by makashi » Tue May 08, 2007 3:31 pm
Claim: If the black square is in square [unparseable or potentially dangerous latex formula], Bob can force the black square to be in square [unparseable or potentially dangerous latex formula] in an adjacent row.
Proof: Group the rows as follows.
Red group: Rows 1 and 2.
Blue group: Rows 3 and 4.
Black group: Rows 5 and 6.
Each time Alice adds a number to a row in column [unparseable or potentially dangerous latex formula], Bob should add a number in column [unparseable or potentially dangerous latex formula] to the other row in the group that is relatively the same compared to the other numbers in that row.
For example, if Alice adds the first number to row 3 column 1, then Bob simply needs to add any rational number to row 4 column 4 which is not already in the grid.
Suppose Alice adds the second number to row 5 in column 3 and that number happens to be larger than the other number in row 5. Then Bob needs to add a number to row 6 in column 6 which is larger than the other number in row 6. We know there is another number in row 6 and also that row 6 column 6 is open by symmetry of the moves.
Suppose Alice adds a number [unparseable or potentially dangerous latex formula] to a row, say row 1, column [unparseable or potentially dangerous latex formula] which already has at least two numbers in that row. Let the largest number in row 1 smaller than [unparseable or potentially dangerous latex formula] be in column [unparseable or potentially dangerous latex formula] and the smallest number larger than [unparseable or potentially dangerous latex formula] be in column [unparseable or potentially dangerous latex formula]. Then Bob shall write a rational number in row 2 column [unparseable or potentially dangerous latex formula] which is larger than the number in row 2 column [unparseable or potentially dangerous latex formula] and smaller than the number in row 2 column [unparseable or potentially dangerous latex formula]. Again, by symmetry, the square in which Bob needs to write his number will be unoccupied and the squares with which Bob needs to make comparisons will be occupied.
By symmetry of the construction, if the black square is in square [unparseable or potentially dangerous latex formula], Bob can force the black square to be in square [unparseable or potentially dangerous latex formula] in an adjacent row.
Therefore, the black squares in any group will not share a vertex under this strategy, and Bob wins.