[gnramires]

I think the b factor is due to the analog representation: if the base has b values, you use on average b/2 voltage for representation. I would think the mean square value in might have been more appropriate though? (as voltage energy losses usually are V^2 / R or C V^2)

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...

[scotty79]

In our computers, to hold b states we are using elements which cost is not proportional to b or b^2 but log2(b).

Which base is optimal then? 2? Any?

[gnramires]

That's true, if you use binary encoding to represent the b states. I were referring to base b representation, in which by definition we use b states to represent log2(b) bits: for example, 4 states to represent bits 00,01,10,11, which could correspond to voltage levels 0V,1V,2V,3V of a variable (analogous for other values of b).

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.

[gnramires]

Correction: I've just realized I've drawn an incorrect conclusion, the energy per bit in the case I've shown is indeed lower than the standard 2 bit encoding, so it's actually a win! Again, for a myriad reasons, I'm not sure you'd want to use that encoding in real life, but it does consume a little under 20% less power in terms of V^2 losses.