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A binary De Bruijn sequence of order $n\ge 1$ is a cyclic binary string in which every length-$n$ binary string occurs exactly once as a consecutive window. It has length $2^n$. For $n=3$, 00010111 has cyclic windows 000, 001, 010, 101, 011, 111, 110, 100; the windows at the end wrap around to the beginning. A linear recording must repeat the first $n-1$ symbols at its end. Finding an Eulerian cycle in the De Bruijn graph on length-$(n-1)$ words constructs one: each edge supplies exactly one required window.

For an alphabet of size $q$, the order-$n$ directed graph has words of length $n-1$ as vertices and words of length $n$ as edges. An edge goes from its prefix of length $n-1$ to its suffix of that length. An Eulerian cycle therefore lists every length-$n$ word once as a cyclic window, producing a De Bruijn sequence.

A cyclic sequence over an alphabet of size $q$ in which every word of length $n$ appears exactly once as a cyclic window. Its length is $q^n$. The De Bruijn graph converts its construction into an Eulerian cycle problem.

False. A connected graph has an Eulerian cycle exactly when every vertex has even degree. In $K_n$ every degree is $n-1$, so it has an Eulerian cycle precisely when $n$ is odd. The complete graph $K_4$ is a counterexample to the statement for all $n\ge 3$.

False. In $K_{r,s}$ the $r$ vertices in one class have degree $s$, while the $s$ vertices in the other have degree $r$. For positive $r,s$, connectedness is automatic, and Euler's criterion requires both $r$ and $s$ to be even. Evenness of their sum alone allows both to be odd. For example $K_{1,3}$ has $r+s=4$ but all four degrees are odd, so it has no Eulerian cycle.