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puzzle-006

MPF - Grid Game - April 6, 2012
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Spiked Math Comic - Math Puzzle Friday - Grid Game


Question 2: What if Randall goes first?

Question 3: What if it is an m x n grid?

Tons of great genie responses from yesterday's comic and even an illustration in the Physics Forums! It's gonna be tough to choose a winner, but on the other hand, I do have a lot of creepy hugs to go around.



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10 Comments

Must the rational numbers be between 1 and m x n ?

Nope, any rational you wish to write down.

Optimal strategy for Randall: write the numbers using Conway's chained arrow notation, then claim Mike can't draw a line because he can't indentify the greatest number.

Hahahahahhahahaah awesome!

1) For a grid with an even number of columns 2n>=4, and m>=3 rows, Randall have a winning strategy :

_first he associates cases by couples, each couple in the same row. In each row, the first case goes with the (n+1)th, the 2nd with the (n+2)th, and so on. The following cases are "forbidden" :
-the n first cases of the 1st row
-the n last cases of the 3rd row
-nth and (n+1)th cases of the 2nd row.

_Each turn, Randall write a number in the case in couple with Mike's. -If neither Mike's case nor the associated case is forbidden, Randall write any rational number.
-If Mike's case if forbidden, Randall write a greater number in the associated case, and a lower one if the associated case is forbidden.

_Of course 2 associated cases never are both forbidden. And, in the end, none of the forbidden cases are the greatest of the row. So, Randall win.

In particular, in a 6*6 grid, Randall win.

This process can easely be generalized for an odd number of column, but an even number of row. (for n and m "great enough", ie >=7 and >=4). Indeed, we just have to forbid the same cases, except we forbid 3 cases and not two in the second row, associate each with a unforbidden case of the same row and then associate randomly all other cases.

As an exemple, here are the forbidden cases of a 6*4 and a 7*4 :
X X X _ _ _ X X X _ _ _ _
_ _ X X _ _ _ _ X X X _ _
_ _ _ X X X _ _ _ _ X X X
_ _ _ _ _ _ _ _ _ _ _ _ _

I just posted my solution in the forum. However, I still have no solution for a 3*5 grid (when Mike begins) and for a 3*5, 3*6, 4*6 and 3*8 (when Randall begins)

You could also do this without having to keep track of larger and smaller numbers by using a symbol for the 'highest number' in a row, and then have a symbol for the 'lower numbers' in the row be a symbol that contains the other symbol. (Confusing, I realize, but let me give an example.)

The first symbol in any row is written as |. If a player wishes to write a higher number in the row, he writes another |, and replaces the current | with a +, by drawing a horizontal line through it. If the player wants to put a lower number in a row, he places a +. At the end of the game, every row will contain 1 |, with the rest being +.

First off relabel Mike as 'A' and Randall as 'B'.

Well, optimal winning solution always goes down to having 2 open spaces in each row. So assuming both A and B know this, A will always win if the grid spaces are odd, and B if the grid spaces are even.

All that happens if B goes first is that A and B are switched.

I think I may have forgotten a case, not sure if it changes anything.

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