I see. In my example a was
sage: type(a)
<type 'sage.rings.integer_mod.IntegerMod_int64'>
John
On 09/11/2007, Robert Bradshaw <[EMAIL PROTECTED]> wrote:
>
> On Nov 9, 2007, at 1:43 PM, John Cremona wrote:
>
> > [Hello Robert -- are you in Bristol yet?]
>
> Just got in.
>
> > 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?
>
> This depends on the size of the modulus. There are three types--
> IntegerMod_int32, IntegerMod_int64, and IntegerMod_gmp. The 32-bit
> one has that code, and only for small p. If your modulus is beg
> enough, a.sqrt?? won't give you that at all.
>
> >
> > 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
> >
> >
>
> >
>
--
John Cremona
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