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Graphs 1

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Graphs 1

Kevin Kauffman

CS 309s

- Pie graphs? Bar graphs? WRONG
- Graphs simply show things and their relationship to one another

- Actors and who they were in movies with (oracle of bacon)
- Family members and who is related to whom (family tree)
- Intersections and the roads that connect them
- Board game spaces and which moves you can make

- Node (vertex): the individual thing we are keepting track of (intersection, family member, actor, game space)
- Edge: the relationship between TWO nodes (road, relationship, game move, shared movie)
- A given set of nodes and edges is called a graph

- Directed/Undirected
- Edges may be one way (“go directly to jail” in monopoly) or both ways (two actors in the same movie)

- Weighted/Unweighted
- Edges may have a value which represents some quality (length of a road between two intersections) or every edge can be valued equally (actors in the same movie)

- Cyclic/Acyclic
- Can you start at one node, and follow edges (any one edge at most once) and get back to the start node (a cycle)?
- Undirected, Acyclic graphs are known as TREES

- 2-D array indicating whether there is an edge between two given nodes
- int[][]
- In unweighted graphs, int[i][j]=1 indicates edge, while 0 indicates no edge
- In weighted graphs, int[i][j] indicates length
- 0,-1, or some huge number may indicate no edge

- O(n^2) storage, O(n^2) iterate through all edges….WASTEFUL

- Only store edges which actually exist (instead of all possible edge)
- For each node, store each node to which it is connected
- Unweighted: ArrayList<ArrayList<Integer>>
- Weighted: Arraylist<HashMap<Integer, Integer>>
- O(E) to iterate through all edges (better) but pain in the neck to initialize and use

- Loads of stuff! (and we’ll get to lots of it throughout the semester)
- Calculate distance between nodes
- Figure out which nodes can be reached from others
- Figure out which edges we can remove and still get to all the other nodes
- ?????

- If we have an unweighted graph (can be either directed or not), how can we figure out the minimum number of edges we must traverse to get from some node s to another node t?
- Ideas?

- Start at s and follow every edge there
- At each node you reach, follow every edge there, recording the distance to each node
- At the end, take the minimum distance to the node
- Problem? Takes O(n!) time to compute!

- We need to come up with a way to only visit a given node once
- If we can visit all the nodes 1 away from s first, then we don’t need to visit them again (because we know we can’t beat that!)
- If we can visit all the nodes 2 away from s next, we don’t need to visit THEM again (since we already know they can’t be 1 away)

- Can we find all nodes 1 away from s?
- Yes…all nodes which have an edge going to s must be at most 1 away from s

- Can we find all nodes 2 away from s?
- Yes… all the nodes (which we haven’t visited) which have an edge from one of the step 1 nodes must be 2 away from s

- By induction, we can find the distance to all the nodes

q.add(s)

while(q isn’t empty)

on=q.poll()

for(edge in on) if (other node unvisited)

distace(other node)=distance(on)+1

q.add(other node)

- We visit all the nodes in order of how far they are away
- Since we directly reference the distance to the node as we pop it off the queue, we don’t have to worry about tracking exactly how far we are away
- Instead of updating distance, we can do other things (perhaps add a “previous” value to get the “turn by turn” of what the path actually is)

- Similar to BFS, but visit all nodes in a given sub-graph before entering another sub-graph (rather than visiting all nodes 1 away first)
- Use a stack instead of a queue

- GNYR 2009 I, Theta Puzzle