This may also end up failing as well, but I though I'd share.
B's plan will be to always pick and even card (from 2 to 48) based only on the even cards that have been picked. C=code number. B=B's card
Case 1. All cards are odd (O1,O2,O3,O4). C=B. A will see a single even card and know C=B
Case 2. One even (O1,O2,O3,E1). B=E1+C mod 100. A will see two even cards, E1 and B. E1-B is too big, but B-E1=C.
Case 3. Two Even (O1,O2,E1,E2). if E1>E2 then B=(E1-E2)+C if E2>E1 then B=(E2-E1)+C. A will see three even cards, E1, E2, and B and will try to decode to find C. There are three choices:
B-(E1-E2) = B-E1+E2 = C
E1-(B-E2) = -B+E1+E2 = not C
E2-(B-E1) = -B+E1+E2 = not C
The unique solution is the correct one, the two incorrect ones come out the same. [Is there an example that doesn't work?]
Case 4. Three even (O1,E1,E2,E3). I haven't tried this yet but I'm thinking if E1>E2>E3 then B=E1-E2+E3+C may work but I haven't checked it yet.
Case 5. Four even (O1,O2,O3,O4). Again I haven't tried this yet but I'm thinking if E1>E2>E3>E4 then B = E1-E2+E3-E4+C may work but I haven't tried it yet.
Posted by Jer
on 2007-05-14 12:59:05