2300+ entries in 0.326s
a111: Logged on 2017-10-31 15:10 asciilifeform: so what does this look like ? walk in with suitcase of benjies, come out with plutoni
^H
^H
^H
^Htitle ?
mircea_popescu:
^ item was in first year of writing. mandatory week spent by 9yos trying to write it out without staining thje paper.
mircea_popescu: i
^i ( ɨi ), short form of to be present indicative ; i-i (ii), article-pronoun manuple.
hanbot:
^ in latest trilema, i mean, mircea_popescu
mod6: (just to see if I get a different value, or non-round value for X
^2)
mod6: ben_vulpes: ah, i seem to remember reading about the X
^2 percentiles in knuth. mine in particular came out pretty round ya. or do you mean "your" in the general sense?
a111: Logged on 2017-10-17 12:46 mircea_popescu: davout
^ BingoBoingo:
^ asciilifeform this suggests 100 Mbps may be eliciting the same response from the connectivity endowed as request for 1U
a111: Logged on 2017-10-14 12:06 apeloyee: it seems I see how to squeeze out one more FZ-sized temporary from FZ_Mod_Exp, besides the
^^ and using a proper barrett; it will exacerbate the above-mentioned physical leakage, though...
http://p.bvulpes.com/pastes/XvDnd/?raw=true <- as usual, not tested.
a111: Logged on 2017-10-14 12:06 apeloyee: it seems I see how to squeeze out one more FZ-sized temporary from FZ_Mod_Exp, besides the
^^ and using a proper barrett; it will exacerbate the above-mentioned physical leakage, though...
http://p.bvulpes.com/pastes/XvDnd/?raw=true <- as usual, not tested.
a111: Logged on 2017-10-08 00:20 asciilifeform:
http://btcbase.org/log/2017-10-07#1722411 << 1 ) ffa is closed form. i.e. it CAN be written as a number of nand gates, with a 'funnel' at the top, to which you present a,b,c, e.g. 4096bit, numbers, and at the bottom in a little cup you get a
^b mod c , and with NO UPWARDS FEEDBACK FLOW of information , i.e. answer comes after same interval of time always, and with strictly downwards signals.
mod6: to understand the thing, one can simply do the: B
^ ((S - 1) & (A
^ B)) on a calc
trinque:
^ lemme know if those look goofy. process that spat them out is brand new.
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.
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).
☝︎ shinohai:
^ I heard the above was edited by sjw on Google translate. It used to be "Take a look at the nigger"
lobbes: !~later tell mircea_popescu
^^ 'help sexpr' and 'help json' also working. lobbesbot has been brought up to spec
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 21:09 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
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: 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 2017-10-05 19:43 asciilifeform: euclidean'd be o(n
^3) yes
mircea_popescu: speaking of
^ asciilifeform , how's the isp thing going ?
apeloyee: but here's an O(n
^2 log n), for a large value of constant.
apeloyee: you gcd is O(N
^3), and so is trial divisiom
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: and in one side only, as obv can't be bigger than (A/N)*4
^K
apeloyee: which is equal to (A/N - 1)*4
^K
apeloyee: A <4
^K, so the above strictly >(A*4
^K)/N - A.
apeloyee: let rewind. A*R = A*(4
^K - B)/N = (A*4
^K)/N - A*B/N; as B <= N, then the previous >= (A*4
^K)/N - A. clear?
apeloyee: A - N*floor(A*R/4
^K) < 2*N <- do you agree with this? (the proof is unnecessarily complicated in that paste)
apeloyee: meaning you divide 4
^K - 1 by the modulus, not 4
^K
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: 4
^K = R*N + B, 0 <= B <= N (line seven). pseudo-remainder.
a111: Logged on 2017-10-05 16:15 asciilifeform: the (unsurprising) surprise is that
^method is wholly absent from the public lit