equal
deleted
inserted
replaced
1549 we say that the regular expression $r$ contains |
1549 we say that the regular expression $r$ contains |
1550 the bitcode $\textit{bs}$, written |
1550 the bitcode $\textit{bs}$, written |
1551 $r \gg \textit{bs}$. |
1551 $r \gg \textit{bs}$. |
1552 The $\gg$ operator with the |
1552 The $\gg$ operator with the |
1553 regular expression $r$ may also be seen as a |
1553 regular expression $r$ may also be seen as a |
1554 machine that does a derivative of regular expressions |
1554 |
1555 on all strings simultaneously, taking |
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1556 the bits by going throught the regular expression tree |
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1557 structure in a depth first manner, regardless of whether |
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1558 the part being traversed is nullable or not. |
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1559 It put all possible bits that can be produced on such a traversal |
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1560 into a set. |
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1561 For example, if we are given the regular expression |
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1562 $((a+b)(c+d))^*$, the tree structure may be written as\\ |
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1563 \\ |
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1564 \\ |
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1565 \\ |
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1566 \\ |
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1567 \\ |
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1568 \\ |
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1569 \begin{center} |
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1570 \begin{tikzpicture} |
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1571 \tikz[tree layout]\graph[nodes={draw, circle}] { |
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1572 * -> |
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1573 {@-> { |
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1574 {+1 -> |
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1575 {a , b} |
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1576 }, |
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1577 {+ -> |
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1578 {c , d } |
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1579 } |
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1580 } |
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1581 } |
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1582 }; |
|
1583 \end{tikzpicture} |
1555 \end{tikzpicture} |
1584 \end{center} |
1556 \end{center} |
1585 \subsection{the $\textit{ders}_2$ Function} |
1557 \subsection{the $\textit{ders}_2$ Function} |
1586 If we want to prove the result |
1558 If we want to prove the result |
1587 \begin{center} |
1559 \begin{center} |