# Math 55 Graph Theory Practice Questions
1. Prove that every simple graph with $n$ vertices and $k$ edges has at least $n − k$ connected components (hint: induction, or contradiction).
3. Let $p\ge 3$ be a prime. Consider the graph $G=(V,E)$ with $V=\{0,1,2,\ldots, (p-1)\}$ and
$$E=\{\{x,y\}:x-y\equiv 2(mod p)\lor x-y\equiv -2 (mod p)\}$$
Show that $G$ is connected. Does $G$ have an Euler circuit? Prove your answer.
3. Let $n\ge 1$ be an integer and let $k\le n/2$. Consider the graph $G=(V,E)$ with $V=\{S\subseteq \{1,2,\ldots,n\}: |S|=k\}$ and
$$E = \{\{S,T\}:|(S-T)\cup (T-S)|=1\}.$$
What are the degrees of the vertices in $G$? Is $G$ connected? For which values of $n$ and $k$ does $G$ have an Eulerian circuit? For which values of $n$ and $k$ is $G$ $2-$colorable? Prove your answers.
4. Do #3 but with the graph defined by $V=\{S\subseteq \{1,2,\ldots,n\}: |S|=k\}$ and
$$E = \{\{S,T\}:|(S-T)\cup (T-S)|=2\}.$$

The official schedule is posted at https://www.jointmathematicsmeetings.org/meetings/national/jmm2023/2270_program_ss94.html#title The Zoom link for virtual attendance/virtual talks is: https://berkeley.zoom.us/j/94982941411?pwd=RWRXeVdoZUNVTmdqK1FIMksrZFllZz09 Friday, January 6 (101 Hynes Convention Center) 1-2pm: Matt Colbrook, The foundations of infinite-dimensional spectral computations (Zoom) 2-3pm: Josué Tonelli-Cueto, Condition-based Low-Degree Approximation of Real Polynomial Systems

1/3/2023Lecture 1: Propositional Logic A proposition is a declarative sentence that is either true or false but not both. A proposition has a truth value which is either $T$ or $F$. A letter used to denote a proposition is called a propositional variable. If $p$ is a proposition, the negation of $p$, denoted $\lnot p$, is the proposition "it is not the case that $p$". The truth value of $\lnot p$ is the opposite of that of $p$. If $p$ and $q$ are propositions, the conjunction of $p$ and $q$ (denoted $p\land q$) is the proposition "p and q." Its truth value is $T$ when both $p$ and $q$ are $T$, and it is $F$ otherwise. If $p$ and $q$ are propositions,he disjunction of $p$ and $q$ is the proposition "p or q". Its truth value is $T$ when at least one of $p$ or $q$ is $T$, and $F$ if both $p$ and $q$ are $F$. If $p$ and $q$ are propositions, the conditional $p\rightarrow q$ is the proposition "if $p$ then $q$". Its truth values are given by the following truth table (which lists its truth values in terms of those of $p$ and $q$): [see Table 5 in Rosen] The converse of $p\rightarrow q$ is $q\rightarrow p$. A compound proposition consists of logical operations (the four listed above) applied to propositions or propositional variables, possibly with parentheses. The truth value of a compound proposition can be mechanically determined given the truth values of its constituents.

12/12/2022Lecture 14 Basic Terminology, Handshaking Theorem A graph is a pair $G=(V,E)$ where $V$ is a finite set of vertices and $E$ is a finite multiset of $2-$element subsets of $V$, called edges. If $E$ has repeated elements $G$ is called a multigraph, otherwise it is called a simple graph. We will not consider directed graphs or graphs with loops (edges from a vertex to itself). Two vertices $x,y\in V$ are adjacent if ${x,y}\in E$. An edge $e={x,y}$ is said to be incident with $x$ and $y$. The degree of a vertex $x$ is the number of edges incident with it. Theorem. In any graph $G=(V,E)$, $$ \sum_{v\in V} deg(v) = 2|E|.$$ Proof. Let $I={(e,v):e\in E,v\in V,\textrm{$e$ is incident with $v$ in $G$}}$ be the set of edge-vertex incidences in $G$. We will count the number of incidences in two ways: (i) Observe that every edge participates in exactly two incidences, and no incidence participates in more than one edge. Thus, the total number of incidences is $|I|=2|E|$.

12/12/2022Prove that every simple graph with $n$ vertices and $k$ edges has at least $n − k$ connected components (hint: induction on $k$). Proof 1. We proceed by induction. Fix $n\ge 1$ and let $P(k)$ be the statement that every simple graph with $n$ vertices and $k$ edges has at least $n-k$ connected components. We will show that $P(k)$ is true for all $k\ge 0$. Basis Step: $P(0)$ says that a graph with no edges has at least $n$ connected components, and this is true since each vertex is its own connected component. Inductive Step: Assume $P(k)$ is true. Let $G$ be a graph with $n$ vertices and $k+1$ edges. Let $e=uv$ be an arbitrary edge of $G$, and let $G'=G-e$ be the graph obtained by deleting $e$. Let $G_1,\ldots,G_\ell$ be the connected components of $G'$ and note that $\ell \ge n-k$ by the inductive hypothesis. Notice that if $u$ and $v$ are in the same connected component of $G'$, then adding it does not change the connected components, so in this case $G$ also has $\ell$ connected components.

11/7/2022
Published on ** HackMD**

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