Completed HW5

hw5/regularization.py

@@ -0,0 +1,93 @@
+import math
+from math import sin, pi
+from random import random
+
+
+def f(x):
+    return sin(pi * x)
+
+
+def generate_training_examples(n=2):
+    xs = [random() * 2 - 1 for _ in range(n)]
+    return [(x, f(x)) for x in xs]
+
+
+def fit_without_reg(examples):
+    """Computes values of w0 and w1 that minimize the sum-of-squared-errors cost function
+
+    Args:
+    - examples: a list of two (x, y) tuples, where x is the feature and y is the label
+    """
+    w0 = 0
+    w1 = 0
+    
+    x1, y1 = examples[0]
+    x2, y2 = examples[1]
+
+    w1 = (y2 - y1) / (x2 - x1)
+    w0 = y1 - w1 * x1
+
+    return w0, w1
+
+
+def fit_with_reg(examples, lambda_hp, learning_rate=0.01, max_iter=1000, tol=1e-6):
+    """Computes values of w0 and w1 that minimize the regularized sum-of-squared-errors cost function
+
+    Args:
+    - examples: a list of two (x, y) tuples, where x is the feature and y is the label
+    - lambda_hp: a float representing the value of the lambda hyperparameter; a larger value means more regularization
+    """
+    w0 = 0
+    w1 = 0
+    for _ in range(max_iter):
+        # Compute the sum of squared errors
+        sse = sum((w0 + w1 * x - y) ** 2 for x, y in examples)
+
+        # Compute the regularization term
+        reg_term = lambda_hp * (w0 ** 2 + w1 ** 2)
+
+        # Compute the total cost function
+        cost = sse + reg_term
+
+        # Compute the partial derivatives of the cost function with respect to w0 and w1
+        d_w0 = 2 * sum(w0 + w1 * x - y for x, y in examples) + 2 * lambda_hp * w0
+        d_w1 = 2 * sum((w0 + w1 * x - y) * x for x, y in examples) + 2 * lambda_hp * w1
+
+        # Update w0 and w1 using gradient descent
+        w0_new = w0 - learning_rate * d_w0
+        w1_new = w1 - learning_rate * d_w1
+
+        # Check for convergence
+        if abs(w0_new - w0) < tol and abs(w1_new - w1) < tol:
+            break
+
+        w0, w1 = w0_new, w1_new
+
+    return w0, w1
+
+
+def test_error(w0, w1):
+    n = 100
+    xs = [i/n for i in range(-n, n + 1)]
+    return sum((w0 + w1 * x - f(x)) ** 2 for x in xs) / len(xs)
+
+def main():
+    print("Starting...")
+    no_reg_errors=[]
+    reg_erros = []
+    exp_size = 1000
+    for _ in range(exp_size):
+
+        train_examples = generate_training_examples(n=2)
+
+        w0, w1 = fit_without_reg(train_examples)
+        w0_r, w1_r = fit_with_reg(train_examples, lambda_hp=1)
+
+        no_reg_errors.append(test_error(w0, w1))
+        reg_erros.append(test_error(w0_r, w1_r))
+
+    print(f"Avg loss, regularization: {sum(no_reg_errors)/(len(no_reg_errors))}")
+    print(f"Avg loss, no regularization: {sum(reg_erros)/(len(reg_erros))}")
+    
+if __name__=="__main__":
+    main()

main.aux

@@ -20,4 +20,6 @@
 \@writefile{toc}{\contentsline {section}{\numberline {4}Homework 4}{15}{}\protected@file@percent }
 \@writefile{toc}{\contentsline {subsection}{\numberline {4.1}Inference in Bayesian Networks}{15}{}\protected@file@percent }
 \@writefile{toc}{\contentsline {subsection}{\numberline {4.2}Gradient Descent}{15}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {section}{\numberline {5}Homework 5}{16}{}\protected@file@percent }
+\@writefile{toc}{\contentsline {subsection}{\numberline {5.1}Regularization}{16}{}\protected@file@percent }
 \gdef \@abspage@last{17}

main.fdb_latexmk

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main.fls

@@ -112,13 +112,6 @@ INPUT .\hw3\motorcade.png
 INPUT .\hw3\motorcade.png
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 INPUT .\hw4\Picture1.png

main.log

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main.tex

@@ -422,52 +422,52 @@ See attached submission.py file
 \subsection{Bayesian Networks}
 \begin{itemize}
     \item[a.] Given the following Bayesian network, write down a factorization of the joint probability distribution P(A,B, C,D, E) as a product of local conditional distributions, one for each random variable.
-    \begin{center}
+          \begin{center}
-        \includegraphics[scale=0.3]{hw3/bayesian_network_1_hw3.png}
+              \includegraphics[scale=0.3]{hw3/bayesian_network_1_hw3.png}
-    \end{center}
+          \end{center}
 
-    $$
+          $$
-    P(A, B, C, D, E) = \ \boxed{P(A) \cdot P(B) \cdot P(C | A, B) \cdot P(D | B) \cdot P(E | C)}
+              P(A, B, C, D, E) = \ \boxed{P(A) \cdot P(B) \cdot P(C | A, B) \cdot P(D | B) \cdot P(E | C)}
-    $$
+          $$
 
     \item[b.] For the same Bayesian network above, list all the random variables that are independent of A, assuming that none of the variables have been observed.
-        \begin{itemize}
+          \begin{itemize}
-            \item B
+              \item B
-        \end{itemize}
+          \end{itemize}
 
     \item[c.] For the same Bayesian network above, assume that E is now observed. List all pairs of random variables (not including E) that are conditionally independent given E.
-        \begin{itemize}
+          \begin{itemize}
-            \item (A, B)
+              \item (A, B)
-            \item (A, D)
+              \item (A, D)
-            \item (B, D)
+              \item (B, D)
-        \end{itemize}
+          \end{itemize}
 
     \item[d.] Consider the following model for traffic jams in a small town, which can be caused by either a car accident or by a visit from the president (and the accompanying security motorcade).
-        \begin{center}
+          \begin{center}
-            \includegraphics[scale=0.3]{hw3/motorcade.png}
+              \includegraphics[scale=0.3]{hw3/motorcade.png}
-        \end{center}
+          \end{center}
 
-        Compute P(Accident = 1 | Traffic = 1) and P(Accident = 1 | Traffic = 1, President = 1). \\
+          Compute P(Accident = 1 | Traffic = 1) and P(Accident = 1 | Traffic = 1, President = 1). \\
 
-        Let:
+          Let:
-            \begin{itemize}
+          \begin{itemize}
-                \item A = event of accident
+              \item A = event of accident
-                \item P = event of president's visit
+              \item P = event of president's visit
-                \item T = event of traffic jam
+              \item T = event of traffic jam
-            \end{itemize}
+          \end{itemize}
 
-        We're given:
+          We're given:
-            \begin{itemize}
+          \begin{itemize}
-                \item P(A) = prior probability of an accident
+              \item P(A) = prior probability of an accident
-                \item P(P) = prior probability of the president's visit
+              \item P(P) = prior probability of the president's visit
-                \item P(T | A) = probability of a traffic jam given an accident
+              \item P(T | A) = probability of a traffic jam given an accident
-                \item P(T | P) = probability of a traffic jam given the president's visit
+              \item P(T | P) = probability of a traffic jam given the president's visit
-            \end{itemize}
+          \end{itemize}
+
+          $$
+              P(\text{Accident} = 1 | \text{Traffic} = 1) \rightarrow P(A | T) = \frac{P(T | A) \cdot P(A)}{P(T)}
+          $$
 
-        $$
-        P(\text{Accident} = 1 | \text{Traffic} = 1) \rightarrow P(A | T) = \frac{P(T | A) \cdot P(A)}{P(T)}
-        $$
-        
 \end{itemize}
 
 \newpage
@@ -479,40 +479,79 @@ See attached submission.py file
 \begin{itemize}
     \item[a - c. ] Refer to bn.py
 
-    \item[d. ] Likelihood weighting gets more accurate as we use more samples. With more samples, the estimated probabilities get closer to the real chances we calculated before. This shows that using a bigger sample size is helpful for getting better results. 
+    \item[d. ] Likelihood weighting gets more accurate as we use more samples. With more samples, the estimated probabilities get closer to the real chances we calculated before. This shows that using a bigger sample size is helpful for getting better results.
 \end{itemize}
 
 \subsection{Gradient Descent}
 
 \begin{itemize}
-    \item[a. ] 
+    \item[a. ]
-    \begin{center}
+          % \begin{center}
-        \includegraphics[scale=0.5]{hw4/Gemini_Chart_Image_5v3caz5v3caz5v3c.png}
+          %     \includegraphics[scale=0.5]{hw4/Gemini_Chart_Image_5v3caz5v3caz5v3c.png}
-    \end{center}
+          % \end{center}
 
 
-    Hinge loss is a reasonable cost function for binary classification because it focuses on creating a good margin between classes. It only penalizes models when their predictions fall close to the wrong class, and ignores classified points even with a small margin. This makes the model prioritize a clear separation between positive and negative classes.
+          Hinge loss is a reasonable cost function for binary classification because it focuses on creating a good margin between classes. It only penalizes models when their predictions fall close to the wrong class, and ignores classified points even with a small margin. This makes the model prioritize a clear separation between positive and negative classes.
 
-    \item[b. ] 
+    \item[b. ]
-    
-    $$
-    \frac{\partial c(h(x), y)}{\partial w_0} = 0
-    $$
 
-    $$
+          $$
-    \frac{\partial c(h(x), y)}{\partial w_1} = \begin{cases}
+              \frac{\partial c(h(x), y)}{\partial w_0} = 0
-        -y, & \text{if } w_0 + w_1x < 1 \\
+          $$
-        0, & \text{if } w_0 + w_1x \geq 1
+
-        \end{cases}        
+          $$
-    $$
+              \frac{\partial c(h(x), y)}{\partial w_1} = \begin{cases}
+                  -y, & \text{if } w_0 + w_1x < 1    \\
+                  0,  & \text{if } w_0 + w_1x \geq 1
+              \end{cases}
+          $$
 
     \item[c. ]
-    
+
-    \begin{center}
+          \begin{center}
-        \includegraphics[scale=0.3]{hw4/Picture1.png}
+              \includegraphics[scale=0.3]{hw4/Picture1.png}
-    \end{center}
+          \end{center}
 
 
 \end{itemize}
 
-\end{document}
+\newpage
+
+\section{Homework 5}
+
+\subsection{Regularization}
+
+\begin{itemize}
+    \item[a.]
+          $$
+              \begin{aligned}
+                  \frac{d}{dw_0} C(w) & = \sum_{i=1}^{n} 2(y_i - (w_0 + w_1x_i))(-1) = \boxed{-2\sum_{i=1}^{n} (y_i - (w_0 + w_1x_i))}      \\
+                  \frac{d}{dw_1} C(w) & = \sum_{i=1}^{n} 2(y_i - (w_0 + w_1x_i))(-x_i) = \boxed{-2\sum_{i=1}^{n} x_i(y_i - (w_0 + w_1x_i))}
+              \end{aligned}
+          $$
+
+    \item[b. ] 
+          $$
+              \begin{aligned}
+                \frac{d}{dw_0} \tilde{C}(w) &= \sum_{i=1}^{n} 2(y_i - (w_0 + w_1x_i))(-1) + 2\lambda w_0 \\
+                &= 2\sum_{i=1}^{n} (w_0 + w_1x_i - y_i) + 2\lambda w_0 ] \\ 
+                &= \boxed{2\sum_{i=1}^{n} (w_0 + w_1x_i - y_i + \lambda w_0)}
+              \end{aligned}
+          $$
+
+          and 
+
+          $$
+              \begin{aligned}
+                \frac{d}{dw_1} \tilde{C}(w) &= \sum_{i=1}^{n} 2(y_i - (w_0 + w_1x_i))(-x_i) + 2\lambda w_1 \\
+                &= \boxed{2\sum_{i=1}^{n} x_i(y_i - (w_0 + w_1x_i)) + 2\lambda w_1}
+              \end{aligned}
+          $$
+
+    \item[c-d.] Refer to regularization.py
+    \item[e. ] Avg loss, regularization: 1.715104320688656 \\
+    Avg loss, no regularization: 0.4384015750703455 
+
+\end{itemize}
+
+\end{document}