330 views

$L_1=\{(xy)^m(yz)^m\;,\;m\geq 1\}$

$L_2=\{a^mb^nc^k\;|\;m>n\;or\;m<n\}$

Which of the following is True?

(a) $L_1$ is CFL, $L_2$ are DCFL                    (b) $L_1$ is DCFL, $L_2$ is CFL

(c) Both $L_1$, $L_2$ are CFLs                       (d) Both $L_1$, $L_2$ are DCFLS

edited | 330 views
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But L2 is not DCFL, it is ncfl actually.
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But Sir a^nb^m |m>n or m<n is DCFL,and adding c^k will make it NDCFL?
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Yes, you are right, missed that k is not in condition

Yes $L_2 =\{a^mb^nc^k\;|\;m>n\;or\;m<n\}$ is DCFL

having DPDA

$L_1=\{(xy)^m(yz)^m\;,\;m\geq 1\}$ is also DCFL

having DPDA

by Veteran (57.1k points)
selected
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Sir , how  L2 is DCFL ?? I think it is NDCFL bcoz m>n or m<n
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push a and for every a pop b , we have to do it for every language... i think there is no ambiguity na  here. so DCFL .
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ok got it sir
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I think here m is != n and therefore, it is dcfl.