first is not regular...because u have to keep count of when w ended and w^r started...its not possible without a stack
second is regular because u can expand x to cover everything leaving just first and last symbol...its regular expression will be like
0(0+1)*0+1(0+1)*1
third is similar to first...not regular u cant keep track of w and w^r

first is not regular...because u have to keep count of when w ended and w^r started...its not possible without a stack

second is regular because u can expand x to cover everything leaving just first and last symbol...its regular expression will be like

0(0+1)*0+1(0+1)*1

third is similar to first...not regular u cant keep track of w and w^r

but for ww^rx we can write the regular expression as 00(0+1)*+11(0+1)* correct me if wrong... using the same logic used as for wxw^r

1st is CFL but not dcfl (non regular)

2nd is regular regular experssion for this is 0(0+1)

^{+}0 + 1(0+1)^{+}13rd is CFL but not dcfl (non regular)

but for ww^rx we can write the regular expression as 00(0+1)*+11(0+1)* correct me if wrong... using the same logic used as for wxw^r

see what are you doing wrong first you should understand why wxw

^{r}when x,w=(0,1)^{+}regular expression is 0(0+1)^{+}0 + 1(0+1)^{+}1 issee if we take w=100 then w

^{r}will be 001 or if w=01 then w^{r }wil be 10for first case 100 x 001

for second case 01 x 10

either string "starts with 1 and ends with 1" or "start with 0 and ends with 0"

or we can manipulate it as " 0 anything 0 + 1 anything 1" because x belongs to (0+1)

^{+}understood now.. thanks

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