4600+ entries in 0.317s
lobbesbot: Logged on 2017-10-11 21:42:21: <diana_coman> !QAuction
2.86mn 48 29173 Flotsam q62
trinque: i.e. if you invoice him it'll be
2 shinohai: YOU ARE HEREBY BANNED FOR JOIN/PART SPAM. If yu wish to recconect plox
2 purchase bouncer from deedbot, thx.
BingoBoingo: Oh, it looks like Hong Kong is waking up
2 hours earlier today
a111: Logged on 2017-10-11 16:53 asciilifeform: proposed item is that if using 1 stream, knowledge of a segment of plaintext gives you something valuable to work with; if using
2+ -- nothing
shinohai: lmao the U.S. futbol team was defeated
2-1 by *Trinidad and Tobago* and will not advance to the World Cup.
jhvh1: mircea_popescu: Bitstamp BTCUSD last: 4779.18, vol: 11788.21139918 | Bitfinex BTCUSD last: 4778.
2, vol: 39777.25429663 | BTCChina BTCUSD last: 4274.3844, vol: 0 | Kraken BTCUSD last: 4780.3, vol: 4840.84169001 | Volume-weighted last average: 4778.58503167
jhvh1: BingoBoingo: Current Blocks: 489300 | Current Difficulty: 1.123863285132E12 | Next Difficulty At Block: 489887 | Next Difficulty In: 587 blocks | Next Difficulty In About: 3 days, 20 hours, 56 minutes, and
2 seconds | Next Difficulty Estimate: None | Estimated Percent Change: None
jhvh1: BingoBoingo: Bitstamp BTCUSD last: 4771.03, vol: 11240.26276269 | Bitfinex BTCUSD last: 4790.
2, vol: 40972.92919445 | BTCChina BTCUSD last: 4282.8318, vol: 0 | Kraken BTCUSD last: 4790.
2, vol: 5483.59331143 | Volume-weighted last average: 4786.46537534
jhvh1: BingoBoingo: Bitstamp BTCUSD last: 4612.17, vol: 9972.22002248 | Bitfinex BTCUSD last: 4619.6, vol: 28296.13901936 | BTCChina BTCUSD last: 4240.5948, vol: 0 | Kraken BTCUSD last: 4614.
2, vol: 4197.98671649 | Volume-weighted last average: 4617.32142572
mircea_popescu: "SGML is a good idea when the markup overhead is less than
2%." << holy shit check him out. that's... a lot narrower than i'd have guessed.
apeloyee: oh noes, the enemy will learn that the first primes are
2,3,5,7,11,..!!!
a111: Logged on 2017-10-07 21:48 apeloyee:
http://btcbase.org/log/2017-10-05#1721485 << alternatively, can *construct* numbers which don't have very small factors. pick a nonzero remainder mod
2, mod 3, ... mod largest-prime-fit-in-your-primorial and find what number of primorial is congruent to it using chinese remainder theorem
a111: Logged on 2017-10-07 19:28 asciilifeform:
http://btcbase.org/log/2017-10-07#1722358 << point was exactly to compare like items. i.e. heathendom does NOT get to 'win' by 'oh hey the hamming weight of exponent is only
2, not 4096, so we only do 4 modexps and not 8192'
a111: Logged on 2017-10-08 00:16 asciilifeform: the ONLY correct method of generating cryptoprimes, is to 1) get N bits from FUCKGOATS
2) determine, in fixed spacetime every single time, whether that string of bits constitutes a usable prime.
apeloyee: right, unclear again. the muliply of N and floor(A*R/4^K) can be calculated mod
2^(K+1)
apeloyee: modulo
2^(K+
2) for classical barrett.
a111: Logged on 2017-10-07 21:25 apeloyee: the multiply-by-approximate quotient in barrett's also needs only the lower part (plus
2 extra bits to the left), and lower part of product can be computed exactly (since rounding is not a problem)
apeloyee:
http://btcbase.org/log/2017-10-07#1722397 << I was unclear. Let A be the number to be reduced mod N, R the approximate reciprocal, K the ffa bitness fitting the modulus, then we know that 0<A - N*floor(A*R/4^K) <
2*N <
2^(K+1). So might as well calculate A - N*floor(A*R/4^K) modulo
2^(K+1).
☝︎ apeloyee:
2 half products out of 3 on the first level of recursion, 4 of 9 on second, and 8 of 27 on third, assuming 64-bit words and unrealistic
2-fold speedup of comba for half-multiply, and no overhead in karatsuba,
a111: Logged on 2017-10-07 21:14 apeloyee:
http://btcbase.org/log/2017-10-07#1722289 << and the point of doing karatsuba is? you do
2 recursive calls to Mul_Karatsuba_TopOnly and one to Mul_Karatsuba. should've simply calculated upper_part(XLo*YHi), upper_part(YLo*XHi) and XHi*YHi
danielpbarron: why is the deposit operation a
2-part thing? couldn't it just encrypt an address to my key with the amount to send?
mircea_popescu: anyway, there's a pile of disinfo and general crap surrounding the events. as an example : on
2-3 dec gorbachev hung out with bush on a soviet ship. on 4th, there was the wasaw pact meeting. gorbachev was well excited of whatever, the new bulgarian (mladenov, his college pal) and generally the western press coverage.
phf: my specific ip was a cloudflare ip 104.28.25.
2. current dig for archive.is resolves to 195.123.218.180, which is a netherlands "mobicom ltd" range. i suspect that archive.is took themselves off cloudflare in the last some months, so now i'm hitting cloudflare proxy servers and they are complaining that the host: is no longer served
a111: Logged on 2017-10-08 00:16 asciilifeform: the ONLY correct method of generating cryptoprimes, is to 1) get N bits from FUCKGOATS
2) determine, in fixed spacetime every single time, whether that string of bits constitutes a usable prime.
a111: Logged on 2017-10-07 21:53 apeloyee: the primorial has to be, say,
2^32 times less than the ffa maxint. then you can add randomnumber*primorial, and such a number is equally likely to any prime from some interval
a111: Logged on 2017-10-07 21:48 apeloyee:
http://btcbase.org/log/2017-10-05#1721485 << alternatively, can *construct* numbers which don't have very small factors. pick a nonzero remainder mod
2, mod 3, ... mod largest-prime-fit-in-your-primorial and find what number of primorial is congruent to it using chinese remainder theorem
a111: Logged on 2017-10-07 21:25 apeloyee: the multiply-by-approximate quotient in barrett's also needs only the lower part (plus
2 extra bits to the left), and lower part of product can be computed exactly (since rounding is not a problem)
a111: Logged on 2017-10-07 21:14 apeloyee:
http://btcbase.org/log/2017-10-07#1722289 << and the point of doing karatsuba is? you do
2 recursive calls to Mul_Karatsuba_TopOnly and one to Mul_Karatsuba. should've simply calculated upper_part(XLo*YHi), upper_part(YLo*XHi) and XHi*YHi
a111: Logged on 2017-10-07 21:09 apeloyee: asciilifeform: turns out a simple, ffa-suitable O(N^
2) algorithm exists for GCD. This is adapted from GMP docs with one extra operation in the loop:
http://p.bvulpes.com/pastes/oupUJ/?raw=true . Note: the code as posted is likely wrong, but I'm sure the idea can be made to work.
a111: Logged on 2017-10-07 21:53 apeloyee: the primorial has to be, say,
2^32 times less than the ffa maxint. then you can add randomnumber*primorial, and such a number is equally likely to any prime from some interval
a111: Logged on 2017-10-07 19:28 asciilifeform:
http://btcbase.org/log/2017-10-07#1722358 << point was exactly to compare like items. i.e. heathendom does NOT get to 'win' by 'oh hey the hamming weight of exponent is only
2, not 4096, so we only do 4 modexps and not 8192'
apeloyee: the primorial has to be, say,
2^32 times less than the ffa maxint. then you can add randomnumber*primorial, and such a number is equally likely to any prime from some interval
☟︎☟︎ apeloyee: *what number has such remainder from division by
2,3, ...
apeloyee:
http://btcbase.org/log/2017-10-05#1721485 << alternatively, can *construct* numbers which don't have very small factors. pick a nonzero remainder mod
2, mod 3, ... mod largest-prime-fit-in-your-primorial and find what number of primorial is congruent to it using chinese remainder theorem
☝︎☟︎☟︎☟︎☟︎☟︎ apeloyee: the multiply-by-approximate quotient in barrett's also needs only the lower part (plus
2 extra bits to the left), and lower part of product can be computed exactly (since rounding is not a problem)
☟︎☟︎ apeloyee: asciilifeform: turns out a simple, ffa-suitable O(N^
2) algorithm exists for GCD. This is adapted from GMP docs with one extra operation in the loop:
http://p.bvulpes.com/pastes/oupUJ/?raw=true . Note: the code as posted is likely wrong, but I'm sure the idea can be made to work.
☟︎ a111: Logged on 2017-10-07 19:28 asciilifeform:
http://btcbase.org/log/2017-10-07#1722358 << point was exactly to compare like items. i.e. heathendom does NOT get to 'win' by 'oh hey the hamming weight of exponent is only
2, not 4096, so we only do 4 modexps and not 8192'
mod6: (fwiw, that machine I just ran it on has Python
2.7.9)
mod6:
2.107 shows extended euclid that yields greatest common divisor d of two integers a and b, but also integers x and y satisfying ax + by = d
mod6:
2.103: FACT If a and b are positive integers with a > b, then gcd(a, b) = gcd(b, a mod b)
mod6: well, if you walk for
2/3h per day, maybe don't think of them as "shoes". perhaps, "proper foot equipment" or something.
phf`: the place to be is of course
http://www.mmts9.ru they do have 1/
2 rack but i haven't tried them, since wasn't sure about our requirements. they can even get you a personal cage. antenna placement on the rooftop so we can put a microwave relay, the works :p
mircea_popescu: ah you can just handle rouble ? send you 10btc, you send out w/e
2.5mn rou or w./e it is ?
jhvh1: asciilifeform: phf was last seen in #trilema
2 days, 21 hours, 4 minutes, and 30 seconds ago: <phf> asciilifeform: that i figured out, the whole contraption reminds me of that ru anti-temper device pdf that you posted recently
jhvh1: mircea_popescu: 156.68421052631578 / 58 =
2.701451905626134
a111: Logged on 2017-10-05 16:42 asciilifeform: i refreshed archive.is of all the major pages
2-3 day before dulap went
a111: Logged on 2017-10-05 19:54 apeloyee: grind until you did (
2*bitness) steps. since remainder can't be more than a/
2.
apeloyee: grind until you did (
2*bitness) steps. since remainder can't be more than a/
2.
☟︎ apeloyee: but here's an O(n^
2 log n), for a large value of constant.
apeloyee: just do some muxes in the end.
2*bitness divisions obv suffice (actually less, but I'm sleepy now)
apeloyee: now let's try generalizing to standard barrett (the error will grow to
2 of course). let L be number of digits in N:
2^(L-1) <= N < L. L is calc'd with the CLZ algorithm
deedbot: asciilifeform updated rating of apeloyee from 1 to
2 << A - N*floor(A*R/4^K) <
2*N
apeloyee: A - N*floor(A*R/4^K) <
2*N <- do you agree with this? (the proof is unnecessarily complicated in that paste)
apeloyee: if N is
2^(K-1), then ordinary quotient won't fit in K+1 bits. but pseudo-quotient (one less the actual quotient) still works.
apeloyee: goddamnit. it's FIXED!!! first sub-word one (which doesn't leak in our model), then by 1 word, then by
2, then 4, 8 and so on
apeloyee: ok, so quotient of division of
2^4096 by
2^4096-1 is 1.
a111: Logged on 2017-10-05 16:06 asciilifeform: compute 'barrettoid' c for the current m : c :=
2^4096 / p . this is done via the knuthian longdividotron, once per modexp.
apeloyee: ok, let m =
2^4096 - 1. then c=1