## Pretty Printing a Tree in a Terminal

December 6, 2016

More often than I’d like, simple tasks turn out to be not that simple. For example, displaying (beautifully) a binary tree for debugging purpose in a terminal. Of course, one could still use lots of parentheses, but that does not lend itself to a very fast assessment of the tree. We could, however, use box drawing characters, as in DOS’s goode olde tymes, since they’re now part of the Unicode standard.

## The infinite radicals of Ramanujan

November 29, 2016

What if I asked you to find the numerical value of

$\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{\cdots}}}}}}$ ?

If you have difficulty figuring out what this thing is, don’t worry. You’re not the only one. This question is one the problems posed by Srinivasa Ramanujan, one of the most brilliant and, well, mysterious mathematicians of all time. In fact, it was head enough that Ramanujan had to give the answer a few months later.

## Tweet time!

November 22, 2016

I’ve been using twitter for about five years, and I wondered if my use of it changed over time, and more precisely, linked to my wake/sleep cycle. That’s fortunately kind of simple to check because you can simply request your whole Twitter archive, delivered as a plain CSV File! Let’s see how we can juice it.

## Square Roots (Part V)

November 15, 2016

Last week we examined the complexity of obtaining $k$ in the decomposition $n=2^k+b$ for some integer $n$. This week, we’ll have a look at how we can use it to speed-up square root extraction. Recall, we’re interested in $k$ because

$2^k \leqslant 2^k+b < 2^{k+1}$,

with $0 \leqslant b < 2^k$, which allows us to get easy bounds on $\sqrt{n}$. Better, we also have that

$\sqrt{2^k} \leqslant \sqrt{2^k+b} \leqslant \sqrt{2^{k+1}}$,

and we know how to compute $\sqrt{2^k}=2\frac{k}{2}$ (somewhat efficiently! Let’s combine all this and see how it speeds up things.

## Square Roots (part IV)

November 8, 2016

In a previous post, we noticed that

$\sqrt{2^k} \leqslant \sqrt{n} = \sqrt{2^k+b} \leqslant \sqrt{2^{k+1}}$,

where $0 \leqslant b < 2^k$, could be used to kick-start Newton's (or another) square root finding algorithm. But how fast can we find $k$ and $b$ in this decomposition?

## Regula Falsi

November 1, 2016

The regular falsi, or method of false position, is a method to solve an equation in one unknown, from an initial “guess”. Guesses are progressively refined, by a rather simple method as we will see, until the exact answer is reached or until convergence is reached. The method is useful when you don’t really know how to divide by arbitrary values (as it was the case in Ancient Times) or when the equation is cumbersome.

## Horner’s Method

October 25, 2016

It’s not like evaluating polynomial is something that comes up that frequently in the average programmer’s day. In fact, almost never, but it brings up a number of interesting questions, such as, how do we evaluate it very efficiently and how much of a simplification in the computation is actually a simplification?

The generic polynomial has the form

$a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_nx^n$.

If we evaluate this naïvely, we end up doing $O(n^2)$ multiply and $O(n)$ additions. Even using the fast exponentiation algorithm, we still use $O(n \lg n)$ multiplies. But we can, very easily, get down to $O(n)$.