# 15-344/Classnotes for Tuesday September 15

We mostly went over Day One Handout today.

 Dror's notes above / Students' notes below

## Lecture Note for September 15

DEFINITION 1 Graph A graph $G = (V,E)$ is a set $V = \{a,b,...\}$ (usually finite, "vertices") along with a set $E = \{(ab),(bc),(bd),...\}$ ("edges") of unordered pairs of distinct elements of $V$.

DEFINITION 2 Incident If an edge $e = (ab)\in{E}$, we say that $e$ is incident to $a$ and $b$.

DEFINITION 3 N-valent In a graph $G = (V,E)$, a vertex $u \in{V}$ is called bivalent if it is incident to precisely two edges and n-valent if incident to precisely n edges, where $n = 0,1,2,..$.

DEFINITION 4 Edge Cover An edge cover for graph $G = (V,E)$ is a subset $C\subset{V}$ such that every edge of $G$ incident to at least one vertex in $C$.

DEFINITION 5 Independent Let $G = (V,E)$ be a graph. A subset $I\subset{V}$ is called independent if whenever $a,b\in{I}$, then $(ab)\notin{E}$.

THEOREM 1 Edge covers are complementary to independent sets. In other words, $C\subset{V}$ is an edge cover if and only if the complementary subset $V-C$ is an independent set.

Proof $(\rightarrow)$ Assume $C$ is an edge cover. I assert that $I = V-C$ is independent. Indeed, if $e=(ab)\in{E}$, then since $C$ is an edge cover, either $a\in{C} \implies a\notin{I}$ or $b\in{C} \implies b\notin{I}$, which implies $e$ does not connect any two elements of $I$.

$(\leftarrow)$ Assume $I=V-C$ is independent. Pick any edge $e = (ab)\in{E}$. As $I$ is independent, $(ab)$ does not connect any two members of $I$. Hence, either $a\notin{I} \implies a\in{C}$ or $b\notin{I} \implies b\in{C}$, which implies $e$ is incident to an element of $C$. QED

DEFINITION 6 Matching is a subset of edges of a graph where no two edges share a common vertex. A matching is said to be perfect if all edges in a graph are matched.