@Gene
As i said in my earlier post right to left diagonal partitions the martix
into 2 equal number of elements. So now the median must be in this
diagonal. Now our focus is on finding median of this diagonal only.
I think this works fine. Can u give some test case for which it fails?
On Sun, Nov 6, 2011 at 3:02 AM, Gene <gene.ress...@gmail.com> wrote:
> Unfortunately this isn't true. See the example I gave earlier:
>
> 1 2 3
> 2 4 5
> 3 4 6
>
> Thje median is 3.
>
> 1 2 2 3 >3< 4 4 5 6
>
> Niether one of the 3's lies on the diagonal.
>
> When you pick any element P on the diagonal, all you know is that
> anything to the right and downward is no less than P and everything to
> the left and upward is no greater. This leaves the upper right and
> lower left rectangles of the matrix unrelated to P.
>
> On Nov 5, 3:51 am, ankit agarwal <ankit.agarwal.n...@gmail.com> wrote:
> > Hi,
> >
> > I think the median will always lie on the diagonal
> > a[n][1] ---- a[1][n]
> > because the elements on the LHS making the upper triangle will
> > always be less than or equal to the elements on the diagonal
> > and the RHS, elements in the lower triangle will be greater than or
> > equal to them.
> >
> > so sort the diagonal and find the middle element, that will be the
> > median.
> >
> > Thanks
> > Ankit Agarwal
> >
> > On Nov 5, 1:29 am, Gene <gene.ress...@gmail.com> wrote:
> >
> >
> >
> > > Here's an idea. Say we pick any element P in the 2D array A and use
> > > it to fill in an N element array X as follows.
> >
> > > j = N;
> > > for i = 1 to N do
> > > while A(i, j) > P do
> > > j = j - 1;
> > > end;
> > > X(i) = j;
> > > end;
> >
> > > This algorithm needs O(N) time.
> >
> > > The elements of X split each row with respect to P. That is, for each
> > > i = 1 to N,
> >
> > > A(i, j) <= P if 0 < j <= X(i), A(i,j) > P if X(i) < j <= N.
> >
> > > Now the strategy is to create two length N arrays a = [0,0,...0]; and
> > > b = [N,N,...]. We'll maintain the invariant that a[i] < Median <= b[i]
> > > for some i. I.e, they "bracket" the median.
> >
> > > We define functions L(a) = sum_i( a(i) ) and R(b) = sum_i( N -
> > > b(i) ). These tell us how many elements there are left and right of
> > > the bracket.
> >
> > > Now reduce the bracket as in binary search: Guess a value P, compute
> > > X. If L(X) >= R(X), set b = X else set a = X.
> >
> > > Keep guessing new P values in a way that ensures we reduce the number
> > > of elements between a and b by some fixed fraction. If we can do
> > > that, we'll get to 1 element in O(N log N) time.
> >
> > > The remaining problem is picking good P's. Certainly the first time is
> > > easy. Just take A(N/2, N/2). This has approximately (at least) N^2/4
> > > elements larger than it and N^2/4 smaller due to the sorted rows and
> > > columns. This is what we need to get O(N log N) performance.
> >
> > > But after the first split, things get trickier. The area between a and
> > > b takes on the shape of a slash / /, so you can't just pick a P that
> > > moves a and b together by a fixed fraction of remaining elements.
> >
> > > Not to worry! You can quickly look up the (at most) N row medians in
> > > the bracket, i.e.
> >
> > > { A(i, (a[i] + b[i] + 1) / 2) | a[i]<b[i] , i = 1 to N }
> >
> > > and use the well known O(N) median selection algorithm to get a median
> > > of this. This has the quality we want of being somewhere roughly in
> > > the middle half of the remaining elements. The logic is the same as
> > > the selection algorithm itself, but in our case the rows are pre-
> > > sorted.
> >
> > > In all, each partitioning step requires O(N), and a fixed fraction
> > > (about 1/2) of the elements will be eliminated from the bracket with
> > > each step. Thus O(log n) steps will be needed to bring the bracket to
> > > size 1 for an overall cost of O(N log N).
> >
> > > I don't doubt that there's a simpler way, but this one seems to work.
> > > Anyone see problems?
> >
> > > On Nov 3, 3:41 pm, sravanreddy001 <sravanreddy...@gmail.com> wrote:
> >
> > > > any better solution than O(N^2) in worst case?
> > > > How do we take advantage of sorting and find in O(N lg N)- Hide
> quoted text -
> >
> > - Show quoted text -
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> To unsubscribe from this group, send email to
> algogeeks+unsubscr...@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>
>
--
Mohit
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
