Translation algorithm - combinations to indices

Question

I have a deck of 52 poker cards represented as array of int - [0, 1, ... 51].

There are 2.598.960 possible combinations of 5 cards. I can generate all combinations like this:

#id #combination
0 - 0, 1, 2, 3, 4
1 - 0, 1, 2, 3, 5
2 - 0, 1, 2, 3, 6
 ...omitted data..
2598957 - 46, 47, 49, 50, 51
2598958 - 46, 48, 49, 50, 51
2598959 - 47, 48, 49, 50, 51

How I can effectively find #id for given #combination and #combination for given #id?

• gbjbaanb: If you generate the list of all combinations, you now have a hash table. Given a set of cards, generate the same hash and do a lookup. The trick is a repeatable way of generating the hash number from a set of cards.
• Worker: @gbjbaanb list or dictionary with indices already in place. I am looking for an algorithm.
• gbjbaanb: If you have no way of calculating a particular row in the table from your cards then you'll just have to loop through them all.

Answer 1

For the first card you have 52 options for the second you have 51 and so on. You need to encode a sequence c1, c2, c3, c4 and c5 where there are 52, 51, 50 ,49, 48 options resp.

You can adjust the numbers by counting how many of the previous numbers are smaller: a5 = c5 - sum(x < c5 for x in [c1, c2, c3, c4])

you can simply encode it as a single integer output =(((a1*51 + a2)*50 + a3)*49 + a4)*48 + a5.

you can extract it by using the modulo operation and integer divide.

int input = //...
a5 = input%48;
input/=48;
a4 = input%49;
input/=49;
a3 = input%50;
input/=50;
a2 = input%51;
input/=51;
a1 = input;

And then the reverse operation is needed to adjust them all back into the range [0..52).

• Worker: Thank you for great tip. I kind of stock with indexing 0,1,2.. Now I just realised that I don't need index. ints are just fine!
• Neil: Or you could multiply by 100 and you'd have an id that you can read as well.
• ratchet freak: @Neil and then you don't have to adjust the individual cards either.
• Neil: The only problem that I see in that is that it leaves spaces in your "id" range.
• ratchet freak: Technically there still is in my answer as it assumes that the order of the cards matter, if you remove that then you have a lot less options for the subsequent cards but the algorithm is more complicated.