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    <p>Lets say you have two filters <code>F1</code> and <code>F2</code>, and an image <code>I</code>. If you pass your image through the two filters, you would get a response that was defined as</p> <pre><code>X = ((I * F1) * F2) </code></pre> <p>Where here I am using <code>*</code> to represent <a href="https://en.wikipedia.org/wiki/Convolution" rel="nofollow noreferrer">convolution</a>.</p> <p>By the associative rule of convolution, this is the same as.</p> <pre><code>X = (I * (F1 * F2)) </code></pre> <p>using commutativity, we can say that </p> <pre><code>X = (I * (F2 * F1)) = ((I * F2) * F1) </code></pre> <p>Of course, this is in the nice continuous domain of math, doing these things on a machine means there will be rounding errors and some data may be lost. You also should think about if your filters are <a href="https://en.wikipedia.org/wiki/Finite_impulse_response" rel="nofollow noreferrer">FIR</a>, otherwise the whole concept of thinking about digital filtering as convolution sorta starts to break down as your filter can't really behave the way you wanted it to.</p> <hr> <p><strong>EDIT</strong></p> <p>The discrete convolution is defined as</p> <p><img src="https://i.stack.imgur.com/qTGZD.png" alt="conv2 uses a straightforward formal implementation of the two-dimensional convolution equation in spatial form"></p> <p>so adding zeros at the edges of you data doesn't change anything in a mathematical sense.</p> <p>As some people have pointed out, you will get different answers numerically, but this is expected whenever we deal with computing actual data. These variations should be small and limited to the low energy components of the output of the convolution (i.e: the edges).</p> <p>It is also important to consider how the convolution operation is working. Convolving two sets of data of length <code>X</code> and length <code>Y</code> will result in an answer that is <code>X+Y-1</code> in length. There is some behind the scenes magic going on for programs like MATLAB and Mathematica to give you an answer that is of length <code>X</code> or <code>Y</code>.</p> <p>So in regards to @belisarius' post, it would seem we are really saying the same thing.</p>
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    1. POaveraging mask and laplacian mask in image processing
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    1. COnicely done thanks!
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    2. COApplying image filters usually require some padding (particularly set for each filter, depending on the values on the image boundary). And that breaks the commutation property.
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      1. USDr. belisarius
    3. CO@belisarius I am not sure I am following how padding (zero or otherwise) would break the commutative property for convolution. Can you provide a proof?
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