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COMP 250 java

创建日期：2024-04-08 15:53:21

所属分类：计算机computer science

标签： COMP 250

java

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COMP 250

General Instructions
• Submission instructions
– Late assignments will be accepted up to 2 days late and will be penalized by 10 points per day.
Note that submitting one minute late is the same as submitting 23 hours late. We will deduct
10 points for any student who has to resubmit after the due date (i.e. late) irrespective of the
reason, be it wrong file submitted, wrong file format was submitted or any other reason. This
policy will hold regardless of whether or not the student can provide proof that the assignment
was indeed “done” on time.
– Don’t worry if you realize that you made a mistake after you submitted : you can submit
multiple times but only the latest submission will be kept. We encourage you to submit a first
version a few days before the deadline (computer crashes do happen and codePost may be
overloaded during rush hours).
– These are the files you should be submitting on codePost:
* DecisionTree.java
Do not submit any other files, especially .class files. Any deviation from these requirements
may lead to lost marks
• To run the code in Eclipse, you will need to use the default package and not have a package
statement. (The reason for this constraint has to do with the use of serialization. We have tried
to make it run in other packages but we not able to at the time of the assignment release. We will
update you if we find a solution.)
• Do not change any of the starter code that is given to you. Add code only where instructed,
namely in the “ADD CODE HERE” block. You should not need to add helper methods. If you
wish to add helper (i.e. private!) methods, you may but only in the DecisionTree class.
• The assignment shall be graded automatically. Requests to evaluate the assignment manually shall
not be entertained, so please make sure that you follow the instruction closely or your code may fail
to pass the automatic tests. Note that for this assignment, you are NOT allowed to import any other
class. Any failure to comply with these rules will give you an automatic 0.
• Whenever you submit your files to codepost, you will see the results of some exposed tests. These
tests are a mini version of the tests we will be using to grade your work. If your code fails those
tests, it means that there is a mistake somewhere. Even if your code passes those tests, it may still
1
contain some errors. We will test your code on a much more challenging set of examples. We highly
encourage you to test your code thoroughly before submitting your final version.
• In a week we will share with you a Minitester class that you can run to test if your methods
are correct. This class is equivalent to the exposed tests on codePost. Please note that these tests
are only a subset of what we will be running on your submissions. We encourage you modify and
expand these classes. You are welcome to share your tester code with other students on Piazza. Try
to identify tricky cases. Do not hand in your tester code.
• Your code will be tested on valid inputs only.
• You will automatically get 0 if your code does not compile.
• Failure to comply with any of these rules will be penalized. If anything is unclear, it is up to you to
clarify it by asking either directly a TA during office hours, or on the discussion board on Piazza.
2
1 INTRODUCTION
Learning Objectives
In this assignment you will be learning about decision trees and how to use them to solve classification
problems. Working on this problem will allow you to better understand how to manipulate trees and how
to use recursion to exploit their recursive structure.
1 Introduction
Congratulations! You have just landed an internship at a startup software company. This company is
trying to use AI techniques – in particular decision trees – to analyze spatial data. Your first task in this
internship is to write a basic decision tree class in Java. This will demonstrate to your new employer that
you understand what decision trees are and how they work.
From a quick web search, you learn that decision trees are a classical AI technique for classifying objects
by their properties. One typically refers to object attributes rather than object properties, and one typically
refers object labels to say how an object is classified.
As a concrete example, consider a computer vision system that analyzes surveillance video in a large store
and it classifies people seen in the video as being either employees or customers. An example attribute
could be the location of the person in the store. Employees tend to spend their time in different places than
customers. For example, only employees are supposed to be behind the cash register.
For classification problems in general, one defines object attributes with x variables and one defines the
object label as a y variable. In the example that you will work with in this assignment, the attributes will
be the spatial position (x1, x2), and the label y will be a color (red or green).
Let’s get back to decision trees themselves. Decision trees are rooted trees. So they have internal nodes
and external nodes (leaves). To classify a data item (datum) using a decision tree, one starts at the root and
follows a path to a leaf. Each internal node contains an attribute test. This test amounts to a question about
the value of the attribute – for example, the location of a customer in a store. Each child of an internal
node in a decision tree represents an outcome of the attribute test. For simplicity, you will only have to
deal with binary decision trees, so the answers to attribute test questions will be either true or false. A test
might be x1 < 5. The answer determines which child node to follow on the path to a leaf.
The labelling of the object occurs when the path reaches a leaf node. Each leaf node contains a label that
is assigned to any test data object that arrives at that leaf node after traversing the tree from the root. The
label might be red or green, which could be coded using an enum type, or simply 0 or 1. Note that, for any
test data object, the label given is the label of the leaf node reached by that object, which depends on the
outcomes of the attribute tests at the internal nodes.
The reason that this document is larger than usual is that decision trees were not covered the pre-recorded
lectures. This document should give you enough information about decision trees for you to do the as-
signment. If you wish to learn more about decision treees, then there are ample resources available on the
web. Steer towards resources that are about decision trees in computer science, in particular, in machine
learning or data mining. For example:
https://en.wikipedia.org/wiki/Decision_tree_learning
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1.1 Creating decision trees 1 INTRODUCTION
Be aware that these resources will contain more information than you need to do this assignment, and so
you would need to sift through it and figure out what is important and what can be ignored. Feel free to use
Piazza to share links to good resources and to resolve questions you might have. The task of understanding
what decision trees are is part of the assignment. The amount of coding you need to do is relatively small,
once you figure out what needs to be done.
1.1 Creating decision trees
To classify objects using a decision tree, we first need to have a decision tree! Where do decision trees
come from? In machine learning, one creates decision trees from a labelled data set. Each data item
(datum) in the given labelled data set has well defined attributes x and label y. We refer to the data set
that is used to create a decision tree as the training set. The basic algorithm for creating a decision tree
using a training set is as follows. This is the algorithm that you will need to implement for fillDTNode()
later.
Data: data set (training)
Result: the root node of a decision tree
MAKE DECISION TREE NODE(data)
if the labelled data set has at least k data items (see below) then
if all the data items have the same label then
create a leaf node with that class label and return it;
else
create a “best” attribute test question; (see details later)
create a new node and store the attribute test in that node, namely attribute and threshold;
split the set of data items into two subsets, data1 and data2, according to the answers to the test
question;
newNode.child1 = MAKE DECISION TREE NODE(data1)
newNode.child2 = MAKE DECISION TREE NODE(data2)
return newNode
end
else
create a leaf node with label equal to the majority of labels and return it;
end
In the program, k is an argument of the decision tree construction minSizeDatalist.
1.2 Classification using decision trees
Once you have a decision tree, you can use it to classify new objects. This is called the testing phase.
For the testing phase, one can use data items from the original data used for training (above) or one can
use new data. Typically when a decision tree is used in practice, the test objects are unlabelled. In the
surveillance example earlier, the system would test a new video and try to classify people as employees
versus customers. Here the idea is that one does not know the correct class for each person. Let’s consider
this general scenario now, that we are given a decision tree and the attributes of some new unlabelled
test object. We will use the decision tree to choose a label for the object. This is done by traversing the
decision tree from the root to a leaf, as follows:
4
2 INSTANTIATING THE DECISION TREE PROBLEM
Data: A decision tree, and an unlabelled data item (datum) to be classified
Result: (Predicted) classification label
CLASSIFY(node, datum) {
if node is a leaf then
return the label of that node i.e. classify;
else
test the data item using question stored at that (internal) node, and determine which child node to go
to, based on the answer ;
return CLASSIFY(child, datum);
end
}
2 Instantiating the decision tree problem
For this assignment, the problem is to classify points based on their position. Each datapoint has an array
of attributes x, and a binary label y (0 or 1). For this section, we will focus on datapoints with only two
attributes.
A graphical representation of example of a data set looks like this. (For the graphs, the attribute value x[0],
is represented as x1 and x[1] as x2.) For those who print out the document in color, the red symbols can
be label 0 and the green symbols can be label 1. For those printing in black and white, the (red) disks are
label 0 and the (green) ×’s are label 1.
Figure 1: x1 is horizontal and x2 is vertical coordinate. Note that the points are intermixed. There is no
way to draw a horizontal or vertical line or any curve for that matter that could split the data.
5
2.1 Finding a good split 2 INSTANTIATING THE DECISION TREE PROBLEM
2.1 Finding a good split
Now that we have an idea of what the data are, let us return to the question of how to split the data into two
sets when creating a node in a decision tree. What makes a ‘good’ split? Intuitively, a split is good when
the labels in each set are as ‘pure’ as possible, that is, each subset is dominated as much as possible by a
single label (and the dominant label differs between subsets). For example, suppose this is our data:
Figure 2: What would be a good split of this data?
Two of many possible splits we could make are shown in Fig. 3. Fig 3-a splits the data into two sets
based on the test condition (x1 < 4), i.e. true or false. (By definition, the green symbol that falls on this
line is considered to be in the right half since the inequality is strict.) This is a good split in that all data
points for which the test condition is false have the same label (green) and all data points for which the
test condition is true have the same label (red), and the labels differ in the two subsets.
The split condition (x1 < 6) in Fig. 3-b is not as good, since the subset for which the condition is true
contains datapoints of both labels.
(a) (b)
Figure 3: Example of different splits on the x1 attribute.
6
2.2 Entropy 2 INSTANTIATING THE DECISION TREE PROBLEM
Splits can be done on either of the attributes. For the example in Fig. 4-a, a good split would be defined
by the test condition (x2 < 4).
The situation is more complicated when the data points cannot be separated by a threshold on x1 or x2, as
in Fig 4-b, however. It is unclear how to decide which of the three splits is best. We need a quantitative
way of deciding how to do this.
(a) (b)
Figure 4: (a) Example of a data set and a split using the x2 attribute, where the two subsets have distinct
labels. (b) An example of a data set in which there is no way to split using either an x1 or x2 value, such
that the two subsets have distinct labels.
2.2 Entropy
To handle more complicated situation, ones needs a quantitative measure of the ‘goodness’ of a split, in
terms of the impurity of the labels in a set of data points. There are many ways to do so. One of the most
common is called entropy.1 The standard definition2 of entropy is:
H = −
∑
i
pi log2(pi) (1)
where p(i) is a function such that 0 ≤ pi ≤ 1 and
∑
i pi = 1. Note that the minus sign is needed to make
H positive, since log2 pi < 0 when 0 < pi < 1. Also, note that if pi = 0 then pi log2(pi) = 0 since that is
the limit of this expression as pi → 0. (Recall l’Hopital’s Rule in Calculus 1.)
For the special case that there are two values only, namely p1 and p2 = 1− p1, entropy H is between 0 and
1, and we can write H as a function of the value p = p1. For a plot of this function H(p), see:
https://en.wikipedia.org/wiki/Binary_entropy_function
1Entropy is an extremely important concept in science. It has its roots in thermodynamics in the 19th century. In the
20th century, “information entropy” was one of the basic for techniques in electronic communication (telephones, cell phones,
internet, etc). In computer science, information entropy is heavily used in data compression, cryptography, and AI.
2Such functions pi are often used to model probabilities, as you will learn if you take MATH 323 or MATH 203 for example.
7
2.3 Using entropy to define a good split 2 INSTANTIATING THE DECISION TREE PROBLEM
In this assignment, we use entropy to choose the best split for a data set, based on its labels. We borrow
the formula for entropy and apply it to our problem as follows:
H(D) = −
∑
y∈L
p(y) log2 p(y) (2)
where
• L is the set of labels, and y is a particular label
• D is a data set; each data point has two attributes and a label i.e. (x1, x2, y)
• H(D) is the entropy of the dataset D
• p(y) is the fraction of data points with label y from dataset D.
Since L consists of only two labels, entropy is between 0 and 1. Entropy is 0 if p(y) takes values 0 and 1
for the two labels. Entropy is 1 if p(y) = 0.5 for both labels. Otherwise it has a value strictly between 0
and 1. See plot in the link above.
2.3 Using entropy to define a good split
During the training phase, when one constructs the decision tree, a node is given a data set D as input.
If D has entropy greater than 0, then we would like to split the data set into two subsets D1 and D2. We
would like the entropy of the subsets to be lower than the entropy of D. The subsets may have different
entropy, however, so we consider the average entropy of the subsets. Moreover, because one subset might
be larger than the other, we would like to give more weight to the larger subset. So we define the average
entropy like this:
H(D1, D2) ≡ w1 ×H(D1) + w2 ×H(D2)
where wi is the fraction of the points in subset i,
wi =
number of datapoints in Di
number datapoints in D, namely D1 +D2
and i is either 1 or 2. Note that w1 + w2 = 1.
NOTE: DO NOT use the formula : w2 = 1 - w1, although correct, this leads to a numerical approximation
error. For each of the weights (w1, w2) use the formula mentioned above separately.
ASIDE: (We mention the following because you will likely encounter it in your reading.) When building
a decision tree, one often considers the difference H(D) − H(D1, D2), which is called the information
gain. For example, one may decide whether or not to split a node based on whether the information gain
is sufficiently large. In this assignment, you will instead use a different criterion to decide whether to split
a node when building the decision tree Your criterion will be based on the number of data items in D, as
will be discussed later.
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2.4 Example 2 INSTANTIATING THE DECISION TREE PROBLEM
2.4 Example
Let us now return to the examples we saw earlier and use entropy to discuss which split is best. Recall the
example of Fig. 4-b. To calculate the entropy of the dataset D before split, note there are two classes with
6 points in each of the classes. So the fraction of points in each class is 0.5 and the entropy is:
H(D) = −1
2
log2
(
1
2
)
− 1
2
log2
(
1
2
)
= 1 (3)
• Split 1 breaks the dataset into two sub datasets, where the subdataset on top contains 8 points (5
red dots , 3 green crosses) and the one below has 4 points (1 red dot, 3 green crosses). Calculating
average entropy H(D1, D2) after the split yields:

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CS 333 CP1406 KIT308 Economics 701 ECON3740 ECE 5759 BFC3340 FINC2012 COMP4500 MATH3871 FIT3080 18-819C Economic ECO202Y5Y PHY100 FINC-780 Econ 100B EC602 IE582 EE450 COMP9517 WRDS 150B ECON 570 LECTURE 2 ECA5101 ECE 4420 ADAD9311 FIT3031 CS 544 MF 703 CS 6140 KIT501 ENGSCI 344 CSC171 IE 332 FINM8007 21S2 COSC 1114 FE545 E471 SIT102 CSC108 MAT301 IAE 101 COP3503 STAT2004 EC 303 CS1026 STAT1203 MATH3171 ISOM5160 IFN657 FIT2014 ACCT2102 DSCI550 STATS 330 ELEC421 COMP3670 QBUS6840 MEES:698B MAST30027 MIET1077 Economics IFT 6758 FIT5122 BUSN5101 CptS 111 HRMT5502 Math2018 PP 312 INFS3200 CZ3005 CSSE1001 SENG 2011 MKTG 553Q BUFN 640 COMP507 ETF5952 DATA 311 ECE4043 BANK3011 finance M3H623544 HW1 INFS 7410 AI6103 STAT3888 INFS3208 ACCT7103 ECOS3013 PSYCH 20B F79MA STAT3006 ENGG952 2X MGSC 7000 ECE4132 MSF 526 MATH 3033 CDS 511 EEEE4120 EEET2465 ECA5103 MATH2070 ELEC3104 BU51019 EEE8147 ENGR20003 COSC 1107 ETF3200 FNCE 435 MSFI 435 ISyE6420 GENG4405 HW ETC5523 IE 7315 MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A PHIL2617 PHYS3116 MIE250 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