Sorry, you are right. And here is the proof of the correctness.

來源: innercool 2010-06-07 19:26:05 [] [博客] [舊帖] [給我悄悄話] 本文已被閱讀: 次 (463 bytes)
回答: Do not think it is going to work...innercool2010-06-07 16:09:12
Let x_i be the number on player i 's head, y_i be his output, e_i be his offset and f_i be his error. We would like to show that there exists some i s.t. f_i = 0.

For every i, we have (all arithmetic modulo n)
y_i = e_i - (x_1+...+x_n -x_i) = x_i - f_i.
Therefore e_i+f_i=x_1+...+x_n. Since {e_i} is the set of congruence modulo n, it follows that {f_i} is also {0,1,...,n-1}; in particular, there exist some j with f_j=0.

Very nice argument.
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