Definitions, definitions and more definitions all about graphs.

I know what a graph is, but just to review - edges E, and vertices (u,v) in V make up a graph. e connects from u to v.

A graph can be:

• Directed: go from vertx v to vertex u with edge e, but you cannot go from u to v on e
• Undirected: how we typically think of graphs the edge connects the two vertices with no direction.
• Tree: connected graph with no cycle
  • Connected is when there are edges to all vertices. We use the edges to fine a path
  • A certain type of path is a cycle that starts at one vertex and end up back at it by following edges

Different types of graphs:

• Transportation networks - algrotihm to find the best route might be an eventual problem, weight with milage, cost of fuel traffic etc.
• Communication networks - problem: find a place in the room to put your wireless router
• information networks: web etc.
• Social networks - still have bad memories from trying to create facebook in 111
• dependency networks

Theorems:

• every n-node tree has exactly n-1 edges: we will prove this in hw ps 3
• let G be an undirected graph of n nodes: Any two fo the following statements
  • G is connected
  • G does not contain a cycle
  • G has n - 1 edges.
• There is a mistake in the above thm written on pg 78. (on instead of). also can't I just use this theorem to prove the one in our home work set?

Already learned what it means to be connected - now looking at what we can do with that

Breadth First Search

• Looks at one node then all the nodes attached to it then all the nodes attached to each of those new nodes, goes by layers
• layer represents how far it is from root node, (also node in layer 5 is two away from node in layer three etc)
• can find shortest path with BFS
• produces a tree with a root at s
• can be made so has order (n+m) - slides from sprenkle explain this well see day 1/27 or 1/25

Theorem 3.3

• for each j >= 1, Lj produuced by BFS consists of all nodes at exactly j distance from s. There is a path from s to t iff t appears in some layer.

Theorem 3.4

• Let T be a breadth-first search tree, let x and y be nodes in T belonging to layers Li and Lj respectively and let (x,y) be an edge of G. Then i and j differ by at most 1.
• proof in book and in notes

Theorem 3.5

• The set R produced at the end of the algorithm is precisely the connected component of G containing s.
• (goes with comment about how connected sets are either disjoint or the same.)
• proof in book.

Deapth First Search

• better for mazes
• go as far as possible that traverse backwards
• longer tree than the wide tree of BFS
• O(m+n)

Theorem 3.6

• More like a lemma…
• for a given recusive call DFS(u), all nodes that are marked Explored between the invocation and end of this recursive call are descendants of u in T.
• use this to prove 3.7

Theorem 3.7

• Let T be a dfs tree, let x and y be nodes in T, and let (x,y) be an edge of G that is not an edge of T. Then one of x or y is an ancestor of the other.

Theorem 3.8

• For any nodes s and t in a graph, their connected components are either identical or disjoint.
• mention this above under theorem 3.5
• proof in book

Representing Graphs

• Adjacency lists, less space and better for sparse graphs O(n+m), possible order n to search for edges.
• adjacency matrix more space O(n^2), constant time access, but if we have to traverse anyways mine as well use adjacency lists?

Queues and Stacks

• not much on these actually in the book in this part, but we have plenty of notes on the powerpoints.

Implementing BFS

• Use an adjacency list structure,
• O(m+n)
• see our notes
• also proof in the book on p. 91, proving the bound time. This could be helpful for future similar problems
• The book references using a queue in class it didn't matter, but now i'm slightly confused…

Implementing DFS

• Need to use a stack. (Last in first off)
• O(m+n)

A little confused about the finding the set of all connected components section, but I think when we talked about htis in class it made sense so I am not going to worry too much.

Bipartite graph - can split nodes into two groups, nodes in group are not connected to each other, only connected to other group. USE BFS to implement. (each layer is a different color, our implementation in class was super helpful.

Theorem 3.14

• I was not going to copy this theorem down, because it is so close to 3.15, but it is the beginning of pi and I cannot resist that.
• If a graph g is bipartite then it cannot contain an odd cycle.

Theorem 3.15

• Let G be a connected graph, and let L1, L2,… be the layers produced by BFS starting at node S. then exactly one of the following two things must be true:
  • There is no edge of G joining two nodes of the same layer. In this case G is a bipartite graph in which the nodes in even numbered layers can be colored red and the notes in odd numbered layers can be colored blue.
  • There is an edge of G joining two nodes of the same layer. It this cane G contains an odd layer cycle and so it cannon be bipartite.

Representatoin - variation of adjacency list from above. Node needs a list to what it points to and what points to it.

Graph search algorithms

• BFS and DFS are very similar to what they are in an undirected graph

Theorem 3.16:

• If u and v are mutually reachable and v and w are mutually reachable then u and w anre mutually reachable.
• similar to the whole connected are the same or disjoint.
• proof in book

Theorem 3.17

• For any two nodes s and t in a directed graph, their strong components are either identical or disjoint.
• proof in book

Directed graph with no cycles - DAG, directed acyclic graph!

• seen in dependency networks
• for example tasks in order, represent each task by a node.
• topological ordering of G is an ordering of its nodes as v1, v2, …. so that for every edge (vi,vj) we have i < j therefore all edges point forward.
• This will be useful for problm set 4

Theorem 3.18

• if G has topological ordering then G is a dag
• see proof in book.
• is this theorem if and only if? need to find out. TWO sentences later we do find out. The converse is true.

theorem 3.20

• see proof in book.
• uses a lemma 3.19

I'm lost a little towards the end of 3.6, but I'm thinking once we get to it in class it will all be clear!

Update: Class did make this easier to understand. I actually remember using djikstra's alogirthm in a math class before!