3. Regular expressions

A = {00, 0000, 00000000 …}

AB = {00, 000, 0000, 00000, …} ← this is the cartesian product between the sets(languages) A and B, where the elements of A come first.

• where the words in the language that have an even length are obtained by combining a word from A with the word ε from B
• and those with an odd length are obtained by combining a word from A with the word 0 from B

A is the language in which the words start with zero and end with one and the number of one is equal to the number of zeros (the same value for n is used).

The notation of n in the definition of B is completely unrelated to the n used to define A.

So, B is the language of words made of sequences of ones, having the length of at least 1, so basically B = L(11*).

A = {01, 0011, 000111, 00001111 …}

B = {1, 11, 111 …}

AB = {011, 00111, 000011111, …, 0111, 001111, 00011111, 0000111111, … }

BA = {101, 10011, 1000111, …, 1101, 110011, …, 11101, 1110011 …}

AB = ∅ (because A is empty, so the cartesian product leads to an empty set)

A* = {ε} (epsilon is always part of Kleene star)

B*= {ε} (epsilon is always part of Kleene star) U {$ 1^n $} U {$ 1^{2n} $} U {$ 1^{3n} $} U … So basically B = L( ($1^n$)* )

$ AB = \{ 0^n 1^n 0^{m+k} \mid m \geq n \geq 1, k \geq 1 \}$ . Deci $ AB = A$ .

Note that the n in the definition of language A is different from the n in in the definition of B, they are independent when used in defining different sets/languages. However, when n is used several times in the definition of one language, such as the 2 times it appears in language A, it is the same value.

$ BA = \{ 0^{(n+k)} 1^n 0^m \mid m \geq n \geq 1, k \geq 1 \}$ . Equivalently: $ BA = \{0^x 1^y 0^z \mid x \geq y\geq 1 \text{ and } z \geq y \geq 1 \}$

We start by defining the regex for a number:

• a number can be a digit from 0 to 9 ⇒ [0-9]
• a number can have several digits, but the first one can't be 0 ⇒ [1-9][0-9]*
• so we have either one of these options: [0-9] U [1-9][0-9]*
• But we can write it in a more concise way: 0 U [1-9][0-9]* ⇐ a number of 1 or more digits that doesn't start with 0, or 0 itself

Having decided on the regex for the number, we can write a regex for expressions

(0 U [1-9][0-9]*) ( ('+' U '*') (0 U [1-9][0-9]*) )*

This can be understood with a clearer/intuitive notation, which however is not exactly formally correct (not a regex):

number ( ('+' U '*') number )*

(1 U ε) ( 0 0* (1 U ε) )* (0 U ε) ( 1 1* (0 U ε) )*

We can start with either 1 or 0. Then we can have any sequences of zeros of any length, but no sequence of ones with a length bigger than 1.

This way we make sure that any sequence of 2 or more zeros precedes any sequence of 2 or more ones.

Then we repeat the same logic on the left, making sure no sequence of 2 or more zeros appear on this side.

Looking at the DFA we can tell state 3 is a sink state. We can simplify the DFA's drawing by not looking at/ignoring it. Let's see what words are accepted:

• ab*ab (when we don't loop on state 1)
• ab*(<a way to leave state 2 and return back to it>)*ab ⇒ ab*(aab*)*ab
• anything that repeats the previous expression several times
• ε (the initial state is also a final state)
• from the previous 2 observations ⇒ we can use Kleene star

So, the regex is:

( ab*(aab*)*ab )*

We the given property implies that the word is composed of repeated sequences of '10' and '01' (so elements at odd positions are different from their left neighbour), possibly followed by a final 1 or 0:

After any prefix of even length with an equal number of 0's and 1's (including $\epsilon$), you can either finish the word, add a single character and then finish the word (both of these keeping the rule true) or add at least 2 characters. In the last case, if the characters are equal the prefix ending at these two characters breaks the rule.

$(01\cup10)^*(0\cup1\cup\epsilon)$