MST (min spanning tree)

1
MST (min spanning tree)
2
Minimum Spanning Trees

• In a weighted, undirected graph, it is a tree
  formed by connecting all of the vertices with
  minimal cost.
• The MST is a tree because its acyclic.
• Its spanning because it covers every vertex.
• And its minimum because it has minimum cost.

2
v1
v2
4
3
1
10
2
7
v4
v3
v5
4
8
6
5
v7
v6
1
2
v1
v2
1
2
v4
v3
v5
4
6
v7
v6
1
3
MSTs proof that greedy works

• Let G be a graph with vertices in the set V
  partitioned into two sets V1 and V2. Then the
  minimum weight edge, e, that connects a vertex
  from V1 to V2 is part of a minimum spanning tree
  of G.
• Proof Consider a MST T of G that does NOT
  contain the minimum weight edge e.
• This MUST have at least one edge in between a
  vertex from V1 to V2. (Otherwise, no vertices
  between those two sets would be connected.)

• Let G contain edge f that connects V1 to V2.
• Now, add in edge e to T.
• This creates a cycle. In particular, there was
  already one path from every vertex in V1 to V2
  and with the addition of e, there are two.
• Thus, we can form a cycle involving both e and f.
  Now, imagine removing f from this cycle.
• This new graph, T' is also a spanning tree, but
  it's total weight is less than or equal to T
  because we replaced e with f, and e was the
  minimum weight edge.

4
MSTs proof that greedy works

• As a spanning tree is created
• If the edge that is added is the one of minimum
  cost that avoids creation of a cycle.
• Then the cost of the resulting spanning tree
  cannot be improved
• Because any replacement edge would have cost at
  least as much as an edge already in the spanning
  tree.
• This is why greedy works!

5
Kruskals Algorithm

• Let V ?
• For i1 to n-1, (where there are n vertices in a
  graph)
• V V ? e, where e is the edge with the minimum
  edge
• weight not already in V, and that does
  NOT
• form a cycle when added to V.
• Return V
• Basically, you build the MST of the graph by
  continually adding in the smallest weighted edge
  into the MST that doesn't form a cycle.
• When you are done, you'll have an MST.
• You HAVE to make sure you never add an edge the
  forms a cycle and that you always add the minimum
  of ALL the edges left that don't.

6
Kruskals
Given Graph G
Edges in sorted order
2
v1
v2
4
3
1
10
v4
v1
4
2
7
v4
v3
v5
1
v7
4
8
v4
v3
6
5
v7
v6
1
v7
v6
5
1
v6
v5
Determine the MST
2
v1
v2
6
v7
2
v1
v2
4
2
3
1
v4
v3
7
v4
v5
2
CYCLE!
v4
v3
v2
v5
v4
3
4
8
6
v4
v7
v6
v6
v1
v2
1
4
10
All Vertices, were done!!
v3
v5
7
Kruskals

• The reason this works
• is that each added edge is connecting between two
  sets of vertices,
• and since we select the edges in order by weight,
  
• we are always selecting the minimum edge weight
  that connects the two sets of vertices.

• Cycle detection
• Keep track of disjoint sets.
• Initially, each vertex is in its own disjoint
  set.
• When you add an edge you are unioning two sets.
• A union cannot happen if the two vertices are
  already in the same set.
• This would create a cycle.

8
Prims Algorithm

• This is quite similar to Kruskal's with one big
  difference
• The tree that you are "growing" ALWAYS stays
  connected.
• Whereas in Kruskal's you could add an edge to
  your growing tree that wasn't connected to the
  rest of it, here you can NOT do it.

• Here is the algorithm
• 1) Set S ?.
• 1) Pick any vertex in the graph.
• 2) Add the minimum edge incident to that vertex
  to S.
• 3) Continue to add edges into S (n-2 more times)
  using the
• following rule
•  
• Add the minimum edge weight to S that is
  incident to S
• but that doesn't form a cycle when added to
  S.

9
Prims
Given Graph G
2
v1
v2
Edges in sorted order
4
3
1
10
2
7
v4
v4
v3
v5
v1
4
4
8
1
6
5
v7
v7
v6
v4
v3
1
v7
v6
5
Determine the MST, using Prims starting with
vertexV1
1
v6
v5
2
v1
v2
6
v7
2
v1
v2
4
2
3
v4
v3
7
1
v4
v5
2
CYCLE!
v4
v3
v2
v5
v4
3
4
8
6
v4
v6
v7
v6
v1
v2
1
4
10
All Vertices, were done!!
v3
v5
10
Huffman Encoding

• Compress the storage of data using variable
  length codes.
• For example, each character in a text file is
  stored using 8 bits.
• Nice and easy because we always read in 8 bits
  for a single character.

• Not the most efficient
• What if e is used 10 times more frequently than
  q.
• It would be more advantageous for us to use a 7
  bit code for e and a 9 bit code for q.

11
Huffman Coding

• Finds the optimal way to take advantage of
  varying character frequencies in a particular
  file.
• On average, standard files can shrink them
  anywhere from 10 to 30 depending on the
  character distribution.

• The idea behind the coding is to give less
  frequent characters and groups of characters
  longer codes.
• Also, the coding is constructed in such a way
  that no two constructed codes are prefixes of
  each other.
• This property about the code is crucial with
  respect to easily deciphering the code.

12
References

• Slides adapted from Arup Guhas Computer Science
  II Lecture notes http//www.cs.ucf.edu/dmarino/
  ucf/cop3503/lectures/
• Additional material from the textbook
• Data Structures and Algorithm Analysis in Java
  (Second Edition) by Mark Allen Weiss
• Additional images
• www.wikipedia.com
• xkcd.com