CS3610 Project 5 solved

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5/5 - (2 votes)

In this day and age, whenever people gear up for a road trip, they turn on their
smart phones and open up Google Maps. Under the directions tab, users type in
the name of their starting location followed by the name of their destination. In
seconds, Google Maps assembles a detailed set of directions covering the shortest route
to the destination city. Such computing speeds are achieved through the use of efficient
mapping algorithms. Unfortunately, how it is these algorithms work is not known
by anyone in the general public. Google has decided to keep the implementations
closed-source and prevent curious computer science students such as yourself from
learning about something awesome. Well, rather than fight a legal battle to have the
licensing rights revised, you are going to get back at the behemoth that is Google
Maps by developing your own competing application. Specifically, here in your humble
beginnings, you will implement a single source shortest path algorithm.
Implementation and Bonus
In order to find a single source shortest path, you must implement the version of
Dijkstra’s algorithm described on Ch12 lect2.ppt powerpoint under blackboard. If you
recall, this algorithm runs in O(V
2 +E) time, where V is the number of vertices in the
graph and E is the number of edges. The O(V
) component refers to the number of
operations carried out to find every minimum distance vertex extracted from the set
of unvisited vertices at the beginning of each iteration of the while loop. The O(E)
component results from comparing and possibly updating the distances of all vertices
adjacent to the minimum distance vertices. In other words, the for loop within the
while loop runs O(E) operations in total.
If your graph is not too dense (meaning the graph does not contain an overwhelming
number of edges), you may want to consider storing the set of unvisited vertices in a
min heap using distance to the source vertex as a key. This will help you find all the
minimum distance vertices extracted at the beginning of each iteration of the while
loop in O(V lg(V )) time as opposed to O(V
) time. Of course, when you now update
the distance of a neighboring vertex in the for loop below, you must also update that
vertex’s position in the min heap. As you already know, bubbling up an element in
a min heap of n elements takes just O(lg(n)) time, but the initial searching for the
element takes O(n) time. In order to avoid the O(n) search, you must implement a
lookup table that returns a vertex’s index in the min heap in O(1) time. If implemented
correctly, the cumulative time complexity of updating the distance values of vertices
in the min heap version of Dijkstra’s would be O(Elg(V )). In other words, using a
lookup table and a min heap, the for loop within the while loop runs in O(Elg(V ))
time as opposed to the O(E) time seen in the version of Dijkstra’s algorithm described
in the previous paragraph. Thus, the total time complexity of this modified Dijkstra’s
1 CS3610
algorithm is O(V lg(V ) + Elg(V )). As said early, it is more advantageous to use the
min heap version if your graph is not overwhelmingly dense.
In this project, the 6 final grade points will be awarded if you successfully implement
the Dijkstra’s algorithm described in lecture. If you successfully implement the min
heap version, you will be awarded the 6 final grade points plus 3 bonus points.
Input is read from the keyboard. The first line of input is the number of test
cases K. Each of the K test cases is written in the following format:
Individual Test Case Format
city 1
city 2
city n
d 11 d 12 … d 1n
d 21 d 22 … d 2n
d n1 d n2 … d nn
The first line of each test case is the number of cities n in the graph. The next n lines
are the names of each city. City names consist only of alphabetic characters. Following
the list of n city names is an nxn distance matrix where each distance d ij is an integer
value in the range [0, 10000] representing the distance of the road connecting city i to
city j. A distance d ij = 0 indicates that there does not exist any road connecting
city i to city j. For this project, all roads will be undirected, which means d ij = d ji
for all cities i and j. As a result, every input distance matrix will be symmetric.
For each test case, output a space delimited list of all the city names in the shortest
path connecting city 1 and city n followed by the integer distance of the path. City
names should be listed in order of when they are to be visited starting from city 1
and ending with city n. If city 1 has multiple shortest paths to city n, just output
one of them. Also note that in every test case, there will always be a path connecting
city 1 to city n.
Sample Test Cases
Use input redirection to redirect commands written in a file to the standard input, e.g.
$ ./a.out < input1.dat.
Input 1
0 1 2 0
1 0 5 6
2 5 0 7
0 6 7 0
Output 1
Akron Athens Cleveland 7
Turn In
Email your source code to with the subject “CS3610 Project 5”.
If you have multiple files, package them into a zip file.
Total: 100 pts.
• 10/100 – Code style, commenting, general readability.
• 05/100 – Compiles.
• 05/100 – Follows provided input and output format.
• 80/100 – Successfully implemented Dijkstra’s algorithm.
• 30/100 – Bonus points for heap implementations.