Integer root extracting

[Torbjorn Granlund]

Hans Aberg <haberg-1 at telia.com> writes:

  > Completely factoring a is of course not always possible.
  
  I'm not sure what you mean here: a is an integer, in fact positive in
  my application. So it always has a prime factorization
  p_1^k_1*...*p_n^k_n.
  
I thought you were interested in a usable algorithm, not merely a
theoretically correct method.  I meant thatt complete factorisation of
large integers is in practice not possible, since it would take too much
time.

  > A more tractable algorithm is the following:
  >
  > for y = floor(log(a)/log(2)) to 2 by -1
  >  If a^(1/y) is an integer, return y
  >
  > The starting value is the largest power y for which a^(1/y) >= 2.
  >
  > Testing if a^(1/y) can be done efficienty by computing mod some small
  > numbers; thus most non-solutions can be quickly rejected.
  
  I am using the prime number decomposition in Haskell, which is very
  fast. But I get some very large primes. For example,
    pi = 2^-48*5*7*11159*2264103847    from pi =
  884279719003555/281474976710656
    e = 2^-51*3*29*283*1483*167639911  from e =
  6121026514868073/2251799813685248
  It does not take that long time to compute pi.
  
  But looking at your code, recurring in log time, it makes me wonder if
  it is faster.

I am quite sure it is faster even for small numbers, and certainly so
for large numbers.  An added feature of my pseudo code is that its
worst-case performance isn't terrible, while factoring takes wildly
different time for different numbers.

-- 
Torbjörn