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chinese emporer

There was no posted solution, so here goes:

Solution

Sage 1 can only guess for one case:
- when ?BB can be seen, Sage 1 must have W.  (Note: if this happened, Sage 2 would see W?B, and Sage 3 would see WB?)
- otherwise, Sage 1 must pass.

Sage 2 can guess for two cases:
- when B?B can be seen, Sage 2 must have W.  (Note: if this happened, Sage 3 would see BW?)
- when W?B can be seen, Sage 2 must have W, otherwise Sage 1 would have seen ?BB and guessed.  (Note: if this happened, Sage 3 would see WW?)
- otherwise, Sage 2 must pass.

Sage 3 will always know:
- when BB? can be seen, Sage 3 must have W.
- when BW? can be seen, Sage 3 must have W, otherwise Sage 2 would have seen B?B and guessed.
- when WB? or WW? can be seen, Sage 3 must have W, otherwise one of the other sages would have guessed.

So, in tabular form:

Result..    Sage sees...        ...and guesses...
    1    2    3    1    2    3
WBB    ?BB    W?B    WB?    W    -    -
WWB    ?WB    W?B    WW?    pass    W    -
BWB    ?WB    B?B    BW?    pass    W    -
BBW    ?BW    B?W    BB?    pass    pass    W
BWW    ?WW    B?W    BW?    pass    pass    W
WBW    ?BW    W?W    WB?    pass    pass    W
WWW    ?WW    W?W    WW?    pass    pass    W

David Woodruff
Tuesday, February 19, 2002

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