stromgo

7 ms·

31.
by stromgo 11y ago
The document recommends "intptr_t" over "long" for system-dependent types, and I've considered doing this but I've been put off by the printing type of %"PRIdPTR" instead of %ld. I wonder if on modern
32.
by stromgo 11y ago
Your Game 2 is poorly defined. What does it mean to "guess correctly" when you're asked to guess twice? If you systematically take the first (or the last) answer, then it's equivalent to being awaken only once on Tails.
33.
by stromgo 11y ago
"Matter contained in space contained in space" might be a problem, but "matter contained in space and space existing directly" should be ok, especially if your alternative is "matter existing directly".
34.
by stromgo 11y ago
For more info about the number of seconds in a solar day, I would rather recommend the table at http://en.wikipedia.org/wiki/Solar_time#Apparent_solar_time . The Leap_second article discusses a much smaller variation (
35.
by stromgo 12y ago
> Most discovered planets to date are... It's possible to do better than lump all of the discovered planets together and complain that the set is biased. For example, Kepler can detect a planet iff there is a chance alignment of i
36.
by stromgo 12y ago
Many such examples can be spotted in the table: [] == 0 == [[]] [1] == true == [1] ... There are even two examples with self-equality (x == x && y == y && z == z): "0" == false == "" "0" == 0 ==
37.
by stromgo 12y ago
The center of mass is the point where the force of gravity would pull you if the force increased proportionally with distance . In reality the force decreases as 1/r^2, which changes things. Think about the more extreme case of a barb
38.
by stromgo 12y ago
Your orbital period of 11.1 hours at 30 km looks wrong. The video shows about 7/8 of a 30-km orbit over 14 days (19 Nov - 3 Dec). The orbital period formula gives 14.6 days (2 * 3.14159265 * sqrt(30000^3/(6.67384e-11 * 10^13)) &#x
39.
by stromgo 12y ago
10e-3 m/s^2 is hard to believe, as the lander would have to be ~250m away from the center of mass [1]. More likely it is 2500m away from the center of mass, giving 6.67e-11 * 1.0e13 / 2500 ^ 2 = 0.0001 m/s^2, so about 1 gram.
40.
by stromgo 12y ago
He mentioned the "exact antipode on Earth", so he's probably thinking of an effect similar to antipodal chaotic terrain [1]. The fact that sound was only audible up to 3000 miles in radius is not enough to rule it out, you ne
41.
by stromgo 12y ago
It's not so clear. Here, as long as you can afford a near-optimal SHA256 processor, more computing power doesn't help. In the "successive squaring" method (see the essay posted by gwern), as long as you can afford a near
42.
by stromgo 12y ago
Looks like you're actually computing the full product. The only optimization you're doing is forgetting the low order bits as soon as you're done with them. The author is asking to keep them, which sounds reasonable. You init
43.
by stromgo 12y ago
Let's say that the full product in your example is 1XXXXYZZZZ To round it we need: (1) the X bits for our answer (2) the Y bit for rounding (3) the logical-or of the Z bits for rounding Ok so there might be a trick to compute (3
44.
by stromgo 12y ago
My example has nothing to do with decimal rounding. It's an example of a product (6172293634027511 * 7059478094414279) whose result is extremely close to the fence between two representable numbers (4837593847918366.5 * 2^53). If the 1
45.
by stromgo 12y ago
How do you correctly round, say, 6172293634027511 * 7059478094414279 / 2^53 = 4837593847918366.50000000000000011102 without essentially computing the full 106-bit product?