Yes, Charles gets a positive answer in a finite number of rounds.
Let's call A the number chosen by Arthur and B the one from Bert. Charles
writes then two numbers: S (Smallest) and L (Largest), possibly S = L.
Let M = A+B. Arthur and Bert must find out if M = S or M = L, this way
they could find the others number.
It is clear that anyone knowing M, deducts his own number and he can
guess the others number.
Round 1: Arthur knows A, S and L, if he sees that S < A, it follows
that M = L. If S = L, Arthur also knows M. On the other hand,
if A <= S < L, then Arthur can't know M. So, if Arthur answers
"YES", everything ends and if he answers "NO", Bert knows
A <= S < L and we go to round 2.
Round 2: Bert knows B, S and L and he knows A<= S. If S < B,
then Bert knows that M = L. Also, as A + B <= S + B, then
M = S if L > S+B. So either Bert says "YES", everything
ends here or Bert answers "NO" and Arthur now knows we are
in a situation where L - S <= B <= S, which means that
L <= 2S. We are now going to round three.
Round 3: Arthur knows A, S and L and he knows that
L - S <= B <= S. He also sees that A + L - S <= M <= A + S.
Arthur answers "NO" if, and only if, A + L + S <= S and
L <= A+L that is: L - S <= A <= 2S - L ( and Arthur answers
"YES" in any other case). Suppose Arthur said "NO". This
situation implies that 2L <= 3S and we go to round 4.
Round 4: Bert knows B, S and L and he knows that
L - S <= A <= 2S - L or that L - S + B <= 2S - L + B.
Bert answers "NO" if, and only if, L - S + B <= S and
L <= 2S - G + B that is 2 (S - L) <= B <= 2S - L.
Suppose Bert said "NO". This situation implies that
3L <= 4S and we go to round 5.
Continuing the above way of thinking, it is shown that
if Arthur answers "NO", it follows that 4L <= 5S and
following rounds with negative answers lead to
5L <= 6S, 6L<= 7S,...(n-1)L <= nS. But there will
always exist a n which is large enough for this not to
be true (Because s < L, as S = L has been eliminated
in round 1)
Charles question willl be answered "YES" before the
n th round. |
