David Atahualpa

[David Atahualpa @datahualpalima]

@mhmck Some will come - the ship that was already coming for weeks before Trump lied that he 'just had a phone call' with Vlad. Then Trump can make a big fuss about it, and even if no more ships come MAGA will be convinced they would (except for evil Ukraine)

[David Atahualpa @datahualpalima]

@Michael33896472 @ukraine_map Mediazona: goes out, does all the hard work of compiling publicly available data, obituaries, social media posts, etc, builds a public interface so everyone can cross-check everything.
Michael: 'grrr propaganda ukronazis grrr'

[David Atahualpa @datahualpalima]

@PopatGuru @engineers_feed go ahead! How would you do it?

[David Atahualpa @datahualpalima]

@dosahm8086 @whitherapathy @Math_files If (A)^4 +(B)^4 +(C)^4 = (D)^4

then

(nA)^4 +(nB)^4 +(nC)^4 = (nD)^4 for any n

[David Atahualpa @datahualpalima]

@whitherapathy @Math_files sure. At the very least multiply each of those four integers by any integer n

[David Atahualpa @datahualpalima]

@RollinsCori @evgen1232007 what would 'critical' look like? 10%?

[David Atahualpa @datahualpalima]

@amazingmap In Portugal: alperce
In Brazil: damasco [ie, the fruit from Damascus]

[David Atahualpa @datahualpalima]

@Schmoepfel3000 @engineers_feed sure, wise guy. But that chance is only 1 in a 1000 (or 1 in a 100 if each prisoner chooses one bottle and sips). If nobody dies after 10hrs (as is 99% likely), all you know is those 10 bottles are safe. The point is finding a scheme that *always* gets you 999 safe bottles

[David Atahualpa @datahualpalima]

@HKUFBE so much for 'two systems' though, huh?

[David Atahualpa @datahualpalima]

@PopatGuru @engineers_feed In that case: to be *sure* 2 prisoners won't ever die, any given bottle can be sipped by at most 1 of them (since it might be poisoned). If you split the set evenly, so each sips 100 bottles, 1 dies and the others live, but you only secure 900 bottles. Any other split does worse

[David Atahualpa @datahualpalima]

@PopatGuru @engineers_feed Fair enough,but you're introducing a new condition, namely "how many bottles can the king identify as safe *while killing at most one prisoner* ?". That's a different math problem to the one proposed

[David Atahualpa @datahualpalima]

@mfa_russia Despite punishments MUCH worse than "bans, restrictions & fines", people in Crimea, Luhansk, Donetsk, Zaporizhzhia and Kherson keep speaking Ukrainian. You evil nazi fucks.

[David Atahualpa @datahualpalima]

@PopatGuru @engineers_feed Sure, but only if the poisoned bottle happened to be the one labeled 1111111111, something that would happen with a chance of 1 in 1024, assuming codes are chosen at random. But the question requires keeping the guests safe, not the prisoners unharmed

[David Atahualpa @datahualpalima]

@Peter05122415 @engineers_feed As @eogfelton said, all prisoners have the same chance of dying, in fact exactly 50% if the poisoned bottle's code was chosen at random from all 1024 possible codes. The coding scheme treats all 10 code positions, hence all prisoners, equally.

[David Atahualpa @datahualpalima]

@DrJMMHall @engineers_feed read it again, Dr Hall

[David Atahualpa @datahualpalima]

@eogfelton @Peter05122415 @engineers_feed Yes, each will take ~500 sips, but they can all die if the poisoned bottle happens to be coded 1111111111. If its code is chosen at random, this is unlikely (1 chance in 1024). Most likely (~25% chance) the code has 5 1's, so 5 die. About 21% chance that 4 die, ditto for 6, etc

[David Atahualpa @datahualpalima]

@mikeromanov91 @engineers_feed since nearly every computer in the world uses binary encoding, including those running malware, weapons control software, dictatorship spyware etc, pretty sure murdering (an expected) 5 prisoners in the name of a good party is rather low on the evil scale

[David Atahualpa @datahualpalima]

@BDiggetyDawg @engineers_feed you wouldn't believe how thirsty those guys were