| A distracting puzzle [message #26698] |
Mon, 17 September 2001 13:58  |
John-David T. Smith
Messages: 384
Registered: January 2000
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Senior Member |
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Given a polygon defined by the vertex coordinate vectors x & y, we've
seen that we can compute the indices of pixels roughly within that
polygon using polyfillv(). You can run the code attached to set-up a
framework for visualizing this. It shows a 10x10 pixel grid with an
overlain polygon by default, with pixels returned from polyfillv()
shaded.
You'll notice that polyfillv() considers only integer pixels, basically
truncating any fractional part of the input polygon vertices (you can
see this by plotting fix([x,x[0]]), etc.). For polygons on a fractional
grid, this error can be significant.
The problem posed consists of the following:
Expand on the idea of the polyfillv algorithm to calculate and return
those pixels for which *any* part of the pixel is contained within the
polygon, along with the fraction so enclosed.
For instance, the default polygon shown (invoked simply as
"poly_bounds"), would have a fraction about .5 for pixel 34, 1 for
pixels 33 & 43, and other values on the interval [0,1] for the others.
Return only those pixels with non-zero fractions, and retain polygon
vertices in fractional pixels (i.e. don't truncate like polyfillv()
does).
JD
pro poly_bounds,x,y,N=n
if n_elements(n) eq 0 then n=10
if n_elements(x) eq 0 then begin
x=[1.2,3,5.3,3.2] & y=[1.3,6.4,4.3,2.2]
endif
window,XSIZE=500,YSIZE=500
;; Set up the plot region, etc.
plot,[0],[0],XRANGE=[0.,n],YRANGE=[0.,n], XMINOR=-1,YMINOR=-1, $
XTICKS=n,YTICKS=n,POSITION=[.05,.05,.95,.95],TICKLEN=0,/NODA TA
p=polyfillv(x,y,n,n)
for i=0,n_elements(p)-1 do begin
xp=p[i] mod n
yp=p[i]/n
polyfill,[xp,xp,xp+1,xp+1],[yp,yp+1,yp+1,yp],COLOR=!D.N_COLO RS/2
endfor
oplot,[x,x[0]],[y,y[0]]
for i=0,n-1 do begin
plots,i,!Y.CRANGE
plots,!X.CRANGE,i
for j=0,n-1 do begin
plots,i+.5,j+.5,PSYM=3
xyouts,i+.1,j+.1,strtrim(i+j*n,2)
endfor
endfor
end
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| Re: A distracting puzzle [message #26727 is a reply to message #26698] |
Tue, 25 September 2001 09:54  |
John-David T. Smith
Messages: 384
Registered: January 2000
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Senior Member |
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Stein Vidar Hagfors Haugan wrote:
>
> If what's being sought here is only to distinguish which pixels have *some*
> area inside the polygon and which do not, wouldn't it be sufficient to check
> the corners? I.e., in a continuum of pixel coordinates, given corners with
> coordinates [0,0], [1,0], [1,1], [0,1], it can be checked whether each of
> those are inside versus outside any defined polygon. If one or more of the
> corners is inside, then some area is also inside..
>
> I have included some simple-minded routines I wrote some years ago to check
> whether a point is inside or outside a polygon...
Thanks Stein Vidar. Your method would seem to provide the answer for
the boolean question; however, my intent was to provide a list of pixels
which are at least partly inside the polygon, *along with* a list of
their fractional areas included. I came up with a solution I call
polyfillaa, which is a direct replacement for polyfillv.
inds=polyfillaa(x,y,sx,sy,AREAS=a)
returns the pixel indices, along with the clipping areas if desired. It
performs a straightforward form of polygon clipping. The "aa" is for
anti-aliasing, which is basically what it does. It works quite well,
but is very slow, thanks to a surplus of looping. In general it returns
more pixels than polyfillv, which neglects pixels with small areas
inside, and (erroneously, I feel) truncates polygon points to integer
pixels.
I may document it and put it up somewhere soon, but I'm embarrassed by
all the for loops. We'll see.
JD
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| Re: A distracting puzzle [message #26814 is a reply to message #26727] |
Wed, 26 September 2001 01:53 |
Martin Downing
Messages: 136
Registered: September 1998
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Senior Member |
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Ah go on JD, show your code - then the rest of us can decide whether we
could do better without reinventing the wheel!
Martin
"JD Smith" <jdsmith@astro.cornell.edu> wrote in message
news:3BB0B6D0.43C7859F@astro.cornell.edu...
> Stein Vidar Hagfors Haugan wrote:
>>
>> If what's being sought here is only to distinguish which pixels have
*some*
>> area inside the polygon and which do not, wouldn't it be sufficient to
check
>> the corners? I.e., in a continuum of pixel coordinates, given corners
with
>> coordinates [0,0], [1,0], [1,1], [0,1], it can be checked whether each
of
>> those are inside versus outside any defined polygon. If one or more of
the
>> corners is inside, then some area is also inside..
>>
>> I have included some simple-minded routines I wrote some years ago to
check
>> whether a point is inside or outside a polygon...
>
> Thanks Stein Vidar. Your method would seem to provide the answer for
> the boolean question; however, my intent was to provide a list of pixels
> which are at least partly inside the polygon, *along with* a list of
> their fractional areas included. I came up with a solution I call
> polyfillaa, which is a direct replacement for polyfillv.
>
> inds=polyfillaa(x,y,sx,sy,AREAS=a)
>
> returns the pixel indices, along with the clipping areas if desired. It
> performs a straightforward form of polygon clipping. The "aa" is for
> anti-aliasing, which is basically what it does. It works quite well,
> but is very slow, thanks to a surplus of looping. In general it returns
> more pixels than polyfillv, which neglects pixels with small areas
> inside, and (erroneously, I feel) truncates polygon points to integer
> pixels.
>
> I may document it and put it up somewhere soon, but I'm embarrassed by
> all the for loops. We'll see.
>
> JD
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