KU Math 725: Lecture 1/26/2026
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Paths, Trails, Walks, and Cycles
[definition=%counter%] Let $x, y \in V(G)$. An $x,y$ walk in $G$ is an alternating sequence of vertices and edges
\[x = v_0, e_0, v_1, e_1, \ldots, v_{n-1}, e_{n-1}, v_n = y\]where $v_i$ and $v_{i+1}$ are the endpoints of $e_i$ for all $i$. The length of the walk is the number of edges in the walk (in the above, that would be $n$). $x$ and $y$ are the endpoints while the other vertices are internal.
If $G$ has no parallel edges, we just list vertices.
A walk is closed if $v_0 = v_n$. A trail is a walk with no repeated edges. A path is a walk with no repeated vertex. A circuit is a closed trail and a cycle is a closed path. [/definition]
[example=%counter%] If you start at vertex $x$ and don’t move at all, that is called the trivial walk from $x$ to itself. [/example]
When we want to denote an $x,y$ walk, we sometimes write $xWy$ or $x\xrightarrow{W} y$. Note that it is obvious that if $xWy$ and $yW’z$ are walks, then $xW''z$ obtained from $xWyW’z$ is a $x,z$ walk.
[proposition=%counter% (1.15 in Reuven’s notes)] If $G$ has an $x,y$ walk, then $G$ has an $x,y$ path. [/proposition]
[proof] Let $xWy$ be a walk. If some vertex $z$ occurs more than once, then $xWy$ has the form
\[xW'zW''zW'''y\]where $W’$ and $W'''$ could be trivial but $W''$ is not. So then
\[xW'zW'''y\]is a strictly shorter walk. Keep repeating this process until no further shortening possible. By definition, this means the walk is a path and we note that this termination happens in finitely many iterations because the length decreases by at least one for each iteration. [/proof]
Connectivity
[definition=%counter%] Two vertices of $G$ are connected if there is a path in $G$ between them. The graph $G$ is connected if every pair of vertices is connected.
The connected components are the maximal connected subgraphs of $G$ and we denote the number of connected components by $c(G)$. [/definition]
[proposition=%counter% (1.17 in Reuven’s notes)] The relation “$u$ is connected to $v$” is an equivalence relation. [/proposition]
[proof] Obvious. Note that transitivity is concatenation of paths. [/proof]
[proposition=%counter% (1.18 in Reuven’s notes)] Let $G$ be a connected graph on $n$ vertices. Then the vertices can be labeled
\[v_1, \quad\ldots, \quad v_n\]so that every induced subgraph $G_j = G[v_1, \ldots, v_j]$ is connected for $1 \le j \le n$. In addition, $v_1$ may be chosen arbitrarily. [/proposition]
[proof] Choose $v_1$ arbitrarily. Obviously $G_1 \cong K_1$ is connected. Now assume that we have constructed $G_j$. Take any vertex $x$ in $V(G) \setminus V(G_j)$. By connectivity of $G$, there is a path $v_1 \to x$ in $G$. Take $v_{j+1}$ to be the first vertex in that path that is not in $G_{j}$. By construction, $v_{j+1}$ is adjacent to one of the vertices of $G_j$ so $G_{j+1}$ is connected. [/proof]
[corollary=%counter% (1.19 in Reuven’s notes)] If $G$ is connected then $e(G) \ge n(G) - 1$. More generally, $c(G) \ge n(G) - e(G)$. [/corollary]
[proof] The inductive process in Proposition 1.18 proves the first inequality inductively. For the second inequality, it suffices to show that
\[e(G) \ge n(G) - c(G).\]For each connected component $C_i$, we have that
\[e(C_i) \ge n(C_i) - 1.\]Summing over $i$ gives the desired formula. [/proof]
[proposition=%counter% (1.20 in Reuven’s notes)] Let $a \in E(G)$. If $a$ belongs to a cycle in $G$, then $c(G - a) = c(G)$. If $a$ does not belong to a cycle, then $c(G - a) = c(G) + 1$. In this latter case, $a$ is called a cut-edge (or co-loop or bridge). [/proposition]
[proof] If $a$ belongs to a cycle in $G$, then removing $a$ does not disconnect $c(G)$. If $a$ does not belong to a cycle, then $a$ does disconnect $G$ into two connected components. [/proof]
[definition=%counter%] A cut vertex is a vertex such that $c(G - v) \ge c(G)$. [/definition]
Acyclicity and Trees
[definition=%counter%] A graph is acyclic (or a forest) if it has no cycles. A connected forest is called a tree. [/definition]
[remark=%counter%] Acyclic $\iff$ every edge is a cut-edge. [/remark]
[proposition=%counter% (1.24 in Reuven’s notes)] A graph is acyclic $\iff$ $c(G) = n(G) - e(G)$. In particular, every tree $T$ has $n(T) - e(T) = 1$. [/proposition]
[proof] Let $c = c(G), e = e(G), n = n(G)$. Start with vertex set $V(G)$ and no edges. This is trivially acyclic and $c = n - e$. Now we add in edges of $G$ one-by-one. Each time we do this, $e$ increases by $1$ and $c$ may or may not decrease by 1. If $c$ ever remains constant, then Proposition 1.20 implies we created a cycle in $G$, which contradicts acyclicity. So every edge is a cut-edge which implies this inductive process of building up $G$ does so maintaining acyclicity at all steps.
[/proof]