Top-Down Parsing/Exercise 3: Difference between revisions

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(New page: = Problem = Consider the following grammar, where '''<tt>F</tt>''' is the initial symbol and '''<tt>{a,b,c,d,e}</tt>''' is the set of terminal symbols: O -> a G -> F c | O c d | (eps) ...)
 
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= Problem =
#REDIRECT [[ist:Top-Down Parsing/Exercise 3]]
 
Consider the following grammar, where '''<tt>F</tt>''' is the initial symbol and '''<tt>{a,b,c,d,e}</tt>''' is the set of terminal symbols:
 
O -> a
G -> F c | O c d | (eps)
F -> G b | O c F e
 
# Examine the grammar and rewrite it so that an LL(1) predictive parser can be built for the corresponding language.
# Compute the FIRST and FOLLOW sets for all non-terminal symbols in the new grammar and build the parse table.
# Show the analysis table (stack, input, and actions) for the parsing process of the '''<tt>acbec</tt>''' input sequence.
 
= Solution =
 
Something to keep in mind at all times: '''never eliminate the rules corresponding to the initial symbol'''
 
There are two ways of handling the mutual recursion: either expand F in G or G in F.
 
== Expanding G in F ==
 
Initial grammar:
 
O -> a
G -> F c | O c d | (eps)
F -> G b | O c F e
 
Eliminating the singularity (O->a):
 
G -> F c | a c d | (eps)
F -> G b | a c F e
 
Expanding G in F:
 
G -> F c | a c d | (eps)
F -> F c b | a c d b | b | a c F e
 
Eliminate left recursion in F:
 
G -> F c | a c d | (eps)
F -> a c d b F' | b F' | a c F e F'
F' -> c b F' | (eps)
 
Factor F prefixes and expand it in G:
 
G -> a c F'' c | b F' c | a c d | (eps)
F -> a c F'' | b F'
F' -> c b F' | (eps)
F'' -> d b F' | F e F'
 
Note that we still have common prefixes to factor in G and a non-terminal left-corner in F''. The next step will handle those cases:
 
G -> a c G' | b F' c | (eps)
G' -> F'' c | d
F' -> c b F' | (eps)
F'' -> d b F' | a c F'' e F' | b F' e F'
 
F was eliminated because it became unreachable from the start symbol. The grammar still has a non-terminal left corner in G', so we need to eliminate it.
 
G -> a c G' | b F' c | (eps)
G' -> d b F' c | a c F'' e F' c | b F' e F' c | d
F' -> c b F' | (eps)
F'' -> d b F' | a c F'' e F' | b F' e F'
 
Factoring prefixes in G', we get to the final version:
 
G -> a c G' | b F' c | (eps)
G' -> d G'' | a c F'' e F' c | b F' e F' c
G'' -> b F' c | (eps)
F' -> c b F' | (eps)
F'' -> d b F' | a c F'' e F' | b F' e F'
 
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Latest revision as of 22:19, 5 December 2018