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 >>>>>>> >>>>>>> >>>>>> >>>>> >>>> >>> >> >

**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: Re: Re: Finding the Number of Pythagorean Triples below a bound***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: Re: Finding the Number of Pythagorean Triples below a bound**

**Re: Re: Re: Re: Finding the Number of Pythagorean Triples below a bound**

**Re: Re: Re: Finding the Number of Pythagorean Triples below a bound**

**Re: Re: Re: Re: Finding the Number of Pythagorean Triples below a bound**