Favorite numeral system?

[muon2]

What's with the votes for base 11-15? I've worked a lot with alternate bases, and this isn't a range that comes up much. At best there is base 12, but that only includes the concepts of dozen and gross (a dozen dozen), but nothing beyond that so it's limited to smaller counts.

[muon2]

Base 3 doesn't come up much, but there is an argument that it is the most efficient numeral system.

Suppose that the cost of a number is related to the product of the number of digits in a number times the number of choices for a digit (the base). For example the number 134₁₀ has a cost of 3*10=30, 10000110₂ has a cost of 8*2=16, so it's more efficient to store that number in binary than decimal. This model of cost is actually not unreasonable given our knowledge of information storage.

The number of digits in an arbitrary number k is equal to the logarithm of a number in the base n it is represented by (rounded down).  Eliminating the rounding factor I can write the cost in terms of the natural logarithm (ln) as

C(k,n) = (ln k/ln n) * n.

The cost as a function of the base is minimized when the derivative of the cost function is zero.

dC(k,n)/dn = 0

Finding the derivative reduces this equation to

ln n - 1 = 0

The solution is n = e, the base of the natural logarithm. e is an irrational number approximately equal to 2.718 and can't be used as a base for discrete counting numbers. However, the closest integer to e is 3, so that is the most efficient integer base. To use my earlier example 134₁₀ = 11222₃ (81+27+2*9+2*3+2*1). The cost of 11222₃ is 5*3 = 15, so it comes up slightly better than binary.

[muon2]

It's a reasonable question. I haven't seen a proof that 3 is better than 2 but consider this table of costs. As I look at the beginning base 3 is better than base 2 more often than the reverse. I suspect the trend continues as base 3 pulls away as the more economical choice.

range	base 2 cost	base 3 cost
81-127	14	15
128-242	16	15
243-255	16	18
256-511	18	18
512-728	20	18
729-1023	20	21
1024-2047	22	21
2048-2186	24	21
2187-4095	24	24
4096-6580	26	24
6581-8191	26	27
8192-16383	28	27
16384-19682	30	27
19683-32767	30	30
32768-59048	32	30
59049-65535	32	33
65536-131071	34	33
131072-177146	36	33
177147-262143	36	36
262144-524287	38	36
524288-531440	40	36
531441-1048575	40	39

edit: I checked 2³⁰ - 1 = 1,073,741,823 takes 30 digits in base 2 (cost = 60) and 18 digits in base 3 (cost = 54) so my supposition seems to hold.

[muon2]

...my supposition seems to hold.

seems so, thus far.  I guess what we want to demonstrate is something like:

 lim  C_n=3  <  lim  C_n=2
n→∞               n→∞

Or, more broadly,

 lim  C_n=3  <  lim  C_{n∈{N ∀ n≠3}}
n→∞               n→∞

Proof by the Second Principle of Mathematical Induction, perhaps, assuming that you can demonstrate conclusively for 2 and 4? 

Well, it's an interesting subject.  I hadn't encountered the idea of the cost of expressing numbers before.  The next time I'm at a cocktail party during which the subject of favorite bases comes up, I'll be able to suggest 3 in a way that makes me look really smart.

The cost grows without limit because it grows as ln(k). If I look at C/ln(k) = n/ln(n) that gives me a measure of the generalized cost independent of the specific number. For base 2 it is 2.89 (same for base 4) and for base 3 it is 2.73. That adds more credence to 3 as the most efficient base.

[muon2]

...my supposition seems to hold.

seems so, thus far.  I guess what we want to demonstrate is something like:

 lim  C_n=3  <  lim  C_n=2
n→∞               n→∞

Or, more broadly,

 lim  C_n=3  <  lim  C_{n∈{N ∀ n≠3}}
n→∞               n→∞

Proof by the Second Principle of Mathematical Induction, perhaps, assuming that you can demonstrate conclusively for 2 and 4? 

Well, it's an interesting subject.  I hadn't encountered the idea of the cost of expressing numbers before.  The next time I'm at a cocktail party during which the subject of favorite bases comes up, I'll be able to suggest 3 in a way that makes me look really smart.

The cost grows without limit because it grows as ln(k). If I look at C/ln(k) = n/ln(n) that gives me a measure of the generalized cost independent of the specific number. For base 2 it is 2.89 (same for base 4) and for base 3 it is 2.73. That adds more credence to 3 as the most efficient base.

[muon2]

As k goes to infinity so does the cost so that doesn't converge. The cost normalized to the log is what I gave above, but that is only correct for a generalized cost that truly goes as ln k. The actual cost goes as floor(log_nk) which is why I drew up my table. The easiest proof is probably to find an integer j such that for all integers k>j, C(k,2) >= C(k,3).