This post comes out of the blue, nearly 2 years since my last one. I realize I’ve been lazy, so here’s hoping I move from an inertia of rest to that of motion, implying, regular and (hopefully) relevant posts. I also chanced upon some wisdom while scrolling through my Twitter feed:

Don’t worry about how many people you think will want to read your blog posts. Writing has value beyond its audience. It can help to crystallize your thinking and also serve as a milestone in your learning journey.

This blog post in particular was meant to be a reminder to myself and other R users that the much used lm() function in R (for fitting linear models) can be replaced with some handy matrix operations to obtain regression coefficients, their standard errors and other goodness-of-fit stats printed out when summary() is called on an lm object.

Linear regression can be formulated mathematically as follows:
,

is the outcome variable and is the data matrix of independent predictor variables (including a vector of ones corresponding to the intercept). The ordinary least squares (OLS) estimate for the vector of coefficients is:

The covariance matrix can be obtained with some handy matrix operations:

given that

The standard errors of the coefficients are basically and with these, one can compute the t-statistics and their corresponding p-values.

Lastly, the F-statistic and its corresponding p-value can be calculated after computing the two residual sum of squares (RSS) statistics:

– for the full model with all predictors

– for the partial model () with the outcome observed mean as estimated outcome

I wrote some R code to construct the output from summarizing lm objects, using all the math spewed thus far. The data used for this exercise is available in R, and comprises of standardized fertility measures and socio-economic indicators for each of 47 French-speaking provinces of Switzerland from 1888. Try it out and see for yourself the linear algebra behind linear regression.

I love probability and have solved countless problems on probability ever since I learned math

…and yet I’ve never coded up probabilistic models!

The assignments and project work for this course are to be implemented in Python!

You don’t need to have prior experience in either probability or inference, but you should be comfortable with basic Python programming and calculus.

WHAT YOU’LL LEARN
– Basic discrete probability theory
– Graphical models as a data structure for representing probability distributions
– Algorithms for prediction and inference
– How to model real-world problems in terms of probabilistic inference

The course started on September 12, is 12-weeks long and is structured in the following manner:

Week 1 (9/12 – 9/16): Introduction to probability and computation
A first look at basic discrete probability, how to interpret it, what probability spaces and random variables are, and how to code these up and do basic simulations and visualizations.

Week 2 (9/19 – 9/23):Incorporating observations
Incorporating observations using jointly distributed random variables and using events. Three classic probability puzzles are presented to help elucidate how to interpret probability: Simpson’s paradox, Monty Hall, boy or girl paradox.

Week 3 (9/26 – 9/30):Introduction to inference, structure in distributions, and information measures
The product rule and inference with Bayes’ theorem. Independence: A structure in distributions. Measures of randomness: entropy and information divergence. Mutual information.

Week 4 (10/3 – 10/7):Expectations, and driving to infinity in modeling uncertainty
Expected values of random variables. Classic puzzle: the two envelope problem. Probability spaces and random variables that take on a countably infinite number of values and inference with these random variables.

Week 5 (10/10 – 10/14):Efficient representations of probability distributions on a computer
Introduction to undirected graphical models as a data structure for representing probability distributions and the benefits/drawbacks of these graphical models. Incorporating observations with graphical models.

Week 6 (10/17 – 10/21):Inference with graphical models, part I
Computing marginal distributions with graphical models in undirected graphical models including hidden Markov models..

Week 7 (10/24 – 10/28):Inference with graphical models, part II
Computing most probable configurations with graphical models including hidden Markov models.

Week 8 (10/31 – 11/4):Introduction to learning probability distributions
Learning an underlying unknown probability distribution from observations using maximum likelihood. Three examples: estimating the bias of a coin, the German tank problem, and email spam detection.

Week 9 (11/7 – 11/11):Parameter estimation in graphical models
Given the graph structure of an undirected graphical model, we examine how to estimate all the tables associated with the graphical model.

Week 10 (11/14 – 11/18):Model selection with information theory
Learning both the graph structure and the tables of an undirected graphical model with the help of information theory. Mutual information of random variables.

Week 11 (11/21 – 11/25):Final project
Final project assigned

Week 12 (11/28 – 12/2):Final project

I’m SO taking this course. Hope this interests you as well!

A key aspect of the Quick Sort algorithm is how the pivot element is chosen. In my earlier post on the Python code for Quick Sort, my implementation takes the first element of the unsorted array as the pivot element.

However with some mathematical analysis it can be seen that such an implementation is O(n^{2}) in complexity while if a pivot is randomly chosen, the Quick Sort algorithm is O(nlog_{2}n).

To witness this in action, one can measure the work done by the algorithm comparing two cases, one with a randomized pivot choice – and one with a fixed pivot choice, say the first element of the array (or the last element of the array).

Implementation

A decent proxy for the amount of work done by the algorithm would be the number of pivot comparisons. These comparisons needn’t be computed one-by-one, rather when there is a recursive call on a subarray of length m, you should simply add m−1 to your running total of comparisons.

3 Cases

To put things in perspective, let’s look at 3 cases. (This is basically straight out of a homework assignment from Tim Roughgarden’s course on the Design and Analysis of Algorithms). Case I with the pivot being the first element. Case II with the pivot being the last element. Case III using the “median-of-three” pivot rule. The primary motivation behind this rule is to do a little bit of extra work to get much better performance on input arrays that are nearly sorted or reverse sorted.

Median-of-Three Pivot Rule

Consider the first, middle, and final elements of the given array. (If the array has odd length it should be clear what the “middle” element is; for an array with even length 2k, use the k^{th} element as the “middle” element. So for the array 4 5 6 7, the “middle” element is the second one —- 5 and not 6! Identify which of these three elements is the median (i.e., the one whose value is in between the other two), and use this as your pivot.

This file contains all of the integers between 1 and 10,000 (inclusive, with no repeats) in unsorted order. The integer in the i^{th} row of the file gives you the i^{th} entry of an input array. I downloaded this file and named it QuickSort_List.txt

You can run the code below and see for yourself that the number of comparisons for Case III are 138,382 compared to 162,085 and 164,123 for Case I and Case II respectively. You can play around with the code in an IPython / Jupyter notebook here.

I have been doing Gilbert Strang’s linear algebra assignments, some of which require you to write short scripts in MatLab, though I use GNU Octave (which is kind of like a free MatLab). I was trying out this problem:

To solve this quickly, it would have been nice to have a function that would give a list of permutation matrices for every n-sized square matrix, but there was none in Octave, so I wrote a function permMatrices which creates a list of permutation matrices for a square matrix of size n.

For example:

The MatLab / Octave code to solve this problem is shown below:

I found myself stuck on this problem recently. I must confess, I lost a couple of hours trying to get to figure the logic for this one. Here’s the problem:

I’ve written 2 functions to solve this problem. The first one I used for smaller N, say N < 30 and the second one for N > 30. The second function is elegant, and it relies on the mathematical property that if a number N is not divisible by 3, it could either leave a remainder 1 or 2.

If it leaves a remainder 2, then subtracting 5 once would make the number divisible by 3. If it leaves a remainder 1, then subtracting 5 twice would make the number divisible by 3.

We subtract 5 from N iteratively and attempt to divide N into 2 parts, one divisible by 3 and the other divisible by 5. We want the part that is divisible by 3 to be the larger part, so that the associated Decent Number is the largest possible. This explanation might seem obtuse, but if you get pen down on paper, you’ll understand what I mean.

I’ve enrolled in Stanford Professor Tim Roughgarden’s Coursera MOOC on the design and analysis of algorithms, and while he covers the theory and intuition behind the algorithms in a surprising amount of detail, we’re left to implement them in a programming language of our choice.

And I’m ging to post Python code for all the algorithms covered during the course!

The Karatsuba Multiplication Algorithm

Karatsuba’s algorithm reduces the multiplication of two n-digit numbers to at most single-digit multiplications in general (and exactly when n is a power of 2). Although the familiar grade school algorithm for multiplying numbers is how we work through multiplication in our day-to-day lives, it’s slower () in comparison, but only on a computer, of course!

Here’s how the grade school algorithm looks: (The following slides have been taken from Tim Roughgarden’s notes. They serve as a good illustration. I hope he doesn’t mind my sharing them.)

…and this is how Karatsuba Multiplication works on the same problem:

A More General Treatment

Let and be represented as -digit strings in some base . For any positive integer less than , one can write the two given numbers as

,

where and are less than . The product is then

where

These formulae require four multiplications, and were known to Charles Babbage. Karatsuba observed that can be computed in only three multiplications, at the cost of a few extra additions. With and as before we can calculate

which holds since

A more efficient implementation of Karatsuba multiplication can be set as , where .

Example

To compute the product of 12345 and 6789, choose B = 10 and m = 3. Then we decompose the input operands using the resulting base (B^{m} = 1000), as:

12345 = 12 · 1000 + 345

6789 = 6 · 1000 + 789

Only three multiplications, which operate on smaller integers, are used to compute three partial results:

We get the result by just adding these three partial results, shifted accordingly (and then taking carries into account by decomposing these three inputs in base 1000 like for the input operands):

result = z_{2} · B^{2m} + z_{1} · B^{m} + z_{0}, i.e.