@Modeling Expert:  Can you please explain your algo with an example...

especially calculating the area of rectangles using min depth and its
traversal through the matrix...

On Tue, Oct 5, 2010 at 7:24 PM, Modeling Expert <[email protected]
> wrote:

> For square matrix , link suggested by Amit would work. Finding a sub-
> rectangle is tougher.
>
> I would go this way; First scan from bottom right till top left and
> for every Non-zero member in matrix create this pair ( width ,
> depth ).
>
> Width is  how many continuous 1's ( including itself ) are there to
> its right,
> Depth is  how many continuous 1's ( including itself ) are there down
> wards
>
> now select a point P(i,j) , which has value Pair ( W , D ) it means
> W-1 columns to right and D-1 rows below are '1' , Note: it doesn't
> give any info about diagonal
>                                    entries to P , and we will see ,
> we actually don't need them.
>
>  Now think about all possible rectangles which could be made when P is
> left top corner
>  so go columnwise from i to  i + 1 , i +2 ,  i + (W-1) points ( call
> it P' point ) , at every step, see the minimum depth for all points
> between P and P' and keep calculating area by width X minDepth
>
>  e.g. P has width = 3, depth = 4 and co-ordinates (i,j ) then go like
> this
> Area for A[i,j] and  P'[i, j+1 ]  = 2 X MinDepth( P, P')   //A1
> Area for A[i,j] and  P"[i, j+2 ]  = 3 X MinDepth( P, P', P")   //A2  :
> Note MinDepth is Minimum Depth of all three points so far
>  Take Max of A1,A2 and you would have max rectangle which could be
> created by having P as top left corner.
>
> Now , keep traveling P throughout this Matrix.
>
> There are few optimisation. A point with width W and depth D can AT
> MAX has a rectangle with AREA MAX = W * D so next time consider points
> only whose W * D > Max Area calcuated so far.E.g if you have already
> got a point which has sub rectangle with area = 12 no need to look for
> points which has such (W,D) pairs = ( 2,3) , (4,2) , (3,3)  etc .
>
> Time complextiy :
> First time reverse traversal O(m * n )  /// m Rows , n Columns
>   Second time , MAX AREA calculation for each points , worse case
> o(n) as we need to travel only column wise as explained above
>   Above action to be done for m*n points so total O( mn * n ) =
> O(mn^2)
>
> So total Time = O ( m * n^2 )
>
> Time complexity :O ( 2 * m * n ) as W and D pair for each entry has to
> be maintained
>
>
> Suggestions, Clarifications, Modifications ?
> -Manish
>
>
>
>
>
>
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to [email protected].
> To unsubscribe from this group, send email to
> [email protected]<algogeeks%[email protected]>
> .
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>
>


-- 
Regards,
Ankit Babbar

-- 
You received this message because you are subscribed to the Google Groups 
"Algorithm Geeks" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to 
[email protected].
For more options, visit this group at 
http://groups.google.com/group/algogeeks?hl=en.

Reply via email to