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Demarcation or scoping of examinations and assessment
School of Computing Short Learning Programmes
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August 25, 2006 01:43PM
Registered: 13 years ago
Is basically two questions in one. where L = below
S -> XaX | bX | XY | a
X -> XbX | bS | /\
Y -> YaS | /\
From what I can see above, the language accepts all words that have atleast one 'a' or one 'b' or can be empty. (Somebody plse correct me here)
1. Write down the grammer that will produce L* (Closure) ?
To me, L* is all the words in L and the empty word.
But L already accept an empty word, because X is nullable, Y is nullable, which in turn S is nullable because of production S->XY, hence
L* = L in this case. Is this correct ????????
2. Prove it in both directions.
If this answer is the same to the tut201 provided, then I know it.
COuld a Lecturer plse give his/her comments ...
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