A row of 2021 balls is given. Pasha and Vova play a game, taking turns to perform moves; Pasha begins. On each turn a boy should paint a non-painted ball in one of the three available colors: red, yellow, or green (initially all balls are non-painted). When all the balls are colored, Pasha wins, if there are three consecutive balls of different colors; otherwise Vova wins. Who has a winning strategy?
Problem
Source: VI Caucasus Mathematical Olympiad
Tags: game, combinatorics
elcinmusazade
14.03.2021 15:42
Pasha wins.for n=5 or 7 easy bash shows that Pasha always wins,For 2021: take 5 (7) points from middle There are even number of points outside of it.If Vova colors outside (inside) then pasha Pasha also colors outside(inside),Pasha starts with algorithm which he wins for n=5 (7),And there is no algorithm for outside,Eventually Vova will color inside with same algorithms with case n=5 (7).I hope you understood
bin_sherlo
20.05.2024 23:26
Answer: Pasha wins. Proof: Let the colours be $A,B,C$. Pasha colours the fourth cell into $A$. If Vova colours a ball with number $\geq 5,$ then Pasha colours the first ball into $B$. Vova cannot colour second or third ball since Pasha wins otherwise. But there are $2018$ balls left. After Vova's moves, Pasha colours any ball with number $\geq 5$. Thus Vova must colour the second or third ball which provides Pasha to win. If Vova colours a ball with number $\leq 3,$ then Pasha colours seventh ball into $B$ and applies the same progress as desired.$\blacksquare$