Consider Microsoft Excel's column-numbering system. Columns are "numbered" A, B, C, ... , Y, Z, AA, AB, AC, ... where A is 1.
The column system is similar to the base-10 numbering system that we're familiar with in that when any digit has its maximum value and is incremented, its value is set to the lowest possible digit value and the digit to its left is incremented, or a new digit is added at the minimum value. The difference is that there isn't a digit that represents zero in the letter numbering system. So if the "digit alphabet" contained ABC or 123, we could count like this:
(base 3 with zeros added for comparison)
base 3 no 0 base 3 with 0 base 10 with 0
----------- ------------- --------------
- - 0 0
A 1 1 1
B 2 2 2
C 3 10 3
AA 11 11 4
AB 12 12 5
AC 13 20 6
BA 21 21 7
BB 22 22 8
BC 23 100 9
CA 31 101 10
CB 32 102 11
CC 33 110 12
AAA 111 111 13
Converting from the zeroless system to our base 10 system is fairly simple; it's still a matter of multiplying the power of that space by the value in that space and adding it to the total. So in the case of AAA with the alphabet ABC, it's equivalent to (1*3^2) + (1*3^1) + (1*3^0) = 9 + 3 + 1 = 13.
I'm having trouble converting inversely, though. With a zero-based system, you can use a greedy algorithm moving from largest to smallest digit and grabbing whatever fits. This will not work for a zeroless system, however. For example, converting the base-10 number 10 to the base-3 zeroless system: Though 9 (the third digit slot: 3^2) would fit into 10, this would leave no possible configuration of the final two digits since their minimum values are 1*3^1 = 3 and 1*3^0 = 1 respectively.
Realistically, my digit alphabet will contain A-Z, so I'm looking for a quick, generalized conversion method that can do this without trial and error or counting up from zero.
Edit
The accepted answer by n.m. is primarily a string-manipulation-based solution. For a purely mathematical solution see kennytm's links:
