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--- lfa:2022:lab10-cfl_2 [2022/12/10 13:50]
alexandra.udrescu01
+++ lfa:2022:lab10-cfl_2 [2022/12/29 00:59] (current)
alexandra.udrescu01
@@ Line 6: / Line 6: @@
   - describe L(G) \\
   - algoritmically construct a PDA that accepts the same language \\
+  - run the PDA on the given inputs \\
   - is the grammar ambiguous? If yes, write a non ambiguous grammar that generates the same language \\
-**10.1.1** \\
-$ S \leftarrow aS | aSb | \epsilon $
-**10.1.2** \\
+**10.1.1** input: aaaabb \\
-$ S \leftarrow aAA  \\ A → aS | bS | a $
+$ S \leftarrow aS | aSb | \epsilon $ \\
+<hidden Solution>
+The start symbol of the PDA is S. \\ The PDA will only have one state q and it will accept via empty stack. \\
+For each nonterminal/rule $ A \leftarrow \gamma $ add a transition **q ---$(\epsilon, A/ \gamma)$--➤ q** and for each terminal c add  **q ---$(c, c/ \epsilon)$--➤ q**
+Thus, our PDA has the following transitions looping on state q:
+  * $ \epsilon, S/aS $
+  * $ \epsilon, S/aSb $
+  * $ \epsilon, S/\epsilon $
+  * $ a, a/\epsilon $
+  * $ b, b/\epsilon $
+Input: aaabb \\ (aaabb, q, S) => (**a**aabb, q, **a**Sb) => (aabb, q, Sb) => (**a**abb, q, **a**Sbb) => (abb, q, Sbb) => (**a**bb, q, **a**Sbb) => (bb, q, Sbb) => (bb, q, bb) => (b, q, b) => ($\epsilon$, q, $\epsilon$) \\
+\\
+Is the grammar ambiguuous? yes, because there exist 2 different left-derivations for word aaabb \\ S => aSb => aaSbb => aaaSbb => aaabb \\ S => aS => aaSb => aaaSbb => aaabb \\
+\\
+The accepted language is $ L(G) = \{a^{m}b^{n} | m \ge n \ge 0\} $ \\
+\\
+Repaired grammar: \\ $ S \leftarrow aS | A \\ A \leftarrow aAb | \epsilon $
+</hidden>
+**10.1.2** input: xayxcayatabcazz \\
+$ S \leftarrow a | xAz | SbS | cS  \\ A \leftarrow SyS | SyStS $
+<hidden Solution>
+The  PDA has the following transitions looping on state q:
+  * $ \epsilon, S/a $
+  * $ \epsilon, S/xAz $
+  * $ \epsilon, S/SbS $
+  * $ \epsilon, S/cS $
+  * $ \epsilon, A/SyS $
+  * $ \epsilon, A/SyStS $
+  * $ a, a/\epsilon $
+  * $ b, b/\epsilon $
+  * $ c, c/\epsilon $
+  * $ x, x/\epsilon $
+  * $ y, y/\epsilon $
+  * $ z, z/\epsilon $
+  * $ t, t/\epsilon $
+Input: xayxcayatabcazz \\ (xayxcayatabcazz, q, S) => (xayxcayatabcazz, q, xAz) => (ayxcayatabcazz, q, Az) => (ayxcayatabcazz, q, SySz) => (ayxcayatabcazz, q, aySz) => (yxcayatabcazz, q, ySz) => (xcayatabcazz, q, Sz) => (xcayatabcazz, q, xAzz) => (cayatabcazz, q, Azz) => (cayatabcazz, q, SyStSzz) => (cayatabcazz, q, cSyStSzz) => (ayatabcazz, q, SyStSzz) => (ayatabcazz, q, ayStSzz) => (yatabcazz, q, yStSzz) => (atabcazz, q, StSzz) => (atabcazz, q, StSzz) => (atabcazz, q, atSzz) => (tabcazz, q, tSzz) => (abcazz, q, Szz) => (abcazz, q, SbSzz) => (abcazz, q, abSzz) => (bcazz, q, bSzz) => (cazz, q, Szz) => (cazz, q, cSzz) => (azz, q, Szz) => (azz, q, azz) => (zz, q, zz)  => (z, q, z) => ($\epsilon$, q, $\epsilon$)\\
+\\
+Is the grammar ambiguuous? yes because of word ababa that has 2 different left-derivations \\ S => SbS => abS => abSbS => ababS => ababa \\ S => SbS => SbSbS => abSbS => ababS => ababa \\
+\\
+It is hard to directly explain the language in this form. Another form may be easier. Let's relabel the terminals: a => bool; b => and; c => not; x => if; y => then; z => fi; t => else.\\
+The grammar becomes: $ S \leftarrow bool | if  A  fi | S  and  S | not  S  \\ A \leftarrow S  then  S | S  then  S  else  S $\\
+The language generated can be described as the language of boolean expressions (considering 'bool' is either a variable or a literal) with the operations 'and', 'not', 'if-then' and 'if-then-else'.
+\\
+Why is it ambigous? The 'and'/b operator does not define its associativity, and the operators 'and'/b and 'not'/c do not have a clear precedence rule.\\
+To fix this grammar we will use the following conventions: ababa == (aba)ba and caba == (ca)ba \\
+Repaired grammar: \\
+$ S \leftarrow TbS | T  \\ T \leftarrow cT | xAz | a \\ A \leftarrow SyS | SyStS $ \\
+</hidden>
+**10.1.3** input: aaabbbbbccc \\
+$ S \leftarrow ABC \\ A \leftarrow aA | \epsilon \\ B \leftarrow bbB | b \\ C \leftarrow cC | c $
+<hidden Solution>
+The  PDA has the following transitions looping on state q:
+  * $ \epsilon, S/ABC $
+  * $ \epsilon, A/aA $
+  * $ \epsilon, A/\epsilon $
+  * $ \epsilon, B/bbB $
+  * $ \epsilon, B/b $
+  * $ \epsilon, C/cC $
+  * $ \epsilon, C/c $
+  * $ a, a/\epsilon $
+  * $ b, b/\epsilon $
+  * $ c, c/\epsilon $
+Input: aaabbbbbccc \\ (aaabbbbbccc, q, S) => (aaabbbbbccc, q, ABC) => (aaabbbbbccc, q, aABC) => (aabbbbbccc, q, ABC) =>
+=> (aabbbbbccc, q, aABC) => (abbbbbccc, q, ABC) => (aabbbbbccc, q, ABC) => (aabbbbbccc, q, aABC) => => (abbbbbccc, q, ABC) => => (abbbbbccc, q, aABC) => (bbbbbccc, q, ABC)
+=> (bbbbbccc, q, BC) => (bbbbbccc, q, bbBC) => (bbbbccc, q, bBC) => (bbbccc, q, BC) => (bbbccc, q, bbBC) => (bbccc, q, bBC) => (bccc, q, BC) => (bccc, q, bC)
+=> (ccc, q, C) => (ccc, q, cC) => (cc, q, C) => (cc, q, cC) => (c, q, C) => (c, q, c) => ($\epsilon$, q, $\epsilon$) \\
+\\
+Is the grammar ambiguuous? no \\
+\\
+The accepted language is $ L(G) = \{a^{m}b^{2n + 1}c^{p+1} | m,n,p \ge 0\} $ \\
+</hidden>
-**10.1.3** \\
-$ S \leftarrow ABC \\ A \leftarrow aA | \epsilon \\ B \leftarrow bbB | b \\ C \leftarrow cC | c $
 ===== 10.2. Lexer Spec =====
@@ Line 22: / Line 115: @@
   * NO_CONSEC_ONE: $ (1 | \epsilon)(01 | 0)* $
 Separate the following input strings into lexemes:
+  * 010101
+<hidden>
+Although the entire string is matched by PAIRS and NO_CONSEC_ONE, PAIRS is defined first, thus it will be the first picked. \\
+PAIRS "010101"
+</hidden>
   * 1010101011
+<hidden>
+First we have a maximal match on "101010101" for regex NO_CONSEC_ONE. The remaining string, "1", is matched by both ONES and NO_CONSEC_ONE, but ONES  is defined first. \\
+NO_CONSEC_ONE "101010101" \\ ONES "1"
+</hidden>
   * 01110101001
+<hidden>
+PAIRS "01" \\
+ONES "11" \\
+NO_CONSEC_ONES "0101001"
+</hidden>
   * 01010111111001010
+<hidden>
+PAIRS "010101" \\
+ONES "11111" \\
+NO_CONSEC_ONES "001010"
+</hidden>
   * 1101101001111100001010011001
+<hidden>
+ONES "11"
+PAIRS "01101001" \\
+ONES "1111" \\
+NO_CONSEC_ONES "0000101001" \\
+PAIRS "1001"
+</hidden>