Description

You might be familiar with converting a given number in binary (base, b = 2) into its decimal equivalent. This question requires you to generalise it to any base b, and the program should be written in Python.

Given a string of a fractional number N_b in base b, convert it into its decimal equivalent N_D. You need to check if the input is a valid number and get rid of leading zeros from the input.

-999999999  < = N_D  <=  999999999,  N_D ∈ ℝ

2  <=  b  <=  36 Example-
For Nb = HELLO.PY	&	b = 35;	output, ND = 26137359.742041.
For Nb = 00101.101	&	b = 2;	output, ND = 5.625.
For Nb = GJDGXR	&	b = 36;	output, ND = 999999999.
For Nb = -4G	&	b = 17;	output, ND = -84.
For Nb = HELLO.PY	&	b = 10;	output, “Invalid Input”.

Important Note: You are NOT allowed to use built in functions like int() etc.

Q2. (Intro to ML)

In this question, we will code a 1-D linear regression problem. We will provide pseudocode here, the students are required to transform it into a python code. Use numpy library for array/vector/matrices etc.

The data files are train.csv and test.csv.

File structure:

The structure of both files are similar. Train.csv has n_train = 10^4 rows and test.csv has  n_test = 10^3 rows, each row corresponding to one data point. Each row has two values separated by comma. The first value is feature and second value is label of the data point.

Example-

If one row of train.txt is :

4,7

Then feature, x = 4 and label, y = 7.

Pseudo-code Step-1:

• Read files train.csv
• Create vector – X_train (dim – n_train x 1) and vector – y_train (dim – n_train x 1)
• Add a column to X_train so that its dimension becomes n_train x 2. First column of X_train should be all 1 and 2nd column is the same as before adding extra column.

Example –

X_train = [

2

3

4

]

New X_train = [

1    2

1    3

1    4                                         ]

Step-2:

Generate a 2-D vector w (dim: 2 x 1)  initialised randomly with floating point numbers.

Step-3:

Plot y vs x using matplotlib where x is the feature and y is the label read from the file train.csv.

Consider x’ = [1 x] (prepending 1 to x to generate 2-dimensional x-vector) On the same figure plot the line w^T*x’ vs x.

Your figure should have a dot corresponding to each datapoint (x,y) and a straight line on the plot corresponding to w^T*x’.

Step-4:

Set w_direct = (X_train^T * X_train)^(-1)* X_train^T*y_train

X_train is the n_train x 2 matrix defined earlier and y_train is the corresponding label vector.

Plot y vs x using matplotlib where x is the feature and y is the label read from the file train.csv.

Consider x’ = [1 x] (prepending 1 to x to generate 2-dimensional x-vector) On the same figure plot the line w_direct^T*x’ vs x.

Your figure should have a dot corresponding to each datapoint (x,y) and a straight line on the plot corresponding to w_direct^T*x’.

Step-5: (Training)

w – 2-dim vector initialised earlier (step-2)

Loop: for nepoch =  1 to N   (N is the number of pass through the data (~10), play with it to find

best fit)

Loop : for j = 1 to n_train

w ← w – eta*(w^T*x’ – y)*x’   (eta = 0.0001 students can change this

value)

If j%100 == 0

Then plot y vs x as earlier and use current value of w to plot

w^T*x’  vs x

Step-6:

Finally redraw the plot as earlier with latest value of w.

 

Note:- Don’t use test.csv for training

Step-7: (Evaluation)

• Read files test.csv
• Create vector – X_test (dim – n_test x 1) and vector – y_test (dim – n_test x 1)
• Add a column to X_test so that its dimension becomes n_test x 2. First column of X_test should be all 1 and 2nd column is the same as before adding extra column.
• Let y_pred1 = X_test*w (w is the final value after doing step 5) – Calculate root mean squared error between y_pred1 and y_test.
• Let y_pred2 = X_test*w_direct
• Calculate root mean squared error between y_pred2 and y_test.

Derivation of Updates (optional reading)

In the loop of step 5 we did the following w ← w – eta*(w^T*x’ – y)*x’

The objective of the above step is to reduce the least squared error.

Let error = 0.5(w^T*x’-y)^2

So, derivative w.r.t. w gives:  (w^T*x’-y)x’

We need to descend down the gradient to reach the minima, so we subtract eta times the above derivative from w to gradually reach minima. We set eta to small value because otherwise, w may overshoot the minima. w_direct can also be computed on the same line by setting derivative to zero (Google it!).