A while ago, Martin Capoušek drew my attention on Kahan’s summation algorithm and ask how it compared to the other methods I presented then.
Well, let’s find out.
A while ago, Martin Capoušek drew my attention on Kahan’s summation algorithm and ask how it compared to the other methods I presented then.
Well, let’s find out.
I’ve visited the median problem a couple of times already. The median is the value in the middle of a sorted list of values. If the list is already sorted, that’s not much of a problem. If it isn’t, then we must use some efficient way to do it.
During last marking season (at the end of the semester), I had, of course, to grade a lot of assignments. For some reason, every semester, I have a good number of students that write code like they just don’t care. I get code that looks like this:
int fonction (int random_spacing)^M{ ^M int niaiseuses; for (int i=0;i<random_spacing; i++){ { { std::cout << bleh << std::endl; }} } }
There’s a bit of everything. Random spacing. Traces of conversions from one OS to another, braces at the end of line. Of course, they lose points, but that doesn’t make the code any easier to read. In a previous installment, I proposed something to rebuild the whitespaces only. Now, let’s see how we can repair as many defects as possible with an Emacs function.
Quite a while ago, while discussing Monte Carlo Integration with my students, the topic of choosing sample locations came up, and we discussed low-discrepancy sequences (a.k.a. quasi-random sequences). In a low-discrepancy sequence, values generated look kind of uniform-random, but avoids clumping. A closer examination reveal that they are suspiciously well-spaced. That’s what we want in Monte Carlo integration.
But how do we generate such sequences? Well, there are many ways to do so. Some more amusing than other, some more structured than others. One of the early example, Halton sequences (c. 1964) is particularly well behaved: it generates 0, 0.5, then 0.25 and 0.75, then 0.125, 0.375, 0.625, and 0.875, etc. It does so with a rather simple binary trick.