A spacefilling curve is also known as a quadtree-representation of a
plane. It's subdivide a plane in continuous 4 cells. My linear
implementations gives at each step one leaf. What do you mean with
unweigthed und BFS (Breadth-First-Search, perhaps?). How can you add 8
vertices to the queue? It makes only sense to add 4 to queue with a
recursive approach!? A chessboard can be viewed as a graph of  2^n -
(n-1) edges. Sorry, I don't know the solution.

On Aug 25, 6:11 am, SHRISH MISHRA <[email protected]> wrote:
> The question is asking for a shortest path of some sort. If we consider each
> cell as a vertex, then the edges will be the possible knight moves. For
> example, the cell (2,3) would have an edge to (and from) the cell (4,4),
> assuming that there are no pieces occupying either cell. This graph is
> unweighted, so the most natural algorithm to use is BFS. Having realized
> this, the graph itself is not necessary anymore; we can just use the board
> directly. S is the source, and is the first cell/vertex added to the BFS
> queue. At each step, at most 8 more vertices will be added to the queue. BFS
> will stop when T is found or when the queue is empty (no path to T was
> found).
>
> Great Minds Discuss Ideas
> Average Minds Discuss Events
> Small Minds Discuss People
> Shrish Chandra Mishra
> CSE NIT,Allahabad
>
>
>
> On Wed, Aug 25, 2010 at 8:25 AM, Chi <[email protected]> wrote:
> > This is a spacefilling-curve. Optimal solution is a Hilbert-Curve or a
> > Morton-Curve. I have written one:
> >http://sourceforge.net/projects/hilbert-curve/files.
>
> > On Aug 25, 4:27 am, Algo geek <[email protected]> wrote:
> > > find the minimum no of moves by which a KNIGHT reaches the destination
> > > from source in a 8x8 chess borad...
>
> > > **you will be given the starting position and ending position ..
>
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