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Which of the following language is $CFL$?

a. $\{a^mb^nc^n\;|\;m!=n\}$

b. $\{a^mb^nc^k\;|\;if\,(m=n)\,then\,(n!=k)\}$

c. $\{a^mb^nc^k\;|\;m>n\;or\;n<k\}$

d. None of these
in Theory of Computation by Junior (995 points)
edited by | 3.1k views
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Ans given is c Need explanation
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DIAGRAM

IN THE OPTION 'C' CONSIDER EITHER M>N (OR) N>K

1ST CASE: CONSIDER M>N THEN LANGUAGE IS L={a^mb^n/m>} IGNORE THE 'C' TERMS THEN IT IS DPDA

2ND CASE:CONSIDER N<K THEN LANGUAGE IS L={b^nc^k/m>} IGNORE THE 'A' TERMS THEN IT IS DPDA

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^^ it look like something in my hand writing
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yes ,for better understanding i just paste the image

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what does this mean:

         q2      (b,z0,z0)   $\rightarrow$  q4 mean in the above diagram ?

+1

@amsar that was solution for something else

$L =\{a^mb^nc^k \;|\;m\neq n\}$

So $q_2$ to $q_4$ was for $m<n$

2 Answers

+12 votes
Best answer
a. $\{a^mb^nc^n \mid m!=n\}$

We need to ensure #a $\neq$ #b, but also #b = #c. Both of these conditions are not restricted to any finite value and hence cannot be checked simultaneously using a PDA. So, $L$ is a CSL.

b. $\{a^mb^nc^k \mid \text{if }(m=n)\text{ then }(n\neq k)\}$

We have 2 condition to be checked here - either $m\neq n$ or $n \neq k$ in either case we accept a string and reject otherwise. Both of these checks are again not restricted to any finite values but we need not check them together as the conditions are separated by "OR". Hence this can be done using an NPDA making this language a CFL but not DCFL.
c. $\{a^mb^nc^k\mid m>n\text{ or }n<k\}$

Similar to 'b', is CFL but not DCFL.
by Veteran (422k points)
selected by
+1
For B). It is " if (m=n) then (n!=k) "

.i.e, if P then Q

P -> Q <=> (P)' $\cup$ Q, rt ?
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yes..
+1
But, sir u have written " m≠n or n==k " .

Won't it be m≠n or n$\neq$k
+2
Thanks, corrected..
+1
Hence both option b , c is correct , Beautiful explanation Sir, Thank you!
+4 votes

for cfl if there is one comparison at one time then its a CFL.

a) $a^mb^nc^n$ | m!=n 

here two comparison first b= c and a !=b & c so not a cfl 

b) $a^mb^nc^k$  | if(m=n) then n!=k
here first we have to compare a=b if it is yes then n!=k 
we can't do it using one stack since we have already popped a using b to match m=n so we cant match the n to k.

c) $a^mb^nc^k$  | m>n or n<k  
here at one time only one compare if m>n then we dont have to check for n<K or vice versa and it can be done by NPDA.

so option c 

by Boss (16.3k points)
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@Umang , can u check this line - 

    we can do it using one stack since we have already popped a using b so we cant match the c to k

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enlightenedits can't typing mistake.

+1
We having nothing left in stack after ensuring  m=n,  so we can't ensure n! =k (or n=k)  later.
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For option c if m=5 and n= 4 then how it will work??
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