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General Discussion / Re: Maths thread.
« on: January 24, 2020, 05:41:33 am »
I corrected my previous result to a worst case of 2n + 3 turns for n > 3. n = 3 requires 10 steps to solve.
The trivial case was outlined above, let's focus on the special case now. In most cases, the 1st player will actually know that they are WB after the first round, because if they weren't it would be a trivial case. In these cases, the first player states their identity and one more passing is required to communicate the player with the doublet their identity. This leads to a worst case scenario of 2n + 2 if the player with the doublet is in position 2 and must pass in the second round.
Now to the remaining 7 cases, which are always the same for all n > 3. They are all players having WB, except for player 1 and n which may also have doublets. This results in 7 cases because they may not have the same doublet. In all these 7 cases, the first player must pass in the second round followed by (n - 2) players who all hold WB and using selective passing to communicate the doublets to player 1 and n. Because in the worst case three options are available for player 1, this requires up to 2 players passing. Combined with the passing of the first player in the second round, we get 2n + 3 turns.
The trivial case was outlined above, let's focus on the special case now. In most cases, the 1st player will actually know that they are WB after the first round, because if they weren't it would be a trivial case. In these cases, the first player states their identity and one more passing is required to communicate the player with the doublet their identity. This leads to a worst case scenario of 2n + 2 if the player with the doublet is in position 2 and must pass in the second round.
Now to the remaining 7 cases, which are always the same for all n > 3. They are all players having WB, except for player 1 and n which may also have doublets. This results in 7 cases because they may not have the same doublet. In all these 7 cases, the first player must pass in the second round followed by (n - 2) players who all hold WB and using selective passing to communicate the doublets to player 1 and n. Because in the worst case three options are available for player 1, this requires up to 2 players passing. Combined with the passing of the first player in the second round, we get 2n + 3 turns.