To me it feels a little bit like Levin's universal search ( in that it keeps ticking things off an infinite list. If the right answer's there, then we'll get to it eventually. The clever bit is ensuring

I couldn't recall what the identity function is, which is why I used (+0) instead. It's id.

Oh, dear. Day 13 looks like it's going to be painful, I will have to think about what's the right way to approach this in Haskell. Do I finally need a monad?

I'm not sure what the uses cases are for multi-auth that are satisfied by shipping keys around. If you have a third party to store N keys, why couldn't they store just 1 key? That is, if your model is

I mean, seriously, Haskell, WTF? Who need a maximum that is right-associative and lazy? maximum :: forall a . Ord a => t a -> a maximum = fromMaybe (errorWithoutStackTrace "maximum: empty

Voted for Yes, I understand and it makes sense to me..

The problem is here: If( buffer[0] == 0 ); break; That If with a capital letter is not the same as if with small letters. So the compiler thinks you're calling a function named If. (The semicolon on the

@stemq is actually a better home, but it seems doubly-redundant to ask a question there just to answer it myself (two separate posts) when I already answered it once on a different site. But the volume

I like Stellaris but I was disappointed that I have to buy an expansion in order to build megastructures. I was looking forward to my Dyson sphere. (Hm, I think I bought the expansion when it went on sale

Lot.

I am somewhat boggled that a mathematics PhD is being asked to do arithmetic as part of a hiring process.

This signage amused me: Whose contact number?

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Looks like it to me too, well done. But I disagree that the the two humans are necessarily all that humble. :)

One idea is replace all the individual constraints: A has to be in [0, 1, 2, 3, 4, ... ] B has to be in [0, 1, 2, 3, 4, ... ] (A,B) has to be in [(0,1), (2,3), (4,5) ... ] C has to be in [0, 1, 2, 3, 4,

You beat me to it by about 20 seconds. :)

But we invented trigonometry (and algebra) so we didn't have to be clever about geometry any more! :) I actually started out with a triangle dissection but I couldn't figure out the trick showing that

The polygon has area 7/2 * s^2 * cot( pi /14), where s is the side length. The rectangle has height s * cot( pi /14), so area s^2 * cot(pi/14) We can get both these from the formula for the length of the

Thanks for the pointer; I became a backer on Kickstarter.

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