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lfa:2022:lab10-cfl_2 [2022/12/12 00:27] mihai.udubasa change exercise 2 to another, more explainable one |
lfa:2022:lab10-cfl_2 [2022/12/29 00:59] (current) alexandra.udrescu01 |
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| - run the PDA on the given inputs \\ | - run the PDA on the given inputs \\ | ||
| - is the grammar ambiguous? If yes, write a non ambiguous grammar that generates the same language \\ | - is the grammar ambiguous? If yes, write a non ambiguous grammar that generates the same language \\ | ||
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| **10.1.1** input: aaaabb \\ | **10.1.1** input: aaaabb \\ | ||
| $ S \leftarrow aS | aSb | \epsilon $ \\ | $ S \leftarrow aS | aSb | \epsilon $ \\ | ||
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| <hidden Solution> | <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. \\ | The start symbol of the PDA is S. \\ The PDA will only have one state q and it will accept via empty stack. \\ | ||
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| Repaired grammar: \\ $ S \leftarrow aS | A \\ A \leftarrow aAb | \epsilon $ | Repaired grammar: \\ $ S \leftarrow aS | A \\ A \leftarrow aAb | \epsilon $ | ||
| </hidden> | </hidden> | ||
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| **10.1.2** input: xayxcayatabcazz \\ | **10.1.2** input: xayxcayatabcazz \\ | ||
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| $ S \leftarrow a | xAz | SbS | cS \\ A \leftarrow SyS | SyStS $ | $ S \leftarrow a | xAz | SbS | cS \\ A \leftarrow SyS | SyStS $ | ||
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| <hidden Solution> | <hidden Solution> | ||
| The PDA has the following transitions looping on state q: | The PDA has the following transitions looping on state q: | ||
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| $ S \leftarrow TbS | T \\ T \leftarrow cT | xAz | a \\ A \leftarrow SyS | SyStS $ \\ | $ S \leftarrow TbS | T \\ T \leftarrow cT | xAz | a \\ A \leftarrow SyS | SyStS $ \\ | ||
| </hidden> | </hidden> | ||
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| **10.1.3** input: aaabbbbbccc \\ | **10.1.3** input: aaabbbbbccc \\ | ||
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| $ S \leftarrow ABC \\ A \leftarrow aA | \epsilon \\ B \leftarrow bbB | b \\ C \leftarrow cC | c $ | $ S \leftarrow ABC \\ A \leftarrow aA | \epsilon \\ B \leftarrow bbB | b \\ C \leftarrow cC | c $ | ||
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| <hidden Solution> | <hidden Solution> | ||
| The PDA has the following transitions looping on state q: | The PDA has the following transitions looping on state q: | ||
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| The accepted language is $ L(G) = \{a^{m}b^{2n + 1}c^{p+1} | m,n,p \ge 0\} $ \\ | The accepted language is $ L(G) = \{a^{m}b^{2n + 1}c^{p+1} | m,n,p \ge 0\} $ \\ | ||
| </hidden> | </hidden> | ||
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| ===== 10.2. Lexer Spec ===== | ===== 10.2. Lexer Spec ===== | ||
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| Separate the following input strings into lexemes: | Separate the following input strings into lexemes: | ||
| * 010101 | * 010101 | ||
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| <hidden> | <hidden> | ||
| Although the entire string is matched by PAIRS and NO_CONSEC_ONE, PAIRS is defined first, thus it will be the first picked. \\ | Although the entire string is matched by PAIRS and NO_CONSEC_ONE, PAIRS is defined first, thus it will be the first picked. \\ | ||
| - | NO_CONSEC_ONE "010101" | + | PAIRS "010101" |
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| </hidden> | </hidden> | ||
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| * 1010101011 | * 1010101011 | ||
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| <hidden> | <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. \\ | 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" | NO_CONSEC_ONE "101010101" \\ ONES "1" | ||
| </hidden> | </hidden> | ||
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| * 01110101001 | * 01110101001 | ||
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| + | <hidden> | ||
| + | PAIRS "01" \\ | ||
| + | ONES "11" \\ | ||
| + | NO_CONSEC_ONES "0101001" | ||
| + | </hidden> | ||
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| * 01010111111001010 | * 01010111111001010 | ||
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| + | <hidden> | ||
| + | PAIRS "010101" \\ | ||
| + | ONES "11111" \\ | ||
| + | NO_CONSEC_ONES "001010" | ||
| + | </hidden> | ||
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| * 1101101001111100001010011001 | * 1101101001111100001010011001 | ||
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| + | <hidden> | ||
| + | ONES "11" | ||
| + | PAIRS "01101001" \\ | ||
| + | ONES "1111" \\ | ||
| + | NO_CONSEC_ONES "0000101001" \\ | ||
| + | PAIRS "1001" | ||
| + | </hidden> | ||