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SYSTEM SOFTWARE [10CS52] Unit-7COMPAILER DESIGN 10CS63 VTU NOTES UNIT-1SYSTEM SOFTWARE [10CS52] Unit-8COMPAILER DESIGN 10CS63 VTU NOTES UNIT-2CD notes for 6th semCOMPAILER DESIGN 10CS63 VTU NOTES UNIT-4COMPAILER DESIGN 10CS63 VTU NOTES UNIT-5Packet SniffersSystem Software and Operating Systems Lab Manual[10csl58]Compiler Design - 06CS63 VTU notes
A lexical analyzer uses the following patterns to recognize three tokens T1, T2, and T3 over the alphabet {a,b,c}. T1: a? (b|c) ∗a T2: b? (a|c) ∗b T3: c? (b|a) ∗ c Note that ‘x?’ means 0 or 1 occurrence of the symbol x. Note also that the analyzer outputs the token that matches the longest possible prefix.
String to be generated :- bbaacabc
The regular expressions :-
T1 : (b+c) * a + a ( b + c ) * a
T2 : (a+c)* b + b ( a + c ) * b
T3 : (b + a) * c + c ( b + a ) * c
lets see which regular expression generations the longest possible prefix
String = bbaacabc
T1 : bba
T2 : bb
T3 : bbaac
So , as we can see T3 is generating the longest possible prefix here
Now , again in the string (abc) also need to be generated , lets check which regular expression generates (abc).
T1 : end with a , but we want to end with c – so abc not possible
T2 : end with b , but we want to end with c – so abc not possible
T3:- (b + a ) * c – ending with c and also b and a can be generated inside the closure – so , yes T3 can generate abc.
Therefore total string (bbaacabc) is generated using T3T3.
So, Answer is option (D) T3T3
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