Regular expression at least 2 out of 3 consecutive characters should be 1

Question

How can I build a regular expression that, using only the concatenate, union and star operations, over the alphabet {0,1}, describes the language "Every three consecutive characters contain at least two 1, and the input has length at least 3"? For instance 110011, 0101 and 11 should be refused. I was thinking on using the logic from this (incomplete) DFA, but I can't figure out how to get a regular expression that follows the rule. Thanks!

Answer 1

Assume that string $s$ is in $L$. We will look at the last two characters of the string:

• 00, impossible, because this string could not be in $L$.
• 01, next character must be 1, new last two characters is 11.
• 10, next character must be 1, new last two characters is 01.
• 11, next character may be 1 or 0, new last two characters is 10 or 11.

Note that no matter your current state, you will always pass through the 11 state within two steps. Assuming that $s$ ended with 11, we get the following loop:

(1|(011))*

This will be the middle section of any string in $L$. All we need to do now is handle possible prefixes, making sure not to allow any with size < 3:

(111|011|1011|11011)

And finally, possible suffixes (note the empty union to express that the suffix is optional):

(|0|01)

Now all that's left is to concatenate them:

(111|011|1011|11011)(1|(011))*(|0|01)