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ThudnBlunder
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 Anti-magic Matrix   « on: Aug 29th, 2004, 1:51pm » Quote Modify

For which values of n does there exist an nxn matrix A such that:

1) All entries of A are from {-1,0,1}
2) The row sums and the column sums are all pairwise distinct?

 « Last Edit: Aug 29th, 2004, 8:46pm by ThudnBlunder » IP Logged

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Grimbal
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 Re: Anti-magic Matrix   « Reply #1 on: Aug 29th, 2004, 3:55pm » Quote Modify

can a row sum equal a column sum?

probably not, or it would be too easy.
 « Last Edit: Aug 29th, 2004, 3:57pm by Grimbal » IP Logged
ThudnBlunder
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 Re: Anti-magic Matrix   « Reply #2 on: Aug 29th, 2004, 8:46pm » Quote Modify

on Aug 29th, 2004, 3:55pm, Grimbal wrote:
 can a row sum equal a column sum?

No, there are 2n different row and column sums.

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Grimbal
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 Re: Anti-magic Matrix   « Reply #3 on: Aug 30th, 2004, 4:26am » Quote Modify

For an even n, I found a pattern:
::
+ + + + + +
+ + + + 0 -
+ + + + - -
+ + 0 - - -
+ + - - - -
0 - - - - -
::
 « Last Edit: Aug 30th, 2004, 4:26am by Grimbal » IP Logged
Aryabhatta
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 Re: Anti-magic Matrix   « Reply #4 on: Aug 31st, 2004, 10:57am » Quote Modify

Only for even n can we have anti-magic squares:

Let p be the missing sum. Then it is clear the the total of the entries of the matrix is -p/2 and p must be even.

We can assume p [le] 0. So the sums 1,2,...n appear in the matrix.

Now consider the rows and columns whose sums are 1,2,...n.
rearrange these rows and columns so that the rows are pushed to left and columns to the top.

Suppose there are r rows and c columns.

The matrix looks like
Code:
 [        ][               ] [ B1     ][      B2       ] --------------------- [         ][               ] [         ][               ]       [  B3     ][      B4       ] [         ][               ]

B1 is the intersection of the r rows (B1 and B3) and c columns (B1 and B2).

So we have that sum of the r rows + c columns
= n(n+1)/2.
Let |Bi| be the sum of entries of block Bi.
Then we have

n(n+1)/2 - |B1| + |B4| = total = -p/2

Now r + c = n. So B1 and B4 are blocks of size rc.
therefore we have that

-p/2 = n(n+1)/2 - |B1| + |B4| [ge] n(n+1)/2 - 2rc
Now rc [le] (r+c)[sup2]/4= n[sup2]/4.
So we have
-p/2 [ge] n(n+1)/2 - n[sup2]/2 = n/2.

This means that p [le] -n.
Which is possible only if p = -n. (as p is one of -n,...,0,1,...n)
Since p is even, we must have that n is even too.

Grimbal's pattern completes the proof that for even n it is possible to do it.

 « Last Edit: Aug 31st, 2004, 11:00am by Aryabhatta » IP Logged
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