For the odd case, is there a solution? Proof in either direction? Thanx

Pratik (https://pratikpoddarcse.blogspot.com)

]]>I would suggest the term “pinwheel” for a set consisting of every other slice (from what is left). For example, if eight slices are left, then slices 1,3,5,7 form a pinwheel, as do slices 2,4,6,8. As you observed, whichever player gets to choose from among an even number of slices can guarantee him- or herself either pinwheel.

Unfortunately for Bob, I can show that in a pizza with 2n+1 slices, there is always a slice Alice can remove such that neither pinwheel in the remaining 2n slices consists of more than half the (original) pizza. So, if Bob has a strategy, it must be more complicated than choosing a pinwheel.

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