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#1
Here's an object I'm going to make?
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I'm conflicted about which thread to put this in, because yes it is an object, but it hadn't had progress made on it yet, hence it is an idea; whatever lets get started I'm making a probe, that shifts to compensate due to magnitude, so it can display any number, but I'm making it in base 10 this time How I will attempt to do this is by creating the numbers like so 8.888888 * 10^88 I'm using these number sizes because 10=2^n; n=3.32, 3.32 bits per decimal digit 24(mantissa)/3.32=7.22891566265 (source (http://stackoverflow.com/questions/7150035/calculating-bits-required-to-store-decimal-number)) But I'm rounding down in mantissa because the last digit has inaccuracies so 7 significands will be the final mantissa range for base 10 and for exponents I'm using 2 digits because 2^127*10^-39 < 1 so 39 powers of 10 enough to overcompensate for 2^127 so 8.888888 * 10^88 has the accuracy to display almost any number to full accuracy (disregarding last digit of inaccuracies) For calculation what I will do is Positive Exponents: X >= 10^32; Y=X*10^-32, N=X X >= 10^16; Y=X*10^-16, N=X X >= 10^8; Y=X*10^-8, N=X X >= 10^4; Y=X*10^-4, N=X X >= 10^2; Y=X*10^-2, N=X X >= 10^1; Y=X*10^-1, N=X I use >= because you cant equal or exceed base value in mantissa By the end if previous X >= 10^1, then result; X < 10^1 Negative Exponents X < 10^-32; Y=X*10^32, N=X X < 10^-16; Y=X*10^16, N=X X < 10^-8; Y=X*10^8, N=X X < 10^-4; Y=X*10^4, N=X X < 10^-2; Y=X*10^2, N=X X < 10^-1; Y=X*10^1, N=X I use < because value can = lower exponent value (1 = 10^-1, but 1 = 10^0 in display range) By the end if previous X < 10^-1, then result; X >= 10^-1 Then the last thing to do is standard probe detection for the mantissa, and it will be finished! Here is an example of how it will work 78,979,430,000 = Display[7.897943 * 10^9] Wish me luck! Giveaway soon, so you can see how the logic works! (logic help from coolman100) | 2014-08-12 14:23:00 Author: amiel445566 Posts: 664 |
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