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# 5.1. Identifying Non-regular Languages¶

## 5.1.1. Identifying Non-regular Languages¶

We have now spent a lot of time time looking at a bunch of ways of describing languages. But they are all pretty much the same, in that they all desribe the same set of languages: Regular Languages. And we have hinted numerous times that not all languages are regular. So, finally, we come to grips with (1) seeing some actual non-regular languages, and (2) using tools that help us to prove that a given language is non-regular.

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We are about to formalize the proof in the slideshow into a tool for proving some languages to be nonregular. But first, let’s explore the relationship of loops in DFAs to regular languages a bit more.

First, we know that loops don’t always cause a problem. In fact, there is a simple relationship between DFAs with or without loops, and languages that are infinite or finite. That is, a finite language is accepted by a DFA with no loop. Its not possible that the language accepted would be finite, since we can’t control the number of times that the machine goes around the loop. That is, the machine can accept all strings in the finite language, but it has to accept more strings as well. Conversely, any infinite language must be accepted by a DFA with one or more loops.

Next, consider that we can use a DFA to “count” a finite number of things. For example, we can make a DFA that never has three consecutive a’s, or one that has at most three a’s in the string. So long as we want to count a fixed number (or maximum number) of things, we are OK. And of course, such a machine can have a loop. What cannot happen is for the loop to affect the counting process. So, the series of states that counts to three a’s can include a loop to process an arbitrary number of b’s, because that will not disrupt the count of a’s. But if the loop includes a’s, then we “lose count” of how many a’s we have seen, because we can’t control the number of times through the loop (or more precisely, we must accept strings that add characters processed an arbitrary number of times through the loop).

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## 5.1.3. The Pumping Lemma¶

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How To Use the Pumping Lemma to prove L is not regular: Proof by Contradiction

• Assume L is regular.

• Therefore $L$ satisfies the pumping lemma.

• Choose a long string $w \in L$, $|w| \ge m$. Remember that $m$ relates to the number of states in the machine, with $m$ being great enough that it has to trigger at least one loop. Which string we pick is critical. We must pick a string that will yield a contradiction. And we succeed with the proof if we find any such string, even if there exist other stings that don’t let us succeed.

• Show that, for our string, there is NO division of $w$ into $xyz$ (we must consider all possible divisions) such that $|xy| \le m$, $|y| \ge 1$ and $xy^iz \in L$ for all $i \ge 0$.

• If we show that there is NO possible division, then we have a contradiction!

• $\Rightarrow L$ is not regular.

Unfortunately, the pumping lemma is one-way: For (some) languages we can use the pumping lemma to prove that they are not regular. But we cannot use the pumping lemma to help us prove that a language is regular. And the pumping lemma is not a universal solution for determining that a language is non-regular. Its just a tool in the toolbox.

## 5.1.4. Some Pumping Lemma Examples¶

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Example 5.1.1

Use the Pumping Lemma to prove that the language $L = \{a^mb^n \mid n+m$ is odd $\}$ is non-regular.

But wait! $L$ is a regular language! For example, it is not hard to create a DFA that maintains states for whether we have seen an even number of symbols so far or an odd number. If the language is indeed regular, you should find it impossible to use the pumping lemma to prove it non-regular. In particular, not all values of $m$ should have a value for $w$ that can be decomposed into $xyz$ such that it is pumpable (that is, $|xy| \le m$, $|y| \ge 1$, and $xy^iz \in L$ for all values of $i \ge 0$). The intuition here is that for any suitably long string, we can always find a decomposition that lets us pump the $y$ substring.

Let’s start by considering the value $m = 1$. For string $w = abb$, it turns out that the only legitimate decomposition yields $y = a$, which cannot be pumped. But this does not mean that the language is non-regular. We can’t just pick our favorite value value for $m$, the Pumping Lemma demands that this condition be true for all other legitimate values of $m$ as well.

In particular, for any value of $m \ge 2$, $w$ has to be at least 3 symbols long (since it has to be of odd length to be in the language). And in this case, we can always decompose the string such that $y$ has either two a’s or two b’s. Which means that it can be pumped any number of times (or deleted), and the resulting string is still of odd length, and therefore is in the language.

Example 5.1.2

Let’s look at some more languages that are easily shown to be non-regular by the Pumping Lemma. In particular, consider these languages:

• $L = \{a^ncb^n | n > 0\}$

• $L = \{a^nb^{n+s}c^s | n,s > 0\}$

• $\Sigma = \{a, b\}, L = \{w \in \Sigma^* | n_a(w) > n_b(w)\}$. (Remember that $n_a(w)$ means the number of a’s in $w$.)

For each of these languages, we can use the same strategy that we used in the examples of $L = \{a^nb^n\}$ and $L = \{ww^R | w \in \Sigma^*\}$. Namely, we pick a string with at least $m$ leading a’s, and show that since this results in $y$ being some number of a’s, it cannot be pumped.

Now let’s look at an example that is not so easy, because we cannot use that simple strategy. This means that we have to pick a string $w$ that will lead to a number of cases for the decomposition into $xyz$ that we will have to get through.

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## 5.1.5. The Pumping Lemma Adversary Game¶

Here is an adversary argument way of looking at this. Your goal is to establish a contradiction (to prove the language is not regular), while the opponent tries to stop the proof. The moves in the game are:

1. The opponent picks $m$.

2. We pick string $w$ in $L$ of length equal or greater than $m$. We are free to chose any $w$, so long as $w\in L$ and $|w|\ge m$.

3. The opponent chooses the decomposition $xyz$, such that $|xy|\le m,|y|\ge1$. The opponent will make the choice that is hardest for us to win the game.

4. We try to pick $i$ so that the pumped string $w_i=xy^iz$ is not in $L$. If we can always do this, we win ($L$ is not regular).

As we see, the adversary games are role based game where we seek to prove the language is non-regular. The adversary seeks to stop us.

Consider the Pumping Lemma definition again:
Let $L$ be an infinite regular language. There exists a constant $m > 0$ such that any $w \in L$ with $|w| \ge m$ can be decomposed into three parts as $w=xyz$ with:
$|xy| \le m$
$|y| \ge 1$
$xy^iz \in L$ for all $i\ge 0$

To connect the adversary game to the pumping lemma proof, we divide proof into steps as follows:

In the pumping lemma proof we write
There exists a constant $m > 0$ [$=$ Adversary picks a value for $m$.]
such that any $w \in L$ with $|w| \ge m$ [$=$ WE pick our choice for $w$.]
can be decomposed into three parts as $w = xyz$ [$=$ Adversary picks $xyz$] (but they are required to meet the length criteria on $xy$ and $y$)
… such that $xy^iz \in L$ for all $i \ge 0$ [$=$ WE pick a value for $i$.]

In the adversary game below, there is a list of languages to chose from. Some of these are regular, some are non-regular. If you think that the language is non-regular, then you should choose to let the computer go first. This makes the computer pick $m$, and then you pick a string $w$ that lets you complete the proof, and so you win.

However, if you think that a language is regular, then you should choose to go first (that is, you pick the value for $m$, which is effectively picking the machine to recognize the language within the context of the proof). In that case, you want to make moves that stop the proof (you will make an effective decomposition for $xyz$ for whatever string $w$ the computer picks), and so you win.

## 5.1.6. Using Closure Properties to Prove L is Not Regular¶

Sometimes we are unable prove that a language is non-regular by using the pumping lemma. Maybe we can’t find the right string to use initially, or we can’t figure out an argument for why there is a contradiction. Either way, it helps to have other tools to prove languages are non-regular. So here is another tool that we might be able to use.

Recall that regular languages are closed under certain operations. For example, a regular language that is the union of two known regular languages is itself regular. This is an example of using closure properties to prove that a language is regular.

In a similar way, we can use closure properties to show that a language is not regular. The approach is to use certain operations to derive a language that we already know is non-regular.

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Let’s do one more quick example. Prove that $L_1 = \{a^nb^na^n\ |\ n > 0\}$ is non-regular by using closure operations.

• Assume that $L_1$ is non-regular, and derive a contradiction.

• The goal is to try to construct $\{a^nb^n | n > 0\}$ which we know is not regular.

• Note that trying to intersect with $\{a^{*}b^{*} \}$ does not help us, because the intersection is just the empty set.

• Let $L_2 = \{a^{*}\}$. $L_2$ is regular, since we are defining this using a regular expression.

• Now define $L_3 = L_1 \backslash L_2 = \{a^nb^na^p\ |\ 0 \le p \le n, n > 0\}$. This is using the right quotient operation, which we know is closed for regular languages. In this case, we are just trimming some number of a’s from the end of $L_1$. Sometimes we trim all of the a’s from the end of $L_1$.

• By closure under intersection, $L_4 = L_3 \cap \{a^{*}b^{*}\} = \{a^nb^n\ |\ n > 0\}$ is regular. Note that we had to do the step with the trimming of letters, because simply intersecting $L_1$ with $a^*b^*$ does not give us what we want.

• We already proved that $L_4$ is not regular. Contradiction.

• $\Rightarrow L_1$ is not regular.

## 5.1.7. Questions to Think About¶

To review what we now know: There are languages that are regular, and there are languages that are nonregular. Regular languages can be represented in any of several interchangeable ways. Some nonregular languages can be proved such using tools like the Pumping Lemma, and closure properties.

These facts should lead us to ask some broader questions. In particular, is every language either regular or nonregular? And if so, can we always decide, for every language, whether it is regular or nonregular?

Remember what a language is: It is simply a set of strings. Most sets of strings are infinite, in that there are many more infinite sets of strings than there are finite sets of strings. (Is this claim really true? Does that actually make sense? There are an infinite number of finite sets of strings.) An important point is that a language is not just those sets of strings that have a description as, for example, a RegEx (of course not, since not all languages are regular). A language is not even just those sets that can be described in English, or a mix of English and math notation.

We will come back to these and similar questions later in the book. They relate to issues of Turing decideable vs. Turing acceptable languages, P vs. NP, and what questions about languages are decideable vs. undecideable. By the end of this book, we should have some answers to these questions, and a better understanding of our limits to what can be known about languages.