> 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.
Is this some weird attempt at humour, or meant as a serious statement? I mean the mention of e of course.
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).
To me this seems like the weak point in the argument.
Simple example, base 1 is obviously inefficient once counting past 1. Similarly base 1 million is inefficient (until you're counting in the bazillions)
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.
In the example you gave, I'm assuming by 'physical element' you mean 'placeholder' or digit. Storing more numbers in a single placeholder seems like you're just getting efficiency for free, but that's not how information works. You have to come up with a unique symbol for one million numbers (0 - 999,999). Which you have to pay for.
base 64 is a more realistic example. With 64 characters per digit, it may seem like it's more efficient, since you require less digits to express the same number as base 10 or 2, but you still have to encode 64 unique characters, and it ends up being less efficient. That doesn't mean less efficient is worse. It just means it is more specialized and used for different things. For example, base64 gets used when you want to encode information in a small amount of space. Otherwise, for the actual storage and computation of data, lower bases are preferred, and base 64 is still obviously stored as binary.
For what it's worth, the base integer with the best radix economy is 3, followed by 2.
I see that radix economy makes some sense if you assume that the "cost" of the "element" is not fixed but linearly proportional to the number of different states it needs to hold. It was apparently linear for the computers built with triodes but log2(n) to hold n states seems more realistic in electronics. Maybe little more for error resiliency.
Repeating the calculation (for k bits and b values) using square values, you get E = b^2 * log_b(k) and dE/db = 2b/ln(b) - b^2 / (ln(b))^2 / b = b/ln(b) * (2 - 1/ln(b)). Setting dE/db to 0, we get 2 - 1/ln(b) = 0, ln(b) = 1/2,
b_opt = sqrt(e) = 1.6487...
Which base is optimal then? 2? Any?
In this case, the average voltage used would be the average of 0, 1, 2^2, 3^2, which is 0+1+...+(b-1)^2 / b ~ b^2 (proportional to b^2). I showed that in this case in theory the "optimal" base would be sqrt(e), but I can't think of simple representations of a variable with less than 2 states (you could have say 2 variables representing less than 4 states, for example the 3 pairs { 0V/1V, 1V/0V, 0V/0V }, but that seems a tad complicated![1]), so the least non-trivial integer base is just 2.
I think in reality this kind of argument doesn't apply to real computers in a straightforward way, because there are many other factors at play than what I would call a 'dynamic power consumption' associated with the square voltage. There are things like leakage currents (that consume some fixed power per transistor) and other effects that significantly complicates things, so that such simple claims of being "optimal" don't apply (so I agree with the other comment about avoiding absolute claims!). But they can inspire designs and we can see if there are any advantages in more realistic cases.
[1] Further analysis: In our example, the 3 voltage pairs have mean square voltage 1/3. If we instead use the pairs { 0V/0V, 0V/1V, 1V/0V, 1V/1V }, the mean squared voltage is 1/2. In the first case, we encode log2(3)/2 bits of information per voltage value, in the second, 1 bit per value. So the energy per bit is (energy per value)/(bits per value) = (1/3)/(log2(3)/2) = 0.420... for the first case, and 0.5 for the second case. Unfortunately, it's a loss!
As an exercise, I believe that if we delete the state 1V/1V/.../1V of a k-voltage tuple, for large enough k-tuple, we come out ahead (probably not by much), although again that's not necessarily useful in practice.
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.
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.
So no, some people really clamor for all of us to switch to ternary logic.
“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
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.
> 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.