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Back to Shortest Path Problem. On to Course Summary and Evaluation.

**Held** Tuesday, May 9, 2000

**Overview**

Today we conclude our discussion of graphs.

**Notes**

- Reminder: Attendance Wednesday is mandatory!
- I"ll bring carrots
- Those who don't attend will receive negative extra credit

**Contents**

**Summary**

- Course topics, Revisited
- Object-Oriented Programming and Program Design
- Algorithms: Analysis, Common, Design
- Data Structures: Design issues, Common
- Java Programming

- Time for official course evaluation

- What's the running time of this algorithm?
- We'll need to talk about the number of nodes in the graph (
*n*) and the number of edges (*m*). - During initialization, we visit each node once, which should be
O(
*n*) in any reasonable implementation. - In the main loop, we remove each node from
*Est*once, so there are O(*n*) repetitions of the main loop. - We look at each edge twice (one for each node connected to the
edge), so the ``update distances'' part takes O(
*m*) across the whole algorithm. (*m*is the number of edges) - How long does it take to find the smallest distance in
*Est*? For the structures we've learned, O(*n*). - We've find the smallest distance O(
*n*) times. - So, the running time is O(
*n*^2 +*m*). - Since
*m*is in O(*n*^2), the running time is O(*n*^2). - That's a significant improvement over the previous algorithm.

- Can we improve this running time? In particular, can we speed up
the determination of the smallest distance in
*Est*? - We might consider using a Heap, since we're doing a lot
of
*removeSmallest*.- However, we do change priorities of objects, and Heaps aren't
designed to easily support that operation. (How might you support
a
`changePriority`

method?) - A data structure that you might learn in 301 (if you take 301), the Fibonacci heap, provides this method efficiently.

- However, we do change priorities of objects, and Heaps aren't
designed to easily support that operation. (How might you support
a

- The algorithm, as given, finds the length of the shortest path,
rather than the actual shortest path.
*How do we find the shortest path?* - It turns out a rather minor change to the algorithm makes it easy to find the shortest path. When we update the distance to a node, we also keep track of the node that led to that distance (the prior node).

- The
*Traveling Salescritter Problem*is, in effect, a variant of shortest path. Instead of finding the shortest path between two nodes, you find the shortest path (or cycle) that visits every node in the graph. - Our ``brute force'' solution works here.
- List all paths
- Find the shortest

- That's an O(n!) algorithm
- Surprisingly, no one has found a substantially better algorithm
(in terms of big-O analysis).
- In fact, there are a class of problems with no known algorithms
that run in time less than O(2
^{n}). - It's an open problem as to whether a better solution can exist.

- In fact, there are a class of problems with no known algorithms
that run in time less than O(2

- An interesting variant of shortest path and traveling salesperson is
the
*minimum spanning tree*(MST) problem. - The MST of an undirected graph is a set of edges that span the
graph (permit one to get from each node to every other node).
- It is minimum in the sense of having the smallest sum of edge weights.

- Note that the MST is a tree since there is no benefit to including a cycle (the extra edges can only add cost).
- It is possible to solve the MST problem by a greedy algorithm.
- Actually, I've been told that there are a variety of greedy algorithms that solve the MST problem.

- Here's one version:
- Separate the nodes into two groups: those in the MST and those not in the MST. Initially one node is in the MST (it doesn't matter which one).
- Repeatedly pick the smallest edge between a node in the MST and a node not in the MST and add it to the MST.
- If there are any nodes left in the set of nodes not in the MST, then there is no MST.

- Here's another one. Is it likely to be successful?
- Order the edges from smallest to largest
- Add an edge as long as it doesn't form a cycle

- In a directed graph, we can treat the edges as giving an ordering.
- For example, if there is an edge from A to B then we can say that A is less than B.

- Unlike typical orderings, this isn't necessarily complete. That is,
there can be pairs of elements that are essentially unordered.
- Even if there are edges from A to C and B to C, we still don't know anything about the relationship of A and B.

- In
*topological sort*, you assign numbers to the elements of the graph in such a way that- No two elements have the same number
- If there is a path from X to Y, then X has a lower number than Y.

- How do we implement graphs? There are many techniques, which may depend on our intended use of the graph.
- The simplest one is to use some sort of
`Node`

structure, and to treat the graph as a collection of nodes.- Each node will need a list of nodes for its neighbors.
- We will probably store the nodes in a hash table for each access (indexed by label).

- If we can number the nodes, we can use an
*adjacency matrix*: if there is an edge from A to B, then M[A,B] is true.- In a weighted graph, we give M[A,B] the weight of the edge. We use some special value (perhaps the largest integer or a negative integer) for ``not there'').

- We can make a list of all the edges in the graph.
- Which implementation should we use? It depends a lot on the algorithm(s)
we intend to use.
- If we need all the neighbors of a node, we want the simple ``collection of nodes'' implementation.
- If we need all the edges of the graph, we use the list of edges.
- If we need to quickly determine whether there is an edge between two nodes, we use the adjacency matrix.

Tuesday, 18 January 2000

- Created as a blank outline.

Tuesday, 9 May 2000

- Filled in the details.

Back to Shortest Path Problem. On to Course Summary and Evaluation.

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