CS70MT1

a graph G is a tree if it is...

- connected and acyclic - connected and |vertices| = |edges| + 1 - connected such that deleting any one edge disconnects it - connected with no cycles, and the addition of any edge creates a cycle

A planar graph can be colored with ___ colors

4

Cycle (or circuit)

A sequence of edges that starts and ends at the same vertex - an extension of a path, so we assume no repeat vertices in between the first/last one

for any planar graph, e <= 3v - 6. why?

Assuming no parallel edges, every face has at least 3 sides (so si = 3 for all i in the sum of faces). Since the sum of faces = 23, we can say 3f = 2e. Solve for f and plug into Euler's formula for 3v - 6

definition of a connected directed graph

For every pair of vertices U and V, both (u,v) and (v,u) are in the edge set

An undirected graph G has a Eulerian tour iff

G is even degree and connected. Touchpoints of the proof: - assuming a Eulerian tour, we prove connected: every vertex attached to an edge must be on the tour, and therefore be connected to all other connections (remember a Eulerian tour can still exist with isolated vertices). - assuming Eulerian tour, we prove even degree: in order to visit every edge exactly once, we must always leave a vertex via a different edge than we entered from - thus each vertex needs 2e edges attached to it (this even works for the start vertex, because we must return to it via a different edge than we departed from) - assuming even degree and connected, we prove eulerian tour: every time we traverse a new edge, we will have an odd number of unvisited options for exiting it (except the start edge). Since we assumed even degree, this means we will have a 'way out' until we get to the first vertex. Thus, a tour without repeat edges is guaranteed. All of these tours, assuming shared vertices, can be spliced together - find a tour, remove all the edges associated with it. find another tour, recursively. The edges will eventually all deplete, and we know this because even degree. inductive proof.

Euler's formula

IN PLANAR GRAPHS, v + f = e + 2; proof by induction - bc: e = 0, v = f = 1 cases: - tree: f = 1, e = v - 1 - not a tree: delete any edge to decrease e and f by 1. This makes the formula true for the smaller graph, so it was true for the larger one (remember this assumes the graph is planar)

K3,3 passes the e <= 3v - 6 planar equation but is not planar. How do we prove its nonplanarity

K3,3 has no triangles because this necessitates the connection of two vertices in the same group. With no three sided faces, the minimum number of sides per face becomes 4, meaning the number of sides as a whole is >= 4f. Plugging into the side equation, 4f = 2e must hold for a planar graph. This means f = 1/2 e. Plugging into euler's formula provides v + 1/2 e = e + 2, e <= 2v - 4, which is an equation that does not hold for K3,3 (9 <= 12 - 6)

Graph theory: U x V (when U and V are sets) denotes..

The cartesian product of U and V, meaning all tuples (u,v) for which the first element is in U and the second element is in V

For any planar graph, e <= 3v - 6. What planar graph properties does this provide

This bounds the number of edges relative to the number of vertices: planar graphs are sparse, and adding edges to a graph with a fixed number of vertices can potentially take it from planar to nonplanar. important: this can be used to prove a graph is nonplanar, but not to prove a graph is planar (this is NOT an iff scenario). Counterexample: K3,3 passes this equation (9 <= 3(6) - 6) but is not planar

Union vs intersect

Union (U): all values in at least one of the two sets Intersect (upside down U): all values in both sets

Eulerian tour

a Eulerian walk that starts and ends at the same point - every edge visited exactly once - intermediate vertices ok - a eulerian tour CAN be a cycle but need not be

a graph is connected if ..

a path exists between any two distinct vertices

Walk

a sequence of edges that can repeat vertices/edges (paths are walks, but a walk might not be a path) - repeat edges are ok unless it's a Eulerian walk

Tour

a sequence of edges that start and end at the same vertex and can repeat intermediate ones (walk is to tour as path is to cycle) - all cycles are tours, but a tour need not be a cycle

Eulerian walk

a walk that uses each edge exactly once - note that unlike paths, walks can repeat vertices. Only the EDGES must be visited once - in a normal walk, repeat edges are ok

Sides formula for planar graphs

assuming a graph is planar, the sum of the number of sides on every face = 2e (each edge is counted twice if it borders 2 faces. it cannot border more than 2 faces)

Any graph that does not contain any odd length cycles is

bipartite - pick a vertex and make it one color, color all its neighbors the other color, and we can 2-color it in this way

planar graph

can be drawn in 2d without crossing edges - some planar graphs have representations that involve crossings. all planar means is that there exists at least one drawing of the graph with no crossings.

what does it mean for a graph to be dual to another

given a graph G, put a node on any face, and then draw an edge between faces of G that have an edge of G between them. The result is also a planar graph - can make problems of coloring edges into problems of coloring vertices

bipartite graph

graph where vertices are split into two groups, and edges only go between the two groups, not within one group. Formally, the vertices are the union (both sets) of set L (1) and R(2), and the edge set is a subset of L x R: edges such that the start vertex is in group 1 and the end vertex is in group 2

Path

in an undirected graph, a sequence of edges - assume all vertices visited are unique - the assumption of unique vertices gives way to the notion of unique edges

5 color thm proof

induction on the number of vertices - given a planar graph, a node of degree 5 or less must exist (by e <= 3v - 6) - if degree 4 or less, done - if degree 5, all neighbors are different colors - review this

depth of a tree

maximum distance from root to leaf (root's distance to self is zero, so this is essentially len(longest path) - 1

complete graphs

maximum number of edges, denoted by K(n)

How many edges are there in a complete graph Kn

n(n-1) / 2

faces of planar graphs

the number of faces can be specific to a drawing. All graphs have one infinite face, and all others are finite

Set notation: A \ B denotes

the set of all values in set A but not in set B