The Python 🐍 problem-solving bootcamp πŸš€ is starting soon. Join the second cohort now!

Progress is great and new things are always exciting... but that doesn't mean old things don't have any value!

The Python 🐍 problem-solving bootcamp is starting soon. Join the second cohort now!

A picture of an abacus
Photo by Crissy Jarvis on Unsplash

Very recently I watched a Youtube video posted by my friend MathGurl, in which she explained the ancient Egyptian multiplication method. At the same time, and for no particular reason, I remembered Haskell, so I decided to implement the method. It was not a big feat of programming, but I did enjoy relearning the basics of Haskell I once knew. You can find the file with the implementation in this GH file or right here:

(The implementation only works for non-negative integers)

The method is quite simple and works because of the binary expansion of a number. Basically, if you want to calculate \(a \times b\), either \(b\) is even or odd. If \(b\) is even, just cut \(b\) in half and duplicate \(a\) to compute \((2a)\times(\frac{b}2)\). If \(b\) is odd, then \(ab = a + (2a)\times\frac{b-1}2\). Another way of thinking about this is by writing \(b\) in the form

\[ b = 2^{k_1} + 2^{k_2} + \cdots + 2^{k_n}\]

and then having

\[ a\times b = a(2^{k_1} + 2^{k_2} + \cdots + 2^{k_n}) = a2^{k_1} + a2^{k_2} + \cdots + a2^{k_n}.\]

If I muster the courage to do it, I might also redo the functions in the post about Kleene recursive functions, in Haskell...

Take your Python 🐍 skills to the next level πŸš€

I write about Python every week. Join +16.000 others who are taking their Python 🐍 skills to the next level πŸš€, one email at a time.

Previous Post Next Post

Blog Comments powered by Disqus.