Homework Assignment 5 Solution

Hand-in Instructions




This homework assignment includes two written problems and a programming problem in Java. Hand in all parts electronically to your Canvas assignments page. For each written question, submit a single pdf file containing your solution. Handwritten submissions must be scanned. No photos or other file types allowed. For the programming question, submit a zip file containing all the Java code necessary to run your program, whether you modified provided code or not.




Submit the following three files (with exactly these names):







<wiscNetID-HW5-P1.pdf

<wiscNetID-HW5-P2.pdf




<wiscNetID-HW5-P3.zip







For example, for someone with UW NetID crdyer@wisc.edu the first file name must be:

crdyer-HW5-P1.pdf




Late Policy




All assignments are due at 11:59 p.m. on the due date. One (1) day late, defined as a 24-hour period from the deadline (weekday or weekend), will result in 10% of the total points for the assignment deducted. So, for example, if a 100-point assignment is due on a Wednesday and it is handed in between any time on Thursday, 10 points will be deducted. Two (2) days late, 25% off; three (3) days late, 50% off. No homework can be turned in more than three (3) days late. Written questions and program submission have the same deadline. A total of three (3) free late days may be used throughout the semester without penalty. Assignment grading questions must be discussed with a TA or grader within one week after the assignment is returned.







Collaboration Policy




You are to complete this assignment individually. However, you may discuss the general algorithms and ideas with classmates, TAs, peer mentors and instructor in order to help you answer the questions. But we require you to:




not explicitly tell each other the answers



not to copy answers or code fragments from anyone or anywhere



not to allow your answers to be copied



not to get any code from the Web
12+12+1+1+13

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Problem 1. Probabilities [15 points]




Suppose three documents 1, 2 and 3 contain only five words “ I”, “am”, “we”, “are” and “groot”. The following table summarizes the number of occurrences of each word in each document. Suppose one document is chosen at random (each document with equal probability), then one word in that document is chosen at random (each word with equal probability within that document).




Document / Word
I
am
we
are
groot
1
12
12
1
1
13
2
17
17
0
0
17
3
14
14
2
2
15









Note that the table can be used to compute conditional probabilities. For example, P(Word = groot | Document = 1) =













[2] What is the probability that the word is “groot”, i.e., P(Word = groot)?



[2] Suppose the randomly chosen word is “we”. What is the probability that it comes from document 1, i.e., P(Document = 1 | Word = we)?
[2] Suppose the randomly chosen word starts with an “a”. What is the probability that it comes from document 2, i.e., P(Document = 2 | Word = am or Word = are)?



Now suppose document 1 is chosen with probability P(Document = 1) = 16, document 2 is chosen with probability P(Document = 2) = 13, and document 3 is chosen with probability P(Document =




= 12. Then, one word in that document is chosen at random (each word with equal probability within that document).



[3] What is the probability that the word is “groot”, i.e., P(Word = groot)?



[3] Suppose the randomly chosen word is “we”. What is the probability that it comes from document 1, i.e., P(Document = 1 | Word = we)?
[3] Suppose the randomly chosen word starts with an “a”. What is the probability that it comes from document 2, i.e., P(Document = 2 | Word = am or Word = are)?






Show your calculations and round all final answers to 4 decimal places.





































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Problem 2. Using a Bayesian Network [20 points]




Use the Bayesian Network below containing 6 Boolean random variables to answer the following questions using inference by enumeration. For the problems, Spring = S, Rain = R, Worms = Wo, Wet = We, Mowing = M, and Birds = B. Give your answers to 4 decimal places. Show your work.




[4] What is the probability that it rains? (P(R))



[4] What is the probability that the grass is wet? (P(We))



[4] Given that it is spring, how likely is there to be someone mowing? (P(M | S))



[4] Given that it is spring, how likely is there to be a bird on the lawn? (P(B | S))



[4] If there are birds on the lawn, but there are no worms out, how likely is it that it is spring? (P(S | B, ¬Wo))


























































































The CPTs are defined as follows:

P(S) = 0.25



P(R | S) = 0.7, P(R | ¬S) = 0.3



P(Wo | R) = 0.7, P(Wo | ¬R) = 0.2



P(We | R, Wo) = 0.12, P(We | R, ¬Wo) = 0.25, P(We | ¬R, Wo) = 0.1, P(We | ¬R, ¬Wo) = 0.08



P(M | We) = 0.02, P(M | ¬We) = 0.42



P(B | S, Wo) = 0.8, P(B | S, ¬Wo) = 0.4, P(B | ¬S, Wo) = 0.4, P(B | ¬S, ¬Wo) = 0.4















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CS 540 Fall 2019










Problem 3. A Naïve Bayes Classifier for Sentiment Analysis [65 points]




One application of Naïve Bayes classifiers is sentiment analysis, which is a sub-field of AI that extracts affective states and subjective information from text. One common use of sentiment analysis is to determine if a text document expresses negative or positive feelings. In this homework you are to implement a Naïve Bayes classifier for categorizing movie reviews as either POSITIVE or NEGATIVE. The dataset provided consists of online movie reviews derived from an IMDb dataset: https://ai.stanford.edu/~amaas/data/sentiment/ that have been labeled based on the review scores. A negative review has a score ≤ 4 out of 10, and a positive review has a score ≥ 7 out of 10. We have done some preprocessing on the original dataset to remove some noisy features. Each row in the training set and test set files contains one review, where the first word in each line is the class label (1 represents POSITIVE and 0 represents NEGATIVE) and the remainder of the line is the review text.




Methods to Implement




We have provided code for you that will open a file, parse it, pass it to your classifier, and output the results. What you need to focus on is the implementation in the file NaiveBayesClassifier.java If you open that file, you will see the following methods that you must implement:




Map<Label, Integer getDocumentsCountPerLabel (List<Instance trainData)



This method counts the number of reviews per class label in the training set and returns a map that stores the (label, number of documents) key-value pair.

Map<Label, Integer getWordsCountPerLabel(List<Instance trainData)



This method counts the number of words per label in the training set and returns a map that stores the (label, number of words) key-value pair.

void train(List<Instance trainData, int v)



This method trains your classifier using the given training data. The integer argument v is the size of the total vocabulary in your model. Store this argument as a field because you will need it in computing the smoothed class-conditional probabilities. (See the section on Smoothing below.)




ClassifyResult classify(List<String words)



This method returns the classification result for a single movie review.




In addition, you will need to implement a k-fold cross validation function as defined in the

CrossValidation.java:




double kFoldScore(Classifier clf, List<Instance trainData, int k, int v)



This method takes a classifier clf and performs k-fold cross validation on trainData. The output is the average of the scores computed on each fold. You can assume that the number of instances in trainData is always divisible by k, so the size




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CS 540 Fall 2019










of each fold will the same. We will use the accuracy (number of correctly predicted instances / number of total instances) as the score metric.




We have also defined four class types to assist you in your implementation. The Instance class is a data structure holding the label and the movie review as a list of strings:




public class Instance {




public Label label;

public List<String words;




}







The Label class is an enumeration of our class labels:




public enum Label {POSITIVE, NEGATIVE}







The ClassifyResult class is a data structure holding each label’s log probability (whose values are described in the log probabilities section below) and the predicted label:




public class ClassifyResult {




public Label label;

public Map<Label, Double logProbPerLabel;




}







The only provided files you need edit are NaiveBayesClassifier.java and CrossValidation.java but you are allowed to add extra helper class files if you like. Do not include any package paths or external libraries in your program. Your program is only required to handle binary classification problems.




Smoothing




There are two concepts we use here:

Word token: an occurrence of a given word



Word type: a unique word as a dictionary entry



For example, “the dog chases the cat” has 5 word tokens but 4 word types; there are two tokens of the word type “the”. Thus, when we say a word “token” in the discussion below, we mean the number of words that occur and NOT the number of unique words. As another example, if a review is 15 words long, we would say that there are 15 word tokens. For example, if the word “lol” appeared 5 times, we say there were 5 tokens of the word type “lol”.




The conditional probability P(w|l), where w represents some word type and l is a label, is a multinomial random variable. If there are |V| possible word types that might occur, imagine a |V|-sided die. P(w |l) is the likelihood that this die lands with the w-side up. You will need to estimate two such distributions: P(w|Positive) and P(w|Negative).










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CS 540 Fall 2019










One might consider estimating the value of P(w|Positive) by simply counting the number of tokens of type w and dividing by the total number of word tokens in all reviews in the training set labeled as Positive, but this method is not good enough in general because of the “unseen event problem,” i.e., the possible presence of words in the test data that did not occur at all in the training data. For example, in our classification task consider the word “foo”. Say “foo” does not appear in our training data but does occur in our test data. What probability would our classifier assign to P(foo|Positive) and P(foo|Negative)? The probability would be 0, and because we are taking the sum of the logs of the conditional probabilities for each word and log 0 is undefined, the expression whose maximum we are computing would be undefined.




What we do to get around this is pretend we actually did see some (possibly fractionally many) tokens of the word type “foo”. This goes by the name Laplace smoothing or add-δ smoothing, where δ is a parameter. The conditional probability then is defined as:

( ∣)=
( ) +






||+∑
∈
( )









where l ∈ {Positive, Negative}, and Cl(w) is the number of times the tokens of word type w appears in reviews labeled l in the training set. As above, |V| is the size of the total vocabulary we assume we will encounter (i.e., the dictionary size). Thus, it forms a superset of the words used in the training and test sets. The value |V| will be passed to the train method of your classifier as the argument int v. For this assignment, use the value δ = 1. With a little reflection, you will see that if we estimate our distributions in this way, we will have




( ∣ ) = 1. Use the equation above for P(w|l) to calculate the conditional probabilities in your implementation.













Log Probabilities




The second gotcha that any implementation of a Naïve Bayes classifier must contend with is underflow. Underflow can occur when we take the product of a number of very small floating-point values. Fortunately, there is a workaround. Recall that a Naïve Bayes classifier computes

( ) = arg max � ( ) � ( ∣ )�

=1




where l ∈ {Positive, Negative} and wi is the ith word type in a review, numbered 1 to k. Because maximizing a formula is equivalent to maximizing the log value of that formula, f(w) computes the same class as

( ) = arg max �log ( ) + � log ( ∣ )�

=1




What this means for you is that in your implementation you should compute the g(w) formulation of the function above rather than the f(w) formulation. Use the Java function log(x) which




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CS 540 Fall 2019










computes the natural logarithm of its input. This will result in code that avoids errors generated by multiplying very small numbers.




This is what you should return in the ClassifyResult class: logProbPerLabel.get(Label.POSITIVE) is the value log ( ) + ∑ =1 log ( ∣ ) with l = Positive and logProbPerLabel.get(Label.NEGATIVE) with l = Negative. The label returned in this class corresponds to the output of g(w). Break ties by classifying a review as Positive.




Testing




We will test your program on multiple training and test sets, using the following command line format:




java SentimentAnalysis <mode <trainFilename [<testFilename | <K]







where trainingFilename and testFilename are the names of the training set and test set files, respectively. mode is an integer from 0 to 3, controlling what the program will output. When mode is 0 or 1, there are only two arguments, mode and trainFilename; when the mode is 2 the third argument is testFilename; when mode is 3, the third argument is K, the number of folds used for cross validation. The output for these four modes should be:




Prints the number of documents for each label in the training set
Prints the number of words for each label in the training set
For each instance in test set, prints a line displaying the predicted class and the log probabilities for both classes



Prints the accuracy score for K-fold cross validation



In order to facilitate debugging, we are providing sample training set and test set files called train.txt and test.txt in the zip file. In addition, we are providing the correct output for each of the four modes based on these two datasets, in the files mode0.txt, …, mode3.txt




For example, the command




java SentimentAnalysis 2 train.txt test.txt




should train the classifier using the data in train.txt and print the predicted class for every review in test.txt




The command




java SentimentAnalysis 3 train.txt 5 should perform 5-fold cross-validation on train.txt




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CS 540 Fall 2019










You are not responsible for handling parsing of the training and test sets, creating the classifier, or printing the results on the console. We have written the main class SentimentAnalysis for you, which will load the data and pass it to the method you are implementing. Do NOT modify any IO code.




As part of our testing process, we will unzip the file you submit, remove any class files, call javac *.java to compile your code, and then call the main method SentimentAnalysis with parameters of our choosing. Make sure your code runs and terminates in less than 20 seconds for each provided test case on the computers in the department because we will conduct our tests on these computers.




Deliverables

Hand in your modified versions of NaiveBayesClassifier.java and CrossValidation.java. Also submit any additional helper .java files required to run your program (including the provided ones in the skeleton). Do not submit any other files including the data files (i.e., no .class or .txt files). All the .java files should be zipped as <wiscNetID-HW5-P3.zip for submission to Canvas.

























































































































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