2400+ entries in 0.313s
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
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.
mod6: <+asciilifeform> mod6, diana_coman , whoever else was tuned in for ffa
^^^ latest . << got it, thanks!
trinque:
^ for some curious reason this thing's behaving strangely dropped on a bsd
mircea_popescu: suppose there's a function that does hashing over a domain consisting of 2
^100 possible distinct values.
mircea_popescu: so notrly. only thing left for him is to joestack before they
^ a111: Logged on 2017-09-20 23:12 apeloyee: longmod procedure in linked pdf says "2
^b < m <2
^(b+1)" on page 9
apeloyee: longmod procedure in linked pdf says "2
^b < m <2
^(b+1)" on page 9
☟︎ mircea_popescu: yes, but as long as you do rsa, a
^ b mod m = either a
^ b or else, a
^ b - m. that's the complete story, there are no other cases. there's specifically no k * m parameter to be explored there.
mircea_popescu: fine. and the reason your mod does not reduce to "At most one substraction op" is "because i believe the product of two prime numbers 2kb each can in fact be less than 2
^4095". that it ?
mircea_popescu: so then! calculating a
^ b mod 2
^4096 is 0.26s ; calculating a
^ b mod m is 51.3s. that it ?
mircea_popescu: soooOOOOooooOOOO.... calculating a
^ b is 0.26s ; calculating a
^ b mod m is 51.3s.
mircea_popescu: sooo.... calculating a
^ b is 0.26s ; calculating a
^ b mod m is 51.3s.
mircea_popescu: so, calculating a
^ b is 0.26s ; calculating a
^ b mod m is 51.3s.
mircea_popescu: asciilifeform so to try and extract actual reality from this : a
^ b mod m takes 0.26 seconds in koch writing ; same a
^b mod m takes 51.3 on your box, notwithstanding a
^ b without any modding takes 1s ?
a111: Logged on 2017-09-17 11:10 mircea_popescu: (pro tip : since the modulus is 4096 bits, all powers of two up to 2
^4095 are unchanged by modding.)
mircea_popescu: (pro tip : since the modulus is 4096 bits, all powers of two up to 2
^4095 are unchanged by modding.)
☟︎ kanzure: 10
^14 kg of wafer-grade silicon vs 1 kg dna
kanzure: anyway the concern is that there's not enough pure silicon available to make enough flash memory for future projected demands of like 2
^28 bits in ~20 years.
kanzure: sorry, that was the wrong number. 10
^14 bits is today's write demand.
kanzure: i was in dc yesterday attending a dna synthesis meeting organied by iarpa. they want 10
^14 data write with dna.
a111: Logged on 2017-09-13 14:34 asciilifeform:
^ http://ulady.ru << exactly what it looks like
apeloyee: standard barrett requires a*b<n
^2, yes. hence I specified the double-precision reciprocal.
apeloyee: what woud be wrong with barrett reduction if you calculate a sufficiently precise reciprocal of modulus N,i.e. (maxint+1)
^2/N?
apeloyee: would O(N
^2) modular multiplication be too slow?
mircea_popescu: consider the number 97. is is 1100001. they do mp_mod (2
^6, 2
^5, 2
^0) ; you can do (2
^6, 2
^5, 0* 2
^4, 0* 2
^3,0* 2
^2,0* 2
^1,2
^0). the list method will sitll work, but this time in constanttime.