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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 8,817 of 10,432    |
|    Roman Pearce to jacob navia    |
|    Re: How big is "infinite" precision?    |
|    22 Jun 15 19:35:55    |
      From: rpearcea@gmail.com              On Monday, June 22, 2015 at 11:17:49 AM UTC-7, jacob navia wrote:       > I am building an integer package with unlimited precision for my        > compiler system lcc-win.       >        > I have limited the size of the numbers to 65536 numbers of 31 bits each.       >        > That gives numbers of 2 031 616 bits, that can represent numbers up to       > pow(2, 2031616) --> 2.596 E+157 826, i e. numbers with more than 157        > thousand digits.       >        > Is that too small?       >        > You use these kind of numbers here. Have you ever used numbers that need        > to be bigger than that?       >        > Each number at that precision takes 248Kbytes. 4096 numbers take a GB of        > memory.       >        > I had the number of digits in a 32 bit number that gives really        > "unlimited" precision but I considered that a waste. I am wrong?       >        > Thanks for your input              In computer algebra at least it is not uncommon to have massive integers. For       example, we might multiply dense multivariate polynomials by packing their       coefficients into integers and doing a single fast integer multiplication.        This type of application        may be irrelevant for your compiler. A better question might be - what       algorithms you intend to implement? Because if you are going to implement an       FFT multiplication then you will probably find 2^16 words to be needlessly       restrictive.              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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