1. - 1-1/injective
2. - onto/surjective
Suppose are finite sets and
(i) If is 1-1 $$\Rightarrow
is a set with weight function
where are rational numbers.
- Coefficient of
,
Addition:
Equality:
Multiplication:
are fps satisfying
- Inverse
Let be fps.
If - Composition
- Negative Binomial Theorem
Let .
has inverse
Let be fps.
If there exists fps where
Let be fps with . Let .
Let be fps s.t. and
Let
- Sum lemma
Let be sets with weight functions:
:
- Product Lemma
Geometric Series:
-
Composition of Geometric Series:
A composition of n is a tuple of positive integers
s.t.
is the number of parts of the composition.
Let be sets of binary strings.
is ambiguous if:
There exist and s.t.
is unambiguous if
Expression with several concatenation/union operations is unambiguous if 1 of the concatenations or unions is.
is ambiguous if there are distinct k-tuples and the concatenations are equal.
is unambiguous
1.
2. is unambiguous for each
Unambiguous Expressions for set of all binary strings:
1.
2.
3.
Theorem 2.6.1:
are sets of unambiguous binary strings.
(i)
(ii)
(iii)
are polynomials where and
where
Partial Fractions Theorem:
Theorem 3.1.4:
There exist polynomials s.t.
Theorem 3.2.1:
Let satisfy the linear recurrence:
Define ,
There exists polynomial s.t.
for recurrence:
Theorem:
Roots:
Multiplicities:
Then there exists polynomials s.t. and
1. A graph is isomorphic to itself (reflexive)
2. If (symmetric)
3. If and (transitive)
Every pair of vertices is an edge.
# of edges =
Graph where every vertex has degree .
# of edges =
Graph where there exists partition of vertices s.t. each edge
of G joins 1 vertex of with a vertex in .
Complete Bipartite Graphs ()
If there is a -walk in there is a -path in .
If there exists -path and -path there exists -path.
If every vertex in has has a cycle.
Length of graph's shortest cycle.
If has no cycles girth =
- Measure of graph density
- High girth = lower average degree (usually)
Connected if there is a -path for any pair of vertices .
Let
If -path exists for each is connected.
Let .
Cut induced by x in is the set of all edges in with exactly 1 end in x.
is disconnected
There exists a nonempty proper subset of
s.t. the cut induced by is .
Closed walk that uses every edge of exactly once.
- Connected unless it has isolated vertices
is connected.
has an Eulerian circuit every vertex in has even degree.
An edge (or cut-edge) in if has more components than .
If is a bridge in and is the component containing
has exactly 2 components
(hence has 1 more component than $G$)
Moreover, and are in different components of .
(hence in )
An edge is a bridge of is not in any cyce of .
Connected graph with no cycles
# of edges = by Theorem 5.1.5
- Bipartite by Lemma 5.1.4
Graph with no cycles.
# of edges = , with vertices and components
Every tree with vertices has leaves.
is connected has a spanning tree.
If is connected with vertices and edges is a tree.
If any of 2 the following 3 conditions hold is a tree:
1. is connected
2. has no cycles.
3. has edges.
If is a spanning tree of and is an edge in that is not in
has exactly 1 cycle.
Moreover, if is an edge in
is a spanning tree of .
If is a spanning tree of and is an edge in
has 2 components ( is a bridge in T).
If is an edge in the cut induced by the vertices of 1 component
is a spanning tree of .
is bipartite has no odd cycles.
Given connected graph and weight function on edges
, find minimum spanning tree in
whose total edge weight in minimized.
If 2 vertices are adjacent in The same 2 vertices are not adjacent in .
A graph is planar Each component is planar
A face of a planar embedding is a connected region on the plane.
2 faces are adjacent $\iff$ The faces share 1 edge in their boundaries.
Let be a planar graph with planar embedding where is the set of all faces.
$$\sum\limits_{f \in F} deg(f)=2
In a planar embedding, an edge is a bridge
The 2 sides of are in the same face
Let be a connected planar graph.
Let = # of vertices in .
Let = # of edges in .
Let = # of faces in a planar embedding of .
Note: If has components
Result: All planar embeddings of a graph have the same number of faces.
A connected planar graph is platonic if it has a planar embedding
where every vertex has same degree () AND
every face has same degree ()
Let be a planar graph with vertices and edges.
If there is a planar embedding of where every face has degree
If contains a cycle
In any planar embedding of , every face boundary contains a cycle.
Let be a planar graph with vertices
Let be a bipartite planar graph with 3 vertices and edges.
Edge subdivision of is obtained by replacing each edge of
with a new path of length
A graph is planar
Graph does not have an edge subdivision of or as a subgraph.
If is a set of size ("colours")
s.t.
Note: A k-colouring does not need to use all colours.
is 2-colourable is bipartite
(complete graph) is n-colourable, and not k-colourable for any
Every planar graph has a vertex of degree
Dual Graph Results L7-8()
Key: Colouring faces is equivalent to colouring vertices of dual graph.
The dual graph is planar.
1. If is connected
2. A vertex in corresponds to a face in of the same degree.
3. A face in corresponds to a vertex in of the same degree.
4. The dual of a platonic graph is platonic.
5. If we draw a closed curve on the plane the faces are 2-colourable.
If is any matching of and is any cover in
$$\Rightarrow
If is a matching of and is a cover of where $$
Let be a matching.
Let be an augmenting path.
Let be a new matching from P.
Notes:
- P is always odd length (L8-2 for proof)
- P always starts and ends in different parts of the bipartition.
If a matching has an augmenting path is not a maximum.
A bipartite graph with edges and maximum degree
has a matching of size
Let .
Neighbour set of is set of all vertices adjacent to at least 1 vertex in .
A bipartite graph with bipartition has a matching that saturates all vertices in A
$$\iff \space \forall \space D \subseteq A,
If is a -regular bipartite graph with
has a perfect matching.
The edges of a -regular bipartite graph can be partitioned into perfect matchings.
{1,1,3}={1,3,1} {1,1,3}={1,3,1} f(x1)=f(x2)⇒x1=x2 ∀y∈T,∃x∈S,f(x)=y f:S→T w:S→{0,1,2,...} Φs(x)=σ∈S∑xw(s) Φs(x)=k≥0∑akxk C(x)=c0+c1x+c2x2+...+k≥0∑∞ckxk (c0,c1,...) [xn]C(x)=cn A(x)=k=0∑∞akxk B(x)=k=0∑∞bkxk A(x)+B(x)=k=0∑∞(ak+bk)xk A(x)=B(x)⟺ak=bk A(x)B(x)=n=0∑∞(j=0∑najbn−j)xn A(x),B(x) A(x)B(x)=1 ightarrow(1−x)k1=k=0∑∞xk ightarrow[x0]A(x)B(x)=1 A(x),B(x) b0=0⇒A(B(x))=j=0∑∞aj(B(x))j (1−x)k1=n=0∑∞(k−1n+k−1)xn A(x)=j=0∑∞ajxj [x0]A(x)=a0=0 A(x),C(x) [x0]A(x)=0⇒ A(x)B(x)=C(x) A(x),C(x) B(x)=A(x)C(x) ightarrow[xn]B(X)=bn=a01(cn−j=0∑n−1an−jbj) A(x),B(x) [x0]A(x)=0 [x0]B(x)=0 ightarrow(A(B(x)))−1=A−1(B(x)) S=A∪B,A∩B=∅,w:S→{0,1,...} ΦS(x)=ΦA(x)+ΦB(x) α:A→{0,1,...} β:B→{0,1,...} w:A×B→{0,1,...} w((a,b))=α(a)+β(b) ightarrowΦA×B(x)=ΦA(x)ΦB(x) j=0∑∞xj=1−x1 [x0]A(x)=0⇒j=0∑∞(A(x))j=1−A(x)1 (a1,a2,...,ak) a1+a2+...+ak=n a1,a2∈A b1,b2∈B a1b1=a2b2 A∩B=∅ Ak∩Aj=∅ S={0,1}∗ S={0}∗({1}{0}∗)∗ S={0}∗({1}{1}∗{0}{0}∗)∗{1}∗ ΦAB(x)=ΦA(x)ΦB(x) ΦA∪B(x)=ΦA(x)+ΦB(x) ΦA∗(x)=1−ΦA(x)1 {cn}n≥0 cn+q1cn−1+...+qkcn−k=0 g(x)=1+q1x+...+qkxk C(x)=n=0∑∞cnxn ightarrow deg(f)<k C(x)=g(x)f(x) G1≅G2⇒G2≅G1 G1≅G2 G2≅G3⇒G1≅G3 ightarrow deg≥2⇒G ightarrow v∈V(G)⇒G x⊆V(G) ightarrow ightarrowG ightarrowG ightarrow ightarrow ightarrowT−e ightarrow w:E(G)→ ightarrown−m+s=2 ightarrow n−m+s=1+C ightarrowm≤d−2d(n−2) ightarrow ightarrowm≤3n−6 ightarrowm≤2n−4 ightarrowf:V(G)→C f(u)=f(v) ∀uv∈E(G) ightarrow ightarrow(G∗)∗=G ightarrow ightarrowM′=[M∪(E(P)∖M)] ∖(E(P)∩M) ightarrow Y∪(A∖X) D⊆V(G) ightarrow deg(f)<deg(g) g(x)=(1−r1x)e1...(1−rkx)ek r_1, ..., r_k \in \C, e_1, .., e_k \in \Z^+
ightarrowg(x)f(x)=(1−r1x)c1,1+... +(1−r1x)e1c1,e1+(1−r2x)c2,1+... +(1−rx)e2c2,e2+...+(1−rkx)ck,1+...+(1−rkx)ekck,ek P1,...,Pk deg(Pi)<ei ightarrow[xn]g(x)f(x)=P1(n)r1n+P2(n)r2n+...+Pk(n)rkn h(x)=qk+qk−1x+...+q1xk−1+xk cn+q1cn−1+...+qkcn−k=0 r_1, r_2, ..., r_L \in \C
e1,e2,...,eL≥1 ightarrow P1,...,PL deg(fj)=ej−1 cn=P1(n)r1n+...+PL(n)rLn