• Disjoint cycle

  • theorem: if (connected) $G=(V,E)$, $\forall i\in V,2|d(i)$
    • edge set $E$: disjoint union of cycles $c_1,c_2,\dots ,c_t$
  • proof: by induction
    • how do we define the size of graph?
    • (strong) induction on no. of edges: $m$
    • base case $m=0,1$: works
    • induction step: $G=(V,E)$, $m=|E|$
      • for a vertex in cc
        • its neighbor's degree, also even
          • i.e. there exists at least another path to continue
        • however, it can't continue forever as $G$'s finite
        • at some point, you will
      • thus, there must be at least one cycle
    • connected graph without cycle: tree
      • but tree: at least with two leaves $⟺2\nmid d(i)$
        • against initial assumption
      • it is connected, and it shouldn't be tree -> it has a cycle
      • cycle $C\in G$
    • $G^{\prime }:=$ $G$ after removal of all edges of $C$
      • all vertex in $C$: loses degree of $2$

• Eulerian circuit

  • Eulerian circuit: closed walk passing every edge exactly once
    • graph: Eulerian if it contains an Eulerian circuit
  • theorem: for $G$ connected, $G$ is Eulerian iff $\forall i\in V,2|d(i)$
    • Eulerian -> $\forall i\in V,2|d(i)$
      • let $v_0,v_1,\dots ,v_k$: closed walk covering all edges once
        • $v_k=v_0$
        • same vertex can appear multiple times in the middle, however
          • as it enters & leaves the node, it must have even degree
          • same for $v_0$
    • $\forall i\in V,2|d(i)$ -> Eulerian
      • let $G$ a cycle connected to multiple Eulerian circuit / graph
      • staring for a vertex cycle, it can have Eulerian walk on Eulerian circuit, and come back to starting point
    • 👨‍🏫 prove base case for yourself
  • Hamiltonian cycle: cycle visiting all vertices
    • 👨‍🏫 solution: an NP-hard problem

• • theorem: $G$ graph w/ $m$ edges
    • $G$ w/ subgraph $H$ $m_H\ge m/2$
    • there exists a bipartite $H$
  • proof idea: consider all subgraph of $G$
    • and consider: largest bipartite subgraph
    • claim: $|E_H|\ge m/2$
  • proof
    • claim, for $\forall c\in V$
      • $d_H(v)\ge \frac{d_G(v)}{2}$
    • if it's not the case, then we can make $H$ largest
      • contradiction: as $H$ must be the largest subgraph
    • theorem: $2\cdot |E_H|={\sum }_{v\in V}{d}_H(v)$
      • ${\sum }_{v\in V}{d}_H(v)\ge {\sum }_{v\in V}{d}_G(v)/2=m$

• Graphic sequence

  • given a seq. $d_1\le d_2\le \cdots \le d_n$
    • $\forall d_i\in [0,n-1]$
    • is the graph w/ this degree sequence?
  • theorem: $d_1,\dots ,d_n$: degree seq of $G$ (not necessarily simple)
    • iff $\sum d_i$ is even
  • if $\forall 2|d_i$: trivial

graph TD
    a_0((a_0))
    a_1((a_1))-->a_1
    a_2((a_2))-->a_2

- and similarly go on

• otherwise
  • no. of odd terms: must be even
  • then: we can have edge between such vertex,
    • and $\forall 2|d_i$
    • which is solvable by above case
• what if we want it to be a simple graph?
  • seq. $d_1,\dots ,d_n$: graphic if $\exists$ a graph based on it
• Havel-Hakimi algorithm
  • choose vertex w. highest degree
• theorem: a seq $A$ $d_1\le d_2\le \dots d_n$ is graphic
  • iff $d_1,d_2,\dots ,d_{n-d_n}-1,\dots ,d_{n-2}-1,d_{n-1}-1$
  • $B$ is graphic -> $A$ graphic: trivial
• $A$ graphic -> $B$ is graphic
  • vertex $n$: connected to red vertices
    • yet, we want it connected to blue vertices
  • then: find a vertex $w$ connected to desirable vertex
    • i.e. swap neighbors, or 2-switch
  • $w$ can always be found:
    • they never have the same degree
• graphic iff Havel-Hakimi algorithm works

• Tree and unique path

  • $G$: tree iff there is a unique path between every pair od vertices
  • tree: no cycle, $n-1$ edges, connected
  • suppose ${\pi }_1,{\pi }_2$: both $(u,v)$-paths and ${\pi }_1\ne {\pi }_2$
    • implies: closed $(u,v)$-walk -> cycle

graph LR
    u((u))
    v((v))
    int_1(( ))
    int_2(( ))
    int_1-->int_2
    u--π1-->int_1
    int_2--π1-->v
    u--π2-->int_1
    int_2--π2-->v

• $G$: has no cycles
  • suppose
• thus: initial statement can be used as fourth definition of tree

• Tree

  • connected graph $G$: tree iff all edges of $G$: cuts
  • tree -> all edge being cut
    • $e\in E_G$
    • as there is only one, unique $(u,v)$-path, removing $e$ disconnects $u,v$
      • thus $e$ is a cut
  • all edges being cut -> tree
    • no edge appears in a cycle
    • there is no cycle
  • proof 2 (more independent)
    • suppose $C$: cycle in G
    • then an edge in cycle: not

• Connect

  • theorem: every connected graph $G=(V,E)$: w/ subgraph $T=(V,E_T)$
    • s.t. $T$ is a tree
  • $E\ge n-1$
    • if $E=n-1$: already a tree
    • else: eliminate all edges that are not cut, one by one
  • proof 2:
    • choose vertex $v$
    • remove all edges except the shortest path from $v$ to all other vertex
    • properties of tree
      • connected
      • having $n-1$ edges
        • d
      • no cycles either
        • the level (distance from $v$): goes only up and up
        • can't form a cycle