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Which of the following languages are CFL?
$$L_1= \left \{ 0^n 1^m \mid n \leq m \leq 2n \right \} \\[1em] L_2 =\left \{ a^i b^j c^k \mid i=2j \text{ or } j=2k \right \}$$

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@Shaik Masthan Brother can you explain L1 in term of the stack. Thankyou

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we can think like, n ≤ m as L1 and m ≤ 2n as L2

finally we want L1.L2,

cfl's are closed under concatenation.

Both are CFL.

CFG for $L_1: \quad \left \{ S \to 0S1 \mid 0S11 \mid \varepsilon \right .$

CFG for L_2 :\quad \left \{ \begin{align} S &\to X \mid Y \\[1em] X &\to Xc \mid P \\ P & \to aaPb\mid \varepsilon\\[1em] Y &\to aY \mid Q \\ Q & \to bbPc\mid \varepsilon \end{align}\right .

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can you please tell me the idea of stack? i.e if i want to check if they are cfl with STACK ,then how should we approach?
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@Uddlpto

to check with stack we need to check how many conditions we have to check in the given language

like for L2 here u need to check i=2j or j=2k....but u dont need to check them simultaneously so we can use one stack at a time to check the conditions...suppose instead of this u were given to check i=2j"and" j=2k..this would not have been possible with one stack...these two checking have to be done together..so u need two stacks here...
overall concept is as we know when we put a stack with FA it works like a memory to remeber things...as u keep on increasing the no of stacks...u start moving up the language hierarchy...and when this memory becomes infinte ..we enter into TM...hope it helps...
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idea for pda of first one?
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1) To accept the input we will push "X" for "0" and when we  will encounter "1" sim ply pop it.

In case M=N  it will be same as 0^N 1^N or M>N then pop every "1" after "0".

2) here we required two memory comparison 1st for i=2j and j =2k , so this is not CFL.

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in second case there is OR condition between two comparision,, so need not to  check both the conditions at once,,,a NPDA can accept it....

correct me if m wrong
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oh yes ,, yeah it will be CFL ,, i just overlooked it .Thanx For Correction :)
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@Abhishek: Your idea for the PDA for $L_1$ is wrong too. $L_1$ is NCFL.
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@Pragy how can we make pda for L1