1. Quiz 4-26-’07
Search

2. Dijkstra’s Shortest Path Algorithm:
Consider the following algorithm to find the shortest distance between two cities (S  G)
A. Maintain a list of cities C that you have visited so far.
Cache the total path-cost g(c) and the predecessor city p(c) for every city c in C.
B. Maintain a list of neighboring cities F (the fringe) that are not in C:
Cache the total path-cost g(f) and the predecessor city p(f) for every city f in F.
Cities in F are ordered according to their total path cost g(f)
C. At every iteration of the algorithm, starting at S, visit the city in the fringe that has the lowest path-cost
Add it to C and remove it from F.
D. Add all new neighboring cities (including their g(f) and p(f)) into the fringe
that are not already in C. (They could already be in F though with a different predecessor)
E. If more than 1 copy of a city is in the fringe, only retain the one with lowest g(f). Delete all other copies.
F. If you reach G, reconstruct the path from SG and report g(G) and the path.
1. [2pts] Work out the first 3 steps of Dijkstra’s algorithm and uniform cost search (UC with repeated states).
This means: visit and expand the first 3 nodes starting at 0.
Use the graph to the right. “0” is the source and “6” is the goal.
2. [2pts] Is this algorithm an instance of a informed or an uninformed search
algorithm? (explain)
3. [2pts] Is this algorithm an instance of a tree search or a graph search
4. [2pts] Is this algorithm complete? (explain)
5. [2pts] Is this algorithm optimal? (explain – you may describe it as
an instance of some algorithm in the book and refer to results
described in the book)

3. Solution Notation for Dijkstra: node (total path cost, predecessor)
Notation for Uniform cost: node (total path cost)
Dijkstra
Nodes visited and expanded: 0 (0, null), 1 (1.1, 0), 3 (3.5, 1)
Uniform cost
Nodes visited and expanded: 0 (0), 1 (1.1), 0 (2.2)
Uninformed (since does not use any estimate of the cost from node to a goal)
Graph search (since the algorithm maintains a list of already expanded nodes)
Yes, it is complete (guaranteed to find a solution)
Yes, it is optimal (because it is uniform cost search with a detection of repeated states)