Question
Even only 52 cards, the permutationIndex
where I describe in Explanations section, would be a huge number; it is
I really don't get your question, but I interpret it like this: you want to calculate the permutationIndex
of a sequence of 52 cards. A given permutation index maps one-to-one to a sequence of cards. Since there are 52! possible arrangements of 52 cards, you'll need at least 226 bits, or 29 bytes. So, your permutationIndex
will already be very big!
Since your permutation index is already 29 bytes long, some extra bytes won't make much of a difference and make the solution a lot easier.
For example, you could map each letter of the Latin alphabet to a card. Given that we have 26 lower case letters, 26 upper case letters, we have lo and behold 52 letters available to represent the 52 cards.
abcdefghijklm nopqrstuvwxyz ♥ A234567890JQK ♦ A234567890JQK ABCDEFGHIJKLM NOPQRSTUVWXYZ ♣ A234567890JQK ♠ A234567890JQK
Now you can make a string of 52 letters. Each unique letter string represents a unique permutation of 52 cards. With this you can:
Each character in a string is represented (in C#) as a 16-bit Unicode value, but for 52 cards you would only need 6 bits. So you have some more options to choose a representation:
Representations 3 and 4 require quite some clever bit fiddling to get the 6 bits for a specific card out of the sequence. I would recommend representation 2, which preserves most of the advantages mentioned above.
When you are using a binary representation instead of a character string representation, then you can create an enum with a unique value for each card, and use that:
public enum Cards : byte
{
HeartsAce
HeartsTwo
// ...
HeartsTen
HeartsJack
HeartsQueen
HeartsKing
DiamondsAce
DiamondsTwo
// ...
SpadesTen
SpadesJack
SpadesQueen
SpadesKing
}