✎ lfa:2024:lab03 [books]

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====== 3. Regular Expressions ======

===== 3.1. Natural Language / DFA $ \rightarrow $ Regex conversion =====

For each of the exercises from DFA Seminary 1 write a regex describing the same language.

**3.1.1.** $ L=\{w \in \{0,1\}^* \text{ | w contains an odd number of ones} \} $

{{:lfa:2022:lfa2022_lab2_ex1.png?400|}}

<hidden 3.1.1.>
$math[0^*10^*(10^*10^*)^*]
</hidden>


**3.1.2.** The language of binary words which contain **exactly** two ones

{{:lfa:2022:lfa2022_lab2_ex2.png?550|}}

<hidden 3.1.2.>
$math[0^*10^*10^*]
</hidden>

**3.1.3.** The language of binary words which encode odd numbers (the last digit is least significative)

{{:lfa:2022:lfa2022_lab2_ex3.png?300|}}

<hidden 3.1.3.>
$math[(0 \cup 1)^*1]
</hidden>


**3.1.4.** The set of all binary strings having the substring 00101

{{:lfa:2022:ex_8_dfa.png?400|}}

<hidden 3.1.4.>
$math[(0 \cup 1)^*00101(0 \cup 1)^*]
</hidden>

===== 3.2. Formation rules (concatenation, reunion, Kleene star) =====

**3.2.1.**

$math[A=\{ 0^{2k} \mid k \geq 1 \}]

$ B = \{0, \epsilon \}$
\\
$ AB = ? $
\\

<hidden>
A = {00, 0000, 00000000 ...}

AB = {00, 00**0**, 0000, 0000**0**, ...} <- 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
</hidden>



**3.2.2.**

$math[A = \{ 0^n 1^n \mid n \geq 1 \}]
\\
$ B = \{ 1^n \mid n \geq 1 \} $
\\
$ AB = ? $ \\ $ BA = ? $

<hidden>
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 = {01**1**, 0011**1**, 00001111**1**, ..., 01**11**, 0011**11**, 000111**11**, 00001111**11**, ... } 

BA = {**1**01, **1**0011, **1**000111, ..., **11**01, **11**0011, ..., **111**01, **111**0011 ...}
</hidden>

**3.2.3**

$math[A = \{ 0^n 1^n 0^m \mid m \geq n  \geq 1 \}]
\\
$ B = \{ 0^n \mid n \geq 1 \} $
\\
$ AB = ? $ \\ $ BA = ? $


<hidden>
$math[AB = \{ 0^n 1^n 0^{m+k} \mid m \geq n  \geq 1, k \geq 1 \}]. Deci $math[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.

$math[BA = \{ 0^{(n+k)} 1^n 0^m \mid m \geq n  \geq 1, k \geq 1 \}]. Equivalently: $math[BA = \{0^x 1^y 0^z \mid x \geq y\geq 1 \text{ and } z \geq y \geq 1 \}]
</hidden>

**3.2.4.**

$ A = ∅ $
\\
$ B = \{ 1^n \mid n \geq 1 \} $
\\
$ AB = ? $
\\
$ A^* = ? $
\\
$ B^* = ? $
\\


<hidden>

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( ($1^n$)* )

</hidden>