158500+ entries in 0.095s
mircea_popescu: alf : "probvlem is mod is not distributive" me : "it can be made, cheaply" alf : "no, it can not" me : "here" alf : "oh, i did
that back in july".
then wtf are you griping about.
mircea_popescu: this approach of "i have a girlfriend and i am blind
to all else" doesn't work with girlfriends, or anything else.
mircea_popescu: that i can believe.
though i expect
the above is actually cheaper
than adding
teh numbers first and modding after.
mircea_popescu: so you won't have many passes of it. just as many as elements in list at
the most.
mircea_popescu: yes, but
this is very cheap here. because
the elements in list are < mod
mircea_popescu: this will always
take
the same ops no matter what elements are in
the original list.
mircea_popescu: repeat 2 until you have populated a list of equal length, and return
the correct element from it.
mircea_popescu: for simplicity, input list limited
to 2 elements, but expansion obvious.
mircea_popescu: and
this is potentially recursive, in
that if you have a 500 bit number with 300 ones in it, you do
the mod for 500
terms which are all a power of 2,
throw 200 away, keep
the other 300 and add
them.
☟︎ mircea_popescu: that is not my concern! if
there IS a mod,
then yo ucan apply it
to
the
terms rather
than add
them first and apply
to result, is all i'm saying.
a111: Logged on 2017-09-12 20:54 asciilifeform: and it happens B^2
times
mircea_popescu: you understand, a mod x + b mod x + c mod x may be > x, but never by more
than op count * x.
mircea_popescu: you write by hand a function which
takes a list with a promise none of
the items on it exceed a mod, and returns
the mod of
the sum of
the sum of
the elements, in constant
time.
mircea_popescu: just write it all out by hand,
the constanttime mod distributivetor.
mircea_popescu: that small cost can be slightly higher and constant
time.
mircea_popescu: i am
talking about how mod is distributive
to addition "at a small cost".
mircea_popescu: asciilifeform you understand you need AT MOST a single pass of knuth ? because it may exceed
the mod but never by more
than 3x ?
mircea_popescu: it is distributive in
this sense at a minimum cost (tm).
rothbart: having some
trouble decrypting
the message deedbot sent me
rothbart: can I just register a new public key with deedbot, without revoking
the old one?
mircea_popescu: ie, all
the
trivial polynomials of
the mod, see ? x, x^2 + x, x ^ 3 + x etc etc
mircea_popescu: asciilifeform if you maintain a list of
the mod and it squares
mircea_popescu: rothbart you're perhaps
too new
to know
this, but
the PREVIOUS
time
there was "miner consensus" it
turned out
the miners
that "voted" and "supermajority" didn't give
the slightest shit about
the whole
thing, and did NOT actually implement what
they were misrepresented as having supported.
mircea_popescu: but whatever, i don't mind making money out of mit's "blockchain of
the future", like i didn't mind fleecing "ripple" or cashing btc crash. free bitcoin will continue for as long as usg can draw breath, i'm not against. let
them lose what
they can't invest.
rothbart: anyone, at any point, can claim all segwit outputs - all
they need
to do is solve
the 80bit hash or whatever?
mircea_popescu: (always and everywhere,
the stupid poor.
to
them it makes sense,
they've nothingh
to lose anyway)
rothbart: how is it
that
the power rangers are all BLIND
to
this?
mircea_popescu: the
t1 wreckers may yell all
they want
this "is not right". but in bitcoin longest chain prevails, and so
the story ends
mircea_popescu: now
then. at
time
t2, any of
those involved, or any
third party, simply SPENDS
that bitcoin.
mircea_popescu: oin chain. on
the basis of
these mystery strings, OTHER PARTIES, which ARE IN NO WAY BOUND
TO
THIS, alloocate wholly imaginary bitcoins
to
the sort of imbeciles who buy into
this scheme (always and everywhere,
the stupid poor.
to
them it makes sense,
they've nothingh
to lose anyway)
rothbart: does
the attacker just build upon
the main chain,
then; sending all
the segwit outputs
to himself?
mircea_popescu: let's drop
the math for a moment and delve. at
time
t0, bitcoin works. at
time
t1, some wreckers under "public pressure" as discussed well in
http://trilema.com/2013/digging-through-archives-yields-gold/ attempt
to attack
this bitcoin
that works, by producing an alt-bitcoin,
that does not work.
the specific way in which
the alt-bitcoin
thatr does not work "works" is by deeding (exactly like deedbot) some strings into
the bitc
rothbart: I've read
the logs, still confused about
this
rothbart: wouldn't he have
to redo all
that PoW since segwit wen't active on his fork?
rothbart: as in,
the attacker would be doing a chain rewrite in order
to keep
the segwit outputs on his fork?
mircea_popescu: whole fucking point of segwit is
to
try and
take pow away.
rothbart: I've been
trying
to grok
the segwit "theft" incentive - as
the bounty grows, so does
the PoW defending it - doesn't
this keep
the segwit outputs safe?
☟︎ mircea_popescu: rothbart if you have it in a portable data format, can just feed it into
trb
rothbart: asciilifeform: will I need
to redownload
the blockchain on
trb?
mircea_popescu: "what do you mean
this problem is hard, i have a half baked item in my head i pompously call abstraction in which it is EASY!!!"
mircea_popescu: i
tried
the haskell approach
to "Solving" problems, whadda ya want from me.
mircea_popescu: 1. multiply x and y ; 2. count bits of result ; 3. count bits of modulus ; 4. multiply modulus with count2 - count3 and
test if larger
than result. if not, substract. if yes, multiply with count2-count3-1 and do
the same. repeat until result smaller
than modulus.
mircea_popescu: ok, how slow is
the following heuristicized approach :
a111: Logged on 2017-09-12 18:23 asciilifeform: you have
the following primitives : multiplier ( 2 N-bit numbers -> 1 2N-bit number ); adder ( ditto ); subtractor ( ditto ); muxer ( 2 N-bit numbers, 1 single-bit number, yields one or
the other N-bit depending on
that single bit ) ; logical ops (and/or/xor/not)