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I'm trying to solve the following question about "Harris Corner Detection".

Consider the following image:

enter image description here

As the first step of the "Harris Corner Detection", we should compute the derivatives using the differentiation kernels shown above. No normalization (division by 2) is needed.

What I did: $$ I_{x}=\frac{\partial I(x,y)}{\partial x}=I(x+1,y)-I(x-1,y)=\begin{bmatrix}4 & 9 & 12\\ 8 & 11 & 14\\ 10 & 15 & 16 \end{bmatrix}-\begin{bmatrix}0 & 1 & 4\\ 0 & 5 & 7\\ 4 & 9 & 12 \end{bmatrix}=\begin{bmatrix}4 & 8 & 8\\ 8 & 6 & 7\\ 6 & 6 & 4 \end{bmatrix} $$ But the solution should be (source):

enter image description here

I'm a bit confused about spatial derivative calculation. I took the formula from here. What am I missing?

I quite don't get how to calculate: $$ M=\sum_{x,y}w(x,y)\begin{bmatrix}I_{x}^{2} & I_{x}I_{y}\\ I_{x}I_{y} & I_{y}^{2} \end{bmatrix} $$ I'm also a bit confused about terminology. Is $I_x$ actually that matrix or is it some value? function?

Is it possible to show how can I calculate $M$ using the following formula I provided above?

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    $\begingroup$ Your $I_x$ is actually moving in the $y$ axis (vertical axis) not the $x$ axis (horizontal axis). That will make the answer match the blue figures in the image. $\endgroup$
    – Peter K.
    Mar 16 at 17:41

1 Answer 1

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What you should do instead to calculate $I_x$:

$$ \begin{align} I_{x}(x,y)&=\frac{\partial I(x,y)}{\partial x}\\ &=I(x+1,y)-I(x-1,y)\\ &= \begin{bmatrix}5& 7 & 11\\ 9 & 12 & 16\\ 11 & 14 & 16 \end{bmatrix}-\begin{bmatrix}1 & 0 & 5\\ 1 & 4 & 9\\ 3 & 8 & 11 \end{bmatrix}\\ &=\begin{bmatrix} \bf\color{blue} 4 & \bf\color{blue}7 & \bf\color{blue}6\\ \bf\color{blue}8 & \bf\color{blue}8 & \bf\color{blue}7\\ \bf\color{blue}8 & \bf\color{blue}6 & \bf\color{blue}5 \end{bmatrix} \end{align} $$ which matches the expected output.

Checking on your source, it seems that $M$ is mis-transcribed:

Harris corner detector equation

Here, the $W$ is a window into the image, not a weighting as is written in the equation for $M$ in the original post. This just means that $x$ and $y$ are varied to capture the window and the sum of the four different items is computed.

To calculate this $H$, all you need is the $I_x$ above and the equivalent in the $y$ direction: $$ \begin{align} I_{y}(x,y)&=\frac{\partial I(x,y)}{\partial y}\\ &=I(x,y+1)-I(x,y+1)\\ \end{align} $$

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