• Graph Theory

  • 👨‍🏫 geometry: doesn't matter:
    • only thing matters: whether edge / path from point $i$ to $j$ exists
    • e.g. can $A$ send message to $B$?
  • graph : $G=(V,E)$
    • tuple of vertices & edge
    • $V$: set of vertices
      • mostly: $V$ are finite (sometimes not)
    • $E$: a finite set of edges
      • every $e\in E$: of form $\{u,v\}\subseteq v$
  • two graphs with same $V,E$: equivalent
    • i.e. we on't care about
  • loop: an edge $\{u,u\}$
    • i.e. edge where start = end
    • 👨‍🏫 stupid idea if you think of it as computer network
  • simple graph: graph without loop
  • if $\{u,v\}\in E$: $u,v$: neighbors / adjacent
  • complete graph: when every element is neighbor of every other one
    • complement: $\rightleftarrows$ empty graph
    • $K_n$: complete graph on $n$ vertices
  • complement: $\bar{G}=(V,\bar{E})$
    • $\{u,v\}\in \bar{E}⟺\{u,v\}\notin E$

---
        title: G
        ---
        graph LR
        1((1))
        2((2))
        3((3))
        4((4))
        1<-->2
        1<-->3
        1<--/-->4
        2<-->3
        2<--/-->4
        3<-->4

---
        title: G'
        ---
        graph LR
        1((1))
        2((2))
        3((3))
        4((4))
        1<--/-->2
        1<--/-->3
        1<-->4
        2<--/-->3
        2<-->4
        3<--/-->4

• $K_n$: has $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ vertices
• $H=(V_H,E_H)$: subgraph of $G$ if:
  • $V_H\subseteq V_G\wedge E_H\subseteq E_G$

• Isomorphism

  • labels on vertices: somewhat meaningless
    • 👨‍🎓 : as they are just names
  • isomorphism: when you can re-label one graph to show another
  • rigorously: for $G=(V_G,E_G)$, $H=(V_H,E_H)$
    • isomorphism: $f:V_G\to V_H$
      • s.t. $f$: one-to-one & onto
    • $\forall u,v\in V_G,\{u,v\}⟺\{f(u),f(v)\}\in V_H$
  • degree of $u$:
    • $d(u)=|\{v|\{u,v\}\in E\}|$
    • i.e. count of neighbors
    • if $d(u)=0$: $u$ is isolated verted
  • theorem: ${\sum }_{v\in V}d(v)=2|E|$
    • basically: the idea of shake-hands
  • say, we add edges from $\varnothing$
    • when you add $v$,
  • adjacent matrix
    • $n:=|V|$
    • $m:=|E|$
    • $n\times n$ matrix $A$ where $$A_{ij}=\{\begin{array}{llc}1& & \{i,j\}\in E\\ 0& & \text{else}\end{array}$$
    • for simple graph: all $A_{ii}=0$
      • as no loop exists
  • if: we want to find all neighbors-of-neighbors:
    • can be given by: $A^2$
    • $A_{ij}^2={\sum }_k{A}_{ik}\cdot A_{kj}$
    • 👨‍🏫 can be further expanded to multiple networks
      • will be used more often in 3711
  • walk: seq. of vertices & edges
    • $v_0,e_0,v_1,e_1,\dots e_{k-1},v_k$
    • s.t. every $e_i=\{v_i,v_{i+1}\}$
    • for simple graph: it doesn't matter
      • 👨‍🏫 however, will be useful later
        • when there exists more than one edge between two vertices
  • trail: walk without repeated edges
  • path: work without repeated vertex
  • path ⊆ trail, but not the other way

---
    title: G
    ---
    graph LR
    1((1))
    2((2))
    3((3))
    4((4))
    5((5))
    1<-->2
    2<-->3
    1<-->3
    4<-->5
    5<-->3
    4<-->3

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• closed trail: when a trail begins & ends at the same edge
  • aka circuit
• cycle: substitute of closed path
  • $v_0,e_0,v_1,e_1,\dots e_{n-1},v_0$
  • and no vertices other than $v_0$ repeats
  • and $v_0$ only appears in the beginning & end
• theorem: every $\{u,v\}$ walk: contains a $\{u,v\}$ path:
  • weaker: if there is a walk $u\to v$, then also path $u\to v$
• proof of weaker theorem:
  • consider: let $\pi$: (one of) the shortest walk from $u\to v$
  • claim: $\pi$ is a path
    • if $\pi$ is not a path
    • => there exists repetition of vertex $w$
    • however, then we can skip the cycle of $w\to w$, and go directly to $v$
    • which, provides a shorter walk
      • contradiction
• proof of stronger theorem:
  • can be proven w/ similar idea

• Königsberg bridges

  • exists a (multi) graph

graph LR
    1((1))
    2((2))
    3((3))
    4((4))
    1<-->2
    1<-->2
    1<-->3
    2<-->3
    2<-->4
    2<-->4
    3<-->4

• is there a closed walk that goes through all edges exactly once?
• solved by Euler: impossible
  • all the vertex: with odd number of degree
  • no vertex w/ even number of degree (essential for comming-back)

• Connectivity

  • vertex $u$ connected to $v$ iff $\exists (u,v)$-path
    • not necessarily adjacent
  • a graph $G$: connected if every pair of vertices are
  • 👨‍🏫 internet is a connected graph
    • not that they are connected in practice...
  • connected component (cc): a maximal connected subset of vertices
  • theorem: a graph with $n$ vertices and $m$ edges: has at least
    • $n-m$ camps
  • proof:
    • empty graph w/ $n$ component
    • adding an edge: decreases no. of cc at most by $1$ $$\{\begin{array}{llc}1& & \text{if it connects to outside of cc}\\ 0& & \text{if it connects within cc}\end{array}$$
  • corollary: every connected graph has $m\ge n-1$
  • tree: connected graph with $n-1$ edges
  • theorem: following three are equivalent
    1. $G$: connected and $m=n-1$ (is a tree)
    2. $G$: connected and has no cycles
    3. $m=n-1$ and $G$ has no cycles
    • 👨‍🏫 like a triangle / triple iff
  • proof:
    • $(i)⟹(ii)$: start w/ empty graph; add edges
      • at least $n-1$ edges must be added to make it connected
      • every edge $e_i$: reduced no. of cc
      • however, to be a cycle: there must be an edge within cc
        • i.e. not reducing no. of cc
    • $(ii)⟹(i)$:
      • you can't have more than $n-1$ edges without making a cycle
      • you can't have less than $n-1$ edges to make it connected
      • thus: it has exactly $n-1$ edges
    • thus $(i)⟺(ii)$
    • $(ii)⟹(iii)$:
      • you can't have more than $n-1$ edges without making a cycle
      • you can't have less than $n-1$ edges to make it connected
      • thus: it has exactly $n-1$ edges (same)
    • $(iii)⟹(i)$:
      • you can't have more than $n-1$ edges without making a cycle
      • you can't have less than $n-1$ edges to make it connected
      • thus: it has exactly $n-1$ edges (same)
  • theorem: an edge $e\in E$: a cut iff
    • $e$: not part of any cycle
    • 👨‍🏫 if you forget what is iff: it's logical error
      • ⚠️ leads to an F!!!
  • proof: assume $e$: a cut
    • of $u,v$
    • s.t. removal of $e$ will make two cc disconnected
    • suppose: $e$ in cycle $C$
    • if there exists a cycle: there must exist another edge connected two cc
    • assume $\forall a,b$: there is an $(a,b)$-path
      • if the path doesn't include $e$: then removal of $e$ will not diconnect the components
      • if the path includes $e$: it can still round its way through $C$
  • proof: assume $e$ is not in any cycle: prove that $e$ is a cut
    • for it to not be a cut: there must exist $\pi$, another path connecting components