We need to have this talk sooner or later guys...
I will spare you my personal opinion and will come right to it:
from en.wikipedia.org/wiki/Radix_economy
If both b and N are positive integers, then the radix economy E ( b , N ) [math]{\displaystyle E(b,N)}[/math] is equal to the number of digits needed to express the number N in base b.
[math]{\displaystyle {E(b,N)}\approx {b\ {\log _{b}(N)}}={b{\ln(N) \over \ln(b)}}}[/math]
Then for a constant N, E ( b , N ) [math]{\displaystyle {E(b,N)}}[/math] will have a minimum at e. Meaning e is therefore the base with the lowest average radix economy.
And since 2 / ln(2) ≈ 2.89 and 3 / ln(3) ≈ 2.73, it follows that 3 is the integer base with the lowest average radix economy.
Now give me one reason why humanity should not switch to three-valued logic!
>>8114835
Because that's not how transistors work
>The operation of rounding to the nearest integer is identical to truncation, or simply deleting everything to the right of the radix point.
This is just one of the reasons why we should implement ternary logic at all scales, down to transistor level.
>>8114854
I agree, converting binary machines to ternary logic, will not be as easy as it is on paper orthough a few attempts have made it to realization in the past (SETUN) but I'm more curious about nowaday possibilities.
>>8114835
You said it yourself, OP: switching to base e is the most efficient. There's no need to round to 2 or 3.
>>8114953
I know you're trolling, but 3/ln(3) - e is pretty damn small.
When Mathematics meets Physics...
"The radix economy of a number in a particular base (or radix) is the number of digits needed to express it in that base, multiplied by the base"
Why is it multiplied by the base?
>>8115329
I'm no mathematician but it seems to have consistency. A static increment so we can get a ratio when we compare two integer(!) bases.
one can change the type of increment to logarithmic or faculty or fibonacci or whatever; there are some wierd bases out there like ß-base or [math]\varphi[/math]-base, but I'm more concerned with integer bases.
>>8115357
I meant, within the context of computers, why multiply by the base? Also, have you seen this: https://en.wikipedia.org/wiki/Ternary_computer
>>8115423
lol you didn't seriously post a wiki link back to me
I'm a bit further into the matter already so this aint news to me
>>8115423
optical computers are the key to implementing ternary on the hardware level
>>8115329
You could choose a very large b, like b=10000. Then most everyday numbers would be single digit numbers, with a few rare two-digit numbers here and there. Surely very short and efficient!
There is just this tiny drawback that you now have to design ten thousand visually distinct digits, memorize them, teach them to your children, and teach them how to do addition and multiplication with them.
So obviously there is some cost associated with a larger b. The formula simply assumes that this cost is linear with the size of b, and therefore multiplies with b.
That assumption of linearity seems a bit naive: humans seem to have not much problems memorizing and recognizing ten different digits, so using ternary seems needlessly frugal and verbose; if you’re dealing with computer hardware, on the other hand, it boils down to the question how you express this shit with transistors. So for that use case, binary it is.
>>8115532
I literally asked if you had seen it, smug retard
>>8115590
>That assumption of linearity seems a bit naive
Exactly. That is my point.
Considering how computers are made... Suppose that I could make a commercial, competitive ternary computer today, it is hard to prove that it is as far as a base goes within computers. So with this consideration, I could go ahead and build a machine with bigger bases, and there is not linearity to it, and it just boils down to Physics and Engineering.
So with all that formally said, OP was making a joke, I presume.
>>8115329
Just my thoughts on this, without knowing anything about radix economy
imagine a number z represented as an mxn matrix, where m is the base and [math] n = \left \lfloor log_m(z) \right \rfloor [/math]
[math] z=\sum _{k=0}^n a_k m^k [/math]
The representation matrix would have a 1 in the i-th row, and the j-th column, when [math] a_i = j [/math] and zeroes everywhere else.
The amount of space to store such a matrix would be [math] m\cdot n = m \left \lfloor log_m(z) \right \rfloor [/math]
