unless you don’t mean the most recent Virtuous
Yeah, that’s her.
Are we not talking about the same Alice?
Doubtful (Alice has been permanently banned for much longer)
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there’s a different Alice who’s permanently banned
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who is not to my knowledge the same person as PlagueSimp
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she cannot escape these vigilant eyes (or can she)
I wonder if Alice vanity searches this forum to see if she still gets mentioned
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They continue alternately choosing one of the integers that has not yet been chosen, until the nth turn, which is forced and ends the game. Bob wins if the parity of {k : the number k was chosen on the kth turn} matches his goal.
I don’t see how game-end vs. end of each turn changes things?
Like, Alice always has a strategy at n > 1 to prevent k from being drawn on the kth turn.
What’d she get banned for?
she was getting pushed in a mafia game so she told the player pushing her to slit their wrists
(she had an extensive previous history of toxicity, that’s obviously not okay regardless but it’s not like it was a one-off thing)
Ah.
Eeks.
mathfia. is this anything
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Benguined mentioned that solving this game would make for a better Putnam problem than the actual A6 Putnam problem, and that sent us down this train of thought.
Alice can prevent K from being drawn on the Kth term*, but that’s not always Alice’s wincon.
If N is even, Bob decides the set will be of even parity. Then, Bob imagines all the numbers in pairs, where each pair is {2a, 2a-1}. Alice goes first, and Bob always guesses the remaining number in the pair. Bob always wins with this strategy.
*actually idk if that’s always true tbhtbhtbhttb. maybe?
Oh, is k just the final number that gets drawn?
no
lets say N=4
Alice chooses 2, so Bob chooses 1. Then Alice chooses 3, so Bob chooses 4.
Our set is {3, 4}
oh also now that I think about it more, when N is odd, Alice can always force the parity to be odd
The game starts soon I just feel bad interrupting these nerds
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What stops Bob picking even, Alice picking 2, Bob picking 1, Alice picking 4, and Bob picking 3?
k never appears in the sequence.