Description
1 Overview
In this project, you will implement topological sorting in a directed acyclic graph. In addition,
your implementation will:
represent the graph as an adjacency matrix
always choose the vertex with the lowest ID when given a choice (e.g., when iterating through
the neighbors of a vertex and when selecting a vertex to start)
throw an exception if it encounters a cycle
nary header file for the project. The provided
header is just to get you started–you are allowed to change it however you like; however, your
solution must be compatible with the provided driver file. You are strongly encouraged to add
more data members and member functions to the class defined in the header.
2 Submitted code
You should submit a zip archive containing two files: digraph.h and digraph.cpp, which define
and implement a DigraphMatrix class. This class must have a constructor that accepts a string (or
const string&), and it must have a function topologicalSort that accepts no parameters and
returns a vector of integers.
The constructor must be able to construct the graph based on the
name of an input file to read. (See section 3 for a description of the graph file format.) Moreover,
it must represent the corresponding graph using an adjacency matrix.
The topologicalSort function should return the vertices of the vertices of the graph, in topologically sorted order if the directed graph does not contain a cycle. If the directed graph does
contain a cycle, though, topologicalSort should throw an exception.
The type of exception it
throws (and its what function) are not important, but it must be an std::exception object or
one that inherits from std::exception.
3 Input file
The input graph is described in edge list format. The first line of the file has the number of vertices
and edges of the graph, separated by a space (n and m, respectively).
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The following m lines have two nonnegative integers each. These integers, u and v, will be in
the range 0 to n − 1, and they represent the endpoints of an edge in the graph. Specifically, they
represent an edge from u to v.
4 Example 1
Consider running the topological sorting algorithm described in class on the graph below:
We start at the unexplored vertex with the lowest ID, vertex 0. We mark 0 discovered and
consider its undiscovered (outgoing) neighbors. Its only neighbor is 1, so we recurse on 1 and
mark it as discovered. The undiscovered neighbors of 1 are 2, 4, and 7.
Since we should consider
neighbors in sorted orde, we recurse on 2 and mark it discovered. The undiscovered neighbors of 2
are 5 and 7, so we recurse on 5 and mark it discovered. The only neighbor of 5 is 7, so we recurse
on 7 and mark it discovered.
Vertex 7 has no outgoing neighbors, so we mark it explored and add it to our output array (7).
When we return to 5, it has no other neighbors, so we mark it explored and add it to the array (7,
5). Returning to 2, we see that 7 has already been explored, so we mark 2 explored and add it to
the array (7, 5, 2).
Returning to 1, the next neighbor of 1 is 4, which we have not visited yet, so
we recurse on 4 and mark it discovered. The only undiscovered neighbor of 4 is 3, so we recurse on
3 and mark it discovered. The neighbors of 3 are 2 and 7, both of which are explored, so we mark
3 as explored and add it to the array (7, 5, 2, 3).
Returning back to 4, we’ve explored both of its
neighbors, so we mark 4 explored and add it to the array (7, 5, 2, 3, 4). When we return to 1, we
see that the last neighbor is 7, which is explored, so we mark 1 explored and add it to the array
(7, 5, 2, 3, 4, 1). We return to 0, but its only neighbor was 1, so we mark it explored and add to
the array (7, 5, 2, 3, 4, 1, 0).
Since we have returned from exploring 0, we need to find the unexplored vertex with the lowest
ID to start at. At this point, we’ve explored every vertex except 6, so we discover 6. All three of
the neighbors of 6 are explored, so we mark 6 as explored and add it to the array (7, 5, 2, 3, 4,
1, 6). Since all of the vertices of the graph have been marked explored, DFS ends. Reversing the
array yields the final answer (6, 0, 1, 4, 3, 2, 5, 7). (Due to the structure of the graph, this is the
only correct answer.)
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5 Example 2
Consider running the topological sorting algorithm described in class on the graph below:
We start with vertex 0 and mark it discovered. The only neighbor of 0 is 1, so we recurse on 1
and mark it discovered. The neighbors of 1 are 4 and 7, so we recurse on 4 and mark it discovered.
Vertex 4 has no outgoing neighbors, so we mark it explored, add it to the list (4), and return to
1. Vertex 1 continues to its neighbor 7, which we mark as discovered. Vertex 7 has no outgoing
neighbors, so we mark it explored and add it to the list (4, 7).
When we return to 1, we have
explored all outgoing neighbors, so we mark 1 explored and add it to the list (4, 7, 1). We then
return to 0, which has no other neighbors, so we mark 0 explored and add it to the list (4, 7, 1,
0). The next unexplored vertex is 2, so we start there and mark it discovered. Vertex 2 has no
neighbors, so we mark it explored and add it to the list (4, 7, 1, 0, 2).
The next unexplored vertex
is 3, so we start there and mark it discovered. The only undiscovered neighbor of 3 is 6, so we
recurse on 6 and discover it. Vertex 6 has no undiscovered neighbors, so we explore it, return, and
explore 3 (4, 7, 1, 0, 2, 6, 3). Since we have not yet discovered vertex 5, we explore it, and as 4 has
already been explored, we mark 5 explored and add it to the list (4, 7, 1, 0, 2, 6, 3, 5). Reversing
this list yield the final answer (5, 3, 6, 2, 0, 1, 7, 4).
6 Example 3
We will trace the topological sorting algorithm on the graph below:
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We start at 0 and mark it discovered. The only neighbor of 0 is 1, so we mark it discovered.
Vertex 1 has no outgoing neighbors, so we mark it explored, add it to the list (1), and return to
0. Since 0 has no other neighbors, we mark it explored and add it to the list (1, 0). We then
start again at 2.
After marking 2 as discovered, we see that its only neighbor, 1, has been marked
explored, so we mark 2 explored and add it to the list (1, 0, 2). We continue on to vertex 3. We
mark 3 as discovered, and since it has no neighbors, we mark it explored and add it to the list (1,
0, 2, 3). When we continue to 4, we mark it as discovered and continue on to 5. We mark vertex
5 discovered and continue to vertex 6, which we mark as discovered and continue to vertex 7. At
vertex 7, we see that vertex 5 is a neighbor, which creates a cycle (5, 6, 7, 5). At this point, your
code should throw an exception.
Note that the graphs formed by removing vertex 4 or reversing the edge (4, 5) would also have
cycles. You may wish to trace the topological sorting algorithm on one of these graphs, as well
7 Grading
Your code will be evaluated based on whether it produces the correct output for some number of
test cases. You may use any development environment that you like when developing the code;
however, it will be compiled and run using g++ in a Linux environment. Code that does not compile
will not receive substantial credit, so be sure that your code can be compiled using g++.
The development environment used in class is VS Code (https://code.visualstudio.com/
download). This IDE includes useful features like syntax highlighting and an integrated terminal
window that you can use to invoke a compiler.
Windows users can download MinGW (https://sourceforge.net/projects/mingw-w64) to
use g++, though you will need to add the MinGW bin directory to the Windows path in order for
the terminal to recognize the command g++ (or gdb or any of the other tools provided by MinGW).
8 Important note
You are allowed to use any code posted for this course on Canvas, and you are allowed to discuss
this project with the course TAs or professor. However, you are not allowed to use code from the
internet or your peers (besides your partner). Your code will be run through plagiarism-detection
software, and violators will be dealt with accordingly.
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