scapp

Hmmm, it could definitely be clearer. It looks like it's using the bulk density for "sugar" (so including the air between the crystals) and the actual density for "sucrose".

It's somewhat a coincidence. In the paper, the dimension has to be larger than this constant K which is shown to be no larger than 1728. This is a fairly crude estimate that comes from multiplying a bunch of small numbers together, so the fact that it ends up being 12³ is not unimaginable. Then taking the dimension to be one larger than this estimate on K, we get 1729 = 12³ + 1³ (= 9³ + 10³).

However, 1728 isn't the minimum possible value for K. With more precise estimates, sketched in the paper, we can bring it down to 8 + ε. So assuming that this finer estimate works, using a 9 dimensional Fourier transform would also make the algorithm be O(n log(n)).

This is the easiest of the paradoxes mentioned in this thread to explain. I want to emphasize that this proof uses the technique of "Assume P, derive contradiction, therefore not P". This kind of proof relies on knowing what the running assumptions are at the time that the contradiction is derived, so I'm going to try to make that explicit.

Here's our first assumption: suppose that there's a set X with the property that for any set Y, Y is a member of X if and only if Y doesn't contain itself as a member. In other words, suppose that the collection of sets that don't contain themselves is a set and call it X.

Here's another assumption: Suppose X contains itself. Then by the premise, X doesn't contain itself, which is contradictory. Since the innermost assumption is that X contains itself, this proves that X doesn't contain itself (under the other assumption).

But if X doesn't contain itself, then by the premise again, X is in X, which is again contradictory. Now the only remaining assumption is that X exists, and so this proves that there cannot be a set with the stated property. In other words, the collection of all sets that don't contain themselves is not a set.

Nothing wrong with starting at 1 for induction, but yes, having an additive monoid is nice (still get a multiplicative monoid with N*)

*Bill Odenkirk. Bob is his brother

Nice! I wondered if someone would beat that TAS level anyway.

It's always fun to make fun of cranks. Thanks for linking that. The author really needs to find the right statement of what they call the Nested Interval Theorem. I cracked up at the complete misuse of it in the "Interval Argument for the Rationals" section

Kind of disappointing that it doesn't have the victory screen. But man, my fingers still remember that combo you have to do for those 4 wide areas.

2 line explanation of where e comes from here: to represent a number x in base b, you need roughly log_b(x) digits. If you weight that by the number of different digits in base b, you get b * log_b(x) = b * log(x)/log(b) = b/log(b) * log(x) (where now the log is any base you care to choose).

So assuming x > 1, this is minimized precisely when b/log(b) is minimized. The derivative of b/log(b) is (log(b) - log(e))/log(b)^2, so this is zero when b = e. (The second derivative is 1/(e log(e)) > 0, so this is a minimum).

There's a way to make it always appear above other windows. Let me see if I can find that again...

Ah, it's in the options menu on the task manager itself (alt + O to access the menu without a mouse).