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CYK Parser Von Carla und Cornelia Kempa

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Overview Top-downBottom-up Non-directional methods Unger ParserCYK Parser

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Cocke Younger Kasami -method

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Recognition phase

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Example grammar Number(s) Integer | Real Integer Digit | Integer Digit Real Integer Fraction Scale Fraction . Integer Scale e Sign Integer | Empty Digit 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Empty ɛ Sign + | -

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Example Sentence: 32.5e+1 1. concentrate on the substrings of the input sentence

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Building the recognition table

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32.5e +1 is in the language What problems can we already see in this example?

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Another complication: Ɛ - rules Input : 43.1

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The ɛ - Problem Shortest substrings of any input sentence : ɛ -substrings We must compute R ɛ the set of non-terminals that derive ɛ R ɛ = { Empty, Scale }

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Non- empty substrings of the input sentence Input : z = z 1 z 2 z 3 z 4 ….z n Compute the set of Non-Terminals that derive the substring of z starting at position i, of length l.

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Terminology (also on the handout) i index we are starting at l length of this substring R s i,l set of Non-Terminals deriving the substring s i, l S i, 0 = ɛ Set of Non- Terminals that derive ɛ : R s i,0 = R ɛ

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S i, l = z i z i+1 …… z i+ l-1

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The set of Non- Terminals deriving the substring s i, l : R s i, l 1.) substrings of length 0 S i, 0 = ɛ and R s i, l = R ɛ 2.) short substrings 3.) longer substrings (say l = j ) All the information on substrings with l < j is available

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Check each RH-side (Right-Hand -side) in the grammar to see if it derives s i, l L A1 ….Am S i, l ( divided into m segments (= possibly empty)) A1 first segment of s i, l A2 second segment of s i, l …. ….

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A 1 ….Am s i,l So A1 first part of s i,l (let´s say A1 has to derive a first part of s i, l of length k) A1 s i, k A1 is in the set R s i,k

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A 1 ….Am s i,l Assuming this A2…Am has to derive the rest: A2 … Am Si+k, l-k This is attempted for every k

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Problems with this Approach 1) Consider A2…Am m could be 1 and A1 a Non-terminal We are Dealing with a unit- rule A1 must derive the whole substring s i, l and thus be a member of R s i, l But that´s the set we are computing right now …

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Solution to this problem A1 s i, l Somewhere along the derivation there must be a first step not using a unit rule A1 B … C * s i, l C is the first Non-Terminal using a non-unit-rule in the derivation

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Solution cont. At some stage C is added to Rs i, l If we repeat the process again and again At some point B will be added and in the next step A1 will be added We have to repeat the process again and again until no new Non-Terminals are added to R s i,l

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Problem 2 Ɛ -rules Consider all but one of the A t derive Ɛ B A1 A2 A3 A4 A5 …. At B and A1 - t are Non-Terminals A2 – At derive Ɛ So what stays is : B A1 A unit-rule

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We have computed all the Rs i,l If S is a member of Rs 1, n the start symbol derives z (=s 1, n) (the input string)

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CYK recognition with a grammar in ****- form: What are the Restrictions we want to have on our grammar ?

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Useful Restrictions No ɛ - rules No unit-rules Limit the length of the right- hand side of each rule, say to two What we get out of this: A a A BC Where a is a terminal and ABC are Non- Terminals

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Chomsky-Normal-Form… (… not only to annoy students ) Perfect grammar for CYK

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How CYK works for a grammar in CNF R ɛ is empty R s i, 1 can be read directly from the rules (A a) A rule A BC can never derive a single terminal

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Procedure Iteratively (as before) : 1) Fill the sets R s, 1 directly 2) Process all substrings of length 1 3) Process all substrings of length 2 4) Process all substrings of length l For the first step we use the rules of the form A a For all the following steps we have to use the rules of the form: A BC

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CYK and CNF Question the CYK-Parser has to answear is: Does such a k exist?

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Answearing this question is easy: Just try all possibilities no problem since you are a computer ;-) Range : from 1 to (l-1) All the sets R s i,k and R s i+k, l-1 have already been computed at this point

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Transform our sample CF-grammar into Chomsky Normal Form Overview 1) eliminate ɛ -rules 2) eliminate unit-rules 3) remove non-productive non-terminals 4) remove non –reachable non-terminals 5) modify the rest until all grammar rules are of the form A a, A BC

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Our number grammar in CNF Number(s) 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Number(s) Integer Digit Number(s) N1 Scale´ | Integer Fraction N1 Integer Fraction Integer 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Integer Integer Digit Fraction T1 Integer T1 . Scale ´ N2 Integer N2 T2 Sign T2 e Digit 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 Sign + | -

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Building the recognition table Input : Our example grammar in CNF input sentence: 32.5 e + 1

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Building the recognition table 1) bottom-row : read directly from the grammar (rules of the form A a ) 2) Check each RHS in the grammar

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Check each RHS of the grammar Two Ways: Example: 2.5 e ( = s 2, 4) 1) check each RHS e.g N1 Scale´ 2) compute possible RH-Sides from the recognition table

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How this is done 1) N1 not in R s 2, 1 or R s 2, 2 N1 is a member of R s 2, 3 But Scale´ is not a member of R s 5, 1 2) R s 2, 4 is the set of Non- Terminals that have a RHS AB where either: A in R s 2, 1 and B in R s 3, 3 A in R s 2, 2 and B in R s 4, 2 A in R s 2, 3 and B in R s 5, 1 Possible combinations: N1 T2 or Number T2 In our grammar we do not have such a RHS, so nothing is added to R s 2, 4.

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Recognition table

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Recognition table (well-formed substring table)

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Computing R s i, l: follow the arrows V and W simultaneously A BC, B a member of a set on the V arrow, C a member of a set on the W arrow

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Comparison This process is much less complicated than the one we saw before Why?

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Conclusion »This process is much less complicated Reasons: 1) We do not have to repeat the process again and again until no new Non- Terminals are added to R s i,l (The substrings we are dealing with are really substrings and cannot be equal to the string we start with)

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Reasons cont. 2) We only have to find one place where the substring must be split into two A B C Here !

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Result of the algorithm we have seen so far: Complete collection of sets R s i, l These sets can be organized in a triangular table:

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Cost of CYK - algorithm Operations dependent on n, the number of input symbols: (n * ( n+1) ) / 2 substrings to be examined For each substring : n-1 different k- positions as the worst case

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Cost of CYK – algorithm cont. All other operations are independent of n The algorithm works in a time at most proportional to n ³ That´s far more efficient than exhaustive search (time exponential in the length of the input sentence)

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