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    <p>The accuracy of the Hough space should be dependent on two main factors.</p> <p>The accumulator maps onto Hough Space. To loop through the accumulator array requires that the accumulator divide the Hough Space into a discrete grid.</p> <p>The second factor in accuracy in Linear Hough Space is the location of the origin in the original image. Look for a moment at what happens if you do a sweep of \theta for any given change in \rho. Near the origin, one of these sweeps will cover far less pixels than a sweep out near the edges of the image. This has the consequence that near the edges of the image you need a much higher \rho \theta resolution in your accumulator to achieve the same level of accuracy when transforming back to Cartesian.</p> <p>The problem with increasing the resolution of course is that you will need more computational power and memory to increase it. Also If you uniformly increase the accumulator resolution you have wasted resolution near the origin where it is not needed.</p> <p>Some ideas to help with this.</p> <ol> <li>place the origin right at the center of the image. as opposed to using the natural bottom left or top left of an image in code.</li> <li>try using the closest image you can get to a square. the more elongated an image is for a given area the more pronounced the <em>resolution trap</em> becomes at the edges</li> <li>Try dividing your image into 4/9/16 etc different accumulators each with an origin in the center of that sub-image. It will require a little overhead to link the results of each accumulator together for rectangle detection, but it should help spread the resolution more evenly.</li> <li>The ultimate solution would be to increase the resolution linearly depending on the distance from the origin. this can be achieved using the</li> </ol> <hr> <pre><code> (x-a)^2 + (y-b)^2 = \rho^2 </code></pre> <hr> <pre><code>circle equation where - x,y are the current pixel - a,b are your chosen origin - \rho is the radius once the radius is known adjust your accumulator resolution accordingly. You will have to keep track of the center of each \rho \theta bin. for transforming back to Cartesian </code></pre>
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