There are 4 red cards and 4 blue cards, the moderator first take any two, and then in a, B, C three people put any two cards on their foreheads,
A, B, c three people can see the other two people's hands on the forehead, after reading let them guess what color on their forehead card,
A said don't know, B said don't know, C said don't know, then a said to know.
For advice on how to reason, A is how to know.
If you use the program, how to achieve it.
Idea: The purpose is to derive a color, because a first look B, C, you should first assume B, C color and then deduce a
Analysis: The cards that may appear on the head are BB, RR, RB (blue, red)
A: Do not know the description B, C in the color added not equal to 4 of the card
B: Do not know the description of A, C color added not equal to 4 of the card
C: Do not know the description of A, b color added not equal to 4 of the card
Process:1> B:RR (BB) c:rr (BB) A must know, so does not meet the requirements
2>
B:RR (BB) C:BB (RR) A don't know, because B don't know, C don't know so a can only take RB
3> B:RR c:rb C don't know->a is not rr a if BB->c according to A, B judge can know oneself for RB, but C don't know, so exclude BB a=rb
B:BB C:RB empathy can be verified A=RB
B:RB C:RR empathy can be verified A=RB
B:RB C:BB empathy can be verified A=RB
4> b:rb C:RB If A=RR b=rb C guess if oneself =bb then B can't not know oneself for RB so C can guess C=RB
If A=BB B=RB the same C can guess yourself =rb
Because a excludes RR BB, it only takes a=rb.
Conclusion: RR and Bb,a cannot be present in both RR and BB