quartz/content/notes/graphs.md

title	aliases	tags
graphs		cosc201 datastructure

Represents a set of things with relationships between them.

• a set of vertices
• some edges between them
• edges on some graphs have weights
• edges on some graphs are directed

Some graphs are named e.g,

Implementation

should:

• be static
• say if two vertices are connected in O(1)
• say what the neighbors of a vertice are in O(1)

Hashset

array wwith element to each verttex, and each element of the array is a TreeSet of its neighbors.

graph is built up by specifying size, then edges.

add edge(int v, int w){
	neighbors[v].add(w);
	neighbors[w].add(v);
}

we could also define a vertex class and use a HashSet\<vertex>

There are many options

Traversals

Breadth first traversal

Stating from v, initialise an empty queue a, and add a marker to each vertex (or make a set of seen vertices).

1. add the start vertex to a queue and mark it as seen
2. while the queue is not empty
3. visit the first ellemnet of a qeue removing ti
4. do someting with it (line 3.5)
5. unseen neighbors neighbors are added to the queue marking them as seen
6. repeat

Depth first

init empty stack, s, and add a marker to each vertex (or make a set of seen vertices)

1. add the start vertex to a stack and mark it as seen
2. while the stack is not empty
3. visit the first ellemnet of a qeue removing ti
4. do someting with it (line 3.5)
5. unseen neighbors neighbors are added to the stack marking them as seen
6. repeat

Cost of traversal

we measure complexity both in terms of veritces V and edges E.

for each vertex

• it occurs in some neghbor list ofot he first time at which point it is added to the list, and marked as seen
• then at soome point it is removed
• so O(1) per vertice so O(V)

for each edge:

• edge arise implicityly in the neihbor list
• each one appreas twice, once per endpoint
• when one appreas is causes a constant amount of work
• also O(1) per edge so O(E)

so O(V + E)