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DATE 2017-03-01

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MESSAGE
DATE 2017-03-27
FROM Ruben Safir
SUBJECT Subject: [Learn] Fwd: Re: hough transform - Lecture 9
From learn-bounces-at-nylxs.com Mon Mar 27 14:09:02 2017
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------

Hello Ruben,
Wow, notes from 2005. I haven=E2=80=99t looked at them for several years. I
retired in 2014.

The algorithm is OK. Note the comment at the top of the slide which says
to divide the parameter space into a set of discrete points (ai, bi).
These will be the points that are involved in the score-keeping and one
of them will be the winner. For example, you might divide the parameter
space into something like an mxn grid. The algorithm has the job of
picking which one of this set is the winner.

In the algorithm the value of b is only temporary. It is a number that
matches ai, but (ai, b) is not likely to be one in the score-keeping
set. You need to find the (ai, bi) that is closest to (ai, b) and give
its count an increment. This gives some tolerance for =E2=80=9Cnoise=E2=80=
=9D in the
values of the (xk, yk) points.

When all the counting is done, pick the (ai, bi) that has the largest count.

There are lots of design trade-offs for a particular problem domain, and
they are related to the how closely you want to quantize the parameter
space. More (ai, bi) points gives you better resolution but makes the
algorithm slower. Also, it is necessary to worry about practical
problems like line gaps, etc.

Best wishes,
Prof Rhody

> On Mar 20, 2017, at 5:43 PM, Ruben Safir wrote:
> =

> Good afternoom Dr Rhody
> =

> I have been studying your wonderful notes on the Hough Transform. They ar=
e the best review of this material that I've seen, but there is one area th=
at has me befuddled. Perhaps there is an error in this part of the notes. =
specifically I am referring to the 9th lecture notes
> =

> https://www.cis.rit.edu/class/simg782/lectures/lecture_09/lec782_05_09.pdf
> =

> Create a set of counters, one for each parameter pair, initialized to zer=
o.
> For each point (x n , y n )
> For each a i compute b =3D =E2=88=92a i x n + y n
> Find the point (a i , b j ) closest to (a i , b)
> Increment counter for (a i , b j )
> =

> I'm struggling to understand how this can work. Once we have a valid a,b=
then we need increment it and walk to the next x,y . After all the X,Y's a=
re covered, the a,b with the largest incremented value is the valid line(s)
> =

> =

> Please let me know how I am wrong in this
> =

> Sincerely
> Ruben Safir MS Comp Sci (candidate 2017)
> =


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