[Guest]

you are playing a game with your friend on a 7x7 chess board.

you begin first and put a "0"(zero) on any square on the board, then your friend puts a "1"(one);

next you put zero, then your firend puts one...

game is going on so until all squares are marked with zero's and one's.

afterthat sum of the numbers in each row and column are notted;

i)odd numbers of those summed numbers go to you,

ii)even numbers of them go to your friend,

you and your friend add your numbers,the one whose collected number is higher than the others wins the game.

Question:

assuming that you and your friend play the game without any mistake/fault(i.e. both are expert on this game),

what can be the maximum number that you can get?

[Guest]

I think that this question is beautifically succinct, and just now I am at a loss as to how to approach an answer.

[Guest]

It all comes down to how your friend plays the game, you have no say in the outcome. as long as he places even number of 1's he gets the point. Since there are 49 squares and you go first that means you have 25 spots and he has 24. Whoever gets to place the 1's will be the winner every time.

The maximun number you can hope for is 1.

[Guest]

'lets start with the 3 by 3 case.

for the first move, no matter where 0 plays, 

it removes his options for the next turn by 4, 

as if he plays on the same row or column, 

1 will get that row on the turn after. 

therefore it really doesn't matter where 0 starts.

let's say he starts in the center.

   |   |

---|---|---

   | 0 |

---|---|---

   |   | 

1 has the same problem, wherever he plays, 

on the next turn he can't play in the same row/column,

otherwise 0 will get that row. however, 

he can definitively limit both players

by playing on the same row or column as 0. 

the question is, does he want to do so? lets try and see.

   |   |

---|---|---

   | 0 | 1

---|---|---

   |   | 

0 now has basically 2 choices, the 2 corners opposite 1, or the two corners same 1.

let's try same 1 first.

   |   | 0

---|---|---

   | 0 | 1

---|---|---

   |   | 

now 0 for the next move has an interesting one, 

he can give both players a point by playing top middle.

1 can prevent this of course by playing either on that spot, 

top left corner, or bottom side, but why do so?

   |   | 0

---|---|---

   | 0 | 1

---|---|---

 1 |   | 

this leaves 0 with no good options. 

while he certainly can take point, 

this would give 1 two points. no other move however is much better.

 1 | 0 | 0    1 wins 4 to 2

---|---|---

 0 | 0 | 1

---|---|---

 1 | 1 | 0


lets back track. is the other choice for 0 any better?

 0 |   |

---|---|---

   | 0 | 1

---|---|---

   |   | 


here, now 1 has the option of giving both players a point.

let's say he does so.

 0 | 1 | 0

---|---|---

   | 0 | 1

---|---|---

   |   | 1


once again, 0 has no good move.

 0 | 1 | 0  1 wins 4 to 2

---|---|---

 0 | 0 | 1

---|---|---

 1 | 0 | 1

it seems nothing for 0 wins, as long as 1 makes no obvious mistakes.

for the 5x5 and 7x7 you have an odd number of rows, an odd number of columns, and an even total.

both players want their total on a particular row/column to be odd.

1 by going second has the advantage. he can chose to have either his 1's to be even/even or odd/odd.

0 must have even/odd. therefore 1 is guaranteed at least 1 point right off the bat.

basically all 1 has to do is play knight's moves and he wins.

Edited November 10, 2009 by phillip1882

[Guest]

I think the question is clear enough.

assume your friend plays the game to get maximum points as you play so.

but remember that you first begin the play

kkehoe5;

total number is 48,

if you get 1 rest is 47?

47 can be obtained from even numbers?

[Guest]

I think the question is clear enough.

assume your friend plays the game to get maximum points as you play so.

but remember that you first begin the play

kkehoe5;

total number is 48,

if you get 1 rest is 47?

47 can be obtained from even numbers?

Sorry I read it as you got a point if you won the row/column not the actual sum.