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Given a language $L$, define $L^i$ as follows:

$$L^0 = \{ \varepsilon \}$$

$$L^i = L^{i-1} \bullet L \text{ for all } I >0$$

The order of a language $L$ is defined as the smallest $k$ such that $L^k = L^{k+1}$. Consider the language $L_1$ (over alphabet O) accepted by the following automaton.

The order of $L_1$ is ____

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0
Is it $2$?
+2

L= eps + 0(00)*

L2 = 0*

L3 = L2.L = 0*.(eps+0(00)*) = 0*

since, L2=L3

order = 2

+1
Can order be zero?

$L_1=\{\epsilon +0(00)^*\}$

Now that is given language L,  we have find order of it.

$L^0=\{\epsilon\}$

$L^1 =L^0.L=\{\epsilon \}.\{\epsilon +0(00)^*\}=\{\epsilon+0(00)^*\}$

$L^2=L^1.L=\{\epsilon+0(00)^*\}.\{\epsilon+0(00)^*\}$

$=\{\epsilon+0(00)^*+0(00)^*0(00)^*\}$

$L^2=\{0^*\}$

$L^3=L^2.L=\{0^*\}.\{\epsilon+0(00)^*\}$

$=\{0^*+0^*0(00)^*\}$

$L^3=\{0^*\}$

Order of L is 2 such that $L^2=L^{2+1}$
answered by Veteran (55.4k points)
selected
L^1 = {null, 0, 000, 00000,.....}
L^2= L.L =  {null, 0, 000, 00000,.....}.{null, 0, 000, 00000,.....} = 0*
L^3 = L^2.L =  0* U {null, 0, 000, 00000,.....} = 0*

L^2=L^3
So Order would be 2.
answered by Veteran (54.1k points)
edited
Ans: 2
answered by Active (1.2k points)

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