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I need to do many binary calculations of the form:

10.1^(1/11) base 2
10.001^(1/11) base 2
10.0001^(1/11) base 2
10.00001^(1/11) base 2
10.000001^(1/11) base 2
...

where 1/11 base 2 = 1/3 base 10.

I don't want to use bin2dec and dec2bin for binary. I want to do calculations like in DEC, without worrying about the cumbersome functions. So how can I do the binary calculations in Octave?

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You should label your radices: (10.1 base 2)^(1/3 base 10) (if that's correct) so your question is more clear. – Dennis Williamson Oct 5 '09 at 23:49
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2 Answers

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The long answer is, it's possible, but you'll have to a) program it yourself, or b) get someone to program it for you. This method of writing fractional base-2 just isn't used in computing.

Programming a function to convert this representation of fractional binary numbers into standard IEEE754 double-precision floating point binary isn't impossible, so if it's an absolute must-have, it is doable.

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Yes, but Masi's fundamental problem here is trying to get octave to interpret "10.1" as the binary 2 + a bit, as opposed to the decimal 10.1. The 10.1 absolutely has to be encapsulated as a argument to some functiont, but that is exactly what masi is trying to avoid - those "cumbersome functions" - so now you're right back where you started, a function to convert the number from it's hybrid binary format to a format understandable by octave.. – DaveParillo Oct 9 '09 at 3:59
correct (see my comments under your answer). this answer is clarifying exactly what he must do to accomplish what he wants to accomplish it -- despite being exactly the way he doesn't want to accomplish it. – quack quixote Oct 9 '09 at 4:31
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up vote 1 down vote

The short answer is you can't. If I read your example right and 10.1 is supposed to be a binary floating point number?, then bin2dec is not really appropriate unless your number are all strings. I'm unaware of the concept of 'floating point binary' in octave. I think you'll have to manage the mantissa & exponent manually yourself if you want to do that.

If you're set on all bit arithmetic, have you tried the bit manipulation functions? bitset, etc.?

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now that sounds like a conversion problem: just need to find a way to auto-convert his string to some binary-float format that Octave will understand. – quack quixote Oct 8 '09 at 22:38
~quack: everything is now in Binary, conversion is not my problem. I want to change base in the environment, not needing to worry about conversion all the time. – Masi Oct 8 '09 at 22:57
Masi, computers don't read floating point numbers in that kind of binary. they use IEEE754 or other representations. – quack quixote Oct 8 '09 at 23:04
eg your "10.1 base 2" = 4.5 base 10 = 0x40900000 in IEEE754 single-precision floating point format. (or does it? did i get the conversion to base 10 right?) see babbage.cs.qc.cuny.edu/IEEE-754/Decimal.html – quack quixote Oct 8 '09 at 23:06
oik? regards. xxxxxxx – Xavierjazz Oct 9 '09 at 3:38
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