Analyzing and Comparing Regular Expression Languages

We are given three regular expressions: $r$, $s$, and $t$. Our goal is to determine the relationship between the languages generated by these regular expressions, denoted as $L(r)$, $L(s)$, and $L(t)$. Let's analyze each regular expression individually.

Understanding Each Regular Expression

1. Regular Expression $r = a(b + a)*$
   • This expression starts with the character 'a'.
   • It is followed by zero or more occurrences of either 'b' or 'a'. The term $(b + a)*$ represents any string composed of 'a's and 'b's, including the empty string ($\epsilon$).
   • Therefore, $L(r)$ consists of all strings that begin with 'a' followed by any sequence of 'a's and 'b's.
   • Examples of strings in $L(r)$: 'a' (a followed by $\epsilon$), 'aa', 'ab', 'aaa', 'aab', 'aba', 'abb', ...
   • Mathematically, $L(r) = \{a\} \cdot (\{a, b\})^*$.
2. Regular Expression $s = a(a + b)+$
   • This expression also starts with the character 'a'.
   • It is followed by one or more occurrences of either 'a' or 'b'. The term $(a + b)+$ represents any non-empty string composed of 'a's and 'b's.
   • Therefore, $L(s)$ consists of all strings that begin with 'a' followed by a non-empty sequence of 'a's and 'b's.
   • Examples of strings in $L(s)$: 'aa', 'ab', 'aaa', 'aab', 'aba', 'abb', ... (Notice 'a' is not in $L(s)$ because it must be followed by at least one character).
   • Mathematically, $L(s) = \{a\} \cdot (\{a, b\})^+$.
3. Regular Expression $t = aa*b$
   • This expression starts with the character 'a'.
   • It is followed by zero or more occurrences of 'a' ($a*$).
   • Finally, it is followed by the character 'b'.
   • Therefore, $L(t)$ consists of strings that start with 'a', followed by zero or more 'a's, and end with a 'b'.
   • Examples of strings in $L(t)$: 'ab' ($a^1 b$), 'aab' ($a^2 b$), 'aaab' ($a^3 b$), ...
   • Mathematically, $L(t) = \{a\} \cdot \{a\}^* \cdot \{b\} = \{a^n b \mid n \ge 1\}$. Note: Some definitions might include $n=0$ for $a^0 b$, but the $a$ at the beginning makes it $a^1 a^* b$. Let's re-evaluate: $aa^*b$ means 'a' followed by zero or more 'a's followed by 'b'. This yields 'ab' (a followed by $\epsilon$ followed by b), 'aab' (a followed by 'a' followed by b), 'aaab' (a followed by 'aa' followed by b), etc. This corresponds to $a^k b$ where $k \ge 1$. So strings are 'ab', 'aab', 'aaab', ...
   • Mathematically, $L(t) = \{a^k b \mid k \ge 1\}$.

Comparing the Languages L(r), L(s), and L(t)

Comparing L(r) and L(s)

• $L(r)$ contains strings starting with 'a' followed by any string from $\{a, b\}^*$. This includes the empty string.
  • $L(r) = \{a, aa, ab, aaa, aab, aba, abb, ...\}$
• $L(s)$ contains strings starting with 'a' followed by any string from $\{a, b\}^+$. This includes any non-empty string.
  • $L(s) = \{aa, ab, aaa, aab, aba, abb, ...\}$
• Notice that any string in $L(s)$ is a string starting with 'a' followed by a non-empty sequence of 'a's and 'b's. Since any non-empty sequence is also a sequence (including empty), every string in $L(s)$ is also present in $L(r)$.
• However, the string 'a' is in $L(r)$ (a followed by $\epsilon$) but not in $L(s)$ (must be followed by at least one character).
• Therefore, $L(s)$ is a proper subset of $L(r)$. This can be written as $L(s) \subset L(r)$ or $L(r) \supset L(s)$ or $L(r) \supseteq L(s)$.

Comparing L(s) and L(t)

• $L(s)$ contains strings starting with 'a' followed by any non-empty string from $\{a, b\}^+$.
  • $L(s) = \{aa, ab, aaa, aab, aba, abb, ...\}$
• $L(t)$ contains strings of the form $a^k b$ where $k \ge 1$.
  • $L(t) = \{ab, aab, aaab, ...\}$
• Let's take any string from $L(t)$, for example, $a^k b$ where $k \ge 1$. This string starts with 'a'. The rest of the string is $a^{k-1}b$. Since $k \ge 1$, $k-1 \ge 0$, so $a^{k-1}b$ is a non-empty string (it contains at least 'b'). This non-empty string is composed only of 'a's and 'b's, so it belongs to $\{a, b\}^+$.
• Thus, every string in $L(t)$ starts with 'a' and is followed by a non-empty string from $\{a, b\}^+$, which means every string in $L(t)$ is also in $L(s)$.
• However, $L(s)$ contains strings like 'aa', 'aaa', 'aba', 'abb', etc., which are not in $L(t)$ (strings in $L(t)$ must end with 'b').
• Therefore, $L(t)$ is a proper subset of $L(s)$. This can be written as $L(t) \subset L(s)$ or $L(s) \supset L(t)$ or $L(s) \supseteq L(t)$.

Concluding the Relationship

• $L(s) \subseteq L(r)$ (specifically $L(s) \subset L(r)$)
• $L(t) \subseteq L(s)$ (specifically $L(t) \subset L(s)$)

• 'a' is in $L(r)$ but not $L(s)$, confirming $L(s) \subset L(r)$.
• 'aa' is in $L(s)$ but not $L(t)$, confirming $L(t) \subset L(s)$.
• 'ab' and 'aab' are in all three, consistent with the subset relationships.

Revision Table: Regular Expression Language Comparison

Additional Information: Regular Expressions and Languages

A regular expression is a sequence of characters that define a search pattern, mainly for use in pattern matching with strings. In the context of formal language theory, regular expressions are used to describe regular languages.

• Regular Language: A language that can be recognized by a finite automaton. Regular expressions are a concise way to represent regular languages.
• Operators:
  • Concatenation ($\cdot$ or implied): Putting strings together. E.g., $ab$ means 'a' followed by 'b'.
  • Union ($+$ or $|$): Means "or". E.g., $a+b$ matches either 'a' or 'b'.
  • Kleene star ($*$): Zero or more repetitions of the preceding element. E.g., $a*$ matches $\epsilon$, 'a', 'aa', 'aaa', etc. $(a+b)*$ matches any string of a's and b's, including the empty string.
  • Kleene plus ($+$): One or more repetitions of the preceding element. E.g., $a+$ matches 'a', 'aa', 'aaa', etc. $(a+b)+$ matches any non-empty string of a's and b's. Note the difference between the '+' used for union and the '+' used for Kleene plus; context usually makes it clear. In $a(a+b)+$, the first $+$ is union, the second is Kleene plus.
• Language of a Regular Expression $R$ ($L(R)$): The set of all strings that the regular expression $R$ can generate or match.
• Subset ($\subseteq$): A set A is a subset of set B if every element in A is also in B.
• Superset ($\supseteq$): A set A is a superset of set B if every element in B is also in A. This is the opposite of the subset relation ($A \supseteq B$ is equivalent to $B \subseteq A$).

Understanding these fundamental concepts is crucial for analyzing and comparing formal languages described by regular expressions.