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java代写-COMP 250-Assignment 3
时间：2021-04-11
Assignment 3
COMP 250 Winter 2021
posted: Friday, Mar. 26, 2021
due: Tuesday, April. 13, 2021 at 23:59
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
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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.
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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:
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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.
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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.
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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.
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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:
4
12
(
−
(
1
4
log2
(
1
4
)
+
3
4
log2
(
3
4
)))
+
8
12
(
−
(
5
8
log2
(
5
8
)
+
3
8
log2
(
3
8
)))
= 0.906 (4)
• Split 2 breaks the dataset into two sub datasets, where the sub dataset on the left has 5 datapoints
(4 red dots, 1 green cross) and the other one has 7 datapoints ( 2 red dots , 5 green crosses). The
average entropy H(D1, D2) after the split is:
5
12
(
−
(
1
5
log2
(
1
5
)
+
4
5
log2
(
4
5
)))
+
7
12
(
−
(
2
7
log2
(
2
7
)
+
5
7
log2
(
5
7
)))
= 0.803 (5)
• Split 3 breaks the dataset into 2 sub datasets, where the sub dataset on the left has 7 datapoints (5 red
dots, 2 green crosses) and the other one has 5 datapoints ( 1 red dot , 4 green crosses). The average
entropy H(D1, D2) after the split is:
7
12
(
−
(
2
7
log2
(
2
7
)
+
5
7
log2
(
5
7
)))
+
5
12
(
−
(
1
5
log2
(
1
5
)
+
4
5
log2
(
4
5
)))
= 0.803 (6)
So, split 2 and 3 have lower average entropy than split 1. It is no problem that split 2 and 3 have the
same average entropy. Either one can be chosen as the ‘best split’. For the assignment, select the first
encountered best split when the check for the best split starts from the first attribute (x[0]) and proceeds
from there and for a given attribute the check starts from the first datapoint and proceeds thereafter.
2.5 Finding the best split
We can use entropy to compare the ‘goodness’ of different splits. The simplest way to define the best
split is just to consider all possible splits and choose the one with the lowest average entropy. In this
assignment, you will take this brute force approach. You will consider each attribute x1 and x2 and each
of the values of that attribute for points in the data set. You will compute the average entropy for splitting
on that attribute value, and you will choose the split that gives the minumum.
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2.5 Finding the best split 2 INSTANTIATING THE DECISION TREE PROBLEM
Data: A Dataset
Result: An attribute and a threshold
FIND BEST SPLIT(data) {
best avg entropy := inf;
best attr := -1;
best threshold := -1;
for each attribute in x do
for each data point in list do
compute split and current avg entropy based on that split;
if best avg entropy > current avg entropy then
best avg entropy := current avg entropy;
best attr := attribute;
best threshold := value;
end
end
end
return (best attr, best threshold)
}
Note the order of the two for loops! If you use the opposite order, you might get a different tree (wrong
answer).Note also that if the minim average entropy you find is equal to the entropy of the input data set,
then there is no point in performing a split. In this case, the node should simply be made a leaf with label
equal to the majority of the labels in the data set.
So, far we have seen how to create the decision tree and what are the intuitions behind the math needed to
get the split to build the tree. Let us consider another example. Consider the data shown below.
A decision tree for this dataset should have splits as shown below.
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2.6 Preventing Overfitting 3 INSTRUCTIONS AND STARTER CODE
Figure 5: Example of an overfit decision tree on the outlier dataset
But does this seem right? Given that all the points around it are of class 1 and the other points of the same
class are far to the left, it might be possible that the datapoint labeled class 2 at the right side of the graph
(6,6) is an outlier or an anomalous reading in the data. Points like these are most of the time, if not always
present in a dataset and care should be taken so that our decision tree does not try too hard to reduce the
impurity of the sub datasets and in the process actively keep splitting the data so as to try to classify the
outliers into purer groups.
This phenomenon described above is called overfitting. The algorithm – in this case the construction of a
decision tree – tries to find a “model” that accounts for all of the data, even the data which might be garbage
for some reason (noise, or due to some error in the program or device that produced the data).
2.6 Preventing Overfitting
There are different ways to prevent overfitting. The method that you will use is called early stopping,
namely stop splitting nodes in the decision tree when the number of data points in a subset is smaller than
some predetermined number. The issue of overfitting is very important in decision trees and in machine
learning in general, but the details are beyond the scope of this assignment.
3 Instructions and starter code
The starter code contains three classes:
Datum.java
This class holds the information of a single datapoint. It has two variables, x and y. x is an array containing
the attributes and y contains the label.
The class also comes with a method toString() which returns a string representation of the attributes
and label of a single datapoint.
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3 INSTRUCTIONS AND STARTER CODE
DataReader.java
This class deals with three things. The method read data() reads a dataset from a csv3, and splits the
read dataset into the training and test set, using splitTrainTestData() file. It also has methods that
deal with reading and writing of “serialized” decision tree objects.
DecisionTree.java
This is the main class which deals with the creation of a decision tree and classification of datapoints using
the created decision tree. You will be implementing some of the methods in this class.
Let us go through the different members and attributes of the class:
• The constructor builds a decision tree by calling the fillDTNode() method on a dataset. It is
given a list of data points and a parameter that specifies the minimum number of datapoints that has
to be present in a given dataset to qualify for a split. This minimum number is used to reduced the
chances of overfitting, as discussed above.
• There is a root node field, called rootDTNode, by which other nodes can be accessed.
• There is a field called minSizeDatalist used to store the minimum number of datapoints that
should be present in the dataset so as to initiate a split.
• There is a subclass class DTNode. This class is used to represent a single node of a decision tree.
There are two types of nodes: the internal nodes which define an attribute and a threshold which
help in classification, and leaf nodes which determine the labels of those data points whose attributes
obey the threshold conditions of the ancestor internal nodes leading up to the leaf node.
The DTNode contains the following members:
– leaf: a boolean variable that indicates whether this node object is a leaf or not.
– label: an integer variable that indicates the label of the node. The label of the node indicates
the class of a datapoint that reaches that particular leaf node after traversing the tree. This is
valid only if the node is a leaf node. The classes for this assignment are simply 0 or 1.
– attribute: The index of the attribute, (i.e. 1, . . . , n) on which the dataset is split in that par-
ticular node. The value of the attribute is one of {1, . . . , n}. The attribute value is meaningful
only for an internal node. The value stored in this field is meaningful only for an internal node.
In the code, the attributes are x[0], x[1], .., x[n-1], hence their indices values are
0, 1, . . . , n− 1, respectively.
– threshold: This holds the value of the attribute at which the split is done. This is also only
meaningful for an internal node.
– left, right These two variables are of type DTNode, and they represent the two children
of an internal node. At the classify stage, the left child leads to a decision tree node that handles
3According to https://en.wikipedia.org/wiki/Comma-separated_values, comma-separated values
(CSV) file is a delimited text file that uses a comma to separate values. A CSV file stores tabular data (numbers and text)
in plain text. Each line of the file is a data record. Each record consists of one or more fields, separated by commas.
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3 INSTRUCTIONS AND STARTER CODE
the case that the value of the attribute is less than the threshold, and right handles the case that
the value is greater than or equal to the threshold. For a leaf node, they are both null.
The class DTNode also contains a few member methods that helps in building the tree.
– fillDTNode(): This is the method that does all the heavy lifting in the entire assignment.
Given a list of data points and a minimum size for splitting (see earlier), this recursive function
creates the entire decision tree. A detailed description is present in the comments of the code.
– findMajority(): Given a list of datapoints D, this method returns a label for that set.
This method is called during training (construction of the decision tree). It can happen that
the size of a dataset falls below the minimum size mentioned in minSizeDatalist, but
still contains datapoints from more than one class. In such a case, a leaf node is created. To
determine the label for this leaf node, the method goes through all the points and finds the
majority label, i.e. the most common label for those points. This label will be used later at the
classification phase, when a new data point reaches that leaf node. When choosing the label
for a leaf node, if there is no majority (a tie), then the label with the smallest value is returned.
– classifyAtNode(): At the testing phase, given a datapoint with only its attributes (no
label), this method uses the label specified by the decision tree leaf node (as was determined at
training time by the findMajority() method).
– equals() : Given another node, this method checks if the tree rooted at the given node is
equal to the tree rooted at the calling node. The definition of ‘equality’ is elaborated in the next
section.
• calcEntropy(): Given a dataset, this function calculates the entropy of the dataset.
• classify(): Given a datapoint (without the label), predicts the label of the datapoint. The only
difference between this method and classifyAtNode() is that classifyAtNode() does
the classification on its member DTNode, whereas for classify() the DTNode is the root of the
created decision tree.
• checkPerformance(): Given a dataset where the datapoints have both attributes and labels,
this method runs the classify() function on all of the datapoints (using only the attributes) and
compares the label that is given by the decision tree with the “ground truth” label for that data point.
The method returns the fraction of datapoints that were predicted wrong, in the form of a string.
• equals(): Given two decision trees, this method checks if the two trees are equal or not. It returns
a boolean value.
Your task
You need to implement three methods from the DTNode class. None of the methods depend on each other.
We suggest that you implement equals() first.
1. DTNode.equals() (30 points)
This method compares two DTNodes. Given another DTNode object, it checks if the tree that is
rooted at the calling DTNode is equal to the tree rooted at DTNode object that is passed as the
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4 DATA
parameter.
Two DTNodes are considered equal if:
(a) a traversal (e.g. preorder) of each of the two trees encounters nodes that are equal;
(b) internal node : the thresholds and attributes should be same;
(c) leaf node : the labels should be same.
Note that the tester you are given uses this equals method to check if the tree you implement in the
second part matches with the actual solution.
2. DTNode.fillDTNode() (50 points)
This method takes in a datalist (i.e. an arraylist of objects of type Datum) and returns the calling
DTNode object as the root of a decision tree trained using the datapoints present in the input datalist.
3. DTNode.classifyAtNode() (20 points)
This method takes in a datapoint (excluding the label) in the form of an array of type double (like
Datum.x) and should return its corresponding label (int).
4 Data
The different datasets used in the assignment are shown below. If you look at the DataReader() class,
you will note that only half of the data in each plot is used in the training. The other half is used to test the
performance of the decision tree. This testing phase is not part of the assignment, but we give you some
performance data so that you can appreciate the differences in the data sets.
For each of the three data sets, we list the performance of the decision tree in classifying the other half
(test set) of the data. We show both the fraction of training points that are misclassified (error) and
the fraction of the test points that are misclassified, and we do so for different values of the variable
minSizeDatalist.
4.1 Highly overlapping data :
Notice that the training error is 0 when the minimum size is 1 and rises as minSizeDatalist increases.
However, the test error is near .5 (50 percent) for all values of minSizeDatalist.
minSizeDatalist : 1 Training error : 0.000 Test error : 0.495
minSizeDatalist : 2 Training error : 0.000 Test error : 0.495
minSizeDatalist : 4 Training error : 0.030 Test error : 0.495
minSizeDatalist : 8 Training error : 0.105 Test error : 0.520
minSizeDatalist : 16 Training error : 0.200 Test error : 0.515
minSizeDatalist : 32 Training error : 0.235 Test error : 0.515
minSizeDatalist : 64 Training error : 0.310 Test error : 0.525
minSizeDatalist : 128 Training error : 0.390 Test error : 0.490
14
4.2 Partially overlapping data : 4 DATA
Figure 6: Plot of data present in data high overlap.csv
4.2 Partially overlapping data :
Figure 7: Plot of data present in data partial overlap.csv
In this case, 10 or 12 of the 200 test data points are misclassified for most values of minSizeDatalist.
The error rises considerably when minSizeDatalist is 128.
minSizeDatalist : 1 Training error : 0.000 Test error : 0.050
minSizeDatalist : 2 Training error : 0.000 Test error : 0.050
minSizeDatalist : 4 Training error : 0.015 Test error : 0.050
minSizeDatalist : 8 Training error : 0.035 Test error : 0.060
minSizeDatalist : 16 Training error : 0.045 Test error : 0.050
15
4.3 Minimal overlapping data : 4 DATA
minSizeDatalist : 32 Training error : 0.045 Test error : 0.050
minSizeDatalist : 64 Training error : 0.075 Test error : 0.050
minSizeDatalist : 128 Training error : 0.255 Test error : 0.245
4.3 Minimal overlapping data :
Figure 8: The data points here have very little overlap. We are loosely calling the overlap “minimal”,
although truly minimal would mean zero overlap. We leave some overlap so that overfitting can potentially
occur.
Note exactly one (of 200) training data points is misclassfied when minSizeDatalist is 4 or more.
The test error is roughly constant. It is slightly larger for small values of minSizeDatalist.
minSizeDatalist : 1 Training error : 0.000 Test error : 0.040
minSizeDatalist : 2 Training error : 0.000 Test error : 0.040
minSizeDatalist : 4 Training error : 0.005 Test error : 0.040
minSizeDatalist : 8 Training error : 0.005 Test error : 0.035
minSizeDatalist : 16 Training error : 0.005 Test error : 0.035
minSizeDatalist : 32 Training error : 0.005 Test error : 0.035
minSizeDatalist : 64 Training error : 0.005 Test error : 0.035
minSizeDatalist : 128 Training error : 0.005 Test error : 0.035
16
学霸联盟
COMP 250 Winter 2021
posted: Friday, Mar. 26, 2021
due: Tuesday, April. 13, 2021 at 23:59
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
3
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.
8
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:
4
12
(
−
(
1
4
log2
(
1
4
)
+
3
4
log2
(
3
4
)))
+
8
12
(
−
(
5
8
log2
(
5
8
)
+
3
8
log2
(
3
8
)))
= 0.906 (4)
• Split 2 breaks the dataset into two sub datasets, where the sub dataset on the left has 5 datapoints
(4 red dots, 1 green cross) and the other one has 7 datapoints ( 2 red dots , 5 green crosses). The
average entropy H(D1, D2) after the split is:
5
12
(
−
(
1
5
log2
(
1
5
)
+
4
5
log2
(
4
5
)))
+
7
12
(
−
(
2
7
log2
(
2
7
)
+
5
7
log2
(
5
7
)))
= 0.803 (5)
• Split 3 breaks the dataset into 2 sub datasets, where the sub dataset on the left has 7 datapoints (5 red
dots, 2 green crosses) and the other one has 5 datapoints ( 1 red dot , 4 green crosses). The average
entropy H(D1, D2) after the split is:
7
12
(
−
(
2
7
log2
(
2
7
)
+
5
7
log2
(
5
7
)))
+
5
12
(
−
(
1
5
log2
(
1
5
)
+
4
5
log2
(
4
5
)))
= 0.803 (6)
So, split 2 and 3 have lower average entropy than split 1. It is no problem that split 2 and 3 have the
same average entropy. Either one can be chosen as the ‘best split’. For the assignment, select the first
encountered best split when the check for the best split starts from the first attribute (x[0]) and proceeds
from there and for a given attribute the check starts from the first datapoint and proceeds thereafter.
2.5 Finding the best split
We can use entropy to compare the ‘goodness’ of different splits. The simplest way to define the best
split is just to consider all possible splits and choose the one with the lowest average entropy. In this
assignment, you will take this brute force approach. You will consider each attribute x1 and x2 and each
of the values of that attribute for points in the data set. You will compute the average entropy for splitting
on that attribute value, and you will choose the split that gives the minumum.
9
2.5 Finding the best split 2 INSTANTIATING THE DECISION TREE PROBLEM
Data: A Dataset
Result: An attribute and a threshold
FIND BEST SPLIT(data) {
best avg entropy := inf;
best attr := -1;
best threshold := -1;
for each attribute in x do
for each data point in list do
compute split and current avg entropy based on that split;
if best avg entropy > current avg entropy then
best avg entropy := current avg entropy;
best attr := attribute;
best threshold := value;
end
end
end
return (best attr, best threshold)
}
Note the order of the two for loops! If you use the opposite order, you might get a different tree (wrong
answer).Note also that if the minim average entropy you find is equal to the entropy of the input data set,
then there is no point in performing a split. In this case, the node should simply be made a leaf with label
equal to the majority of the labels in the data set.
So, far we have seen how to create the decision tree and what are the intuitions behind the math needed to
get the split to build the tree. Let us consider another example. Consider the data shown below.
A decision tree for this dataset should have splits as shown below.
10
2.6 Preventing Overfitting 3 INSTRUCTIONS AND STARTER CODE
Figure 5: Example of an overfit decision tree on the outlier dataset
But does this seem right? Given that all the points around it are of class 1 and the other points of the same
class are far to the left, it might be possible that the datapoint labeled class 2 at the right side of the graph
(6,6) is an outlier or an anomalous reading in the data. Points like these are most of the time, if not always
present in a dataset and care should be taken so that our decision tree does not try too hard to reduce the
impurity of the sub datasets and in the process actively keep splitting the data so as to try to classify the
outliers into purer groups.
This phenomenon described above is called overfitting. The algorithm – in this case the construction of a
decision tree – tries to find a “model” that accounts for all of the data, even the data which might be garbage
for some reason (noise, or due to some error in the program or device that produced the data).
2.6 Preventing Overfitting
There are different ways to prevent overfitting. The method that you will use is called early stopping,
namely stop splitting nodes in the decision tree when the number of data points in a subset is smaller than
some predetermined number. The issue of overfitting is very important in decision trees and in machine
learning in general, but the details are beyond the scope of this assignment.
3 Instructions and starter code
The starter code contains three classes:
Datum.java
This class holds the information of a single datapoint. It has two variables, x and y. x is an array containing
the attributes and y contains the label.
The class also comes with a method toString() which returns a string representation of the attributes
and label of a single datapoint.
11
3 INSTRUCTIONS AND STARTER CODE
DataReader.java
This class deals with three things. The method read data() reads a dataset from a csv3, and splits the
read dataset into the training and test set, using splitTrainTestData() file. It also has methods that
deal with reading and writing of “serialized” decision tree objects.
DecisionTree.java
This is the main class which deals with the creation of a decision tree and classification of datapoints using
the created decision tree. You will be implementing some of the methods in this class.
Let us go through the different members and attributes of the class:
• The constructor builds a decision tree by calling the fillDTNode() method on a dataset. It is
given a list of data points and a parameter that specifies the minimum number of datapoints that has
to be present in a given dataset to qualify for a split. This minimum number is used to reduced the
chances of overfitting, as discussed above.
• There is a root node field, called rootDTNode, by which other nodes can be accessed.
• There is a field called minSizeDatalist used to store the minimum number of datapoints that
should be present in the dataset so as to initiate a split.
• There is a subclass class DTNode. This class is used to represent a single node of a decision tree.
There are two types of nodes: the internal nodes which define an attribute and a threshold which
help in classification, and leaf nodes which determine the labels of those data points whose attributes
obey the threshold conditions of the ancestor internal nodes leading up to the leaf node.
The DTNode contains the following members:
– leaf: a boolean variable that indicates whether this node object is a leaf or not.
– label: an integer variable that indicates the label of the node. The label of the node indicates
the class of a datapoint that reaches that particular leaf node after traversing the tree. This is
valid only if the node is a leaf node. The classes for this assignment are simply 0 or 1.
– attribute: The index of the attribute, (i.e. 1, . . . , n) on which the dataset is split in that par-
ticular node. The value of the attribute is one of {1, . . . , n}. The attribute value is meaningful
only for an internal node. The value stored in this field is meaningful only for an internal node.
In the code, the attributes are x[0], x[1], .., x[n-1], hence their indices values are
0, 1, . . . , n− 1, respectively.
– threshold: This holds the value of the attribute at which the split is done. This is also only
meaningful for an internal node.
– left, right These two variables are of type DTNode, and they represent the two children
of an internal node. At the classify stage, the left child leads to a decision tree node that handles
3According to https://en.wikipedia.org/wiki/Comma-separated_values, comma-separated values
(CSV) file is a delimited text file that uses a comma to separate values. A CSV file stores tabular data (numbers and text)
in plain text. Each line of the file is a data record. Each record consists of one or more fields, separated by commas.
12
3 INSTRUCTIONS AND STARTER CODE
the case that the value of the attribute is less than the threshold, and right handles the case that
the value is greater than or equal to the threshold. For a leaf node, they are both null.
The class DTNode also contains a few member methods that helps in building the tree.
– fillDTNode(): This is the method that does all the heavy lifting in the entire assignment.
Given a list of data points and a minimum size for splitting (see earlier), this recursive function
creates the entire decision tree. A detailed description is present in the comments of the code.
– findMajority(): Given a list of datapoints D, this method returns a label for that set.
This method is called during training (construction of the decision tree). It can happen that
the size of a dataset falls below the minimum size mentioned in minSizeDatalist, but
still contains datapoints from more than one class. In such a case, a leaf node is created. To
determine the label for this leaf node, the method goes through all the points and finds the
majority label, i.e. the most common label for those points. This label will be used later at the
classification phase, when a new data point reaches that leaf node. When choosing the label
for a leaf node, if there is no majority (a tie), then the label with the smallest value is returned.
– classifyAtNode(): At the testing phase, given a datapoint with only its attributes (no
label), this method uses the label specified by the decision tree leaf node (as was determined at
training time by the findMajority() method).
– equals() : Given another node, this method checks if the tree rooted at the given node is
equal to the tree rooted at the calling node. The definition of ‘equality’ is elaborated in the next
section.
• calcEntropy(): Given a dataset, this function calculates the entropy of the dataset.
• classify(): Given a datapoint (without the label), predicts the label of the datapoint. The only
difference between this method and classifyAtNode() is that classifyAtNode() does
the classification on its member DTNode, whereas for classify() the DTNode is the root of the
created decision tree.
• checkPerformance(): Given a dataset where the datapoints have both attributes and labels,
this method runs the classify() function on all of the datapoints (using only the attributes) and
compares the label that is given by the decision tree with the “ground truth” label for that data point.
The method returns the fraction of datapoints that were predicted wrong, in the form of a string.
• equals(): Given two decision trees, this method checks if the two trees are equal or not. It returns
a boolean value.
Your task
You need to implement three methods from the DTNode class. None of the methods depend on each other.
We suggest that you implement equals() first.
1. DTNode.equals() (30 points)
This method compares two DTNodes. Given another DTNode object, it checks if the tree that is
rooted at the calling DTNode is equal to the tree rooted at DTNode object that is passed as the
13
4 DATA
parameter.
Two DTNodes are considered equal if:
(a) a traversal (e.g. preorder) of each of the two trees encounters nodes that are equal;
(b) internal node : the thresholds and attributes should be same;
(c) leaf node : the labels should be same.
Note that the tester you are given uses this equals method to check if the tree you implement in the
second part matches with the actual solution.
2. DTNode.fillDTNode() (50 points)
This method takes in a datalist (i.e. an arraylist of objects of type Datum) and returns the calling
DTNode object as the root of a decision tree trained using the datapoints present in the input datalist.
3. DTNode.classifyAtNode() (20 points)
This method takes in a datapoint (excluding the label) in the form of an array of type double (like
Datum.x) and should return its corresponding label (int).
4 Data
The different datasets used in the assignment are shown below. If you look at the DataReader() class,
you will note that only half of the data in each plot is used in the training. The other half is used to test the
performance of the decision tree. This testing phase is not part of the assignment, but we give you some
performance data so that you can appreciate the differences in the data sets.
For each of the three data sets, we list the performance of the decision tree in classifying the other half
(test set) of the data. We show both the fraction of training points that are misclassified (error) and
the fraction of the test points that are misclassified, and we do so for different values of the variable
minSizeDatalist.
4.1 Highly overlapping data :
Notice that the training error is 0 when the minimum size is 1 and rises as minSizeDatalist increases.
However, the test error is near .5 (50 percent) for all values of minSizeDatalist.
minSizeDatalist : 1 Training error : 0.000 Test error : 0.495
minSizeDatalist : 2 Training error : 0.000 Test error : 0.495
minSizeDatalist : 4 Training error : 0.030 Test error : 0.495
minSizeDatalist : 8 Training error : 0.105 Test error : 0.520
minSizeDatalist : 16 Training error : 0.200 Test error : 0.515
minSizeDatalist : 32 Training error : 0.235 Test error : 0.515
minSizeDatalist : 64 Training error : 0.310 Test error : 0.525
minSizeDatalist : 128 Training error : 0.390 Test error : 0.490
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4.2 Partially overlapping data : 4 DATA
Figure 6: Plot of data present in data high overlap.csv
4.2 Partially overlapping data :
Figure 7: Plot of data present in data partial overlap.csv
In this case, 10 or 12 of the 200 test data points are misclassified for most values of minSizeDatalist.
The error rises considerably when minSizeDatalist is 128.
minSizeDatalist : 1 Training error : 0.000 Test error : 0.050
minSizeDatalist : 2 Training error : 0.000 Test error : 0.050
minSizeDatalist : 4 Training error : 0.015 Test error : 0.050
minSizeDatalist : 8 Training error : 0.035 Test error : 0.060
minSizeDatalist : 16 Training error : 0.045 Test error : 0.050
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4.3 Minimal overlapping data : 4 DATA
minSizeDatalist : 32 Training error : 0.045 Test error : 0.050
minSizeDatalist : 64 Training error : 0.075 Test error : 0.050
minSizeDatalist : 128 Training error : 0.255 Test error : 0.245
4.3 Minimal overlapping data :
Figure 8: The data points here have very little overlap. We are loosely calling the overlap “minimal”,
although truly minimal would mean zero overlap. We leave some overlap so that overfitting can potentially
occur.
Note exactly one (of 200) training data points is misclassfied when minSizeDatalist is 4 or more.
The test error is roughly constant. It is slightly larger for small values of minSizeDatalist.
minSizeDatalist : 1 Training error : 0.000 Test error : 0.040
minSizeDatalist : 2 Training error : 0.000 Test error : 0.040
minSizeDatalist : 4 Training error : 0.005 Test error : 0.040
minSizeDatalist : 8 Training error : 0.005 Test error : 0.035
minSizeDatalist : 16 Training error : 0.005 Test error : 0.035
minSizeDatalist : 32 Training error : 0.005 Test error : 0.035
minSizeDatalist : 64 Training error : 0.005 Test error : 0.035
minSizeDatalist : 128 Training error : 0.005 Test error : 0.035
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