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

Hm. Sorry to have wasted your time. I'd picked up on this bit of the OP

... since A and E are restricted to only two values between them, they must consume those two values;

... and hung my hat on it, but that obviously doesn't apply in the same way to the two examples above.

Question: Would this example also decompose into the (same?) two groups as the above?

my @grid = ( [ 0, 1, 0, 1, 0, 0, ], [ 0, 0, 0, 1, 1, 0, ], [ 0, 1, 0, 0, 1, 0, ], [ 1, 0, 0, 0, 0, 1, ], [ 1, 0, 1, 0, 0, 0, ], [ 1, 0, 0, 0, 0, 1, ], );

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

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Re^4: decomposing binary matrices
by hv (Parson) on Feb 16, 2007 at 16:06 UTC

    Yes: labelling the columns A..F and the rows 1..6, we know variables {B, D, E} must consume values {1, 2, 3} between them, and likewise {A, C, F} must consume {4, 5, 6}. In this example, however, the latter can be further decomposed: C can only be 5, so {A, F] are left with {4, 6}.

    Hugo