CYK Parser

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CYK Parser

• Von Carla und Cornelia Kempa

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Overview
Top-down Bottom-up
Non-directional methods Unger Parser CYK 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 ? ?
• Sign ? -

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Example Sentence 32.5e1

• 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 ?- rules

• Input 43.1

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The ?- Problem

• Shortest substrings of any input sentence
• ?-substrings

We must compute R? the set of non-terminals that
derive ?
R? Empty, Scale
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Non- empty substrings of the input sentence

• Input z z1 z2 z3 z4 .zn
• 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 ?
• Set of Non- Terminals that derive ?
• R s i,0 R ?

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S i, l z i z i1 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 ? and R s i, l R ?
• 2.) short substrings
• 3.) longer substrings (say l j )
• All the information on substrings with
• l lt 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
• (lets 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 A2Am has to derive the rest
• A2 Am ? Sik, l-k

This is attempted for every k
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Problems with this Approach

• 1) Consider A2Am
• 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 thats 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

• ?-rules
• Consider all but one of the At derive ?
• B ? A1 A2 A3 A4 A5 . At
• B and A1 - t are Non-Terminals
• A2 At derive ?
• 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 ?- 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? 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 ik , 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 ?-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, lfollow 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 ( n1) ) / 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
• ? Thats far more efficient than exhaustive
  search (time exponential in the length of the
  input sentence)