Use a sudoku-solving algorithm

[Bartleby2718]

If a value x must be in one of m decks, say Deck 3, and if there are n numbers that can be in Deck 3, what is the probability that x is in Deck 3? It's complicated. Neither 1/m nor 1/n might be the answer. Since a value in a card affects the values in the other decks, you actually have to count the number of possible cases. It's like solving a sudoku. This issue will add on to #94, although not anytime soon.

[Bartleby2718]

Actually, it's just logic, not a sudoku-solving algorithm, but anyway, let me give you a concrete example.
If the rightmost decks, none of which is open, are:

• Deck 7: 10, ?, ?
• Deck 8: 12, ?, ?
• Deck 9: 13, ?, ?

Deck 8 contains at least one 11, and Deck 9 contains at least one 13 besides its delegate.
In fact, the possibilities are:

Case	Deck 8	Deck 9
1	12, 11, 11	13, 13, 12
2	12, 11, 11	13, 12, 13
3	12, 11, 12	13, 13, 11
4	12, 12, 11	13, 13, 11
5	12, 11, 12	13, 11, 13
6	12, 12, 11	13, 11, 13

For done, both Deck 8 and Deck 9 must be opened, so maybe you should look for other numbers first. On the other hand, the big sums can be used to dare some of the biggest decks the opponent has. Finally, there is a guarantee that joker is in neither of Deck 8 and Deck 9.

[Bartleby2718]

Some links on sudoku-solving algorithms:

• http://pi.math.cornell.edu/~mec/Summer2009/meerkamp/Site/Solving_any_Sudoku_II.html
• http://www.sudokuwiki.org/Crooks_Algorithm

The problem is that I have no idea how I should implement this... (because I haven't looked into it yet)

[Bartleby2718]

Another example:

• Deck 1: 2, ?, ?
• Deck 2: 4, ?, ?
  ...
• Deck 9: 13(Joker), 2, 3

In this case, Deck 1 has to be {2, 1, 1} and Deck 2 has to be {4, 4, 3} because it is the only possible case, although it may be evident at first glance.

[Bartleby2718]

What concerns me is that this is clearly a logical but may not worth the computing time.