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Java Forum / General / February 2008do u know ramanujan numbers algorithm
I think about ramanujan numbers and I need to know is there any algorithm for solving ramanujan numbers, please help me >I think about ramanujan numbers and I need to know is there any >algorithm for solving ramanujan numbers, please help me Assuming this is 1729 and friends, then there are a lot of links at: http://mathworld.wolfram.com/TaxicabNumber.html rossum On Feb 27, 7:45 pm, "emre esirik(hacettepe com. sci. and eng.)" <emreesi...@gmail.com> wrote: > I think about ramanujan numbers and I need to know is there any > algorithm for solving ramanujan numbers, please help me Make 4 for loops with int a,b,c,d Check a^2+v^2=c^2+d^2 Whichever number follows print them you will find many such numbers. > On Feb 27, 7:45 pm, "emre esirik(hacettepe com. sci. and eng.)" > <emreesi...@gmail.com> wrote: [quoted text clipped - 6 lines] > > Whichever number follows print them you will find many such numbers. Shouldn't that calculation involve cubes rather than squares? Doesn't the definition of "taxicab" numbers require that the number equal the sums of two /different/ pairs of cubes? Assuming you only want factors between 1 and 100, the suggested algorithm requires 10^8 iterations. At, say, 100 clock cycles per iteration, that's about 10^10 clocks, or 50 seconds on a 2GHz CPU. For a maximum root of 1000 it would take 10^12 iterations or 500,000 seconds, almost six days. It would grow as to the fourth power of the maximum root, a rather big big O, wouldn't you say? To the OP: What exactly do you mean by "solving ramanujan [sic] numbers"? Do you mean finding them? Factoring them? What?  SignatureLew > > On Feb 27, 7:45 pm, "emre esirik(hacettepe com. sci. and eng.)" > > <emreesi...@gmail.com> wrote: [quoted text clipped - 25 lines] > -- > Lew Yes Ramanujm numbers are those numbers which follow this rul a^3+b^3 =c^3+d^3=a Number Then the Number is Called Ramanujm Number. Just follow 4 for loops And test a^3+b^3 =c^3+d^3=a Number And print those number. "emre esirik(hacettepe com. sci. and eng.)" wrote: >>>> I think about ramanujan numbers and I need to know is there any >>>> algorithm for solving ramanujan numbers, please help me >>> Make 4 for loops with int a,b,c,d >>> Check a^2+v^2=c^2+d^2 >>> Whichever number follows print them you will find many such numbers. Lew wrote: >> Shouldn't that calculation involve cubes rather than squares? > Yes Ramanujm numbers are those numbers which follow this rul > > a^3+b^3 =c^3+d^3=a Number > > Then the Number is Called Ramanujm Number. Lew wrote: >> Doesn't the definition of "taxicab" numbers require that the number equal the >> sums of two /different/ pairs of cubes? Shouldn't that be addressed in the algorithm also? > Just follow 4 for loops And test a^3+b^3 =c^3+d^3=a Number And print > those number. What about the O(n^4) algorithmic inefficiency of that approach? Isn't there a better way? Lew wrote: >> Assuming you only want factors between 1 and 100, the suggested algorithm >> requires 10^8 iterations. At, say, 100 clock cycles per iteration, that's [quoted text clipped - 3 lines] >> It would grow as to the fourth power of the maximum root, a rather big big O, >> wouldn't you say? Until the issues of duplicate results and algorithmic inefficiency are addressed, I'd advise the OP to be skeptical of this approach. This all still leaves open the question of what the OP wants: >> To the OP: What exactly do you mean by "solving ramanujan [sic] numbers"? Do >> you mean finding them? Factoring them? What? SignatureLew > "emre esirik(hacettepe com. sci. and eng.)" wrote: > [quoted text clipped - 43 lines] > > - Show quoted text - I have O(n^2) algorithm in mind i will tell it if he really need it. I will take $200 for this solution. But only if He is really in need of this algorithm. Bye Sanny > [...] > I have O(n^2) algorithm in mind i will tell it if he really need it. I > will take $200 for this solution. > > But only if He is really in need of this algorithm. He must be: He's multi-posted his question to half of Usenet. (For suitable values of "half.") SignatureEric.Sosman@sun.com Lew asked: >> What about the O(n4) algorithmic inefficiency of that approach? Isn't there >> a better way? > I have O(n^2) algorithm in mind i will tell it if he really need it. I > will take $200 for this solution. Oh, my! So is it fair to say that providing a bad algorithm was merely a way of marketing the for-pay solution? "Here's a bad answer, and if you pay me I'll give a good answer"? I would advise anyone who thinks of taking a poster up on such an offer to review said poster's other posts to see if it is credible that their offer is worth the suggested cost, speaking, of course, in the most general terms and not about any one particular such (at best) dubious offer. Seriously, Sanny, oh, my! SignatureLew I will have to invest my time in generating the O(n^2) algorithm. For O(n^4) algorithm I gave it by withing in 10 seconds for free. Its true I can give little work for free but to design something that needs time I cant do that for free. If you are ill you go to docter he may give a free advice. But incase he need to do an Xray/ Or other diagnostic Tests then the cost increases. Simmilarly I can devote max 2-3 min for a O(n^2) Solution but if I have to invest 2-3 hours then I atleast ask for $200. I generally take $50-$100 / hr for Coding work. So $200 is reasionable. And I think the Guy is rich enough to pay me that much. Bye Sanny > I will have to invest my time in generating the O(n^2) algorithm. For > O(n^4) algorithm I gave it by withing in 10 seconds for free. [quoted text clipped - 10 lines] > $50-$100 / hr for Coding work. So $200 is reasionable. And I think the > Guy is rich enough to pay me that much. I will bear your philosophy in mind when answering any of your future questions. BugBear > > I will have to invest my time in generating the O(n^2) algorithm. For > > O(n^4) algorithm I gave it by withing in 10 seconds for free. [quoted text clipped - 15 lines] > > BugBear I only ask for advices which you can give in 10-20 seconds I will never ask something that will need you 2-3 hours. Naturally You can see I have helped many people in comp.lang.java.programmer But only when I can help in 1-2 min. I will never waster 2-3 hours for a question and never want any one to devote their precious hours solving my problems. Bye Sanny On Wed, 27 Feb 2008 09:34:53 -0800 (PST), Sanny <softtanks@hotmail.com> wrote, quoted or indirectly quoted someone who said : >Just follow 4 for loops And test a^3+b^3 =c^3+d^3=a Number And print >those number. you could improve slightly on that brute force approach this way: 1. you don't need to find all the permutations, just the ones where a < b, a < c, and c < d (perhaps <=, I am not familiar with the exact rules). That means your loops don't need to start at but at a or c. 2. you can compute an approximation for d as Math.pow( a * a * a + b * b * b - c * c * c , 1./3.); round to int, throw away anything not very close to int, and check the int. 3. Use Jet. It will be clever about optimising the * inside loops. see http://mindprod.com/jgloss/jet.html -- Roedy Green Canadian Mind Products The Java Glossary http://mindprod.com >>Just follow 4 for loops And test a^3+b^3 =c^3+d^3=a Number And print >>those number. > you could improve slightly on that brute force approach this way: > 1. you don't need to find all the permutations, just the ones where > a < b, a < c, and c < d (perhaps <=, I am not familiar with the exact > rules). You could also iterate only over two variables and save the values of a^3+b^3 as keys in a map. Before actually adding it to the map, you'd check for it's previous existence in the map: if it was already there, you've found a solution. Of course, when adding one, you need to store the value a as value for key a^3+b^3, such that when you re-encounter the key, you also know the previous a and (with simple math) b. There's further potential for improvement by casually cleaning up keys in the map that are small enough that re-encountering them can be proven impossible. e.g. because it is smaller than current a's cube. Perhaps, it's even better to iterate the sum a+b in the other loop and the difference (only till sign-change) in the inner loop, so you do not have to decide on any maximum beforehand. Has anyone paid the guy for a O(n^2) solution, yet? ;-) On Feb 29, 2:32 pm, Andreas Leitgeb <a...@gamma.logic.tuwien.ac.at> wrote: > >>Just follow 4 for loops And test a^3+b^3 =c^3+d^3=a Number And print > >>those number. [quoted text clipped - 22 lines] > > Has anyone paid the guy for a O(n^2) solution, yet? ;-) Pay and get the Solution The guy has not responded it means he just ask for curiosity. Any one else interested pay me $200 for the solution. Bye Sanny Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. 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