Q. 1. Marks [10]
Suppose we have a Transition Graph (TG) as follows:
Derive regular
expression for the above transition graph. Briefly explain each and every step.
Answer:
According to Kleenes’ theorem
part II, step 1 introduces new start state connected with the old start states
through NULL transitions as follows:
In step 2, new final state is introduced connected
with both final states through NULL transitions as follows:
Step 3 replaces multiple incoming transition edges by
a single transition labeled by the corresponding regular expression as follows:
In step 4, loops are eliminated as follows:
After applying step 4 to eleminate and bypass states
we get:
Finally
applying step 4 once again we get:
Q. 2. Marks
[10]
Develop a Finite Automaton
(FA) corresponding to the union of above two FAs.
Answer:
Old states
|
New states after
reading
|
|
a
|
b
|
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Z1- = (X1,Y1)
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(X2,Y1) = Z2
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(X2,Y2) = Z3
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Z2 = (X2,Y1)
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(X3,Y1) = Z4
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(X3,Y2) = Z5
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Z3 = (X2,Y2)
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(X3,Y3) = Z6
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(X3,Y2) = Z5
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Z4 = (X3,Y1)
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(X4,Y1) = Z7
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(X3,Y2) = Z5
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Z5 = (X3,Y2)
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(X4,Y3) = Z8
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(X3,Y2) = Z5
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Z6+ = (X3,Y3)
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(X4,Y3) = Z8
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(X3,Y2) = Z5
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Z7+ = (X4,Y1)
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(X4,Y1) = Z7
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(X4,Y2) = Z9
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Z8+ = (X4,Y3)
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(X4,Y3) = Z8
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(X4,Y2) = Z9
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Z9+ = (X4,Y2)
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(X4,Y3) = Z8
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(X4,Y2) = Z9
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