CSCI Compiler Construction

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1 CSCI Compiler Construction Lecture 12 Cocke-Younger-Kasami (CYK) Algorithm Instructor: Hossein Hojjat February 20, 2017

2 Recap: Chomsky Normal Form (CNF) A CFG is in Chomsky Normal Form if each rule is of the form A BC A a where a is any terminal A,B,C are non-terminals B, C cannot be start variable We allow the rule S ɛ if ɛ L 1

3 Recap: Conversion to CNF 1. remove unproductive non-terminals (optional) 2. remove unreachable non-terminals (optional) 3. make terminals occur alone on right-hand side 4. reduce arity of every production to 2 5. remove ɛ-production rules X ɛ 6. remove unit productions X Y 7. unproductive non-terminals 8. unreachable non-terminals 2

4 Approaches to Parsing Top Down (Goal driven) S Start from the start non-terminal S Grow parse tree downwards to match the input word Expr Plus Term Bottom Up (Data Driven) Start from the input word Term num Build up parse tree whose root is the start non-terminal S num 3

5 CYK (Cocke-Younger-Kasami) One of the earliest recognition and parsing algorithms Bottom-up parsing (starts with words) Independently developed by J. Cocke, D. Younger, T. Kasami (CYK) Build solutions compositionally from sub-solutions Based on a dynamic programming approach 4

6 CYK algorithm: Basic idea For a grammar G and a word w: For every substring v 1 of length 1, find all non-terminals A such that A v 1 For every substring v 2 of length 2, find all non-terminals A such that A v 2. For the unique substring w of length w, find all non-terminals A such that A w Check whether S belongs to the last set 5

7 CYK algorithm Standard CYK operates only on context-free grammars in Chomsky Normal Form (CNF) To illustrate the benefits of CNF we use a non-cnf grammar 6

8 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ Sign e + 4 7

9 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ Sign Sign e + 4 We start with substrings of length 1 7

10 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ Sign e + Sign 4 We start with substrings of length 1 Grammar has unit-rules: after finding a non-terminal, we need to repetitively search for other non-terminals that derive the same word Chain of rules: 7

11 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ , Fraction Sign e + Sign 4 7

12 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ , Fraction Scale Sign e + Sign 4 7

13 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ , Fraction, Real Scale Sign e + Sign 4 7

14 Example word: 12.3e+4 Real Real Fraction Scale Fraction. Scale e Sign ɛ ,, Real Fraction, Real Scale Sign e + Sign 4 7

15 Example word: 12.3 Real Real Fraction Scale Fraction. Scale e Sign ɛ , Fraction Sign

16 Example word: 12.3 Real Real Fraction Scale Fraction. Scale e Sign ɛ , Fraction Scale, Real, Real ɛ Sign + 1 ɛ 2 ɛ. ɛ 3 ɛ We neglected ɛ-production rules until now ɛ-rules can be used between any two symbols, at the beginning and at the end 8

17 Substring Recognition s 1,4 s 2,4 word: w = a 1 a 2 a n s 1,3 s 3,4 s 2,3 s s 1,1 1,2 s 2,2 s 3,3 s 4,4 a 1 a 2 a 3 a 4 X i,j set of all non-terminals deriving the substring s i,j Start non-terminal S derives w (= s 1,n ) if and only if S is a member of X 1,n 9

18 Substring Recognition How to check if the RHS of rule A A 1 A 2 A n derives s i,j? A 1 A 2 A n (i < k < j) a i a k 1 a k a j a k+1 We need to check all the possible matching of substrings with A i non-terminals Results of checking subproblem solutions are reusable Subproblem results are computed once and then memoized for later Matching in CNF: A 1 A 2 a i a k 1 a k a j 10

19 Recognition Table X 1,5 X 1,4 X 1,3 X 2,5 X 2,4 X 3,5 X4,5 X 1,2 X 2,3 X 3,4 X 1,1 X 2,2 X 3,3 X 4,4 X 5,5 a 1 a 2 a 3 a 4 a 5 Each row corresponds to one length of substrings Bottom Row: words of length 1 Second from Bottom Row: words of length 2. Top Row: word w 11

20 Example S AB CB grammar: A BA a B BC b intput: babab C AC a B A, C B A, C B b a b a b 12

21 Example S AB CB grammar: A BA a B BC b intput: babab C AC a A, B S A, B S B A, C B A, C B b a b a b 12

22 Example S AB CB grammar: A BA a B BC b intput: babab C AC a S S S A, B S A, B S B A, C B A, C B b a b a b 12

23 Example S AB CB grammar: A BA a B BC b intput: babab C AC a S, A S S S A, B S A, B S B A, C B A, C B b a b a b 12

24 Example S AB CB grammar: A BA a B BC b intput: babab C AC a S S, A S S S A, B S A, B S B A, C B A, C B b a b a b 12

25 Exercise What is the time and space complexity of CYK? Assume size of input word is n, number of non-terminals is r 13

26 CYK Problem Parsers usually parse the original grammar without further modification Grammar reflects the actual structure of language Conversion to CNF can make it difficult to keep the intended structure 14

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