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by lblume
481 days ago
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> Any value representable in n bits is representable in n+1 bits > [1 1 1 1 1 1 1] 2.004e+19728 Does that mean that the 8 bits version has numbers larger than this? Doesn't seem very useful for 10^100 is already infinity for all practical purposes. |
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Yes. Not just a bit larger but an even more ridiculous leap. Notice that you're iterating exponents. That last one (7 bits) was 2^65536 so the next one (8 bits) will be 2^2^65536.
Python objects to 2^65536 complaining that the base 10 string to represent the integer contains more than 4300 digits so it gave up.
> Doesn't seem very useful
By that logic we should just use fixed point because who needs to work with numbers as large as 2^1023 (64 bit IEEE 754). These things aren't useful unless you're doing something that needs them, in which case they are. I could see the 5 or 6 bit variant of such an iterated scheme potentially being useful as an intermediary representation for certain machine learning applications.