[Hello Robert -- are you in Bristol yet?]
I'm puzzled now. My comment on inefficiency and Steffen's were based
on the code
> for i from 0 <= i <= n/2:
> if (i*i) % n == self.ivalue:
> return self._new_c(i)
>
but now when I do a.sqrt?? I do not see that code. Steffen, are you
using 2.8.12?
John
On 09/11/2007, Steffen <[EMAIL PROTECTED]> wrote:
>
> Thx, I am wondering why I did not try the command a.sqrt?? on my own.
>
> However, it seems as the implemented algorithm is not the most
> efficient one. My result from a.sqrt?? from the latest release:
>
> def sqrt(self, extend=True, all=False):
> cdef int_fast32_t i, n = self.__modulus.int32
> if n > 100:
> moduli = self._parent.factored_order()
> # Unless the modulus is tiny, test to see if we're in the
> really
> # easy case of n prime, n = 3 mod 4.
> if n > 100 and n % 4 == 3 and len(moduli) == 1 and moduli[0]
> [1] == 1:
> if jacobi_int(self.ivalue, self.__modulus.int32) == 1:
> # it's a non-zero square, sqrt(a) = a^(p+1)/4
> i = mod_pow_int(self.ivalue, (self.__modulus.int32+1)/
> 4, n)
> if i > n/2:
> i = n-i
> if all:
> return [self._new_c(i), self._new_c(n-i)]
> else:
> return self._new_c(i)
> elif self.ivalue == 0:
> return self
> elif not extend:
> raise ValueError, "self must be a square"
> # Now we use a heuristic to guess whether or not it will
> # be faster to just brute-force search for squares in a c
> loop...
> # TODO: more tuning?
> elif n <= 100 or n / (1 << len(moduli)) < 5000:
> if all:
> return [self._new_c(i) for i from 0 <= i < n if (i*i)
> % n == self.ivalue]
> else:
> for i from 0 <= i <= n/2:
> if (i*i) % n == self.ivalue:
> return self._new_c(i)
> if not extend:
> raise ValueError, "self must be a square"
> # Either it failed but extend was True, or the generic
> algorithm is better
> return IntegerMod_abstract.sqrt(self, extend=extend, all=all)
>
> The easier %4 == 3 case seems to be implemented efficiently, but the
> %4 == 1 not. The algo from Tonelli and Shanks might be a good solution
> here. Any thoughts on other/better algorithm?
>
> Steffen
>
>
>
>
> On 9 Nov., 18:24, "John Cremona" <[EMAIL PROTECTED]> wrote:
> > Use the ?? operator to see the algorithm:
> >
> > sage: a=GF(next_prime(10^6)).random_element()^2;
> > sage: a.sqrt??
> >
> > Type: builtin_function_or_method
> > Base Class: <type 'builtin_function_or_method'>
> > String Form: <built-in method sqrt of
> > sage.rings.integer_mod.IntegerMod_int64 object at 0x99055a4>
> > Namespace: Interactive
> > Source:
> > def sqrt(self, extend=True, all=False):
> > r"""
> > Returns square root or square roots of self modulo n.
> >
> > INPUT:
> > extend -- bool (default: True); if True, return a square
> > root in an extension ring, if necessary. Otherwise,
> > raise a ValueError if the square is not in the base ring.
> > all -- bool (default: False); if True, return *all* square
> > roots of self, instead of just one.
> >
> > ALGORITHM: Calculates the square roots mod $p$ for each of the
> > primes $p$ dividing the order of the ring, then lifts them
> > p-adically and uses the CRT to find a square root mod $n$.
> >
> > See also \code{square_root_mod_prime_power} and
> > \code{square_root_mod_prime} (in this module) for more
> > algorithmic details.
> >
> > EXAMPLES:
> > sage: mod(-1, 17).sqrt()
> > 4
> > sage: mod(5, 389).sqrt()
> > 86
> > sage: mod(7, 18).sqrt()
> > 5
> > sage: a = mod(14, 5^60).sqrt()
> > sage: a*a
> > 14
> > sage: mod(15, 389).sqrt(extend=False)
> > Traceback (most recent call last):
> > ...
> > ValueError: self must be a square
> > sage: Mod(1/9, next_prime(2^40)).sqrt()^(-2)
> > 9
> > sage: Mod(1/25, next_prime(2^90)).sqrt()^(-2)
> > 25
> >
> > sage: a = Mod(3,5); a
> > 3
> > sage: x = Mod(-1, 360)
> > sage: x.sqrt(extend=False)
> > Traceback (most recent call last):
> > ...
> > ValueError: self must be a square
> > sage: y = x.sqrt(); y
> > sqrt359
> > sage: y.parent()
> > Univariate Quotient Polynomial Ring in sqrt359 over Ring
> > of integers modulo 360 with modulus x^2 + 1
> > sage: y^2
> > 359
> >
> > We compute all square roots in several cases:
> > sage: R = Integers(5*2^3*3^2); R
> > Ring of integers modulo 360
> > sage: R(40).sqrt(all=True)
> > [20, 160, 200, 340]
> > sage: [x for x in R if x^2 == 40] # Brute force verification
> > [20, 160, 200, 340]
> > sage: R(1).sqrt(all=True)
> > [1, 19, 71, 89, 91, 109, 161, 179, 181, 199, 251, 269,
> > 271, 289, 341, 359]
> > sage: R(0).sqrt(all=True)
> > [0, 60, 120, 180, 240, 300]
> >
> > sage: R = Integers(5*13^3*37); R
> > Ring of integers modulo 406445
> > sage: v = R(-1).sqrt(all=True); v
> > [78853, 111808, 160142, 193097, 213348, 246303, 294637, 327592]
> > sage: [x^2 for x in v]
> > [406444, 406444, 406444, 406444, 406444, 406444, 406444, 406444]
> > sage: v = R(169).sqrt(all=True); min(v), -max(v), len(v)
> > (13, 13, 104)
> > sage: all([x^2==169 for x in v])
> > True
> >
> > Modulo a power of 2:
> > sage: R = Integers(2^7); R
> > Ring of integers modulo 128
> > sage: a = R(17)
> > sage: a.sqrt()
> > 23
> > sage: a.sqrt(all=True)
> > [23, 41, 87, 105]
> > sage: [x for x in R if x^2==17]
> > [23, 41, 87, 105]
> >
> > """
> > if self.is_one():
> > if all:
> > return list(self.parent().square_roots_of_one())
> > else:
> > return self
> >
> > if not self.is_square_c():
> > if extend:
> >
> > and so on!
> >
> > John
> >
> > On 09/11/2007, Steffen <[EMAIL PROTECTED]> wrote:
> >
> >
> >
> >
> >
> > > Hi, I need to find square roots in GF(prime). I did it like this:
> >
> > > y = sqrt(GF(prime)(ySquare))
> >
> > > So far I am not quite happy with the calculation periods and I would
> > > like to know which algorithm is used. In my case is prime % 4 == 1,
> > > which is the hardest case to find the square root mod p. I am sage
> > > newbie and probably its again only a command that I cant find. I tried
> > > sqrt? and similar things, but only got the information that sqrt is a
> > > symbolic function. Could somebody tell me which algo is implemented or
> > > better how to find the implemented algo.
> >
> > > Cheers, Steffen
> >
> > --
> > John Cremona
>
>
> >
>
--
John Cremona
--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to sage-devel@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/sage-devel
URLs: http://sage.scipy.org/sage/ and http://modular.math.washington.edu/sage/
-~----------~----~----~----~------~----~------~--~---
