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Game of Dodge


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Player 1 has a 6x6 grid With rows numbered 1 - 6

 

Player 2 has a 1x6 grid with columns numbered 1-6.

 

Player One begins by filling in the first horizontal row of her table (row 1) with a run of X’s and O’s. That is, on the first line of her board, she will write six letters – one in each box – each letter being either an X or an O. Player Two looks at Player One’s first row and then places either one X or one O in the first box of her board. So at this point, Player One has filled in the first row of her board with six letters, and Player Two has filled in the first box of her board with one letter. The game continues with Player One noticing what Player Two placed on her board and then writing down a run of six letters (X’s and O’s), one in each box of the second horizontal row of her board, followed by Player Two writing one letter (an X or an O) in the second box of her board. This game proceeds in this fashion, with each player’s moves visible to the other, until all of Player One’s boxes are filled with X’s and O’s; thus, Player One has produced six rows of six marks each, and Player Two has produced one row of six marks. Player One wins if any horizontal row she wrote down is identical to the row that Player Two created. Player Two wins if Player Two’s string is not one of the six strings made by Player One.

 

Would you rather be Player One or Player Two?

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Hints and (deeper) solution

Spoiler

 

See "bias", think "diagonal"

Hint and (deeper) solution

 

Spoiler

 

See "Georg", think "Cantor"

Solution

Spoiler

 

Player 2 can construct a vector V that is guaranteed to be unequal to any of the rows in Player 1's array A, by using the Diagonal Construction pioneered by Georg Cantor.

For each round i, Player 1 fills A(i,j), for j in [1...6] with X or O, at Player 1's choice
Then, to defeat Player 1,  Player 2 shall fill V(i) with the opposite of A(i,i).

By the end of the game, V does not equal any of the rows of A, because it differs with each row by the value on the diagonal. That is, for each i in [1...6], V(i) is not equal to A(i,i).

 

Spoiler

 

 

 

 

 

Edited by CaptainEd
Tried unsuccessfully to nest spoilers further, sigh
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