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I found a paragraph in the book about $SSD$, can't get one thing:

Most commonly, the distance measure is the sum of squared differences. For two images $f(x, y)$ and $g (x, y)$ it is defined as $$ SSD(d_1,d_2) = \sum_{i=-n_1}^{n_1} \sum_{j=-n_2}^{n_2} \big(f(x+i,\,y+j)-g(x+i-d_1,\,y+j-d_2)\big)^2 $$

where the summation extends over the region of size $(2n_1 + 1) \times (2n_2 + 1)$.

I can not get, why does $i$ changes from $-n_1$ to $n_1$, but not from $0$ to $n_1$. The similar about $j$. And why does summation goes over the $(2n_1 + 1) \times (2n_2 + 1)$ but not over the $(n_1) \times (n_2)$

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(0, 0) represents the pixel in question. In order to go both positive and negative in both axes it is required to use both strictly positive (e.g. n<sub>1</sub>) and strictly negative (e.g. -n<sub>1</sub>) offsets. –  Ignacio Vazquez-Abrams Jun 21 '12 at 19:06
But in Matrix, there are no negative offsets oO –  user1448906 Jun 21 '12 at 19:17
As defined, SSD(d1,d2) appears to be an image itself for each d1 and d2: It depends on x and y. This might be correct, but I am unsure. –  willem Jun 21 '12 at 19:26
It seems that $(i, j)$ represents the offset from the point $(x, y)$. So, for example, $(i, j) = (0, 0)$ corresponds to the point $(x, y)$ itself. The collection of $(i, j)$ that you sum over forms a $2$-dimensional array (matrix if you like) that happens to be indexed where $(0, 0)$ is in the middle. This is okay and makes a lot of sense, given the context. You can define $k = i + n_1 + 1$ and see that it ranges from $1$ to $2n_1 + 1$ if you insist. Analogously, for $\ell = j + n_2 + 1$. –  Sammy Black Dec 13 '13 at 22:01
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