4 ms·
A byte is 8 bits long (2^3), so there are 256 distinct values you can store in a byte (2^(2^3) = 2^8 = 256). A kilobyte is 8192 bits long (2^13), so there are
by mbrubeck 4y ago
A byte is 8 bits long (2^3), so there are 256 distinct values you can store in a byte (2^(2^3) = 2^8 = 256).
A kilobyte is 8192 bits long (2^13), so there are 2^8192 distinct values you can store in a kilobyte.
- onlyrealcuzzo 4y agoWith 98% accuracy, a Hyperloglog can count ~4Bn distinct 64-bit integers in ~1.5Kb. Without a data structure, this would take GigaBytes (1M times the space). That's pretty incredible to me. Do the combinatorics of every possible combination of 4B distinct 64-bit integers that that 1.5kb can represent. Mind blowing.
- pca006132 4y agoI think what really fastinates me is how much better we can solve a problem if we allows some approximation/uncertainty.
- onlyrealcuzzo 4y agoIf I did the calculations correctly, that's over 20 Exabytes worth of data that ~1.5kb could represent with ~98% accuracy. 20 Exabytes is probably close to all the information stored in Google...