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

*To*: mathgroup at smc.vnet.net*Subject*: [mg68442] 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 03:46:44 -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>*Sender*: owner-wri-mathgroup at wolfram.com

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

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

**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**

**Re: Precision of arguments to FunctionInterpolation**

**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**