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\section*{Homework 2}

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\begin{enumerate}

\item What is the difference between \emph{basic} regular expressions  

  and \emph{extended} regular expressions?

  \solution{Basic regular expressions are $\ZERO$, $\ONE$, $c$, $r_1 + r_2$,

    $r_1 \cdot r_2$, $r^*$. The extended ones are the bounded

    repetitions, not, etc.

    Maybe explain here how the extended regular expressions

    are defined in terms of the basic ones.

  }

\item What is the language recognised by the regular

  expressions $(\ZERO^*)^*$.

  \solution{$L(\ZERO^*{}^*) = \{[]\}$,

    remember * always includes the empty string}

\item Review the first handout about sets of strings and read

      the second handout. Assuming the alphabet is the set

      $\{a, b\}$, decide which of the following equations are

      true in general for arbitrary languages $A$, $B$ and

      $C$:

      \begin{eqnarray}

      (A \cup B) @ C & =^? & A @ C \cup B @ C\nonumber\\

      A^* \cup B^*   & =^? & (A \cup B)^*\nonumber\\

      A^* @ A^*      & =^? & A^*\nonumber\\

      (A \cap B)@ C  & =^? & (A@C) \cap (B@C)\nonumber

      \end{eqnarray}

      \noindent In case an equation is true, give an

      explanation; otherwise give a counter-example.

      \solution{1 + 3 are equal; 2 + 4 are not. Interesting is 4 where

        $A = \{[a]\}$, $B = \{[]\}$ and $C = \{[a], []\}$ is an

        counter-example.\medskip

        For equations like 3 it is always a good idea to prove the

        two inclusions

        \[

          A^* \subseteq  A^* @ A^*   \qquad

          A^* @ A^* \subseteq A^*

        \]  

        This means for every string $s$ we have to show

        \[

          s \in A^*    \;\textit{implies}\;  s \in A^* @ A^* \qquad

          s \in A^* @ A^* \;\textit{implies}\;  s \in A^*

        \]

        The first one is easy because $[] \in A^*$ and therefore

        $s @ [] \in A^* @ A^*$.

        The second one says that $s$ must be of the form $s = s_1 @ s_2$ with

        $s_1 \in A^*$ and $s_2 \in A^*$. We have to show that

        $s_1 @ s_2 \in A^*$.

        If $s_1 \in A^*$ then there exists an $n$ such that $s_1 \in A^n$, and

        if $s_2 \in A^*$ then there exists an $m$ such that $s_2 \in A^m$.\bigskip

        Aside: We are going to show the following power law

        \[

        A^n \,@\, A^m = A^{n+m}  

        \]  

        We prove that by induction on $n$.

        Case $n = 0$: $A^0 \,@\, A^m = A^{0+m}$ holds because $A^0 = \{[]\}$

        and $\{[]\} \,@\, A^m = A ^ m$ and $0 + m = m$.\medskip

        Case $n + 1$: The induction hypothesis is

        \[ A^n \,@\, A^m = A^{n+m}

        \]

        We need to prove

        \[

        A^{n+1} \,@\, A^m = A^{(n+1)+m}  

        \]

        The left-hand side is $(A \,@\, A^n) \,@\, A^m$ by the definition of

        the power operation. We can rearrange that

        to $A \,@\, (A^n \,@\, A^m)$.   \footnote{Because for all languages $A$, $B$, $C$ we have $(A @ B) @ C = A @ (B @ C)$.}

        By the induction hypothesis we know that $A^n \,@\, A^m = A^{n+m}$.

        So we have $A \,@\, (A^{n+m})$. But this is $A^{(n+m)+1}$ again if we

        apply the definition of the power operator. If we

        rearrange that we get $A^{(n+1)+m}$ and are done with

        what we need to prove for the power law.\bigskip

        Picking up where we left, we know that $s_1 \in A^n$ and $s_2 \in A^m$. This now implies that $s_1 @ s_2\in A^n @ A^m$. By the power law this means

        $s_1 @ s_2\in A^{n+m}$. But this also means $s_1 @ s_2\in A^*$.

      }

\item Given the regular expressions $r_1 = \ONE$ and $r_2 =

      \ZERO$ and $r_3 = a$. How many strings can the regular

      expressions $r_1^*$, $r_2^*$ and $r_3^*$ each match?

      \solution{$r_1$ and $r_2$ can match the empty string only, $r_3$ can

        match $[]$, $a$, $aa$, ....}

\item Give regular expressions for (a) decimal numbers and for

      (b) binary numbers. Hint: Observe that the empty string

      is not a number. Also observe that leading 0s are

      normally not written---for example the JSON format for numbers

      explicitly forbids this. So 007 is not a number according to JSON.

      \solution{Just numbers without leading 0s: $0 + (1..9)\cdot(0..9)^*$;

        can be extended to decimal; similar for binary numbers

      }

\item Decide whether the following two regular expressions are

      equivalent $(\ONE + a)^* \equiv^? a^*$ and $(a \cdot

      b)^* \cdot a \equiv^? a \cdot (b \cdot a)^*$.

      \solution{Both are equivalent, but why the second? Essentially you have to show that each string in one set is in the other. For 2 this means you can do an induction proof that $(ab)^na$ is the same string as $a(ba)^n$, where the former is in the first set and the latter in the second.}

\item Give an argument for why the following holds:

  if $r$ is nullable then $r^{\{n\}} \equiv r^{\{..n\}}$.

  \solution{ This requires an inductive proof. There are a number of

    ways to prove this. It is clear that if $s \in L (r^{\{n\}})$ then

    also $s \in L (r^{\{..n\}})$.

    So one way to prove this is to show

    that if $s \in L (r^{\{..n\}})$ then also $s \in L (r^{\{n\}})$

    (under the assumption that $r$ is nullable, otherwise it would not

    be true).  The assumption $s \in L (r^{\{..n\}})$ means

    that $s \in L(r^{\{i\}})$ with $i \leq n$ holds

    and we have to show that $s \in L(r^{\{n\}})$ holds. 

    One can do this by induction for languages as follows:

    \[

      \textit{if}\; [] \in A \;\textit{and}\; s \in A^n

      \;\textit{then}\; s \in A^{n+m}

    \]  

    The proof is by induction on $m$. The base case $m=0$ is trivial.

    For the $m + 1$ case we have the induction hypothesis:

    \[

      \textit{if}\; [] \in A \;\textit{and}\; s \in A^n

      \;\textit{then}\; s \in A^{n+m}  

    \]  

    and we have to show

    \[

    s \in A^{n+m+1}   

    \]  

    under the assumption $[] \in A$ and $s \in A^n$. From the

    assumptions and the IH we can infer that $s\in A^{n + m}$.

    Then using the assumption $[] \in A$ we can infer that also

    \[

      s\in A \,@\, A^{n + m}

    \]

    which is equivalent to what we need to show $s \in A^{n+m+1}$.

    Now we know $s \in L(r^{\{i\}})$ with $i \leq n$. Since $i + m = n$

    for some $m$ we can conclude that $s \in L(r^{\{n\}})$. Done.

  }

\item Given the regular expression $r = (a \cdot b + b)^*$.

      Compute what the derivative of $r$ is with respect to

      $a$, $b$ and $c$. Is $r$ nullable?

\item Define what is meant by the derivative of a regular

      expressions with respect to a character. (Hint: The

      derivative is defined recursively.)

      \solution{The recursive function for $der$.}

\item  Assume the set $Der$ is defined as

  \begin{center}

    $Der\,c\,A \dn \{ s \;|\;  c\!::\!s \in A\}$

  \end{center}

      What is the relation between $Der$ and the notion of

      derivative of regular expressions?

      \solution{Main property is $L(der\,c\,r) = Der\,c\,(L(r))$.}

\item Give a regular expression over the alphabet $\{a,b\}$

      recognising all strings that do not contain any

      substring $bb$ and end in $a$.

      \solution{$((ba)^* \cdot (a)^*)^*\,\cdot\,a + ba$ \quad Not sure whether this can be simplified.}

\item Do $(a + b)^* \cdot b^+$ and $(a^* \cdot b^+) +

  (b^*\cdot b^+)$ define the same language?

  \solution{No, the first one can match for example abababababbbbb

  while the second can only match for example aaaaaabbbbb or bbbbbbb}

\item Define the function $zeroable$ by recursion over regular

      expressions. This function should satisfy the property

  \[

  zeroable(r) \;\;\text{if and only if}\;\;L(r) = \{\}\qquad(*)

  \]

      The function $nullable$ for the not-regular expressions

      can be defined by 

  \[

  nullable(\sim r) \dn \neg(nullable(r))

  \]

      Unfortunately, a similar definition for $zeroable$ does

      not satisfy the property in $(*)$:

  \[

  zeroable(\sim r) \dn \neg(zeroable(r))

  \]

  Find a counter example?

  \solution{

    Here the idea is that nullable for NOT can be defined as

    \[nullable(\sim r) \dn \neg(nullable(r))\]

    This will satisfy the property

    $nullable(r) \;\;\text{if and only if}\;\;[] \in L(r)$. (Remember how

    $L(\sim r)$ is defined).\bigskip

    But you cannot define

    \[zeroable(\sim r) \dn \neg(zeroable(r))\]

    because if $r$ for example is $\ONE$ then $\sim \ONE$ can match

    some strings (all non-empty strings). So $zeroable$ should be false. But if we follow

    the above definition we would obtain $\neg(zeroable(\ONE))$. According

    to the definition of $zeroable$ for $\ONE$ this would be false,

    but if we now negate false, we get actually true. So the above

    definition would not satisfy the property

    \[

      zeroable(r) \;\;\text{if and only if}\;\;L(r) = \{\}

    \]  

    }

\item Give a regular expressions that can recognise all

      strings from the language $\{a^n\;|\;\exists k.\; n = 3 k

      + 1 \}$.

      \solution{$a(aaa)^*$}

\item Give a regular expression that can recognise an odd 

  number of $a$s or an even number of $b$s.

  \solution{

    If the a's and b's are meant to be separate, then this is easy 

    \[a(aa)^* + (bb)^*\]

    If the letters are mixed, then this is difficult

    \[(aa|bb|(ab|ba)\cdot (aa|bb)^* \cdot (ba|ab))^* \cdot (b|(ab|ba)(bb|aa)^* \cdot a)

    \]

    (copied from somewhere ;o)

    The idea behind this monstrous regex is essentially the DFA

\begin{center}    

\begin{tikzpicture}[scale=1,>=stealth',very thick,

                    every state/.style={minimum size=0pt,

                    draw=blue!50,very thick,fill=blue!20}]

  \node[state,initial]   (q0) at (0,2) {$q_0$};

  \node[state,accepting] (q1) at (2,2) {$q_1$};

  \node[state]           (q2) at (0,0) {$q_2$};

  \node[state]           (q3) at (2,0) {$q_3$};

 \path[->]  (q0) edge[bend left] node[above] {$a$} (q1)

            (q1) edge[bend left] node[above] {$a$} (q0)

            (q2) edge[bend left] node[above] {$a$} (q3)

            (q3) edge[bend left] node[above] {$a$} (q2)

            (q0) edge[bend left] node[right] {$b$} (q2)

            (q2) edge[bend left] node[left]  {$b$} (q0)

            (q1) edge[bend left] node[right] {$b$} (q3)

            (q3) edge[bend left] node[left]  {$b$} (q1);

\end{tikzpicture}

\end{center}

  Maybe a good idea to reconsider this example in Lecture 3

  where the Brzozowski algorithm for DFA $\rightarrow$ Regex

  can be used.

}

\item \POSTSCRIPT  

\end{enumerate}

\end{document}

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