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CS 461 – Oct. 10

CS 461 – Oct. 10. Review PL grammar as needed How to tell if a word is in a CFL? Convert to PDA and run it.  CYK algorithm Modern parsing techniques. Accepting input. How can we tell if a given source file (input stream of tokens) is a valid program? Language defined by CFG, so …

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CS 461 – Oct. 10

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  1. CS 461 – Oct. 10 • Review PL grammar as needed • How to tell if a word is in a CFL? • Convert to PDA and run it.  • CYK algorithm • Modern parsing techniques

  2. Accepting input • How can we tell if a given source file (input stream of tokens) is a valid program? Language defined by CFG, so … • Can see if there is some derivation from grammar? • Can convert CFG to PDA? • Exponential performance not acceptable. (e.g. doubling every time we add token) • Two improvements: • CYK algorithm, runs in O(n3) • Bottom-up parsing, generally linear, but restrictions on grammar.

  3. CYK algorithm • In 1965-67, discovered independently by Cocke, Younger, Kasami. • Given any CFG and any string, can tell if grammar generates string. • The grammar needs to be in CNF first. • This ensures that the rules are simple. Rules are of the form X  a or X  YZ • Consider all substrings of len 1 first. See if these are in language. Next try all len 2, len 3, …. up to length n.

  4. continued • Maintain results in an NxN table. Top right portion not used. • Example on right is for testing word of length 3. • Start at bottom; work your way up. • For length 1, just look for “unit rules” in grammar, e.g. Xa.

  5. continued • For general case i..j • Think of all possible ways this string can be broken into 2 pieces. • Ex. 1..3 = 1..2 + 3..3 or 1..1 + 2..3 • We want to know if both pieces  L. This handles rules of form A  BC. • Let’s try example from 3+7+. (in CNF)

  6. 337  3+7+ ? S  AB A  3 | AC B  7 | BD C  3 D  7 For each len 1 string, which variables generate it? 1..1 is 3. Rules A and C. 2..2 is 3. Rules A and C. 3..3 is 7. Rules B and D.

  7. 337  3+7+ ? S  AB A  3 | AC B  7 | BD C  3 D  7 Length 2: 1..2 = 1..1 + 2..2 = (A or C)(A or C) = rule A 2..3 = 2..2 + 3..3 = (A or C)(B or D) = rule S

  8. 337  3+7+ ? S  AB A  3 | AC B  7 | BD C  3 D  7 Length 3: 2 cases for 1..3: 1..2 + 3..3: (A)(B or D) = S 1..1 + 2..3: (A or C)(S) no! We only need one case to work.

  9. CYK example #2 Let’s test the word baab S  AB | BC A  BA | a B  CC | b C  AB | a Length 1: ‘a’ generated by A, C ‘b’ generated by B

  10. baab S  AB | BC A  BA | a B  CC | b C  AB | a Length 2: 1..2 = 1..1 + 2..2 = (B)(A, C) = S,A 2..3 = 2..2 + 3..3 = (A,C)(A,C) = B 3..4 = 3..3 + 3..4 = (A,C)(B) = S,C

  11. baab S  AB | BC A  BA | a B  CC | b C  AB | a Length 3: [ each has 2 chances! ] 1..3 = 1..2 + 3..3 = (S,A)(A,C) = Ø 1..3 = 1..1 + 2..3 = (B)(B) = Ø 2..4 = 2..3 + 4..4 = (B)(B) = Ø 2..4 = 2..2 + 3..4 = (A,C)(S,C) = B

  12. Finally… S  AB | BC A  BA | a B  CC | b C  AB | a Length 4 [has 3 chances!] 1..4 = 1..3 + 4..4 = (Ø)(B) = Ø 1..4 = 1..2 + 3..4 = (S,A)(S,C) = Ø 1..4 = 1..1 + 2..4 = (B)(B) = Ø Ø means we lose! baab  L. However, in general don’t give up if you encounter Ø in the middle of the process.

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