Last time, we used word rank to try to figure out who could possibly have written Cuckoo’s calling. It didn’t work out so well, but we at least have a nice framework in place. So perhaps we can try a few more ways of turning entire novels into a few numbers.
JP's Blog
Authorship attribution: Part 1
About two weeks ago, the new crime fiction novel Cuckoo’s Calling was revealed to have actually been written by J.K. Rowling under the pseudonym Robert Galbraith. What’s interesting is exactly how they came to that conclusion. Here’s a quote from Time magazine (via Programming Praxis):
A programming puzzle: f(f(n)) = -n
Two Programming Praxis puzzles in a week? Madness! Let’s do it!
This time, the puzzle at first seems rather minimal:
Write a function f so that f(f(n)) = -n for all integers n.
If you haven’t seen this problem before, take a moment to think though it. It’s a neat little problem–a close cousin to a lateral thinking puzzle.
Adventures in optimization, re: Typed Racket
Last Thursday I wrote a post about generating Perlin/simplex noise in Racket. Later that day, I posted to the Racket mailing list asking how I could make it faster. What resulted was a whole sequence of responses (primarily about Typed Racket) and a bit of a rabbit hole that I’m still trying to wrap my head around.
Perlin and simplex noise in Racket
Many games need noise. No, not that noise–this noise. Or better yet, this noise. More seriously, noise in this context refers to psuedo-randomly generated images that can be used for natural looking terrain generation1. Something like this:
Approximating Pi with Buffon's Needle
I’m a bit late for Pi Day, but Programming Praxis had a neat problem on Friday that I wanted to check out:
Suppose we have a floor made of parallel strips of wood, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two strips?
Triangle Trilemma
Four points, a square?) and comes originally from a Google Code Jam problem. The problem is stated simply enough
Accept three points as input, determine if they form a triangle, and, if they do, classify it at equilateral (all three sides the same), isoceles (two sides the same, the other different), or scalene (all three sides different), and also classify it as acute (all three angles less than 90 degrees), obtuse (one angle greater than 90 degrees) or right (one angle equal 90 degrees).
But once you start implementing it, that’s when things get more interesting. 😄
Four points, a square?
Another post from Programming Praxis. This one was originally intended for Friday but they posted it early, so I figured I would go ahead and do the same. The problem is actually deceptively straight forward:
Given four points, do they form a square?
Happy New Year
Yesterday’s post from Programming Praxis asks us to build a very special sort of expression. Using the numbers 10, 9, 8, 7, 6, 5, 4, 3, 2, and 1 in that order along with the operators of multiplication, division, addition, subtraction, and concatenation, find all of the ways that we can write an expression totaling 2013. Here’s one valid solution:
109 - 8 * 7 + 654 * 3 - 2 / 1 = 2013
Narcissistic Numbers
Today’s post from Programming Praxis posits an interesting problem: how can we (efficiently) find all of the narcissistic numbers (in base 10)?