If L2 is DCFL ,how to draw its DPDA?

Question

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

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

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

Comments

But L2 is not DCFL, it is ncfl actually.

---

But Sir a^nb^m |m>n or m<n is DCFL,and adding c^k will make it NDCFL?

---

Yes, you are right, missed that k is not in condition

Answer 1

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 

Comments

Sir , how  L2 is DCFL ?? I think it is NDCFL bcoz m>n or m<n

---

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 .

---

ok got it sir

---

I think here m is != n and therefore, it is dcfl.

Answer 2

Both are DCFLs

• L1’s has CFG as follows:

                   S → xySyz | epsilon

• L2 = { $a^{m}b^{n}c^{k} | m!=n$}

                  Now you can easily think of stack logic.

Answer is D