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> A demonstration that e=2.718 rounded to 3 is the best radix for computation is disproved. Is this some weird attempt at humour, or meant as a serious stateme
by FartyMcFarter 3y ago
> A demonstration that e=2.718 rounded to 3 is the best radix for computation is disproved.
Is this some weird attempt at humour, or meant as a serious statement? I mean the mention of e of course.
- Tomte 3y agoIt‘s a fringe idea put forward by a German cryptocurrency (of course), and loosely based on some old result by Knuth. So no, some people really clamor for all of us to switch to ternary logic.
- FartyMcFarter 3y agoSure, 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.
- woadwarrior01 3y agoNot too fringe of an idea, IMO. The idea of tri-state logic is fairly ubiquitous in analog electronics. Also, was a family of Soviet ternary computers[1] that were fairly competitive with their contemporary binary logic based counterparts. [1]: https://en.wikipedia.org/wiki/Setun https://en.wikipedia.org/wiki/Setun
- Tomte 3y agoOkay, let me more precise: the fringe idea in Iota was to insist on ternary logic implemented in software on binary logic hardware.
- woadwarrior01 3y agoI didn't know what IOTA is and had to look it up. And I agree with your assessment, I suspect their decision was motivated by marketing reasons and not technical reasons.
- int_19h 3y agoSetun is very heavily mythologized as "groundbreaking", but the reality is that it was never a true ternary machine on hardware level. They tried to make it work, but couldn't, so in the end each hardware trit was implemented as a pair of ferromagnetic bits.
- deleted 3y ago[deleted]
- IshKebab 3y agoIt's not a joke. e is theoretically the most efficient base. Of course you can't actually have fractional bases so the most efficient real base (again, theoretically) is 3. This paper is just showing that with real technology it's still less efficient than 2. I don't think anyone was really suggesting that we switch to base 3 hardware so it's a completely academic result, but still interesting.
- scapp 3y ago2 line explanation of where e comes from here: to represent a number x in base b, you need roughly log_b(x) digits. If you weight that by the number of different digits in base b, you get b * log_b(x) = b * log(x)/log(b) = b/log(b) * log(x) (where now the log is any base you care to choose). So assuming x > 1, this is minimized precisely when b/log(b) is minimized. The derivative of b/log(b) is (log(b) - log(e))/log(b)^2, so this is zero when b = e. (The second derivative is 1/(e log(e)) > 0, so this is a minimum).
- remcob 3y agoWhy should we weigh by the number of different digits? I.e. is there an argument why the cost of a single digit is linearly proportional to the number of values it can hold? To me this seems like the weak point in the argument.
- deleted 3y ago[deleted]
- scotty79 3y agoWhy does the number of digits in the base should matter?
- asimpletune 3y agoIt's called radix economy. Different radices have different efficiencies as defined by the ratio of number of digits in a radix alphabet over the number of placeholders required. Simple example, base 1 is obviously inefficient once counting past 1. Similarly base 1 million is inefficient (until you're counting in the bazillions)
- scotty79 3y agoIf you can store one digit of a number base million in one physical element then it's exactly as efficient when it comes to storage of digits as binary. One element per one digit. But it's about 20 times more efficient when it comes to storage of whole numbers because you can store number up to a million in a single physical elements while binary needs 20 elements.
- adrian_b 3y agoNo there has been such a demonstration, which they quote and from which they show the results: S.L. Hurst, “Multiple-Valued Logic - Its Status and Its Future“, IEEE Trans. on Computers, VOL. C-33, No 12, December 1984. According to that demonstration, e would be optimal, but it is impossible because it is not integer. Among integers, 3 is best, followed by 2 and 4, which are equally good. However, as mentioned in the article, the difference between 2 and 3 is very small and that classic demonstration does not take into account the difference in the complexity of the logic gates and registers for binary and ternary logic.