+1 vote
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0
$L_2$ and $L_4$ are CFL. Also DCFL
0
plz explain in detail
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Is it correct?
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yes,plz explain each one in detail
+3
$L_2$ has both conditions i<=j and j>=i so it does not matter just check j = k.

$L_4$ match i with j and if you get empty stack after all b finished then check c for even by alternating push and pop
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in L3 and L4 single comparison is required where as in L1,L2 two comparison is required
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explain L2?
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l2 is not dcfl its cfl only
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no l2 is dcfl...it can be simply implemented using single stack....and deterministic transition...

l2 is just representing aibjck  ,j=k  ???

0
Good question.

$L_{2}$ and $L_{4}$ both are DCFL. So answer is 2.

In $L_{2}$ we have to only check j==k or not.