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Re: Re: Re: Finding the Number of Pythagorean Triples below a bound
*To*: mathgroup at smc.vnet.net
*Subject*: [mg68415] Re: [mg68382] Re: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Fri, 4 Aug 2006 03:59:29 -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>
*Sender*: owner-wri-mathgroup at wolfram.com
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
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