This is just a mind fart of mine, but if you have a bit of a twisted sense of fun, you may enjoy this.
We're all used to counting 1, 2, 3... And those of use that program are used to counting in other systems like binary (0, 1, 10, 11, 100, 101, 111, 1000...), or octal (0, 1, 2, 3, 4, 5, 6 7, 10, 11, 12, 13, 14, 15, 16, 17, 20...) or hexidecimal (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B...) or whatever. But no matter which one you are using, you are still using a base 'n' system, where decimal is base 10, binary is base 2, octal is base 8 and hexadecimal is base 16.
But just for the sake of musing, what about a non-base n system? I'd like to briefly outline a simple musing on base 'prime', where the base is any traditional base, but instead of counting, we uniquely enumerate occurances in a set with primes and set a final value that tells us how many of each element there are. i.e. You end with 2 numbers, the final sum, and the number of items in the original set, and an enumeration (pairing of prime numbers to unique elements from the set). (You don't actually need the number of items in the original set because you have the enumeration, and the number of enumerators is equal to the number of items in the original set.)
The number of elements in your set determines the highest prime that you use as well as the 'prime' base multiplier. The prime base multiplier determines the lowest possible number of the next prime number in your series of primes that you use to enumerate the elements in your set with. This is always given as you know the number of elements in the set.
As a simple example, consider that your set consists of only 2 elements. Your first prime is always 2. The next prime enumerator will be 5. We get this by using the prime base multiplier by the number of elements, in this case 2, i.e. 2*2 = 4. Since 4 is not prime (trivially true), we go to the next higher prime, which is 5.
To count/enumerate our elements, we simply count occurances of non-unique elements in our set and assign them a prime enumerator, then multiply the number of occurances by the prime enumerator. So, consider our set consists of {John, John}. Our prime enumerators are 2 and 5. We go to our first element, 'John', and assign that value to the prime enumerator 2, then advance to our next element, 'John'. We now have 2 occurances of 'John' assigned to the prime enumerator 2, and 2x2=4, so we sum all of our products (product [4] = prime enumerator [2] * occurances of item {John} in the set [2]).
The end result is that we have a single number [4] that combined with the enumerator assignment [2 : John, 5 : Nothing] uniquely tells us how many of each item are in the set.
Now, here's the question: Does the above outlined system work? True or false.
If you need some help, here are the first 5 in the series for sets with 1, 2, 3, 4 and 5 elements:
|
# of Elements
in Set |
First
Enumerator |
Second
Enumerator |
Third
Enumerator |
Fourth
Enumerator |
Fifth
Enumerator |
|
1 |
2 |
- |
- |
- |
- |
|
2 |
2 |
5 |
- |
- |
- |
|
3 |
2 |
7 |
23 |
- |
- |
|
4 |
2 |
11 |
47 |
191 |
- |
|
5 |
2 |
11 |
59 |
307 |
1543 |