|         |      1 \documentclass{article} | 
|         |      2 \usepackage{../style} | 
|         |      3 \usepackage{../graphics} | 
|         |      4  | 
|         |      5 \begin{document} | 
|         |      6  | 
|         |      7 \section*{Complement Sets} | 
|         |      8  | 
|         |      9 Consider the following picture: | 
|         |     10  | 
|         |     11 \begin{center} | 
|         |     12 \begin{tikzpicture}[fill=gray] | 
|         |     13 % left hand | 
|         |     14 \scope | 
|         |     15 \fill (0,0) circle (1); | 
|         |     16 \endscope | 
|         |     17 % outline | 
|         |     18 \draw (0,0) circle (1) (0,1)  node [text=white,below] {$P(s)$} | 
|         |     19       (2,0) node [text=black,above] {$\neg P(s)$} | 
|         |     20       (-2,-2) rectangle (3,2) node [text=black,above] {$\Sigma^*$}; | 
|         |     21 \end{tikzpicture} | 
|         |     22 \end{center} | 
|         |     23  | 
|         |     24  | 
|         |     25 \noindent | 
|         |     26 where $\Sigma^*$ is in our case the set of all strings (what follows | 
|         |     27 also holds for any kind of ``domain'', like the set of all integers or | 
|         |     28 set of all binary trees, etc). Let us assume $P(s)$ is a property that | 
|         |     29 is about strings, for example $P(s)$ could be ``the string $s$ has | 
|         |     30 an even length'', or ``the string $s$ starts with the letter | 
|         |     31 \texttt{a}''. Every such property carves out a subset of strings from | 
|         |     32 $\Sigma^*$, which in the picture above is depicted as a grey | 
|         |     33 circle. This subset of strings is often written as a comprehension like | 
|         |     34  | 
|         |     35 \begin{equation}\label{set} | 
|         |     36 \{s \in \Sigma^*\;|\; P(s) \} | 
|         |     37 \end{equation} | 
|         |     38  | 
|         |     39 \noindent | 
|         |     40 meaning all the $s$ (out of $\Sigma^*$) for which the property $P(s)$ | 
|         |     41 is true. If $P(s)$ would not be true then the corresponding string $s$ | 
|         |     42 would be outside the grey area where $\neg P(s)$ holds. Notice that | 
|         |     43 sometimes the property $P(s)$ holds for every string in | 
|         |     44 $\Sigma^*$. Then the grey area would fill out the whole rectangle and | 
|         |     45 the set where $\neg P(s)$ holds is empty. Similarly, the property $P(s)$ | 
|         |     46 holds for no string in which case the grey circle is empty. | 
|         |     47  | 
|         |     48 Now, we are looking for the complement of the set defined in \eqref{set}. | 
|         |     49 This complement set is often written as | 
|         |     50  | 
|         |     51 \[ | 
|         |     52 \overline{\{s \in \Sigma^*\;|\; P(s) \}} | 
|         |     53 \] | 
|         |     54  | 
|         |     55 \noindent | 
|         |     56 It is the area of $\Sigma^*$ which isn't grey, that is $\Sigma^*$ | 
|         |     57 minus $\{s \in \Sigma^*\;|\; P(s) \}$, \textbf{or} written differently | 
|         |     58 it is the set $\{s \in \Sigma^*\;|\; \neg P(s) \}$. That means it the | 
|         |     59 set of all the strings where $\neg P(s)$ holds. Consequently we have | 
|         |     60 for any complement set the equation: | 
|         |     61  | 
|         |     62 \begin{equation}\label{eq} | 
|         |     63 \overline{\{s \in \Sigma^*\;|\; P(s) \}} = \{s \in \Sigma^*\;|\; \neg P(s) \} | 
|         |     64 \end{equation} | 
|         |     65  | 
|         |     66 \section*{Semantic derivative} | 
|         |     67  | 
|         |     68 Our semantic derivative $Der\;c\;A$ is nothing else than a property that defines | 
|         |     69 a subset of strings (inside $\Sigma^*$). The corresponding | 
|         |     70 property $P(s)$ is $c::s \in A$ because we defined $Der\;c\;A$ as | 
|         |     71  | 
|         |     72 \[ | 
|         |     73 Der\;c\;A \;\dn\; \{s \in \Sigma^*\;|\; c::s \in A\} | 
|         |     74 \] | 
|         |     75  | 
|         |     76 \noindent | 
|         |     77 That means $Der\;c\;A$ is some grey area inside $\Sigma^*$. Obviously | 
|         |     78 which subset, or grey area, we are carving out from $\Sigma^*$ depends on what we | 
|         |     79 choose for $c$ and $A$.\bigskip | 
|         |     80  | 
|         |     81  | 
|         |     82 \noindent | 
|         |     83 Let us see how this pans out in a concrete example. For this let | 
|         |     84 $\Sigma^*$ not be the set of all strings, but only the set of strings | 
|         |     85 upto a length of 3 over the alphabet $\{a, b\}$. That means $\Sigma^*$ | 
|         |     86 (or the rectangle in the picture above) consists of the strings | 
|         |     87  | 
|         |     88 \[ | 
|         |     89   \Sigma^* = \left\{\begin{array}{l} | 
|         |     90     $\ensuremath{[]}$\\ | 
|         |     91     $\ensuremath{[a], [b]}$\\ | 
|         |     92     $\ensuremath{[aa], [ab], [ba], [bb]}$\\ | 
|         |     93     $\ensuremath{[aaa], [aab], [aba], [abb], [baa], [bab], [bba], [bbb]}$ | 
|         |     94     \end{array}\right\} | 
|         |     95 \]  | 
|         |     96  | 
|         |     97  | 
|         |     98 \noindent | 
|         |     99 If we set $A$ to $\{[aaa], [abb], [aa], [bb], []\}$, then $Der\;a\;A$ is the subset | 
|         |    100  | 
|         |    101 \[ | 
|         |    102 Der\;a\;A = \{[aa], [bb], [a]\} | 
|         |    103 \]   | 
|         |    104  | 
|         |    105 \noindent | 
|         |    106 which is given by the definition of | 
|         |    107 $Der\;a\;A \dn \{s \in \Sigma^*\;|\;a::s\in A\}$. Now lets look at what | 
|         |    108 the complement of this set looks like: | 
|         |    109  | 
|         |    110 \begin{equation}\label{Der} | 
|         |    111   \overline{Der\;a\;A} = \left\{\begin{array}{l} | 
|         |    112     $\ensuremath{[]}$\\ | 
|         |    113     $\ensuremath{[b]}$\\ | 
|         |    114     $\ensuremath{[ab], [ba]}$\\ | 
|         |    115     $\ensuremath{[aaa], [aab], [aba], [abb], [baa], [bab], [bba], [bbb]}$ | 
|         |    116     \end{array}\right\} | 
|         |    117 \end{equation} | 
|         |    118  | 
|         |    119 \noindent | 
|         |    120 This can be calculated by ``subtracting'' $\{[aa], [bb], [a]\}$ from | 
|         |    121 $\Sigma^*$. I let you check whether I did this correctly.  According | 
|         |    122 to the equation in \eqref{eq} this should be equal to | 
|         |    123  | 
|         |    124 \[ | 
|         |    125 \overline{Der\;a\;A} = \{s \in \Sigma^*\;|\;a::s\not\in A\} | 
|         |    126 \]   | 
|         |    127  | 
|         |    128 \noindent Let us test in turn every string in $\Sigma^*$ and see | 
|         |    129 whether $a::s$ is in $A$ which we set above to | 
|         |    130  | 
|         |    131 \[\{[aaa], [abb], [aa], [bb], []\}\] | 
|         |    132  | 
|         |    133 \noindent | 
|         |    134 This gives rise to the following table where in the first column are | 
|         |    135 the strings of $\Sigma^*$ and in the second whether $a::s \in A$ holds. The third column | 
|         |    136 is the negated version of the second. | 
|         |    137  | 
|         |    138 \begin{center} | 
|         |    139 \begin{tabular}{r|c|l} | 
|         |    140   s & is $a::s \in A$? & $\neg(a::s \in A) \Leftrightarrow  a::s \not\in A$\\ | 
|         |    141   \hline | 
|         |    142   $[]$    & no & yes\\ | 
|         |    143   $[a]$   & yes& no\\ | 
|         |    144   $[b]$   & no & yes\\ | 
|         |    145   $[aa]$  & yes& no\\ | 
|         |    146   $[ab]$  & no & yes\\ | 
|         |    147   $[ba]$  & no & yes\\ | 
|         |    148   $[bb]$  & yes& no\\ | 
|         |    149   $[aaa]$ & no & yes\\ | 
|         |    150   $[aab]$ & no & yes\\ | 
|         |    151   $[aba]$ & no & yes\\ | 
|         |    152   $[abb]$ & no & yes\\ | 
|         |    153   $[baa]$ & no & yes\\ | 
|         |    154   $[bab]$ & no & yes\\ | 
|         |    155   $[bba]$ & no & yes\\ | 
|         |    156   $[bbb]$ & no & yes\\ | 
|         |    157 \end{tabular}     | 
|         |    158 \end{center} | 
|         |    159  | 
|         |    160 \noindent | 
|         |    161 Collecting all the yes in the third row gives you the set in \eqref{Der}. So it works out in this example. | 
|         |    162  | 
|         |    163 \end{document} | 
|         |    164  | 
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