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POFinding an Image Inside Another Image
 

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    <p>I'm trying to build an application that solves a puzzle (trying to develop a graph algorithm), and I don't want to enter sample input by hand all the time.</p> <p><strong>Edit:</strong> I'm not trying to build a game. I'm trying to build an agent which plays the game "SpellSeeker"</p> <p>Say I have an image (see attachment) on the screen with numbers in it, and I know the locations of the boxes, and I have the exact images for these numbers. What I want to do is simply tell which image (number) is on the corresponding box. </p> <p><a href="http://i46.tinypic.com/3089vyt.jpg">Numbers http://i46.tinypic.com/3089vyt.jpg</a></p> <p>So I guess I need to implement </p> <p><code>bool isImageInsideImage(Bitmap numberImage,Bitmap Portion_Of_ScreenCap)</code> or something like that.</p> <p>What I've tried is (using <a href="http://code.google.com/p/aforge/">AForge</a> libraries)</p> <pre><code>public static bool Contains(this Bitmap template, Bitmap bmp) { const Int32 divisor = 4; const Int32 epsilon = 10; ExhaustiveTemplateMatching etm = new ExhaustiveTemplateMatching(0.9f); TemplateMatch[] tm = etm.ProcessImage( new ResizeNearestNeighbor(template.Width / divisor, template.Height / divisor).Apply(template), new ResizeNearestNeighbor(bmp.Width / divisor, bmp.Height / divisor).Apply(bmp) ); if (tm.Length == 1) { Rectangle tempRect = tm[0].Rectangle; if (Math.Abs(bmp.Width / divisor - tempRect.Width) &lt; epsilon &amp;&amp; Math.Abs(bmp.Height / divisor - tempRect.Height) &lt; epsilon) { return true; } } return false; } </code></pre> <p>But it returns false when searching for a black dot in this image.</p> <p>How can I implement this?</p>
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    1. COSome sudoku or tic tac toe game?
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    2. COHave you seen other questions on the [similar ground](http://stackoverflow.com/questions/2472467/how-to-find-one-image-inside-of-another) ?
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    3. COIt's a game called "spellseeker". But it's not important I just want to build the algorihm to solve such a problem. And yes, actually I found this solution in other questions, but they didn't help me. And forgive me for my lazyness but I really think that there should be an easier way for this :)
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