+1 vote
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$L_2$ and $L_4$ are CFL. Also DCFL
plz explain in detail
Is it correct?
yes,plz explain each one in detail
$L_2$ 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
in L3 and L4 single comparison is required where as in L1,L2 two comparison is required
explain L2?
l2 is not dcfl its cfl only

no l2 is dcfl...it can be simply implemented using single stack....and deterministic transition...

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

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.