I have developed a system to trace the outlines of (images of) objects. Now I want to test whether two independent traces represent a common feature.
Imagine two people (or machines) tracing the outline of a feature in an image, recording it as sequence of vertices. Inaccuracies in recognizing the feature boundaries and in specifying the vertices can be viewed as random errors in vertex positions. The problem is the two traces might use completely different (Cartesian) coordinate systems (set up on two digitizing tablets, for instance). The null hypothesis to test is that they represent a common feature.
This is illustrated below. I drew a figure and recorded the $x$ and $y$ co-ordinates of its vertices in its coordinate system. Let's call this figure $M$. It is represented as a sequence $(x_i,y_i), i=1, 2, \ldots, m$.
Then I drew the same figure in a bigger size in another coordinate system (with no known relationship to the first coordinates system) and recorded the $x$ and $y$ co-ordinates. Let's call this $N$, represented as a sequence $(x_i^\prime, y_i^\prime), i=1, 2, \ldots, n$.
Question: Having these data, how can I test whether the figures $M$ and $N$ represent the same image features even though they have different sizes and co-ordinates?
If the conclusion is that yes, they do represent a common object, then how can I estimate a similarity transformation between $M$ and $N$ so that I can work on formulas or equations to check the results with different figures?
$M$ and $N$ both seem to be cirles with the center $\vec{c}=(7,9)^T$. One has a radius of $1$, the other has a radius of $3$.
Thus, you can translate both sets by $(7,9)^T$ to get two circles which are centered at the origin $M' = M - \vec{c}$ and $N' = N -\vec{c}$. If you take a point in $M'$ and stretch it by $3$, then you end up in $N'$, and you have $$N'=3M'$$
or $$(N - (7,9)^T) = 3(M - (7,9)^T),$$
where it's worth noting that you can't expand the brackets.
Edit: the corresponding equations which describe $M,N$ are
$$M: (x-7)^2 + (y-9)^2 = 1^2, \quad \quad (x-7)^2 + (y-9)^2 = 3^2 $$