4 ms·
I don't think that's right. Assuming we had 5 characters to encode data, with base2 we get 2^5 = 32 possible combinations with base4 we get 4^5 = 1024 possibl
by slayed0 12y ago
I don't think that's right. Assuming we had 5 characters to encode data,
with base2 we get 2^5 = 32 possible combinations
with base4 we get 4^5 = 1024 possible combinations
- epistasis 12y agoBut storage length is the log of the number of possible combinations, so you're back to just double the amount of storage.
- slayed0 12y agoBoth functions (2^x and 4^x) have exponential growth, but their exponential growth is not linearly related. 2*(2^x) != (4^x)
- epistasis 12y agoWe are talking about data, which is the log of the number of combinations, like any measure of information. If this interests you, definitely look into basic information theory, and then move onto coding theory. Take, for example, 32bit integers vs 64bit integers (unsigned for simplicity). Two 32 bit integers can represent exactly the same number of combinations that a single 64 bit number can. Sure, there's an exponential number more combinations in a 64bit integer than a 32 bit, but the number of combinations is not how the storage capacity is measured.