I am not sure, but I think I solved it : We will not allow the first player to pass the second row. We take the numbering of collumns 0 1 2 3 4 5 Now put the numbers 0 1 2 3 4 5 4 5 0 1 2 3 In the first 2 rows of another table. Any time the first player puts a number in one of x, we put a number in the other x so that the relationship (bigger, smaller ) between the same numbers in both rows is same. Because nr are rational, we can do it. Now, obviously, biggest number will not be neighbours

DusT wrote: Now, obviously, biggest number will not be neighbours Obviously? $\begin{array}{cccccc} 0&2&4&6&8&\mathbf{10}\\ 9&11&1&3&5&\mathbf{15} \end{array}&$

Here's a way: Denote columns as c1, c2, c3, c4, c5, c6. Show that Bob has the ability to do one of the following in any row: (the basic strategy for Bob is to always place numbers in the row Alice put hers) #1 Force the black square to be in one of c1, c2, or c3. #2 Force the black square to be in one of c4, c5, or c6. #3 Force the black square to be in one of c1, c2, c5, c6. Now let Bob execute #1 on row 1, #3 on row 2, and #2 on row 3. If, in the second row, black square is in one of c1, c2, then it cannot be connected to row 3. (case 1) If, in the second row, black square is in one of c5, c6, then it cannot be connected to row 1. (case 2) In either case, Bob wins. _: Grid boxes X: Possible locations of black square Case 1: X X X_ _ _ X X _ _ _ _ _ _ _ X X X Case 2: X X X_ _ _ _ _ _ _ X X _ _ _ X X X So in fact, Bob wins in a 6x3 grid.

Your solution is very similar to my one. Only my one is based on more explicit results: 1. Bob can force the maximum in a row to lie in a group of 3 squares he chooses. 2 Bob reaches the maximum in row ! to be on c1,c2,c3, on row 2 on c4,c5,c6, on row 3 on c1,c2,c6. this is his winning strategy

Yeah, obiously! Reply tu lugash 0 2 4 6 8 10 9 11 1 3 5 15 The second array is not sorted in the same way. It should be 0 2 4 6 8 10 11 15 1 3 5 9

Hmm... DusT How do you take care of cases in which Alice places her first number in the first row but the second number the second row, so that Bob cannot always place his number according to 012345 450123 ? For example, let's say the first two moves are as follows: _ _ 2 _ _ _ _ _ _ _ 3 _ _ _ _ _ _ _ (Alice places 2, Bob places 3) _ _ 2 _ _ _ 1 _ _ _ 3 _ _ _ 4 _ _ _ (Alices places 1, Bob places 4) _ _ 2 _ 5 _ 1 _ _ _ 3 _ _ _ 4 _ _ _ Now Alice places 5, but Bob's counterspot is already taken.

Sorry if I explain wrong , but I mean whean Alice puts a number in square x, Bob puts the number in the other x, so that the arrangement of the numbers remains the same Hope no it's clear

Yes, but my point was that what if the other "x" is already taken?

Isn't it obvious that there are only 2 letters of the same type, so when the 1st player takes one, we take the other, so either both are taken, or none is taken. Now, it is solved I think

Isn't this sorta obvious? Bob always makes the last move. He knows which square will be black in every row except the one in which the last empty square is. If he sees that this square will complete the line, he can always choose a sufficiently small number. If it won't, then we can always choose a sufficiently large number.

No, that doesn't work -- given the black squares in the first five rows, there might be more than one possible position for a black square in the sixth row that Alice can use to draw her line.

Er, how? Can you give an example?

I think that the confusion here is that a "line" doesn't mean a straight line. It means a path from the top to the bottom where adjacent squares share a vertex.

If the first square in rows 1 through 5 and the second square in row 6 is colored black, there exists a line (in the Euclidean sense) that connects the top edge to the bottom edge and passes only though black squares.

Is it necessary for each player to choose a rational number??????

I think I am interpreting this problem wrong or this is trivial. Kindly clarify: The grid has 36 (even) number of squares so Bob will make the last move. Clearly he can fill the number such that if the greatest numbers of the other rows lie in that column, he can choose a sufficiently small number. If not, a sufficiently large number suffices.

In the 6 by 6 grid, shade in the first 3 squares in the first and fourth rows, the first, 5th, and 6th squares in the 2nd and 5th rows, and the 3rd, 4th, and 5th squares in the 3rd and 6th rows. If the black squares were located in any of these shaded squares, a path wouldn't exist. Hence, Bob can win if he can force the black squares to be in these shaded squares. To do this, if Alice writes a number in a shaded square, Bob writes a smaller number in an unshaded square in the same row. If Alice writes a number in an unshaded square, but writes a larger number in a shaded square in the same row.

billzhao wrote: Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players. Bob wins no matter what. (Refer to the figure)Bob's strategy is to make sure that no two black squares share a vertice.He puts numbers in any row in an increasing order. $\;\;\;\;\; \bullet $ If Alice decides to follow along,by trying to disrupt Bob's strategy,then Bob starts going for $3 \cdot 3$ squares,there are $14$ of these.He changes the Corner's rather than the middle if Alice has fixed the middle and if Alice hasn't fixed the middle he gors for the middle. Alice can't disrupt it;there are an even number of squares i.e Bob finishes it. $\;\;\;\;\; \bullet $ If Alice decides not to play along then Bob goes for squares of the same parity;this is a clever strategy i.e Alice can't give Bob a last minute surprise;If after some point she decides to play along then Bob is safe;All the odd parity/even parity squares are reeserved so in order to move to the bottom of the board Alice would need to use an square of the parity Bob choose but this is impossible since all of these are white squares.

not sure if this works but any line going through the top and to the bottom has to pass through only one square in at least one row also bob can make sure that his turns are the turns that color squares black so on each of these turns he has two choices for what the black square will be and im pretty sure this implies bob will always win

We claim that Bob always wins. Bob's winning strategy is to always place in the same row as Alice. Consider any row, and we claim that Bob can limit the black square's position to three of these squares. If Alice places a number in one of the squares Bob wants black, then Bob can place a smaller number in a square he wants white, and vice versa. Thus, our claim is true. Bob can limit row 1's to the rightmost three columns, row 2's to the leftmost two columns and rightmost column, and row 3's to the leftmost three columns. Starting from the top row, there is no way to draw a line that gets past row $3,$ so we're done.

Divide the board into 12 "zones", each of which is a set of 3 (not necessarily connected) cells, which we call $A1,A2,B1,B2,C1,C2,D1,D2,E1,E2,F1,F2$, in the following manner: $$\begin{matrix} A1 & A1 & A1 & A2 & A2 & A2\\ B2 & B2 & B2 & B1 & B1 & B1\\ C1 & C1 & C2 & C2 & C2 & C1\\ D1 & D1 & D1 & D2 & D2 & D2\\ E2 & E2 & E2 & E1 & E1 & E1\\ F1 & F1 & F2 & F2 & F2 & F1\\ \end{matrix}$$ Bob can then use the following strategy: If Alice writes a number in a "1" zone, Bob comes up with a number smaller than all numbers on the board (which is always possible) and writes it in a "2" zone of the same letter as Alice's zone. If Alice writes a number in a "2" zone, Bob comes up with a number larger than all numbers on the board (which is again possible) and writes it in a "1" zone of the same letter as Alice's zone. Note that the only time when the largest number in a row can be in a "2" zone is right after Alice writes a number in that zone, but using this strategy Bob will immediately overturn that by placing a larger number in the corresponding "1" zone. Thus, by the end of the process, all black cells are in "1" zones. This clearly prevents Alice from winning, so Bob wins are we are done.

We claim that $\boxed{\text{Bob}}$ is the winner. The key in this problem is that Bob can always prevent Alice from taking any $3$ squares he chooses. Indeed say on one of Alice's turns she plays in a square he wishes to be white. Then he can simply play a higher number in a square that he wishes to be black and so on. Now the finish is obvious. Restrict Alice to the square $(1, 2, 3)$ in row $1$. On row $2$ restrict Alice to the squares $(1, 5, 6)$. Then on the third row restrict her to the square $(4, 5, 6)$. Clearly Alice cannot win.