Math 239 Formula Theorems
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Enumeration
Sets
Cartesian Product
Unordered: Ordered: $$
A \cup B
Binomial Coefficient Sets
Binomial Theorem
$$S_{k,n}={\Omega \subseteq {1, 2, ..., n}:
\Omega
Bijection
Bijection Theorems
1. - 1-1/injective 2. - onto/surjective
Suppose are finite sets and (i) If is 1-1 $$\Rightarrow
Generating Series
Generating Series Theorems
is a set with weight function
$$
Formal Power Series
FPS Operations
where are rational numbers. - Coefficient of
, Addition: Equality: Multiplication:
Inverse/Composition of FPS (not all FPS have inverses)
Inverse/Recurrence Theorems
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
Sum and Product Lemmas for Generating Series
Geometric Series
Let - Sum lemma Let be sets with weight functions: : - Product Lemma
Geometric Series: - Composition of Geometric Series:
Integer Composition
Ambiguous/Unambiguous Binary Strings
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.
Unambiguous/Ambiguous Binary Strings Expressions
Unambiguous Binary String Operations
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)
Partial Fractions Theorems
Homogeneous Linear Recurrences
are polynomials where and where r_1, ..., r_k \in \C, e_1, .., e_k \in \Z^+ 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.
Characteristic Polynomial
for recurrence: Theorem: Roots: r_1, r_2, ..., r_L \in \C Multiplicities: Then there exists polynomials s.t. and
Graph Theory
$$\sum\limits_{v \in v(G)} deg(v)=2
E(G)
Corollary 4.3.3 (Average Vertex Degree)
Isomorphic
Avg vertex degree = $$\frac{2
E(G)
Isomorphic Equivalence Relation
Complete Graphs ()
1. A graph is isomorphic to itself (reflexive) 2. If (symmetric) 3. If and (transitive)
Every pair of vertices is an edge. # of edges =
Regular Graphs (k-regular)
Bipartite Graphs
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 ()
N-cube
Graph with partition (A,B) where all possible edges joining vertex in A with vertex in B. $$
A
Theorem 4.5.2 (Walk, Path)
Corollary 4.6.3 (Path Transitive)
If there is a -walk in there is a -path in .
If there exists -path and -path there exists -path.
Theorem 4.6.4 (Cycle, degree)
Girth
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)
Spanning Cycle/Hamiltonian Cycle
Connectedness
Cycle that uses every vertex in the graph
Connected if there is a -path for any pair of vertices .
Theorem 4.8.2 (Connectedness)
Cuts
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.
Theorem 4.8.5 (Disconnectedness)
Eulerian Circuit/Tour
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
Theorem 4.9.2 (Eulerian Circuit Vertices)
Bridge
is connected. has an Eulerian circuit every vertex in has even degree.
An edge (or cut-edge) in if has more components than .
Lemma 4.10.2 (Bridge Component)
Theorem 4.10.3 (Bridge, Cycle)
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 .
Tree
Forest
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
Lemma 5.1.4 (Tree, Bridge)
Leaf
Every edge in a tree/forest is a bridge.
Tree with vertex degree 1.
Theorem 5.1.8 (Tree Leaves)
Lemma 5.1.3 (Unique Path Tree)
Every tree with vertices has leaves.
There is a unique path between any 2 vertices in a tree.
Theorem 5.2.1 (Connectedness, Spanning tree)
Corollary 5.2.2 (Connected, tree)
is connected has a spanning tree.
If is connected with vertices and edges is a tree.
Corollary (Tree-Graph equivalence)
Corollary 5.2.3 (Spanning Tree, Cycle)
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 .
Corollary 5.2.4 (Spanning Tree, Component)
Bipartite Characterization Theorem
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.
Minimum Spanning Tree (MST) Problem
Prim's Algorithm Theorem
Given connected graph and weight function on edges , find minimum spanning tree in whose total edge weight in minimized.
Prim's algorithm produces a MST.
Complement
Planar
If 2 vertices are adjacent in The same 2 vertices are not adjacent in .
A graph is planar Each component is planar
Face
Boundary
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.
Subgraph of all vertices and edges that touch a face.
Boundary Walk
Handshaking Lemma for Faces
Closed walk once around the perimeter of the face boundary. Degree of face = length of boundary walk
Let be a planar graph with planar embedding where is the set of all faces. $$\sum\limits_{f \in F} deg(f)=2
Lemma L7-1
Jordan Curve Theorem
In a planar embedding, an edge is a bridge The 2 sides of are in the same face
Every planar embedding of a cycle separates the plane into 2 parts: 1 on the inside, 1 on the outside
Euler's Formula
Platonic
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 ()
Lemma 7.5.2
Lemma 7.5.1
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.
Theorem 7.5.3
Corollary 7.5.4
Let be a planar graph with vertices
is not planar.
Theorem 7.5.6
Corollary 7.5.7
Let be a bipartite planar graph with 3 vertices and edges.
is not planar.
Edge Subdivision
Kuratowski's Theorem
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.
K-colouring
Theorem 7.7.2
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
Theorem 7.7.3
6-Colour Theorem
(complete graph) is n-colourable, and not k-colourable for any
Every planar graph is 6-colourable.
Corollary 7.5.4
5-Colour Theorem
Every planar graph has a vertex of degree
Every planar graph is 5-colourable.
Contraction Result L7-7
4-Colour Theorem
If is planar is planar.
Every planar graph is 4-colourable.
Dual Graph Results L7-8()
Lemma 8.2.1
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 $$\Rightarrow
Lemma 8.2.2
Konig's Theorem
If is a matching of and is a cover of where $$
M
Augmenting Path Results L8-2
Lemma 8.1.1
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.
Bipartite Matching Algorithm
Corollary L8-4
Minimum cover =
A bipartite graph with edges and maximum degree has a matching of size
Neighbour Set ( or )
Hall's Theorem
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,
Corollary 8.6.2
Corollary L8-6
If is a -regular bipartite graph with has a perfect matching.
The edges of a -regular bipartite graph can be partitioned into perfect matchings.
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