Csdrhrt

$L_n = {x in Sigma^{*} | exists w, y, z in Sigma^*, x = ywz, w^r = w, |w| = n }$

Informally x is palindrome of length n

where $Sigma = {0,1}$

I'm having a hard time understanding this could someone please explain.

For $n = 1$

$L_1 = (0+1)^*(0+1)^*$ (Why? What does this mean?)

$L_1^c = sigma$ (What would the compliment mean)

For $n = 2$

$L_2 = (0+1)^*(00+11)(0+1)^*$

$L_2^c = (10)^*(epsilon + 1) + (01)^*(epsilon + 0)$

So my attempt on $n = 3$ (the above two were given)

$L_3 = (0+1)^* (000 + 111 + 101 + 010) (0+1)^*$

$L_3^c = (100)^*(epsilon + 1) + (011)^*(epsilon + 0)$

I get how regexes work I just don't understand how $L_1$ is the regex for palindrome of length one. Not sure what the compliment means. Same for $L_2$

$L_n$ is the set of strings in $Sigma^*$ that have a palindrome of length $n$ as a substring. The complement of a language $L^C$ on an alphabet $Sigma$ is the set of strings in $Sigma^*$ that are not in $L$.

For $L_1$, every string of length $1$ is a palindrome, namely $1$ and $0$. Any nonempty string is in $L_1$, so $$
L_1=L((0+1)(0+1)^*)
$$
since any nonempty string is either a $1$ or $0$ followed by some number of $1$'s and $0$'s. The complement of the language is just the empty string (which we denote $epsilon$). $$
L_1^C={epsilon}
$$
For $L_2$, the palindromes of length $2$ are $11$ and $00$, so the language is any string with either $11$ or $00$ preceded and followed by some number of $1$'s and $0$'s. $$
L_2=L((0+1)^*(00+11)(0+1)^*)
$$
The complement of $L_2$ is the set of strings that don't contain $11$ or $00$, so these are just strings of alternating $1$'s and $0$'s. These could begin or end with $0$ or $1$, but the string must alternate characters. $$
L_2^C=L((01)^*(0+epsilon) + (10)^*(1+epsilon))
$$
For $L_3$, the palindromes of length $3$ are $000$, $010$, $101$, and $111$, so$$
L_3=L((0+1)^*(000+010+101+111)(0+1)^*)
$$
The complement of $L_3$, the strings that do not contain $000$, $010$, $101$, or $111$, is more difficult. Consider such a string that begin with $00$. The next character must be $1$ since if it was $0$, the string would contain $000$. Now, the next character for string $001$ must be $1$ since if it was $0$, the string would contain $010$. We can continue this argument to get $00110011...$ for arbitrarily long strings. From this, we can see that if the string has at least $4$ characters, it can be represented as some number of repetitions of $0011$, $0110$, $1100$, or $1001$ followed by a prefix of that string. Otherwise, if it has less than $4$ characters, it is a substring of one of these. $$
L_3^C=((0011)^*(epsilon+0+00+001)+(0110)^*(epsilon+0+01+011) \
+(1100)^*(epsilon+1+11+110)+(1001)^*(epsilon+1+10+100))
$$