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# ASSIGNMENT NO:2

DEAR FELLOW!

LETS START DISCUSSION HERE ...

GOOD LUCK TO ALL...

Fall%202015_CS402_2.doc

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u got your point??? i forgot ur email that'swhy i did not mail you..

is it complete solution?

i think no

ary bhi koi solution upload kar do

Salam,

Given Diagram for Task 1 has been completed using Method 2 (Lecture 16) where we combine the stages when we have multiple transitions for different stages.

My concern, that in given NFA, for stage 3 there are two transitions for letter B, one leading to Stage 2(Final Stage) and other one leading to Stage 4 but in FA given above, from combined stage of (2,3 +), there is only one transition for B that is going to stage 4.

Now either, we combine stage 2,4 or we separate them, there should be transition from stage 3 to stage 2 for letter B.

Any thoughts ??

in this fA of question 2 their is not any initial state why? explain
in my opinion z3 and z6 are not final states m confused plz explain it

I m very sorry but still I could not understand the logic of FA. how to creat an FA and what is the logic to draw an FA. can anyone tell me the basic steps of drawing an FA...?

plz find errors in this nfa to fa
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yai first question ka nfa to fa thk ha bus is main intial stage sy b ki transition empty stat ki traf move krw dain r empty stat py is ka khud a r b ka loop lga lain

Well made FA i must say.

However, my concern that ur FA is accepting string ending in "bb" OR strings ending in even number of "b". while the given NFA doesn't accept strings ending in "bb". solution made by transition table also accepts the strings ending in "bb". So as per me, this is a problem.

Any thoughts ??

but 4 sy 3 r 3 sy 2 bb accepted ha

correct but in order to reach 4, we have to go through 3 that will append one more "b"

and in this way total number of b (in the end of string) will be odd number.

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