### Lecture 10 - Dijkstra's path reconstruction, Minimum Spanning Trees, Huffman Encoding

```Dijkstra’s Algorithm
Announcements

Assignment #2 Due Tonight


Assignment #3 Posted
Announcements
Last Time

Breadth First Search, Depth First Search
◦ Different algorithms for searching all nodes in
a graph.

Topological Sort
◦ Produces an ordering of vertices in a graph,
given certain constrains (A before B, B before
C, etc.)

Dijkstra’s Algorithm
◦ Finds the shortest paths from a source vertex
to all other vertices.
Today

Finish up Dijkstra’s
◦ Path reconstruction

MSTs – Minimum Spanning Tree
algorithms
◦ Prim’s
◦ Kruskal’s
Dijkstra’s Algorithm

This algorithm finds the shortest path from a
source vertex to all other vertices in a
weighted directed graph without negative
edge weights.
5
a
10
2
3
2
b
8
3
16
4
4
12
e
d
c
13
Dijkstra’s Path Reconstruction

Example shown on
the Board
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
it’s acyclic.
◦ It’s spanning because it
covers every vertex.
◦ And it’s minimum because
it has minimum cost.
2
v1
4
1
2
v3
v2
3
7
v4
v5
4
8
5
10
v6
6
v7
1
2
v1
v2
1
v3
2
v4
v5
4
v6
1
6
v7
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.
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!
Kruskal’s Algorithm
Let V = 
For i=1 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.
Kruskal’s
Given Graph G:
Edges in sorted order:
4
v4
v1
4
1
v4
v6
v3
1
2
v1
3
v6
v4
v5
7
4
v3
v5
4
8
6
v7
1
Determine the MST:
v7
4
7
v4
2
v1
3
1
v5
v3
2
v2
CYCLE!
v4
v4
v6
v5
4
8
v1
10
6
v2
v4
v2
3
v6
v2
2
v3
1
v4
5
5
v7
v1
2
v3
v7
2
6
v2
v6
10
v5
1
v7
All Vertices, we’re
done!!
Kruskal’s

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.
edge you are unioning
two sets.
◦ A union cannot
happen if the two
the same set.
 This would create a
Prim’s 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
Prim’s
Given Graph G:
Edges in sorted order:
4
v4
v1
1
v4
v6
2
v1
3
v4
8
v5
7
6
v7
1
4
7
v4
2
v1
3
1
v5
v3
2
v2
CYCLE!
v4
v4
4
v3
v6
v5
4
8
v1
v5
4
Determine the MST, using Prim’s
starting with vertexV1:
v7
v2
v4
10
6
v2
2
v3
v6
v2
3
v6
5
v7
1
1
v4
5
v7
v3
v1
2
v3
4
2
6
v2
v6
10
v5
1
v7
All Vertices, we’re
done!!
Example on the board
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
bits for a single
character.

Not the most
efficient…
◦ What if ‘e’ is used 10
times more frequently
than ‘q’.
◦ It would be more
use a 7 bit code for e
and a 9 bit code for q.
Huffman Coding

Finds the optimal
way to take
character frequencies
in a particular file.
◦ On average, standard
files can shrink them
anywhere from 10% to
30% depending to 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.
code is crucial with
respect to easily
deciphering the code.
Building a Huffman Tree

Example on the
board
References
Slides adapted from Arup Guha’s Computer
Science II Lecture notes:
http://www.cs.ucf.edu/~dmarino/ucf/cop3503/le
ctures/
 Additional material from the textbook:

Data Structures and Algorithm Analysis in Java (Second
Edition) by Mark Allen Weiss
