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Re: Re: Re: Re: Finding the Number of Pythagorean Triples below a bound
*To*: mathgroup at smc.vnet.net
*Subject*: [mg68470] Re: [mg68442] Re: [mg68382] Re: [mg68345] Re: Finding the Number of Pythagorean Triples below a bound
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Sun, 6 Aug 2006 02:57:06 -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> <C56DFF0B-6F0B-4E64-B72A-D805B3C6C063@mimuw.edu.pl> <7AA8804E-73F2-4D7B-BCAB-633CE33CD1D0@mimuw.edu.pl> <F7520FE0-16DB-4F8F-A39E-DF64749B85F0@mimuw.edu.pl> <B45555BF-3093-4487-A3EA-0981783CA75A@mimuw.edu.pl> <B935EC43-291F-4C01-B646-534789B6E908@mimuw.edu.pl> <F7F7DB69-7894-466A-99BC-065835567D5C@mimuw.edu.pl> <200608050746.DAA14183@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
I have succeeded in computing the number of solutions of a^2+b^2==c^2
+2 where c<10^5. This was done by means of the follwoign code:
test1 = Compile[{{m, _Integer}, {k, _Integer}}, Module[
{i = 0, c2, diff, sdiff},
Do [
If[Mod[c^2 + k, 4] != 3, c2 = Floor[Sqrt[(c^2 + k)/2]];
Do [
diff = c^2 + k - b^2;
sdiff = Sqrt[N[diff]];
If [sdiff >= b && sdiff == Round[sdiff], i++];
, {b, 1., c2}]];
, {c, 10.^(m - 1), 10.^m - 1}];
i]]
I got
test1[5,2]//Timing
{5136.24 Second,8929}
To this one should add the outcome of
countTriplesP[4,2]
983
where countTriplesP is defined in my earlier post. So the actual
number of solutions is
8929+983
9912
This means that compiled Mathematica code can quite successfully deal
with the case m=5 of Titus's problem. This computation took 1 hour
and 43 minutes on a 1 Gigahertz PowerBook G4, to which we have to add
about 1.5 minute for solving the 10^4 case. That gives the total time
of 1.45 minutes, which seems to me pretty impressive, and shows, I
think, that Mathematica is quite usable for hard numerical
computations, provided one one is lucky enough to have ones code
compile properly. I really hope some work at WRI is going into making
Compile work better (in more cases and more predictably).
Andrzej Kozlowski
On 5 Aug 2006, at 09:46, Andrzej Kozlowski wrote:
> Daniel Lichtblau has already informed me that running the code below
> on a 64 bit computer will not help at all, so there is no point
> trying :-(
> However, he also suggested a way around the Compile problem with
> integers larger than machine integers :-)
> I am trying it out right now.
>
> Andrzej Kozlowski
>
>
> On 4 Aug 2006, at 21:09, Andrzej Kozlowski wrote:
>
>> I need to add some corrections. There were some some small mistakes
>> in the code I posted earlier, which caused problems with running
>> the compiled code for negative values of k, and which also probably
>> accounted for the slightly incorrect answers which the
>> "approximate" code returned. I attributed the inaccuracy to the use
>> of numerical precision in the test for a number being a perfect
>> square but it seems that the cause was (probably) elsewhere. I
>> don't want to try to make this message too long so I won't bother
>> explaining what I think the mistakes were; but I will give what I
>> think is the correct code. I have decided to separate the code for
>> negative and positive values of k. The code for negative k works
>> also for positive k's but is slightly slower, due to the extra test
>> that needs to be performed. For this reason, and for the sake of
>> greater clarity I have decided to separate the two codes.
>>
>> The code for positive k:
>>
>> countTriplesP = Compile[{{m, _Integer}, {k, _Integer}}, Module[
>> {i = 0, c2, diff, sdiff},
>> Do [
>> If[Mod[c^2 + k, 4] != 3, c2 = Floor[Sqrt[(c^2 + k)/2]];
>> Do [
>> diff = c^2 + k - b^2;
>> sdiff = Sqrt[N[diff]];
>> If [sdiff >= b && sdiff == Round[sdiff], i++];
>> , {b, 1, c2}]];
>> , {c, 1, 10^m - 1}];
>> i]]
>>
>> The code for negative k:
>>
>> countTriplesN = Compile[{{m, _Integer}, {k, _Integer}}, Module[
>> {i = 1, c2, diff, sdiff},
>> Do [
>> If[Mod[c^2 + k, 4] != 3 && c^2 + k ³ 0, c2 = Floor[Sqrt[(c^2 +
>> k)/2]];
>> Do [
>> diff = c^2 + k - b^2;
>> sdiff = Sqrt[N[diff]];
>> If [sdiff >= b && sdiff == Round[sdiff], i++];
>> , {b, 1, c2}]];
>> , {c, 1, 10^m - 1}];
>> i]]
>>
>> Now we get:
>>
>> countTriplesP[1,6]+countTriplesN[1,-6]//Timing
>>
>>
>> {0.000221 Second,3}
>>
>>
>> countTriplesP[3,2]+countTriplesN[3,-2]//Timing
>>
>>
>> {0.95186 Second,282}
>>
>>
>> countTriplesP[4,2]+countTriplesN[4,-2]//Timing
>>
>>
>> {95.2177 Second,2762}
>>
>>
>>
>>
>> Note that these values are consistently less by one form the values
>> obtained by Titus and also by me using my earlier "exact" code, but
>> actually I believe that this disparity was due to mistakes in the
>> earlier code. In any case, if we replace the "numerical" test for a
>> perfect square by the much slower "exact" test, the answers will be
>> the same, so the difference of 1 is certainly not due to the use of
>> numerical precision. Anyway, everything works fast and looks
>> perfectly satisfactory but then there is a snag. I decided to run
>> the code for m=5 and k=2, started it and went out for several
>> hours. When I came back I was rather disappointed to see that it
>> was still running and then I saw the message:
>>
>> countTriplesP[5,2]//Timing
>>
>> CompiledFunction::"cfn" Numerical error encountered at instruction
>> 10; proceeding with uncompiled evaluation.
>>
>> I assume the cause of this is that on 32 bit computers Compile
>> cannot deal with integers larger than 2^32 but we have:
>>
>>
>> c=10^5;
>>
>> Log[2,c^2]//N
>>
>> 33.2193
>>
>> I can't at the moment see any way around this problem except to run
>> uncompiled code (far too long) or try a 64 bit computer.
>> Unfortunately I do not have one and I don't think Titus has one, so
>> if any body has a 64-bit computer with a 64 bit version of
>> Mathematica installed, and a little spare time, I think both of us
>> would like to know how the above code performs in this setting. If
>> it works it will provide a nice bit of advertising for 64 bit
>> computing.
>>
>> Andrzej Kozlowski
>>
>>
>>
>>
>> On 4 Aug 2006, at 12:48, Andrzej Kozlowski wrote:
>>
>>> My latest code has just managed:
>>>
>>> countT[4,2]+countT[4,-2]//Timing
>>>
>>>
>>> {93.9638 Second,2762}
>>>
>>>
>>> That is less than two minutes on a 1 gigahertz computer. The
>>> correct answer is actually 2763 (by my earlier computation using
>>> the exact test) so we have lost one solution but gained more than
>>> 50 fold improvement in performance!
>>> The case m=5 is now certainly feasible, although I am not sure if
>>> I wish my not very powerful PowerBook to be occupied for so long,
>>> as I need to use Mathematica for other tasks. Perhaps I can now
>>> leave this to others.
>>>
>>> Andrzej Kozlowski
>>>
>>> On 4 Aug 2006, at 12:38, Andrzej Kozlowski wrote:
>>>
>>>> I have good news: the code I just posted can be compiled and then
>>>> it becomes really fast ;-)
>>>>
>>>> countT = Compile[{{m, _Integer}, {k, _Integer}}, Module[
>>>> {i = 0, c2, diff, sdiff},
>>>> Do [
>>>> If[Mod[c^2 + k, 4] != 3, c2 = Floor[Sqrt[(c^2 + k)/2]];
>>>> Do [
>>>> diff = c^2 + k - b^2;
>>>> sdiff = Sqrt[N[diff]];
>>>> If [sdiff >= b && sdiff == Round[sdiff], i++];
>>>> , {b, 1, c2}]];
>>>> , {c, 1, 10^m - 1}];
>>>> i]]
>>>>
>>>>
>>>> countT[3,2]+countT[3,-2]//Timing
>>>>
>>>>
>>>> {1.0032 Second,282}
>>>>
>>>> I will try the case m=4 and m=5 and send you the results. I am
>>>> not promising to do this very soon, just in case ;-)
>>>>
>>>> Andrzej Kozlowski
>>>>
>>>>
>>>>
>>>> On 4 Aug 2006, at 12:09, Andrzej Kozlowski wrote:
>>>>
>>>>> Here is the fastest code I have seen so far that works (it is
>>>>> based on the one that Daniel sent you). I have corrected and
>>>>> enhanced and enhanced it. It gives approximate answers for
>>>>> reasons that I explained in earlier postings (use of machine
>>>>> arithmetic to test for perfect integers). Of course it is easy
>>>>> to replace the code by exact code by replacing the numerical
>>>>> test for a perfect square by an exact one.
>>>>>
>>>>>
>>>>> countTriples[m_,k_] := Module[
>>>>> {i=0, c2, diff, sdiff},
>>>>> Do [
>>>>> If[Mod[c^2+k,4]!=3, c2 = Floor[Sqrt[(c^2+k)/2]];
>>>>> Do [
>>>>> diff = c^2+k-b^2;
>>>>> sdiff = Sqrt[N[diff]];
>>>>> If [sdiff>=b&&sdiff==Round[sdiff],i++];
>>>>> , {b,1,c2}]];
>>>>> ,{c,1,10^m-1}];
>>>>> i]
>>>>>
>>>>>
>>>>> countTriples[3,2]+countTriples[3,-2]//Timing
>>>>>
>>>>>
>>>>> {12.3746 Second,282}
>>>>>
>>>>> The correct answer is 283.
>>>>>
>>>>> This code should easily deal with the case m=4 (I have not yet
>>>>> tried it) and I think even m=5 should now be within reach.
>>>>>
>>>>> Andrzej Kozlowski
>>>>>
>>>>>
>>>>>
>>>>> On 4 Aug 2006, at 11:27, Andrzej Kozlowski wrote:
>>>>>
>>>>>> The "improvement" below which I sent a little earlier is wrong
>>>>>> (even though it returned correct answers). Obviously the point
>>>>>> is that c^2+k can't be of the form 4n + 3, but there is no
>>>>>> reason why a^2+b^2-k can't be of that form. Since my code does
>>>>>> not explicitly select c it can't make use of this additional
>>>>>> improvement. A different code, which uses explicit choices of
>>>>>> (say) a and c and tests for b being a perfect square could
>>>>>> exploit this fact and perhaps gain extra speed. It should not
>>>>>> be difficult to write such a code along the lines I have been
>>>>>> using.
>>>>>>
>>>>>> Andrzej
>>>>>>
>>>>>>
>>>>>> On 4 Aug 2006, at 10:47, Andrzej Kozlowski wrote:
>>>>>>
>>>>>>>
>>>>>>> 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
>>>>>>>
>>>>>>>
>>>>>>
>>>>>
>>>>
>>>
>>
>
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