Given the context of L1, free irregular language. This regular language is L2.
Is it possible that L1 U L2 = the correct language? Is it possible that L1 * L2 = the correct language?
I think the second is impossible. But I'm not sure.
I would like to see an example if one of the above statements (or both) is / true.
Is it possible that L1 U L2= ordinary language?
L1 U L2
Yes it is possible.
Simple case: if L 1 is a subset of L 2, then it L1 U L2will be regular ( ), for example: L=L2 1: { | where } andanbnn >= 0L2 = (a + b)*
=L2
anbn
n >= 0
L2 = (a + b)*
is it possible that L1 * L2 = the correct language?
L1 * L2
. - - ( L1 L1 * L2).
: CS 273:
, -
L1 = a ^ nb ^ n, CFL, L2 = Ø, .
, L1.L2 = (a ^ nb ^ n).Ø = Ø,
. , , .
, L1 U L2 = ?
, . :
L1 = {0*1*} () L2 = {0^n1^n |n>=0} ( ).
L1 = {0*1*}
L2 = {0^n1^n |n>=0}
L = L1 U L2 = {0*1*}, , . , , - .
L = L1 U L2 = {0*1*}
, L1·L2 = ?
L1·L2
- - . :
L = L1·L2 = {(0*1*)·(0^n1^n) |n>=0} ( ).
L = L1·L2 = {(0*1*)·(0^n1^n) |n>=0}
, , , L1 L2 Ø, L1·L2 Ø (). , Ø .
L1
L2
Ø
: Geeks for Geeks