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CYK Algorithm

CYK Algorithm. Introduction. Problem: Given a context free grammar and a string s is it possible to decide whether s can be generated by the grammar or not? If the grammar is not in a very special form this is not so efficient.

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CYK Algorithm

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  1. CYK Algorithm

  2. Introduction • Problem: Given a context free grammar and a string s is it possible to decide whether s can be generated by the grammar or not? • If the grammar is not in a very special form this is not so efficient. • If the grammar is in Chomsky Normal Form, we have an elegant algorithm for testing this, the CYK algorithm.

  3. The CYK algorithm • Suppose that we are given a grammar in Chomsky Normal form S → AB A → BB | 0 B → AA |1 • We would like to see if 10110 is generated by this grammar or not.

  4. Substrings of length 1 • Since the only way to produce terminals is by following the rules A → a, just replace every terminal with the variables that produce it. 1 0 1 1 0 B A B B A

  5. Substrings of length 2 Suppose now that we want to see how every substring of length 2 can be generated. This is equivalent with finding ways to produce all the length 2 substrings where terminals are replaced with the variables that represent them. But since every rule is of the form A → BC, it suffices to replace every two consecutive variables with the variables that produce them. 1 0 1 1 0 B A B B A - S A -

  6. Substrings of length 3 • To produce the substring 101 (in 10110) we can either take 1 with 01 or 10 with 1. Here BS cannot be produced by any variable. 10 1 1 0 B A B B A - S A - -

  7. Substrings of length 3 • To produce the substring 101 (in 10110) we can either take 1 with 01 or 10 with 1. Here we don’t have a pair since 10 cannot be produced. 1 01 1 0 B A BB A - S A - -

  8. Substrings of length 3 • To produce the substring 011 (in 10110) we can either take 0 with 11 or 01 with 1. Here AA can be produced by B. 101 1 0 B A B B A - S A - - B

  9. Substrings of length 3 • To produce the substring 011 (in 10110) we can either take 0 with 11 or 01 with 1. Here SB cannot be produced by any variable 1 0 11 0 B A B B A - S A - - B

  10. Substrings of length 3 • To produce the substring 110 (in 10110) we can either take 1 with 10 or 11 with 0. Here we don’t have a pair since 10 cannot be produced by a variable. 1011 0 B A BB A - S A - - B -

  11. Substrings of length 3 • To produce the substring 110 (in 10110) we can either take 1 with 10 or 11 with 0. Here AA can be produced by B 10 110 B A B BA - S A - - B B

  12. Substrings of length 4 • To produce the substring 1011 (in 10110) we can take 1 with 011 or 10 with 11, or 101 with 1. Here BB can be produced by A. 10 1 1 0 B A B B A - S A - - BB A

  13. Substrings of length 4 • To produce the substring 1011 (in 10110) we can take 1 with 011 or 10 with 11, or 101 with 1. Here we don’t have a pair since 10 cannot be produced. 1 011 0 B A B B A - S A - - B B A

  14. Substrings of length 4 • To produce the substring 1011 (in 10110) we can take 1 with 011 or 10 with 11, or 101 with 1. Here we don’t have a pair since 101 cannot be produced. 1 0 11 0 B A B B A - S A - - B B A

  15. Substrings of length 4 • To produce the substring 0110 (in 10110) we can take 0 with 110 or 01 with 10, or 011 with 0. Here AB can be produced by S. 101 1 0 B A B B A - S A - - B B A S

  16. Substrings of length 4 • To produce the substring 0110 (in 10110) we can take 0 with 110 or 01 with 10, or 011 with 0. Here we don’t have a pair since 10 cannot be produced. 1 0110 B A B B A - S A - - B B A S

  17. Substrings of length 4 • To produce the substring 0110 (in 10110) we can take 0 with 110 or 01 with 10, or 011 with 0. Here BA cannot be produced by any variable. 1 0 1 10 B A B BA - S A - - BB A S

  18. Combine previous solutions • In order now to produce the whole string 10110 we can take 1 with 0110 or 10 with 110 or 101 with 10, or 1011 with 0. Here, BS cannot be produced. 1 0 1 1 0 B A B B A - S A - - B B A S -

  19. Combine previous solutions • In order now to produce the whole string 10110 we can take 1 with 0110 or 10 with 110 or 101 with 10, or 1011 with 0. Here we don’t have a pair. 1 0 1 1 0 B A B B A - S A - - B B A S -

  20. Combine previous solutions • In order now to produce the whole string 10110 we can take 1 with 0110 or 10 with 110 or 101 with 10, or 1011 with 0. Here we don’t have a pair. 1 01 1 0 B A B B A - S A - - B B A S -

  21. Combine previous solutions • In order now to produce the whole string 10110 we can take 1 with 0110 or 10 with 110 or 101 with 10, or 1011 with 0. Here, AA is produced by B. 1 0 1 10 B A B BA - S A - - B B A S B

  22. Answer • If the last line contains the start variable S, then there is a way to produce the string else the string cannot be generated. For our example 10110 cannot be generated.

  23. Mechanical way • Now that we show why this method works lets give an easy way to compute the table • Suppose that we are about to fill in the position with the cycle. We take the pairs that the arrows designate 10 1 1 0 B A B B A - S A - - B B A S

  24. Mechanical way • Now that we show why this method works lets give an easy way to compute the table • Suppose that we are about to fill in the position with the cycle. We take the pairs that the arrows designate 10 1 1 0 B A B B A - S A - - B B A S

  25. Mechanical way • Now that we show why this method works lets give an easy way to compute the table • Suppose that we are about to fill in the position with the cycle. We take the pairs that the arrows designate 10 1 1 0 B A B B A - S A - - B B A -

  26. Mechanical way • Now that we show why this method works lets give an easy way to compute the table • Suppose that we are about to fill in the position with the cycle. We take the pairs that the arrows designate 10 1 1 0 B A B BA - S A - - BB A -

  27. Mechanical way • So finally: 10 1 1 0 B A B B A - S A - - B B A S

  28. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1

  29. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB

  30. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB -

  31. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B ABBB - S

  32. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A BBB - S A

  33. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A A

  34. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - -

  35. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A BBB - S A - -

  36. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B

  37. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B

  38. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A BBB - S A - - B -

  39. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S

  40. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A

  41. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 BA B BB - S A - - B S A

  42. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A

  43. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A -

  44. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 BA B BB - S A - - B S A -

  45. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A A

  46. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A A -

  47. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A A -

  48. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A A -

  49. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S AA S

  50. A string that is produced • Run the CYK algorithm for the string 10111 10 1 1 1 B A B BB - S A - - B S A A S The derivation is: S

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