No, that is not the same.
It won't find a square smaller than s.
Don
On Thursday, March 27, 2014 2:56:29 AM UTC-4, atul007 wrote:
>
> @Don : your algo time complexity is O(n^2)
>
> It can be reduced to this :-
>
> // Now look for largest square with top left corner at (i,j)
> for(i = 0; i < n; ++i)
> for(j = 0; j < n; ++j)
> {
> s = Min(countRight[i][j], countDown[i][j]);
> if (countRight[i][j] && countDown[i][j] &&
> (countRight[i+s][j] >= s) && (countDown[i][j+s] >= s) && s>max)
> {
> bestI = i; bestJ = j; max = s;
> }
> }
>
> On 1/25/13, Don <[email protected] <javascript:>> wrote:
> > The worst case I know of is when the matrix is solid black except for
> > the lower right quadrant. In this case, it does break down into O(n^3)
> > runtime. It took about 8 times as long to run n=4000 as it took for
> > n=2000.
> > Don
> >
> > On Jan 24, 10:29 am, Don <[email protected]> wrote:
> >> I'm not sure I understand your case. However, I stated that there are
> >> cases where it is worse than O(N^2). The runtime is highly dependent
> >> on the contents of the matrix. In many cases it takes fewer than N^2
> >> iterations. Occasionally it takes more. On average it seems to be
> >> roughly O(N^2), but again that depends a lot on what is in the matrix.
> >> I got that result by trying different ways of filling the matrix. I
> >> tried things like randomly setting each pixel with various
> >> probabilities, placing random horizontal and vertical segments,
> >> placing random squares, or placing random filled squares. I did all of
> >> those both in black on white and white on black. In all of those
> >> cases, going from n=1000 to n=2000 resulted in a runtime increase of
> >> less than a factor of 4.
> >>
> >> Don
> >>
> >> On Jan 23, 10:33 pm, bharat b <[email protected]> wrote:
> >>
> >>
> >>
> >>
> >>
> >>
> >>
> >> > @Don: the solution is very nice.. But, how can u prove that it is
> >> > O(n^2)..
> >> > for me it seems to be O(n^3) ..
> >>
> >> > Ex: nxn matrix .. all 1s from (n/2,0) to (n/2,n/2).
> >> > all 1s from (n/2,0) to (n,0).
> >>
> >> > On Thu, Jan 17, 2013 at 9:28 PM, Don <[email protected]> wrote:
> >> > > The downside is that it uses a bunch of extra space.
> >> > > The upside is that it is pretty fast. It only does the
> time-consuming
> >> > > task of scanning the matrix for contiguous pixels once, it only
> >> > > searches for squares larger than what it has already found, and it
> >> > > doesn't look in places where such squares could not be. In practice
> >> > > it
> >> > > performs at O(n^2) or better for most inputs I tried. But if you
> are
> >> > > devious you can come up with an input which takes longer.
> >> > > Don
> >>
> >> > > On Jan 17, 10:14 am, marti <[email protected]> wrote:
> >> > > > awesome solution Don . Thanks.
> >>
> >> > > > On Thursday, January 17, 2013 12:38:35 AM UTC+5:30, marti wrote:
> >>
> >> > > > > Imagine there is a square matrix with n x n cells. Each cell is
> >> > > > > either
> >> > > > > filled with a black pixel or a white pixel. Design an algorithm
> >> > > > > to
> >> > > find the
> >> > > > > maximum subsquare such that all four borders are filled with
> >> > > > > black
> >> > > pixels;
> >> > > > > optimize the algorithm as much as possible
> >>
> >> > > --
> >
> > --
> >
> >
> >
>
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
