>I can't say that I understood this reasoning. My problem is
>this: I do not see how, in the last paragraph, you
>distinguish betweeb EW and NS directions. The impression is
>that what applies to one, applies to the other. It's not the
>matter of "vertical or horizontal", but rather of "vertical
>and horizontal".

That's because I forgot to put in a crucial step. Earlier I had already distinguished between the cases max(B)>0 and max(B)=0, and in a similar way I wanted to distinguish between the cases where the net length L = (white length - black length) on the upper and lower sides was L>0 or L=0. In the first case, the grid crossing argument applies and forces the rectangular height to be a multiple of 2 squares. In the second case, there is after all no sign change as the top and bottom sides of the rectangle sweep across a gridline, so the height need not be a multiple of 2. However, in that case, L=0 forces the rectangle width to be a multiple of 2.

There is something strangely asymmetrical about this reasoning, in that the cases L>0 and L=0 are treated by quite different logic, and yet produce a conclusion that is symmetrical upon interchange of vertical and horizontal. It makes me feel that there must be a smoother approach, because (symmetrical problem statement) --> (asymmetrical reasoning) --> (symmetrical conclusion) just doesn't FEEL right to me.

Also, now that you have brought it to my attention, the fact that the same dichotomy is used on the whole rectangle and on its edges seems to invite generalization to n-dimensional boxes on n-dimensional chessboards, perhaps proceeding by induction on the dimension. I'll think about that one some more later.

Sorry for the initial lack of clarity; I had thought through this stage of the argument, but somehow overlooked actually writing it down. Thank you for not being telepathic, and thereby helping me to improve my expository skill.

--Stuart Anderson

Of course you are right. I don't seem to be doing too well with the details lately . . . perhaps I should get in the habit of thinking more and posting less. Most embarrassing.

Fortunately, this does not affect the main thread of the argument, which really only depends on the sign reversals, not on the particular value of the maximum. You do raise an interesting question though: for what class of shapes is it true that there is a uniform bound on the net area? My hunch is that convex shapes will turn out to have a maximum area that never exceeds 1.0. I'll see if I can come up with a concise proof, unless you happen to know of one already, in which case I will only bother about it if I think the proof can be improved in some way.

Thank you for pointing out my mistake!

--Stuart Anderson

If the length of the diagonal is 2n + 1, then the black-white area balance is n, so it may be made arbitrarily large. For example, the square shown above has an area balance of 9.

Thankyou

sfwc
<><

Thankyou,

sfwc
<><

Well, that shoots down my hunch, but this outcome is actually more interesting (to me, at least), since in the context of convolution it'shows that some methods of smoothing are markedly superior to others on a given grid. Using rectangles of the same area, one aligned with the grid and one tilted 45 degrees, convolving with the aligned rectangle would produce much lower amplitude of periodic variation. Hence it is a "better" convolution smoother than the other: same area implies roughly the same computational cost, but the amplitude of variation is much smaller.

This gives me an applicaton idea (which perhaps has already been implemented; I'll have to check): this means that there is a nearly ideal method of removing half-tone dots from photographs (supposing that one had a newsprint or magazine phote which was scanned into a graphic application). Using a very small rectangle (a 1 by 2 square would be minimal I think), half-tone could be removed while leaving the large scale features of the picture virtually unchanged.

Of course, half-tone is usually in a hex grid, but this grid has (ignoring orientation) an equilateral triangle for its unit cell and so a small rhombus should provide an optimal covolution filter for removing half-tone.

Perhaps this was obvious anyway, but I came upon the idea unexpectedly in this context. Just a lot of speculation for now. I'll see if anything comes of it.

--Stuart Anderson