000000 011110 010100 011100 010000 000000 In this case, when i and j are 1, your algorithm will set s = 3. The if statement will fail, and it will never notice that it could have formed a square with s=2.
Don On Sunday, March 30, 2014 9:49:21 AM UTC-4, atul007 wrote: > > @don : According to question we want to find "the maximum subsquare". > can you give me test case for which this wont work? > > > > On Fri, Mar 28, 2014 at 9:41 PM, Don <[email protected] <javascript:>>wrote: > >> 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]> 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] <javascript:>. >> > > -- 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].
