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In the network 200.20.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is____

Which of the following languages are context free?

  1. $L_1=\{a^mb^na^nb^m \mid m,n \geq 1\}$
  2. $L_2=\{a^mb^na^mb^m \mid m,n \geq 1\}$
  3. $L_3=\{a^mb^n \mid m= 2n+1\}$
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mask is 27 so last 5 bit will be host bits here

200.20.11.144=200.20.11.10010000/27

Lowest Address=200.20.11.100 00000/27=200.20.11.128/27

Highest Address=200.20.11.100 11111/27=200.20.11.159/27

Here, Subnet id=200.20.11.128/27

       Direct Broadcasting address=200.20.11.159/27

so last address to give to host is 200.20.11.158/27

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Q49 L1 and L3 are Context Free 

L1 is ambnanbm ,here in one stack we will push first m a's then push n b's,then for each pop of 'b' print 'a' and each pop of 'a' print 'b', so only one stack is required, so CFL

L2 is ambnambn,, here two stacks will be required one stack will not be sufficient for comparison, say we have push first m a's then n b's in stack,then how to print m a's ? so 2 stacks will be required here.So not CFL

L1 is ambn,m=2n+1, is equivalent to a2n+1b,only one comparison so CFL

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