hi guys
i guess it s like the sudoku prob.. for a matrix to be rearrangeable u can t hav any of d ones in d same row or even in d same column.. cause if u re arrange it.. then the diagonal wont b able to get the entry of 1.. hop i could b of any help
aakriti
On 3/18/06, kool_guy <[EMAIL PROTECTE
the matrix
1 1 1
1 1 1
1 1 1
has rank 1, but is still rearrangable, so it seems linear dependence
has not much to do with the problem.
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To po
SPX2 wrote:
> what complexity aj ?
O(ne) where e is the n is the number of rows and e is the number of 1's
in the matrix.
worst case O(n^3).
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Hi..frns...
here algorithms for this problem...
1st assume each row contains only 1.
store the position of 1's in each row in one dimensional array.
and use hash technique ..[ this is for to know whether any row is repeated or not ] if two numbers repeated
this is not rearrang
what complexity aj ?
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its about liniar independence
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hi,
I guess the problem is find a permutation of rows and columns of A,
such that resulting matrix has all diagonal entries as 1 (since any
permutation can be obtained by a sequence of swappings).
1) Column permutations are redundant: i.e.. if A is rearrangable then
A can be converted into d
I think the condition that there is one and only one 1 in every row and
every column is not necessary
since if we have a matrix like:
11 1
11 1
11 1
it is automatically rearrangeable. But I do not have an algorithm for
determining whether a matrix is
rearrangeable.
--
Vijay Venkat Raghavan N a écrit :
> hi,
>
> well am not really sure about this stuff, but i guess if theres a 1 in each
> row and each coloum initially, then it is rearrangeable. i am not able to
> think of a rigorous proof as of now.
>
> -Vijay
>
> On 3/18/06, kool_guy <[EMAIL PROTECTED]> wrote
hi,
well am not really sure about this stuff, but i guess if theres a 1 in
each row and each coloum initially, then it is rearrangeable. i am not
able to think of a rigorous proof as of now.
-VijayOn 3/18/06, kool_guy <[EMAIL PROTECTED]> wrote:
Let A be a "n by n" matrix. A is rearrangeable if t
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