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Re^3: decomposing binary matrices

by BrowserUk (Pope)
on Feb 16, 2007 at 23:11 UTC ( #600542=note: print w/replies, xml ) Need Help??


in reply to Re^2: decomposing binary matrices
in thread decomposing binary matrices

I just ran that example through the code I posted (Note: My code uses letters for columns not rows as you have). The groupings produced:

23415 0000E 000BB 23415 AAAAA CCC0C DDD0D

gives rows 1 & 5 (your A & E), as a group which when subtracted from the remainer leaves rows 2, 3 & 4 (your {BCD}) as a group which is what you are after?

Full output from the unchanged program (except for inserting the new data (first line below)):

# my @grid = ( "AB0001","A0CD02","A0CD03","A0CD04","ABCDE5" ); c:\test>600418-2.pl This input AB0001 A0CD02 A0CD03 A0CD04 ABCDE5 sorted A0CD02 A0CD03 A0CD04 AB0001 ABCDE5 Tranformed looks like this AAAAA 000BB CCC0C DDD0D 0000E 23415 sorted 0000E 000BB 23415 AAAAA CCC0C DDD0D These are the sets where the letters denote the columns of the origina +l matrix (0 mean column not used in this set). And the numbers above, the rows they came from. 23415 0000E 000BB 23415 AAAAA CCC0C DDD0D

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
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Re^4: decomposing binary matrices
by hv (Parson) on Feb 17, 2007 at 13:11 UTC

    Sorry, I transformed the data already, and included more than necessary. Try this instead (in which, again, {B, C, D} yield a separable submatrix):

    my @grid = ( "ABCD1", "A0002", "0BCD3", "0BCD4");

    Hugo

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