Proof Flashcards
in any graph, nb of odd dergee vertices is even
Handshake lemma
simple graph G contains no odd length closed walk => G bipartite
- suppose G connected
- AUB = V
-AinterB = vide
-e = uw
G graph, e cut-edge; no cycle contains e
contradiction
G simple graph, deg(v)>=2 for all v€V, G contains a cycle
P maximum length path
Cayley’s Theorem
p: T->C prüfer code functIon
show bijectivity
G graph n vertices k edges, comp(G)>=n-k
induction on k
G connected on n vertices, |E(G)|>=n-1
G a tree, |V(G)| = |E(G)|+1
every closed walk of odd length contains an odd cycle
exists repetition
Euler theo
=>) contradiction
<=) induction on |E(G)|
element u,v de A^k = nb of walks from u to v of k length
statement : sum of walks of length k-1 form u to w for all w€N(v) = sum of walks of length k from u to v
M maximal, M’ maximum |M|>= |M’|/2
Hall’s Mariage theorem
=>) f: S -> N(S) matching function
prove injectivity
<=) induction on|A|
D digraph, deg(v)>=1 for all v€E => contains a cycle
maximum path
every tournament has an Hamiltonian path
induction on |V(T)|
if H1 and H2 2 strongly connected components then H1 and H2 do not share any vertex
contradiction