• Leaves

  • recall: tree = connected graph w/ $n-1$ edges
  • leaf: a vertex of degree 1
  • theorem: every tree on $n\ge 2$ vertices: w/ at least 2 leaves
  • proof: based on maximum path
    • let $\pi =v_0{v}_1\cdots v_k$: longest path
    • claim: $v_0,v_k$: leaves
    • proof by contradiction: $v_0\&v_k$: not a leaf
      • $v_0,v_k$: cannot be connected to vertex in the same path
        • as it won't be a cycle
      • also: it cannot be connect to another vertex
        • as it is against statement: $v_0,v_k$ is the longest path
  • w/ $V=[n]$, how many graphs $G=(V,E)$ exist?
    • $2^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$
    • 2 choices (yes / no) for total of $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possibilities
    • or: largest set of size $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$, and all (distinct) subset of it
  • how many of above are trees?
    • solution: $n^{n-2}$
    • translate all trees into seq. of length $n-2$
    • translation
      • for each leaf (or: w/ min. neighbors) of tree: write its neighbor, and erase the leaf
      • repeat until only one edge remains
      • e.g.

graph LR
        1((1))
        2((2))
        3((3))
        4((4))
        5((5))
        6((6))
        7((7))
        3-->1
        3-->5
        1-->4
        5-->6
        5-->2
        5-->7

    - gets translated into `5 1 3 5 5`, vertex <span class="katex"><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8389em;vertical-align:-0.1944em;"></span><span class="mord">5</span><span class="mpunct">,</span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord">7</span></span></span></span> remaining

• recovery
  • if a node is originally a leaf: it won't be on a list
    • i.e. 2, 4, 6
    • choose a remaining one, and connect with existing node

graph LR
        5((5))
        7((7))
        7-->5

        6_2((6))
        5_2((5))
        7_2((7))
        7_2-->5_2
        6_2-->5_2

        6_3((6))
        5_3((5))
        7_3((7))
        4_3((4))
        7_3-->5_3
        6_3-->5_3
        4_3-->5_3

        6_4((6))
        5_4((5))
        7_4((7))
        4_4((4))
        2_4((2))
        3_4((3))
        7_4-->5_4
        6_4-->5_4
        4_4-->5_4
        2_4-->3_4

        6_5((6))
        5_5((5))
        7_5((7))
        4_5((4))
        2_5((2))
        3_5((3))
        1_5((1))
        7_5-->5_5
        6_5-->5_5
        4_5-->5_5
        2_5-->3_5
        3_5-->1_5

- forest:  graph consisted of forest
  - HW: show how many forests are in all possible cases

• theorem: every connected graph w/ $n\ge 2$ vertices: w/ at least $2$ non-cut vertices
  • proof: w/ longest path
  • $v_0,\dots v_k$
  • even if $v_0$ are to be cut: its neighbors can still more
  • same for $v_k$
• theorem: every odd closed walk: contains an odd cycle
  • $w=v_0{v}_2{v}_2\cdots v_k$
  • $v_0=v_k$
    • if it is a cycle: problem solution
    • or else: $\exists i,ji\ne k,v_i=v_j$
  • graph: gets divided into two smaller cycle, around $v_i,v_j$
    • and choose an odd graph, and work on it
    • as it is only subgraph of original graph, its size must
• Bipartite graph: $G=(V,E)$
  • $V=V_1\bigsqcup V_2$ (disjoint)
  • s.t. every edge: has endpoint in $V_1$, another in $V_2$
• theorem: $G$ is bipartite iff it has no odd cycles
• proof
  • bipartite -> no odd cycles
    • start in $v_1$
    • taking 1 edge: goes to opposite side
    • taking odd no. of edge: goes to opposite side from starting side
    • i.e. cannot form a cycle, as $V_1\bigsqcup V_2$
    • no cycles!
  • no odd cycles -> bipartite
    • put a vertex $x$ in $V_1$
    • add all its neighbor in $V_2$, opposite side
    • add all their neighbor in $V_3$, opposite side
    • ... repeat
    • there can't be an edge within the same party
      • as that would create an odd cycle
• theorem: assume a $d(i)\ge 2\forall i\in \forall$, then $G$ has a cycle
• proof
  • let $v_0,\dots ,v_k$: a longest path
  • $v_0,v_k$: must be connected to somewhere
  • i.e. within the path -> creating a cycle
  • or: outside the path -> against the claim that $v_0,v_k$ is the longest path
• theorem: in finite $G$, $\forall i\in V,d(i)\le 2$
  • every cc of $G$: either a cycle or path
  • pick an arbitrary $v\in V$
  • let $v,\dots ,v_k$ be entire graph
  • for $v_k$ to create a cycle
    • cannot connect to vertex in the middle: $d(v_i)\ge 3$
    • cannot connect to $v$: it's a cycle
  • if you stuck in a dead end: iterate through the opposite direction ($v_k\to v$)
    • as the graph is finite, we can reach the end
    • and apply the same rule described above
  • 👨‍🏫 continue until this process covers your entire graph
  • 👨‍🎓 a non-cycle non-path graph: must have at least 3 neighbors