Assignment 7 Topological Sort of a Directed Graph

CMPT 280– Intermediate Data Structures and Algorithms
Assignment 7

Total Marks: 57
1 Submission Instructions
• Assignments must be submitted using Moodle.
• Responses to written (non-programming) questions must be submitted in a PDF file, plain text file
(.txt), Rich Text file (.rtf), or MS Word’s .doc or .docx files. Digital images of handwritten pages
are also acceptable, provided that they are clearly legible.
• Programs must be written in Java.
• If you are using IntelliJ (or similar development environment), do not submit the Module (project).
Hand in only those files identified in Section 5. Export your .java source files from the workspace
and submit only the .java files. Compressed archives are not acceptable.

2 Background
2.1 Topological Sort of a Directed Graph
Imagine a directed graph is used to represent prerequisite relationships between university courses. There
is a node in the graph for each course, and a directed edge from course A to course B if course A is a
prerequisite for course B. Here’s what such a graph would look like for some of our courses might look
like:
CMPT 141
MATH 110 CMPT 145
CMPT 270 CMPT 214 CMPT 260
CMPT 280 CMPT 215
Note that this is a graph and not a tree, because CMPT 270 has more than one "parent". But, there are no
cycles in this graph.
If there are no cycles in a directed graph, we can perform an operation called a topological sort. A
topological sort of a graph produces a linear ordering of the nodes such that if there is a directed edge
from node A to node B, then node A appears before node B in the ordering. In the specific case of
the course prerequisite graph, a topological sort produces an ordering of the nodes such that no course
appears before any of its prerequisites – precisely the thing one needs to know in order to take courses in
the right order!
Note that there is not necessarily a unique answer for topological sort. For example, in the graph pictured above, we could take MATH 110 and CMPT 111 in either order (because neither has prerequisites),
so both of the following sequences would be valid topological sorts:
• 111, 110, 115, 270, 214, 260, 280, 215
• 110, 111, 115, 270, 214, 260, 280, 215
Also note that 280 and 215 could be taken in either order as long as they are taken after both 270 and 214,
resulting in more possible topological orderings:
• 111, 110, 115, 270, 214, 260, 215, 280
• 110, 111, 115, 270, 214, 260, 215, 280
The basic algorithm for topological sort, and a variation that is relevant to the specific problem we will
want to solve, is presented with Question 2.
Finally, it is worth noting that you cannot perform a topological sort on a graph that has a cycle. If
there is a cycle, then it is not possible to satisfy the prerequisites for any node in the cycle because each
course in the cycle is a prerequisite of itself.
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3 Your Tasks
Question 1 (42 points):
For this question you will be implementing a k-D tree. We begin with introducing some algorithms
that you will need. Then we will present what you must do.
Helper Algorithms for Implementing k-dimensional Trees
As we saw in class, in order to build a k-D tree we need to be able to find the median of a set of
elements efficiently. The “j-th smallest element” algorithm will do this for us. If we have an array of n
elements, then finding the n/2-smallest element is the same as finding the median.
Below is a version of the j-th smallest element algorithm that operates on a subarray of an array
specified by offsets le f t and right (inclusive). It places at offset j (where le f t ≤ j ≤ right) the element
that belongs at offset j if the subarray were sorted. Moreover, all of the elements in the subarray
smaller than that belonging at offset j are placed between offsets le f t and j − 1 and all of the elements
in the subarray larger than that element are placed between offsets j + 1 and right, but there is no
guarantee on the ordering of any of these elements! The only element guaranteed to be in its sorted
position is the one that belongs at offset j. Thus, if we want to find the median element of a subarray
of the array list bounded by offsets le f t and right, we can call
jSmallest(list, left, right, (left+right)/2)
The offset (le f t + right)/2 (integer division!) is always the element in the middle of the subarray
between offsets le f t and right because the average of two numbers is always equal to the number
halfway in between them.
The j-smallest algorithm is presented in its entirety on the next page.
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Algorithm jSmallest ( list , left , right , j )
list - array of comparable elements
left - offset of start of subarray for which we want the median element
right - offset of end of subarray for which we want the median element
j - we want to find the element that belongs at array index j
To find the median of the subarray between array indices ’ left ’ and ’ right ’ ,
pass in j = ( right + left )/2.
Precondition : left <= j <= right
Precondition : all elements in ’ list ’ are unique ( things get messy otherwise !)
Postcondition : the element x that belongs at offset j , if the subarray were
sorted , is at offset j . Elements in the subarray
smaller than x are to the left of offset j and the
elements in the subarray larger than x are to the right
of offset j .
if( right left )
// Partition the subarray using the last element , list [ right ] , as a pivot .
// The index of the pivot after partitioning is returned .
// This is exactly the same partition algorithm used by quicksort .
pivotIndex := partition ( list , left , right )
// If the pivotIndex is equal to j, then we found the j-th smallest
// element and it is in the right place ! Yay!
// If the position j is smaller than the pivot index , we know that
// the j-th smallest element must be between left , and pivotIndex -1 , so
// recursively look for the j-th smallest element in that subarray :
if j < pivotIndex
jSmallest ( list , left , pivotIndex -1 , j )
// Otherwise , the position j must be larger than the pivotIndex ,
// so the j-th smallest element must be between pivotIndex +1 and right .
else if j pivotIndex
jSmallest ( list , pivotIndex +1 , right , j )
// Otherwise , the pivot ended up at list [j] , and the pivot *is* the
// j-th smallest element and we ’re done .
Notice that there is nothing returned by jSmallest, rather, it is the postcondition that is important.
The postcondition is simply that the element of the subarray specified by left and right that belongs
at index j if the subarray were sorted is placed at index j and that elements between le f t and j − 1
are smaller than the j-th smallest element and the elements between j + 1 and right are larger than
the j-th smallest element. There are no guarantees on ordering of the elements within these parts of
the subarray except that they are smaller and larger than the the element at index j, respectively. This
means that if you invoke this algorithm with j = (right + le f t)/2 then you will end up with the median element
in the median position of the subarray, all smaller elements to its left (though unordered) and all larger elements
to its right (though unordered), which is just what you need to implement the tree-building algorithm!
NOTE: for this algorithm to work on arrays of NDPoint280 objects you will need an additional parameter d that specifies which dimension (coordinate) of the points is to be used to compare points.
An advantage of making this algorithm operate on subarrays is that you can use it to build the k-d
tree without using any additional storage — your input is just one array of NDPoint280 objects and
you can do all the work without any additional arrays — just work with the correct subarrays.
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You may have noticed that jSmallest uses the partition algorithm partition the elements of the
subarray using a pivot. The pseudocode for the partition algorithm used by the jSmallest algorithm
is given below. Note that in your implementation, you will, again, need to add a parameter d to denote
which dimension of the n-dimensional points should be used for comparison of NDPoint280 objects.
// Partition a subarray using its last element as a pivot .
Algorithm partition ( list , left , right )
list - array of comparable elements
left - lower limit on subarray to be partitioned
right - upper limit on subarray to be partitioned
Precondition : all elements in ’ list ’ are unique ( things get messy otherwise !)
Postcondition : all elements smaller than the pivot appear in the leftmost
part of the subarray , then the pivot element , followed by
the elements larger than the pivot . There is no guarantee
about the ordering of the elements before and after the
pivot .
returns the offset at which the pivot element ended up
pivot = list [ right ]
swapOffset = left
for i = left to right -1
if( list [ i ] <= pivot )
swap list [ i ] and list [ swapOffset ]
swapOffset = swapOffset + 1
swap list [ right ] and list [ swapOffset ]
return swapOffset ; // return the offset where the pivot ended up
Algoirthm for Building the Tree
An algorithm for building a k-d tree from a set of k-dimensional points is given below. It is slightly
more detailed than the version given in the lecture slides. It uses the jSmallest algorithm presented
above.
Algorithm kdtree ( pointArray , left , right , int depth )
pointArray - array of k - dimensional points
left - offset of start of subarray from which to build a kd - tree
right - offset of end of subarray from which to build a kd - tree
depth - the current depth in the partially built tree - note that the root
of a tree has depth 0 and the $k$ dimensions of the points
are numbered 0 through k -1.
if pointArray is empty
return null ;
else
// Select axis based on depth so that axis cycles through all
// valid values . (k is the dimensionality of the tree )
d = depth mod k ;
medianOffset = ( left + right )/2
// Put the median element in the correct position
// This call assumes you have added the dimension d parameter
// to jSmallest as described earlier .
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jSmallest ( pointArray , left , right , d , medianOffset )
// Create node and construct subtrees
node = a new kD - tree node
node . item = pointArray [ medianOffset ]
node . leftChild = kdtree ( pointArray , left , medianOffet -1 , depth +1);
node . rightChild = kdtree ( pointArray medianOffset +1 , right , depth +1);
return node ;
Implementing the k-D Tree – What You Must Do
Implement a k-D tree. You must use the NDPoint280 class provided in the lib280.base package of
lib280-asn6 to represent your k-dimensional points. You must design and implement both a node
class (KDNode280.java) and a tree class (KDTree280.java). Other than specific instructions given in
this question, the design of these classes is up to you. You may use as much or as little of lib280 as
you think is appropriate. You’ll be graded in the actual appropriateness of your choices here. You
should aim to make the class fit into lib280 and its hierarchy of data structures, but you should not
force things by extending classes inappropriately. You may use whatever private/protected methods
you deem necessary.
A portion of the marks for this question will be awarded for the design/modularity/style of the implementation of your class. A portion of the marks for this question will be awarded for acceptable
inline and javadoc commenting.
Your ADT must support the following operations:
• Construct a new (balanced) k-D tree from a set of k-dimensional points (it must work for any
k 0).
• Perform a range search: given a pair of points (a1, a2, . . . ak
) and (b1, b2, . . . , bk
), ai <= bi
for all
i = 1 . . . k, return all of the points (c1, c2, . . . , ck
) such that a1 ≤ c1 ≤ b1, a2 ≤ c2 ≤ b2, . . . , ak ≤
ck ≤ bk
.
Note: your tree does not have to have operations that insert or remove individual NDPoints.
In addition, you should write a test program that demonstrates the correctness of your tree. The test
program should consist of two parts:
1. Show that your class can correctly build a k-D tree from a set of points. For k=2, display the set
of k-dimensional points that you used as input (use between 8 and 12 elements), followed by a
graphical representation of the built tree (similar to the toStringByLevel() output in the trees
we’ve done previously). Do this again for one other value of k, between 3 and 5 (your choice).
2. For the second of the two trees you displayed in part 1, perform at least three range searches. For
each search, display the query range, execute the range search, and then display the list of points
in the tree that were found to be in range. A sample test program output is given below.
Implementation and Debugging Strategy
In order to implement the tree-building algorithm kdtree you first need to implement jSmallest
which, in turn requires partition. It is strongly suggested that you implement and thoroughly
test partition before trying to implement jSmallest. In turn, throughly test jSmallest before you
implement kdtree. If you don’t do this, I can tell you from experience that it will be a nightmare to
debug. You need to be sure that each algorithm is correct before implementing the algorithms that
depend on it, otherwise, if you run into a bug it will be very hard to determine in which method in
the chain of dependent methods the bug is occurring. This is a fundamental principle that is crucial
to designing complex software systems. Make sure each piece is correct before relying on it later.
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Sample Output
Input 2 D points :
(5.0 , 2.0)
(9.0 , 10.0)
(11.0 , 1.0)
(4.0 , 3.0)
(2.0 , 12.0)
(3.0 , 7.0)
(1.0 , 5.0)
The 2 D tree built from these points is :
4: -
3: (9.0 , 10.0)
4: -
2: (5.0 , 2.0)
4: -
3: (11.0 , 1.0)
4: -
1: (4.0 , 3.0)
4: -
3: (2.0 , 12.0)
4: -
2: (3.0 , 7.0)
4: -
3: (1.0 , 5.0)
4: -
Input 3 D points :
(1.0 , 12.0 , 1.0)
(18.0 , 1.0 , 2.0)
(2.0 , 12.0 , 16.0)
(7.0 , 3.0 , 3.0)
(3.0 , 7.0 , 5.0)
(16.0 , 4.0 , 4.0)
(4.0 , 6.0 , 1.0)
(5.0 , 5.0 , 17.0)
5: -
4: (5.0 , 5.0 , 17.0)
5: -
3: (16.0 , 4.0 , 4.0)
4: -
2: (7.0 , 3.0 , 3.0)
4: -
3: (18.0 , 1.0 , 2.0)
4: -
1: (4.0 , 6.0 , 1.0)
4: -
3: (1.0 , 12.0 , 1.0)
4: -
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2: (2.0 , 12.0 , 16.0)
4: -
3: (3.0 , 7.0 , 5.0)
4: -
Looking for points between (0.0 , 1.0 , 0.0) and (4.0 , 6.0 , 3.0).
Found :
(4.0 , 6.0 , 1.0)
Looking for points between (0.0 , 1.0 , 0.0) and (8.0 , 7.0 , 4.0).
Found :
(7.0 , 3.0 , 3.0)
(4.0 , 6.0 , 1.0)
Looking for points between (0.0 , 1.0 , 0.0) and (17.0 , 9.0 , 10.0).
Found :
(16.0 , 4.0 , 4.0)
(7.0 , 3.0 , 3.0)
(3.0 , 7.0 , 5.0)
(4.0 , 6.0 , 1.0)
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Question 2 (15 points):
In this question we will once again be looking at a problem related to quests in video games. In many
roleplaying video games, one completes quests to gain experience points. When one’s charater gains
enough experience points, their character advances and grows in power, or gains new abilities. Very
often there are quests that can only be attempted after one or more prerequisite quests have been
completed. We can represent this as a quest prerequisite graph, much like the course prerequisite graph
in Section 2.1, in which each quest is represented by a node, and there is a directed edge from node
A to node B if node A’s quest is a prerequisite to node B’s quest. In this question you will work with
just such a graph, and implement a topological sort algorithm that will output an ordering in which
all of the quests in the graph can be completed.
We can perform the topological sort of the graph using the following algorithm:
Algorithm TopologicalSort ( G )
G is a directed graph .
L = Empty list that will contain the result of the topological sort
S = set of nodes in G with no incoming edges ( i . e . indegree 0)
while S is non - empty do
remove a node n from S
add n to the end of the list L
for each node m with an edge e from n to m do
remove edge e from the graph
if m now has no incoming edges ( i . e . indegree 0) then
insert m into S
if the graph has any edges in it then
throw exception ( the graph had at least one cycle !!!)
else
return L ( a topologically sorted order )
Keeping in mind that each node of the graph G stores information about one quest, we can say that S
stores the set of quests that are "doable" or "available" (those for which all prerequisites have already
been completed) any any partcular moment. Being a set, it stores these quests in no particular order.
But... we are greedy. When we remove a node from S at the top of the while loop, we always want
to remove the quest in S that has the biggest experience point (XP) reward. In this way, we always do
the available quest with the largest experience reward first so as to gain experience more quickly. But
to do this we have to change this algorithm a little. Fortunately, it’s a very easy change. All we have
to do is change S from a set of graph nodes to a heap of quests that are keyed on the quest’s XP
value. Then we are guaranteed to always remove from S the available quest with the highest possible
experience point reward (because it will always be at the top of the heap!)1 The modified algorithm is:
1Note: although the modified algorithm will get us XP faster than if we used the original algorithm, it will NOT guarantee that
we gain the most possible XP with the fewest number of quests. To illustrate this, suppose we have Quests 1, 2, 3, and 4 which are
wroth 50, 500, 500, and 5000 XP respectively, but, quest 1 is a prerequisite to quest 4. Initially, the available quests are 1, 2, and 3
(4 cannot be done without doing 1 first). Following the algorithm with the "set" replaced by a heap, we would end up with quests
2 and 3 being first in the topological order, because they are both worth more experience than quest 1. Then we would have to do
quest 1, and finally quest 4. If we did the quests in this order, over the first 3 quests we would gain 1050 experience. But we could
have done quest 1 first, and then quest 4 immediately, giving us 5050 experience after just two quests. But we didn’t realize this
because we didn’t look ahead... our algorithm is "greedy" in this respect and we took the available quest with the best reward first.
This is fine and this is what we want to do here, I just don’t want you to expect you will always get the "optimum" experience gain
from this algorithm.
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Algorithm TopologicalSort ( G )
G is a directed graph .
L = Empty list that will contain the result of the topological sort
H = heap of quests ( compared by experience value ) whose corresponding
nodes in G have no incoming edges ,
while H is non - empty do
remove a quest n from H
add quest n to the end of the list L
for each graph node m such that there is an edge e from n ’ s graph node to m do
remove edge e from the graph
if m now has no incoming edges then
insert m ’ s quest into H
if the graph has any edges in it then
throw exception ( the graph had at least one cycle !!!)
else
return L ( a topologically sorted order )
You will be provided with the following files as part of an IntelliJ module called QuestPrerequisites-Template:
Quest.java A class for storing information about a quest. Note that it is different from the QuestLogEntry
class from Assignment 4. It contains mostly accessors and mutators for protected instance variables. It also has a toString() method and a compareTo methods that allow quests to be compared by their experience values. You may not modify this class.
QuestVertex.java A class to be used as the vertex class in a graph. It defines a graph node that can
store a reference to a quest. You can use the quest() method of this graph node to obtain the
quest associated with this node. You may not modify this class.
QuestProgression.java This class contains a few static methods that together form a program for
doing a topological sorting of quests in a "quest prerequisite graph". It contains the following
static methods:
readQuestFile: Reads a data file containing information about quests and quest prerequisites
and builds a quest prerequisite graph. This method is finished, and you don’t have to do
anything to it. Note that the graph that is created has nodes that are of type QuestVertex.
hasNoIncomingEdges: A method that determines whether a given node in a given graph has no
incoming edges, i.e., has indegree zero.
questProgression: A method that performs a topological sort of a given quest prerequisite
graph.
main: The main program that loads the quest data, constructs the graph, performs the topological
sort, and displays the result. This method is finished and you don’t have to do anything to
it.
quests16.txt A data file containing information on 16 quests and their prerequisites which is used by
main() as the program input.
You must complete the implementation of the hasNoIncomingEdges and questProgression methods. It is up to you to inspect the methods available in the GraphMatrixRep280 class (and it’s superclasses!) and use them to solve the problem. Because you are not implementing methods within the
graph class, you can only access and modify the graph via its public interface.
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Hints
This program relies on a correspondence between graph nodes and quests. Note that in the graph
ADTs, nodes are referred to by integers, starting from 1. Note also that each Quest instance contains an
ID (you can see this in Quest.java). The quest prerequisite graph is constructed by the readQuestFile
method such that a quest with a particular ID k is always associated with node k of the graph. Thus,
if you have a quest object, you can find its corresponding graph node by looking up the node in the
graph with the quest’s ID. If you have a node ID, you can find its corresponding quest by looking
up the node object (of type QuestVertex) using its ID, and call the quest() method of the resulting
vertex object to obtain the quest stored there.
Sample output is shown below.
Sample Output
If you implement the unfinished methods correctly, the output of the program when given quests16.txt
as input will be:
13 , Discover Peppermint Butler ’ s Secrets , Candy Kingdom , XP : 14550
7 , Find the Ice King ’ s Wizard Eye , Ice Kingdom , XP : 12000
12 , Rescue Marceline from the Nightosphere , The Nightosphere , XP : 25000
10 , Win Wizard Battle , Wizard Battle Arena , XP : 29000
9 , Rescue Wildberry Princess from the Ice King , Ice Kingdom , XP : 4200
3 , Defeat Goliad , Candy Kingdom , XP : 2578
5 , Atone for Shoko ’ s Sins , Finn ’ s Treehouse , XP : 2700
4 , Gain Approximate Knowledge of Many Things , Demon Cat Lair , XP : 1000
8 , Get some pickles from Prismo , Prismo ’ s Home , XP : 150
6 , Make an Amazing Sandwich , Finn ’ s Treehouse , XP : 1900
16 , Watch what Beemo Does When He Is Alone , Finn ’ s Treehouse , XP : 70
1 , Steal a Treat from the Donut Witch ’ s Garden , Blue Plains , XP : 250
14 , Defeat the Ice King ’ s Penguin Army , Candy Kingdom , XP : 50000
11 , Eat Marceline ’ s Fries , Marceline ’ s House , XP : 50
15 , Discover the Ice King ’ s Secret Past , The Past , XP : 3299
2 , Recover the Stolen Items from the Door Lord , The Sandylands , XP : 700
Another test you can do to check whether your program is working is to make a copy of quests16.txt
and remove all of the ordered pairs starting on line 18 of the file. If you do this, then the topological
sort should result in a list of quests sorted in descending order by their experience value.
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4 Files Provided
lib280-asn7: A copy of lib280 which includes:
• solutions to assignment 6;
• the NDPoint280 class in the lib280.base package for representing n-dimensional points for
question 1;
• the GraphMatrixRep280 which you’ll use in question 2;
• QuestPrerequisites-Template — an IntelliJ module with templates for question 2.
5 What to Hand In
KDNode280.java: The node class for your k-D tree from Question 1.
KDTree280.java: Your k-D tree class for Question 1.
a6q2.txt/doc/pdf: The console output from your test program for question 1, cut and paste from the IntelliJ
console window.
QuestProgression.java Your completed QuestProgression class from question 2.
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