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Topic: Antimagic Matrix (Read 1273 times) 

ThudnBlunder
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Antimagic Matrix
« on: Aug 29^{th}, 2004, 1:51pm » 
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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 29^{th}, 2004, 8:46pm by ThudnBlunder » 
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Grimbal
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Re: Antimagic Matrix
« Reply #1 on: Aug 29^{th}, 2004, 3:55pm » 
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can a row sum equal a column sum? probably not, or it would be too easy.

« Last Edit: Aug 29^{th}, 2004, 3:57pm by Grimbal » 
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ThudnBlunder
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Re: Antimagic Matrix
« Reply #2 on: Aug 29^{th}, 2004, 8:46pm » 
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on Aug 29^{th}, 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: Antimagic Matrix
« Reply #3 on: Aug 30^{th}, 2004, 4:26am » 
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For an even n, I found a pattern: :: + + + + + + + + + + 0  + + + +   + + 0    + +     0      ::

« Last Edit: Aug 30^{th}, 2004, 4:26am by Grimbal » 
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Aryabhatta
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Re: Antimagic Matrix
« Reply #4 on: Aug 31^{st}, 2004, 10:57am » 
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Only for even n can we have antimagic 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 31^{st}, 2004, 11:00am by Aryabhatta » 
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