite automata, among other things. The main characteristic of a finite automation is that it only has a finite amount of memory, described by its states.

Now suppose you have a string, which is recognized by a finite automaton, and which is long enough to "exceed" the memory of the automation, i.e. in which states must repeat. Then there is a substring where the state of the automaton at the beginning of the substring is the same as the state at the end of the substring. Since reading the substring doesn't change the state it may be removed or duplicated an arbitrary number of times, without the automaton being the wiser. So these modified strings must also be accepted.

There is also a somewhat more complicated pumping lemma for context-free languages, where you can remove/insert what may intuitively be viewed as matching parentheses at two places in the string.

Answer 5

It's a device intended to prove that a given language cannot be of a certain class.

Let's consider the language of balanced parentheses (meaning symbols '(' and ')', and including all strings that are balanced in the usual meaning, and none that aren't). We can use the pumping lemma to show this isn't regular.

(A language is a set of possible strings. A parser is some sort of mechanism we can use to see if a string is in the language, so it has to be able to tell the difference between a string in the language or a string outside the language. A language is "regular" (or "context-free" or "context-sensitive" or whatever) if there is a regular (or whatever) parser that can recognize it, distinguishing between strings in the language and strings not in the language.)

LFSR Consulting has provided a good description. We can draw a parser for a regular language as a finite collection of boxes and arrows, with the arrows representing characters and the boxes connecting them (acting as "states"). (If it's more complicated than that, it isn't a regular language.) If we can get a string longer than the number of boxes, it means we went through one box more than once. That means we had a loop, and we can go through the loop as many times as we want.

Therefore, for a regular language, if we can create an arbitrarily long string, we can divide it into xyz, where x is the characters we need to get to the start of the loop, y is the actual loop, and z is whatever we need to make the string valid after the loop. The important thing is that the total lengths of x and y are limited. After all, if the length is greater than the number of boxes, we've obviously gone through another box while doing this, and so there's a loop.

So, in our balanced language, we can start by writing any number of left parentheses. In particular, for any given parser, we can write more left parens than there are boxes, and so the parser can't tell how many left parens there are. Therefore, x is some amount of left parens, and this is fixed. y is also some number of left parens, and this can increase indefinitely. We can say that z is some number of right parens.

This means that we might have a string of 43 left parens and 43 right parens recognized by our parser, but the parser can't tell that from a string of 44 left parens and 43 right parens, which isn't in our language, so the parser can't parse our language.

Since any possible regular parser has a fixed number of boxes, we can always write more left parens than that, and by the pumping lemma we can then add more left parens in a way that the parser can't tell. Therefore, the balanced parenthesis language can't be parsed by a regular parser, and therefore isn't a regular expression.

Answer 6

For example, take this language L = aⁿbⁿ.

Now try to visualize finite automaton for the above language for some n's.

if n = 1, the string w = ab. Here we can make a finite automaton with out looping if n = 2, the string w = a²b². Here we can make a finite automaton with out looping

if n = p, the string w = a^pb^p. Essentially a finite automaton can be assumed with 3 stages. First stage, it takes a series of inputs and enter second stage. Similarly from stage 2 to stage 3. Let us call these stages as x, y and z.

There are some observations

1. Definitely x will contain 'a' and z will contain 'b'.
2. Now we have to be clear about y:
   • case a: y may contain 'a' only
   • case b: y may contain 'b' only
   • case c: y may contain a combination of 'a' and 'b'

So the finite automaton states for stage y should be able to take inputs 'a' and 'b' and also it should not take more a's and b's which cannot be countable.

1. If stage y is taking only one 'a' and one 'b', then there are two states required
2. If it is taking two 'a' and one 'b', three states are required with out loops and so on....

So the design of stage y is purely infinite. We can only make it finite by putting some loops and if we put loops, the finite automaton can accept languages beyond L = aⁿbⁿ. So for this language we can't construct a finite automaton. Hence it is not regular.