in reply to Factors
in thread Vampire Numbers
I think what is being suggested here is that instead of
creating all possible pairs of numbers (x,y) from the
digits of a number p and testing if x*y == p, it is more
efficient to factor the number p first and and then test
all possible factorizations to see if they are just a
permuation of the digits of p.
Finidng a quadratic residue x^2=y^2 (mod p) is useful,
because then x^2  y^2 = (x+y)*(xy) = 0 (mod p) and
so (x+y) or (xy) may be factors of p.
In practice, one starts with an x that has the smallest
square larger than p, then computes x^2p to see if it
is a perfect square. If not, increment x and repeat.
This is pretty fast relative to testing by division
from 2 to sqrt(p), but there exist even faster methods
based on the quadratic residue. The Quadratic Sieve,
invented by Pomerance in 1981 is a direct derivative
and the Number Field Sieve is a related method.
Getting back to Vampire numbers, what are the relative
efficiencies for a number p with 2d digits? For the first
method, a naive approach generates all permutations of 2*d
digits and splits each into two ddigit numbers to test.
Thus each 2d digit number requires about (2*d)! == (2*log(p))!
tests, or (2*log p)^(2*log p) by Stirling's approximation.
For the second method, a quadratic sieve has running
time of order exp(sqrt(log(p)*log(log(p)))) to find a
single factor of p. So finding a single factor using QS
is more efficient that testing all factors using the
first method. But I don't know the running time for
finding all the factors using the QS and I don't know the
typical multiplier for the scaling result of the QS method.
Anyone?
Mark


Replies are listed 'Best First'.  

Re: Re: Factors
by gumby (Scribe) on Jun 12, 2002 at 21:45 UTC  
Re: Re: Factors
by YuckFoo (Abbot) on Jun 12, 2002 at 22:30 UTC  
by kvale (Monsignor) on Jun 13, 2002 at 04:51 UTC  
by gumby (Scribe) on Jun 13, 2002 at 15:43 UTC  
by gumby (Scribe) on Jun 15, 2002 at 19:21 UTC 
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