All squares of a $ 20\times 20$ table are empty. Misha* and Sasha** in turn put chips in free squares (Misha* begins). The player after whose move there are four chips on the intersection of two rows and two columns wins. Which of the players has a winning strategy? Proposed by A. Golovanov US Name Conversions: Misha*: Naoki Sasha**: Richard
Problem
Source: Tuymaada 2009, Junior League, First Day, Problem 1
Tags: geometry, geometric transformation, combinatorics unsolved, combinatorics
MellowMelon
21.07.2009 06:48
Naoki and Richard, huh? Is this a thing of translation or did the author make a tribute to artofproblemsolving.com? Richard wins. Partition the columns into pairs. Whenever Naoki plays, play in the same row and the paired column. This forbids any more moves in either of the paired columns, even if the row played on is not empty. So after 20 moves by both players, Naoki has no more moves left without giving Richard a win. Before I realized I misread it, I was trying to solve the problem where making the configuration of four chips was a loss and not a win. This is much harder.
guas
21.07.2009 07:07
MellowMelon wrote: Naoki and Richard, huh? Is this a thing of translation or did the author make a tribute to artofproblemsolving.com? They were Misha and Sasha. Or, if these names are too difficult, Michael and Alexander (as these two people are called in official documents). I know both, they will soon appear and restore their identity by any means however violent.
jgnr
21.07.2009 11:38
MellowMelon wrote: Before I realized I misread it, I was trying to solve the problem where making the configuration of four chips was a loss and not a win. This is much harder. I misread it too. I was really confused when I read your solution.