@Don: what is the point of considering s=2 when we have already found s=3.As question says find "the maximum subsquare". Which is of size 3 and this the expected outcome.
On Mon, Mar 31, 2014 at 11:28 PM, Don <[email protected]> wrote: > 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]> 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]. >>> >> >> -- > 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]. > -- 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].
