Re: Re: Re: Finding the Number of Pythagorean Triples below a bound
- To: mathgroup at smc.vnet.net
- Subject: [mg68425] Re: [mg68382] Re: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sat, 5 Aug 2006 03:46:01 -0400 (EDT)
- References: <eaeqa3$53v$1@smc.vnet.net><200607300848.EAA25171@smc.vnet.net> <eakfgm$rl6$1@smc.vnet.net> <200608020923.FAA28520@smc.vnet.net> <79C36C70-E091-4A82-8EC5-0EDC743D081D@mimuw.edu.pl> <200608031007.GAA15743@smc.vnet.net> <76C59C34-5A55-4318-B0AF-3A572E71B421@mimuw.edu.pl>
- Sender: owner-wri-mathgroup at wolfram.com
On 3 Aug 2006, at 16:42, Andrzej Kozlowski wrote:
>
> On 3 Aug 2006, at 12:07, Andrzej Kozlowski wrote:
>
>> On 2 Aug 2006, at 20:01, Andrzej Kozlowski wrote:
>>
>>>
>>> On 2 Aug 2006, at 11:23, titus_piezas at yahoo.com wrote:
>>>
>>>> Hello all,
>>>>
>>>> My thanks to Peter and Andrzej, as well as those who privately
>>>> emailed
>>>> me.
>>>>
>>>> To recall, the problem was counting the number of solutions to a
>>>> bivariate polynomial equal to a square,
>>>>
>>>> Poly(a,b) = c^2
>>>>
>>>> One form that interested me was the Pythagorean-like equation:
>>>>
>>>> a^2 + b^2 = c^2 + k
>>>>
>>>> for {a,b} a positive integer, 0<a<=b, and k any small integer. I
>>>> was
>>>> wondering about the density of solutions to this since I knew in
>>>> the
>>>> special case of k=0, let S(N) be the number of primitive solutions
>>>> with
>>>> c < N, then S(N)/N = 1/(2pi) as N -> inf.
>>>>
>>>> For k a squarefree integer, it is convenient that any solution is
>>>> also
>>>> primitive. I used a simple code that allowed me to find S(10^m)
>>>> with
>>>> m=1,2,3 for small values of k (for m=4 took my code more than 30
>>>> mins
>>>> so I aborted it). The data is given below:
>>>>
>>>> Note: Values are total S(N) for *both* k & -k:
>>>>
>>>> k = 2
>>>> S(N) = 4, 30, 283
>>>>
>>>> k = 3
>>>> S(N) = 3, 41, 410
>>>>
>>>> k = 5
>>>> S(N) = 3, 43, 426
>>>>
>>>> k = 6
>>>> S(N) = 3, 36, 351
>>>>
>>>> Question: Does S(N)/N for these also converge? For example, for the
>>>> particular case of k = -6, we have
>>>>
>>>> S(N) = 2, 20, 202
>>>>
>>>> which looks suspiciously like the ratio might be converging.
>>>>
>>>> Anybody know of a code for this that can find m=4,5,6 in a
>>>> reasonable
>>>> amount of time?
>>>>
>>>>
>>>> Yours,
>>>>
>>>> Titus
>>>>
>>>
>>>
>>> Here is a piece code which utilises the ideas I have described in
>>> my previous posts:
>>>
>>> ls = Prime /@ Range[3, 10];
>>>
>>> test[n_] :=
>>> Not[MemberQ[JacobiSymbol[n, ls], -1]] && Element[Sqrt[n],
>>> Integers]
>>>
>>> f[P_, k_] := Sum[If[(w =
>>> a^2 + b^2 - k) < P^2 && test[w], 1, 0], {a, 1, P}, {b, a,
>>> Floor[Sqrt[P^2 - a^2]]}]
>>>
>>> g[m_, k_] := f[10^m, k] + f[10^m, -k]
>>>
>>> We can easily confirm the results of your computations,.e.g. for
>>> k=2.
>>>
>>>
>>> Table[g[i,2],{i,3}]
>>>
>>>
>>> {4,30,283}
>>>
>>>
>>> Since you have not revealed the "simple code" you have used it is
>>> hard to tell if the above one is any better. It is however,
>>> certainly capable of solving the problem for m=4:
>>>
>>>
>>> g[4,2]//Timing
>>>
>>>
>>> {4779.39 Second,2763}
>>>
>>> The time it took on my 1 Gigahertz PowerBook was over 70 minutes,
>>> which is longer than you thought "reasonable", so I am still not
>>> sure if this is any improvement on what you already have. The time
>>> complexity of this algorithm seems somewhat larger than exponential
>>> so I would expect that it will take about 6 hours to deal with n=5
>>> on my computer, and perhaps 2 weeks to deal with n=6.
>>>
>>> Andrzej Kozlowski
>>>
>>>
>>>
>>
>>
>> I mistakenly copied and pasted a wrong (earlier) definition of f.
>> Here is the correct one:
>>
>> f[P_, k_] := Block[{w}, Sum[If[GCD[a, b, k] ==
>> 1 && (w = a^2 +
>> b^2 - k) < P^2 && test[w], 1, 0], {a, 1,
>> P}, {b, a, Floor[Sqrt[P^2 - a^2]]}]]
>>
>> The definition of g is as before:
>>
>> g[m_, k_] := f[10^m, k] + f[10^m, -k]
>>
>> Andrzej Kozlowski
>>
>
>
> Below is a faster version of the above code. (It owes a significant
> improvement to Daniel Lichtblau, which I borrowed from him without
> his knowledge ;-))
>
> test[n_] :=
> JacobiSymbol[n,
> Prime[Random[Integer, {2, 20}]]] -1 && Element[Sqrt[n],
> Integers]
>
>
> f[P_, k_] := Block[{w}, Sum[If[GCD[a, b, k] ==
> 1 && (w = a^2 +
> b^2 - k) < P^2 && test[w], 1, 0], {a, 1,
> Floor[Sqrt[(P^2+k)/2]]}, {b, a, Floor[Sqrt[P^2 - a^2+k]]}]]
>
> g[m_, k_] := f[10^m, k] + f[10^m, -k]
>
> The improvement is in the upper bound on a in the sum. Since a is
> the smaller of the two squares whose sum is equal to P^2+k it can't
> be larger than Floor[Sqrt[(P^2+k)/2]].
>
> Note that you can improve the performance by loosing some accuracy
> if you use a cruder test for a perfect square:
>
> test1[n_] := With[{w = Sqrt[N[n]]}, w == Round[w]
> ]
>
> f1[P_, k_] := Block[{w}, Sum[If[GCD[a, b, k] ==
> 1 && (w = a^2 +
> b^2 - k) < P^2 && test1[w], 1, 0], {a, 1,
> Floor[Sqrt[(P^2+k)/2]]}, {b, a, Floor[Sqrt[P^2 - a^2+k]]}]]
>
>
> g1[m_, k_] := f1[10^m, k] + f1[10^m, -k]
>
>
> Let's compare the two cases.
>
> In[7]:=
> g[3,2]//Timing
>
> Out[7]=
> {89.554 Second,283}
>
> In[8]:=
> g1[3,2]//Timing
>
> Out[8]=
> {37.376 Second,283}
>
> So we see that we get the same answer and the second approach is
> considerably faster. However:
>
> In[9]:=
> g[1,6]//Timing
>
> Out[9]=
> {0.008863 Second,3}
>
> In[10]:=
> g1[1,6]//Timing
>
> Out[10]=
> {0.005429 Second,5}
>
>
> The correct answer is 3 (returned by the first method). The faster
> method found two false solutions. This should not matter if you are
> interested only in approximate answers (as you seem to be) but it
> is worth keeping in mind.
>
> Andrzej Kozlowski
I have noticed one more obvious improvement. We can replace test by:
test[n_] :=
Mod[n, 4] =!= 3 && JacobiSymbol[n,
Prime[Random[Integer, {2, 100}]]] -1 && Element[Sqrt[n],
Integers]
and test1 by
test1[n_] := With[{w = Sqrt[N[n]]}, Mod[n, 4] =!= 3 && w == Round[w]
]
We are simply making use of the easily to prove fact that an integer
of the form 4 k + 1 can never be the sum of two squares. There is a
noticeable improvement in the performance of g:
g[3,2]//Timing
{58.0786 Second,283}
However, the performance of g1 seems to actually decline slightly:
g1[3,2]//Timing
{40.8776 Second,283}
However, there are fewer cases of "false solutions":
In[22]:=
g1[1,6]//Timing
Out[22]=
{0.006694 Second,4}
I am still not sure what is the most efficient use of JacobiSymbol in
this kind of problems. In the first version of my code I used a test
involving the first few odd primes, afterwards I switched to using
just one random prime in taken form a certain range. Evaluation of
JacobiSympol[m,p] != -1 is certainly much faster than that of Element
[Sqrt[m],Integers], but it is not clear what is the most efficient
number of primes to use, which are the best primes and whether it is
better to choose them at random or just use a fixed selection.
The numerical test Sqrt[N[n]]==Round[Sqrt[N[n]] is even faster, but
it will sometimes produce "false squares".
Andrzej Kozlowski
- References:
- Re: Finding the Number of Pythagorean Triples below a bound
- From: titus_piezas@yahoo.com
- Re: Re: Finding the Number of Pythagorean Triples below a bound
- From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
- Re: Finding the Number of Pythagorean Triples below a bound