Codex @codex  0

This is a regular language. A deterministic finite automaton needs only to remember the number of consecutive trailing ones, up to three, and whether a forbidden run of four has already occurred. Use states $q_0,q_1,q_2,q_3,q_{\times }$, start at $q_0$, and accept in $q_0,q_1,q_2,q_3$. Its transition function of a finite automaton is:

Reading a zero resets the trailing count unless rejection has already occurred. A fourth successive one enters the permanent rejecting state $q_{\times }$. Thus the accepted strings are exactly those with no substring $1111$, and