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Sure, but what does the number e rounded up to 3 have to do with it?
by FartyMcFarter 3y ago
Sure, but what does the number e rounded up to 3 have to do with it?
- beardyw 3y agoThe golden ratio rounded up is 2. QED.
- Retric 3y agoIt’s not quite nonsense: “100 in decimal has three digits, so its radix economy is 10×3 = 30; its binary representation has seven digits (1100100 2) so it has radix economy 2×7 = 14 in base 2; in base 3 its representation has five digits (10201 3) with a radix economy of 3×5 = 15; in base 36 (2S 36) its radix economy is 36×2 = 72.” https://en.wikipedia.org/wiki/Radix_economy https://en.wikipedia.org/wiki/Radix_economy The idea being that there’s a tradeoff between the cost of each digit vs the number of digits you need. However we use base 2 because the cost of base 3 is more than 50% higher than using base 2.
- ymerej 3y agoBest mic drop.
- wk_end 3y agoThe article explains this: > For computation, radix R = 3 would be more economical than R = 2 because the “optimal” radix would be R = e = 2.718, according to a demonstration presented in [1]. Where [1] is "S.L. Hurst, “Multiple-Valued Logic - Its Status and Its Future“, IEEE Trans. on Computers, VOL. C-33, No 12, December 1984." I can't find that one for free anywhere, but there's your lead.
- downvotetruth 3y agoIEEE754 enters chat - synthwave starts banging
- mrob 3y agoSee https://en.wikipedia.org/wiki/Radix_economy https://en.wikipedia.org/wiki/Radix_economy
- hexomancer 3y agoIf you assume the complexity of an elements grows linearly with the number of states it can assume (e.g. if complexity of a bit is 2a then complexity of a trit is 3a). Then the total complexity of for example representing a number N is log{N}_{B} \times B \times a. Minimizing this with respect to B yields B=e.