25 seat bar riddle
There is a bar with 25 seats in a line. The people there are anti-social so when they walk in the bar, they always try to find a seat farthest away from others. If one person walks in and find there is no seat are adjacent to nobody, that person will walk away. The bar owner wants as many people as possible. The owner can tell the first customer where to sit. all the other customers will pick the farthest possible seat from others.
So where should the first customer sit?
Answer – 9 or 17
Consider a case. If the number of seats is n = (2^k)+1 for some k, then, the maximum possible seating of (n/2)+1 can be achieved by having the first person sit at position 1 or n or ceil(n/2).
This is true since when n=(2^k)+1, then, when someone sits in the first or middle or the last seat then all the other gaps are of length in powers of 2 (as suggested above by Vinay).
This implies that for this problem, with n=25, we need to split it into n1 and n2 such that: n1 = (2^k1)+1 and n2 = (2^k2)+1, for some k1, k2. Note that n1+n2 = n+1 = 26 (since they both share a common seat).
Thus, the only possible values are 9 or 17.
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