Computer Science 3510
Fall 2002
Programming Assignment 3

First version due: Tue, Oct 29
Second version due: Mon, Nov 11

A word to the wise

Begin working on this program right away. Do not wait. Create a refinement plan based on the plan presented below, and follow that plan. Do not attempt to write the entire program and then start testing it.

Get something done for the first pass. Make your first version the best that you can.


Preliminaries

A graph is a collection of vertices connected by a collection of edges. An edge connects exactly two vertices to one another. You can draw a picture of a graph by showing the vertices and the edges connecting them. Here is an example. The vertices are shown as circles with numbers in them and the edges are lines connecting the vertices.

A weighted graph is a graph in which each edge has a number attached to it, called the weight of the edge. Here is a picture of a weighted graph.

A spanning tree of a graph is obtained by deleting as many edges as possible, without making it impossible to get from any vertex to any other by following a sequence of edges. For example, the following is a spanning tree of the above weighted graph.

The weight of a spanning tree is the sum of the weights of its edges. For example, the weight of the above spanning tree is 59. Here is another spanning tree for the same graph. Its weight is 48.

Obviously, some spanning trees have smaller weight than others. A minimal spanning tree is a spanning tree with the smallest possible weight.


Motivation for computing minimal spanning trees

Suppose that we have a collection of towns, and we must build railroad tracks so that it is possible for a train to get from any town to any other town. Our budget for building tracks is small, so we choose to build as little track as possible. One approach to deciding which tracks to build is to construct a weighted graph, where the vertices are the towns, and the weight of the edge from town A to town B is the length of the railroad track that would need to be built to connect towns A and B directly. Then a minimal spanning tree of that graph would be a good choice for connecting the towns, since it is the cheapest way to connect all of the towns using railroad tracks between towns.

A similar problem occurs when a house is being wired for electricity. All of the outlets on a given circuit need to be connected to the circuit panel, and to each other. To connect them with a minimum amount of wire, you might build a graph having a vertex for each outlet and for the circuit panel, with the weight between two vertices being the wiring distance between those vertices. Then get a minimal spanning tree for the graph.

(A minimal spanning tree is not always the best solution to either of these problems. You can often do better by introducing new railroad track junctions or wiring junctions, called Steiner points. For example, if you have three towns in a triangle, you can connect them together by sending tracks to a point in the middle of the triangle. The middle point is a Steiner point. But good placements of Steiner points are difficult to find, so a minimal spanning tree is a reasonable compromise.)


The assignment

For this assignment, you will write a program that reads a description of a weighted graph, and prints the edges that are part of a minimal spanning tree of that graph, and also prints the total weight of the minimal spanning tree.

The input starts with a line that tells how many vertices the graph has. If there are five vertices, then those vertices have numbers 1, 2, 3, 4 and 5. In general, if there are n vertices, then they are numbered 1...n.

Following the first line are the edges, one per line. Each edge line has three integers on it. Line

2 4 50
indicates that there is an edge between vertices 2 and 4, and that its weight is 50. The end of the input is signaled by a line that contains just a 0. An input describing graph

might look like this.

5
1 2  9
1 3 12
2 4 18
2 3  6
2 5 20
3 5 15
0

The output of your program for this input might look like this.

The input graph has 5 vertices, and its edges are as follows.

  vertices    weight
  1   2            9
  1   3           12
  2   4           18
  2   3            6
  2   5           20
  3   5           15
  
A minimal spanning tree uses the following edges.

  vertices    weight
  2   3            6
  1   2            9
  3   5           15
  2   4           18

The total weight of the spanning tree is 48.

Your program is not required to insist that the three numbers that describe an edge are on one line, but it can if it wants to. For example, if your program allows an input like the following, that is acceptable. But your program is not required to allow such inputs. (You will find it easier to allow these additional input forms.)

5
1 2  
9
1 
3 12
2 4 18  2 3  6
2 5 20
3 5 15
0
The only difference between this and the previous input is the placement of line breaks.


Data structures

When designing a program, you need to decide how to represent your data. Typically, you draw diagrams showing arrays, structures, pointers, etc. to understand the data representation.

Once you have decided on a representation, you need to decide whether you want the data structure to be open or closed in your program.

Choosing an open data structure allows all parts of the program that use the data structure to see the exact representation, and to manipulate that representation directly. This has the advantage that you do not need to introduce any special functions to manipulate the data structure, and you have a lot of flexibility in how to use it. This approach has the disadvantage of making the form of data structure nearly impossible to change. The code is also typically less readable than if you use a closed approach. Also, you often end up writing the same code to manipulate the data in more than one place in your program.

When you choose a closed representation, you are creating an abstract data type. Choosing a closed data structure involves hiding the representation inside a class or module, and only allowing the data to be accessed through public functions of the class or module. This typically involves more initial work, but makes your program easier to modify and often makes it easier to write in the long run.


Representing a weighted graph

For this program I will have mercy on you and allow you to use an open representation for graphs. You will find this a little simpler. If you prefer to use a closed representation (a class) that is fine.

Create a type Graph. You can store a graph as an array of edges, where each member of the array is a triple giving two vertex numbers and a weight. A triple is just a structure with three variables in it. Of course, you must also store the number of edges and the number of vertices, in addition to the array. Store those numbers once for the graph, not once for each edge. So the graph itself will be a structure holding the number of vertices, the number of edges and an array of edges.

You may presume that there will be a maximum of 100 edges in the graph, but you must design your program so that it is very easy to change the maximum number of edges to some other number.

Use sensible names for things. If something is an edge, don't call it a graph. Call it an edge. Do not call a vertex and edge, or a graph a vertex. Call things what they are.


An abstract data type

An algorithm for computing a minimal spanning tree is described below. It builds up a spanning tree gradually. During the process, it needs to be able to ask whether two vertices are already connected by the edges that have been selected so far. Here, we describe a tool for keeping track of which vertices are connected.

The tool is an abstract data type for managing connectedness. I will call an object that manages information about connections a connection manager. Initially, a connection manager CM starts with no connections at all. That is, nothing is connected to anything else.

The connection manager allows you to place connections between certain vertices, and to ask if two vertices are connected. If you connect A to B, and then connect B to C, then A and C become connected automatically. It is not necessary to tell the connection manager that A and C are connected.

This abstract data type must be implemented as a class. The operations are as follows.

A constructor with a single integer parameter n.

The constructor builds a connection manager that can handle up to n vertices, numbered 1,...,n.

CM.connect(m,n)

This tells connection manager CM to make a connection between vertices m and n. Here, m and n are positive integers, up to the size that this connection manager can handle.

CM.connected(m,n)

This returns true (1) if m and n are connected, and false (0) if they are not.


Implementation of the abstract data type

Here is how to implement these operations. The idea is to have a leader of each group of connected vertices. For example, if vertices 3, 5 and 8 are connected together, then they might decide that 5 is their leader. Write a function leader(k) that returns the leader of the group that contains vertex k. In the example, leader(3) would be 5, and leader(5) would also be 5. (Vertex 5 is its own leader.) To find out whether two vertices are connected, just check whether they have the same leader.

To implement function leader, keep an array called superior of integers. Ideally, the superior[k] is the leader of the group containing vertex k. Sometimes, however, the superior of a vertex is not its leader. But the leader of k is the same as the leader of the superior of k. By going to superior[k], at least you get closer to the leader.

Suppose that r = superior[k]. It might turn out that r is not a leader at all --- it is not its own leader. Instead, you must take its superior. You continue doing this until you find a vertex that is its own leader. That vertex is the true leader of the group. So to find the leader of k, you would do the following.

  r = k;
  while(superior[r] != r) r = superior[r];
Now r is the leader of k.

When you are told to make a connection between two vertices m and n, what you need to do is

  1. Find the leader r of m and the leader s of n.

  2. Force r to be the superior (and leader) of s, by making superior[s] = r.

Now s is no longer its own leader. All vertices that used to have s as their superior now automatically have r as their leader, since the loop will not stop at s any more. It is very important that you only change the superior of a number that is currently its own leader, or the method will not work.

Keep as much private in the class as possible. There should be no public variables. Do not make a function public unless you need to, because it is one of the operations provided by the abstract data type.

Use the implementation discussed here. Do not invent your own. In particular, be sure than neither the connect nor the connected function performs a scan of every member of the superior array. That is not called for here, and it is much less efficient than what is described here.


Computing minimal spanning trees

Here is a simple algorithm for computing a minimal spanning tree of a weighted graph.

  1. If the array of edges of the graph is called G.edges, then sort the edges in G.edges by their weights, with smaller weights earlier in the array. (You can use function qsort, described below, to do this.)

  2. Create a new connection manager. Initially, there are no connections, so every vertex is its own leader.

    Create an array that will end up holding the edges of the spanning tree. Initially, it has no edges in it, but you gradually add more edges. Also create a count, initially 0, of the total weight of the edges in the spanning tree array.

  3. Scan through the array G.edges in increasing order of weight. (That is, do the following in a loop.) When you see edge (i,j), with weight w, ask the connection manager whether vertices i and j are already connected. If they are connected, then just skip this edge, and go to the next edge. If they are not connected then add edge (i,j) to the spanning tree, add w to the total weight of the tree, and tell the connection manager that i and j are connected. Then do the next edge.

When you are finished, you will have created an array holding the edges of a minimal spanning tree of the original graph, and you will have computed the total weight. Print them all.


Using qsort

There is a standard library function called qsort that implements a variant of the quicksort algorithm for sorting arrays. You should include header file stdlib.h to use qsort. This function is designed to be able to sort any kind of array into any desired order. In order to achieve this degree of generality, qsort needs information about the array and how to sort it.

The prototype for qsort is as follows.

     void qsort(void *base, 
                size_t nel, 
                size_t width,
                int (*compar) (const void *, const void *));
Parameter base is a pointer to the array that you want to sort. Notice that its type is void*. That just means that qsort knows that it is a pointer, but does not know what type of thing it points to.

Parameter nel is the number of elements in the array, and parameter width is the number of bytes occupied by each element. (Type size_t is typically the same as int, and is used for values that describe the size of an array.)

Parameter compar is a function. It is responsible for telling qsort the order in which you want the array sorted. qsort will sort into ascending order, according to what function compar says is ascending order. Function compar takes two parameters, which will be pointers to particular members of the array. compar(A,B) should return
   a negative number if A < B
0 if A = B
a positive number if A > B
according to the order in which you want your array to be sorted.

For example, suppose that you want to sort an array of long integers into descending order. You write a comparison function.

int compareLongInts(const long* A, const long* B)
{
  return *B - *A;
}
Notice that compareLongInts returns a positive number when A < B. This is because you want to sort into descending order, but qsort wants to sort into what it thinks is ascending order. So you tell qsort that 30 < 10. Also notice that the parameters to compareLongInts are pointers. qsort will call this function, and will pass it pointers to the members that qsort wants to compare. qsort will call the comparison function many times.

There is an obvious problem, however. The prototype for function compareLongInts says that its parameters have type const long*. But qsort wants to be passed a function whose parameters have type const void*. To fix this, you can use a cast. A cast is a way of telling the compiler to treat something as if it has a different type. Create a type QSORT_COMPARE_TYPE that you want to cast to, as follows.

   typedef int (*QSORT_COMPARE_TYPE)(const void*, const void*);

To sort an array Arr of n long integers, you use

   qsort((void*)Arr, n, sizeof(long), (QSORT_COMPARE_TYPE)compareLongInts);
We have passed qsort the array to sort, the number of elements, the number of bytes occupied by each element (given by sizeof(long), since each element has type long) and the comparison function, cast to the appropriate type.

Casts are, in general, very dangerous. They are a way of pulling the wool over the eyes of the compiler. This particular kind of cast is safe, and you can use it, but become sloppy with casts at your peril.

You will need to write a comparison function that is appropriate for sorting an array of edges into ascending order according to weight. Function compareLongInts is only an example, and cannot be used to compare edges. An edge is not an integer. Write a function to compare two edges.


Improving the efficiency

There is an easy way to greatly speed up your program for large numbers of edges. The improvement is made in the implementation of the connection manager abstract data type.

There are two modifications to do. One involves a change to the leader function, the other a change to the connect function and to the representation of the abstract data type.

IMPORTANT. These improvements are part of the assignment, and you will lose points for not doing them. But only make these improvements after everything is working correctly without them. You will receive no credit for implementing these improvements if your basic implementation of the connection manager is wrong.

Changing the abstract data type implementation should not involve changing anything else at all. Not a single character outside of the abstract data type should be changed. If you cannot do that, then you have done something wrong.

The first improvement

The first change involves a change to the leader function. After the leader function scans through a chain to find the leader of a vertex, it goes back through the chain and puts the genuine leader in the superior array, for every vertex that was looked at in the chain. That way, subsequent leader computations will go much more quickly. For example, if the superior array contains

  superior[1] = 2
  superior[2] = 4
  superior[3] = 3
  superior[4] = 6
  superior[5] = 3
  superior[6] = 6
  superior[7] = 4
  superior[8] = 5
then computing leader(1) requires chaining through 1, 2, 4, 6. The improvement adds a loop that changes the contents of the array to the following, by installing the correct leader (6) of each of those vertices. (It is just a matter of rescanning through the chain, the same way the chain was scanned the first time.)
  superior[1] = 6
  superior[2] = 6
  superior[3] = 3
  superior[4] = 6
  superior[5] = 3
  superior[6] = 6
  superior[7] = 4
  superior[8] = 5
Notice that we have not scanned the entire array from beginning to end! Superior[8] is still 5, even though the leader of 8 is 3. Only the vertices that were looked at in the original scan have their superior values changed. If you try to change everything in the array, you make the program slower, not faster.

Also notice that it was not just the superior of 1 that was changed. All of the vertices that were examined in the chain have their superiors set to their leaders. For example superior[2] was changed too.

The second improvement

Each vertex that is a leader has a collection of constituents who have it as their leader. For example, in the above array, vertex 3 is a leader, and the constituents of vertex 3 are 3, 5 and 8. So vertex 3 has three constituents, counting itself.

When making a connection, you find two values s and t that are leaders. You can then either change the superior of s to t (so s is no longer a leader) or change the superior of t to s (so t is no longer a leader). The choice of which to do influences the efficiency of the implementation. The best choice is to change the superior of the vertex that has the fewest constituents. That tends to keep the lengths of the superior chains up to the leader short.

Modify the data structure so that each vertex has not only a superior, but also a count of its constituents. A vertex that is not a leader has no constituents. So now the information is an array of structures, where each structure contains a superior and a constituent count.

A picture of the initial array, before any connections have been made, might look like this. Notice that each vertex has one constituent, itself.

  index   superior  numConstituents
    1        1           1
    2        2           1
    3        3           1
    4        4           1
    5        5           1

When making a connection, compare the constituent counts. Change the superior of the value with the fewest constituents. (If they have the same number of constituents, then the choice is arbitrary. For example, change the first one.) If you change things so that the superior of s becomes t, then remember that all of the constituents of s become new constituents of t. For example, if you do connect(3,5) in the above array, you might arbitrarily decide to make the superior of 3 be 5. Then the array looks like this.

  index   superior  numConstituents
    1        1           1
    2        2           1
    3        5           0
    4        4           1
    5        5           2
If you now do connect(5,1), you must change the superior of 1, since it has fewer constituents than 5. The array ends up looking like this.
  index   superior  numConstituents
    1        5           0
    2        2           1
    3        5           0
    4        4           1
    5        5           3
As before, only change the superior of a vertex that is currently a leader. If you now do connect(2,1), you must realize that you are really being asked to connect 2 and 5, since 5 is the representative of 1. Since 5 has more constituents, you change 2, yielding
  index   superior  numConstituents
    1        5           0
    2        5           0
    3        5           0
    4        4           1
    5        5           4
As you can see, this improvement tends to lead to shorter chains of superiors before the leader is found. It is not guaranteed to ensure that the superior is always the leader, though. You still need to do the leader calculation using a loop.

Implementing the program: a refinement plan

Here is a suggestion for how to implement this program. This is a refinement plan. It tells how to produce more and more of the program, gradually refining it until eventually it does the whole job. It is critical that a refinement plan have tests along the way, so that the part written so far is known to work. That way, if you get an error, it is almost always in the part of the program that was most recently written, and finding bugs is relatively easy.

  1. Implement the part of the program that reads in and prints out the input graph. Test it. It is a good idea to use functions. You will need to print other graphs later.

  2. Implement the beginnings of the minimal spanning tree algorithm by sorting the edges and printing them after sorting. Be sure that this is working.

  3. Implement the connection manager ADT. Test it. Testing involves writing a test program that is not part of the finished product of the program. The extra work to write the tester is well worth it. Proceeding beyond this step with a faulty connection manager will cause huge problems later. Your tester can create a connection manager and make a few connections. Then it can ask if a few vertices are connected. Ask some questions where the answer should be yes, and some where the answer should be no. Check that the answers are correct.

  4. Implement the rest of the minimal spanning tree algorithm. Test it. Look at the results. Are they correct? You will have to solve an example by hand to make sure.

  5. Make the program's output look nice, if necessary. Test it.

  6. Implement the improvements to the connection manager. Test your program again. Notice how nice it would be to have an automated tester, so that you can easily retest your program when the improved connection manager implementation is installed.

Keep the program well-commented and well-indented throughout. Make sure every function has a clear contract. Do not plan to add in the comments after the program is finished.

When you do a test, if things don't work, then use the debugger or add some prints to see what is going on. Do not try to work in the dark. Before you try to fix the problem, understand what is wrong. After you have diagnosed the problem, fix it. Do not move on until everything that you have written so far is working.


What to turn in

Turn in all of your source files. There should be, at a minimum, two files for the connection manager (a header file and an implementation file) and one file for the minimal spanning tree program. There might be more, depending on how you implement this program.

The connection manager must be in a separate file from the minimal spanning tree finder. Your program should be well documented, well indented and easy to read. Keep it that way during development. Do not try to work on a sloppy, poorly documented program, unless you have time to burn.

Use the handin utility to turn in your work. Version 1 will be handin assignment 5, and version 2 will be handin assignment 6.


Asking questions

If you have questions, be sure to ask, and do so as soon as possible, so that you have time to deal with the response.