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Most efficient way to identify a circle

rashtastic

New member
Jan 3, 2013
11
A friend's hw problem (Prove any five points in the plane determines a possibly degenerate conic section) led us to a different problem that we found more interesting.

We can identify a circle with three points on the circle, or six parameters $(x_1,y_1,x_2,y_2,x_3,y_3)$

where, keeping order, every circle can be uniquely identified with six reals. So the above tuple is a point in the set of circles on the plane, encoded in a certain way.

We can do this more efficiently by using the center and the radius, $(x_1,y_1,r)$, bringing us down to three parameters to uniquely identify any circle. Can we go any further?

Defining a circle in terms of real parameters involves finding a parameter set $\mathcal{B}=\{(\theta_1,...,\theta_n)\}$ where the set of all circles $\mathcal{C}$ can be mapped injectively into $\mathcal{B}$. We have done this for $\mathcal{B}$ with order 6 and 3. Can we do it for order 2?

Yes... ignoring decimal points and negatives for the moment, we can take the center coordinates and make a new number from interpolating the center coordinates' digits. So (11111,22222) can be (1212121212). This is quite silly but allows us to define an injective function from $\mathcal{C}$ into $\mathcal{B}$ where $\mathcal{B}$ has order 2- one parameter for the combined center, one for the radius. Again, given that you produce rules for decimals, you can then combine the same thing for the radius and create a single parameter that can be used to uniquely identify a circle. Or a single parameter can identify any conic section, or a list of conic sections, and so on. If five points is sufficient to identify a conic section, then ten parameters (two reals per point) can be uniquely mapped into a single real that identifies the conic. Consider that $\mathcal{R}^{10}\sim\mathcal{R}$.

You might notice that somewhere along this line we have basically lost touch with reality.

Can you think of a good argument for why $(x_1,y_1,r)$ is the most efficient way to uniquely express any circle in a way that is geometrically sensible?

Or can you find a two-parameter solution (reals only) that can uniquely identify any circle in a reasonable way?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
There is a natural bijective correspondence $\mathcal{C}\leftrightarrow\mathbb{R}^2 \times\mathbb{R}$ between the set of all circles in the plane and the space $\mathbb{R}^2 \times\mathbb{R}$ (given by taking the point in $\mathbb{R}^2$ to be the centre of the circle, and the point in $\mathbb{R}$ to be its radius). Since any "reasonable" way to parametrise a point in point in $\mathbb{R}^2$ uses two parameters, it follows that you need three parameters to specify a circle.

Notice that if you specify a circle by giving three points on it, that does not give a bijective correspondence between $\mathcal{C}$ and $\mathbb{R}^6$ because there are many triples of points giving rise to the same circle.
 

rashtastic

New member
Jan 3, 2013
11
Notice that if you specify a circle by giving three points on it, that does not give a bijective correspondence between $\mathcal{C}$ and $\mathbb{R}^6$ because there are many triples of points giving rise to the same circle.
Agreed- also there are triples of points giving rise to no circle.

In the three point/six parameter system, given two points, there are constraints on what the third point can be. We might reason that taking advantage of whatever structure exists may allow us to more efficiently encode a circle.

Whereas having any two parameters of $(x_1,y_1,r)$ gives you no information on the third parameter. There is no more structure to take advantage of. The x-coordinate, y-coordinate, and size-coordinate are simultaneously all the information you need and least amount you can get by with. In statistics we would call it "minimally sufficient." I guess this is my hand-waving argument that if I set it up differently, e.g. in polar coordinates, I would still need three parameters to represent the three "free"/independent components.

A parabola should be similar: identifiable by three points/six parameters, but reducible to three free parameters (The coefficients of $ax^2+bx+c$).

Maybe the general idea is "Structure = potential for data reduction. Independent components = information efficiency." Will explore further.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I was just wondering...

Is the function $f: \mathbb R^2 \to \mathbb R$ given by something like $(111.111..., 222.222...) \mapsto 121212.121212...$ bijective?

I don't see why it wouldn't be... although I'm actually sure it's not bijective.
 

rashtastic

New member
Jan 3, 2013
11
I was just wondering...

Is the function $f: \mathbb R^2 \to \mathbb R$ given by something like $(111.111..., 222.222...) \mapsto 121212.121212...$ bijective?

I don't see why it wouldn't be... although I'm actually sure it's not bijective.
You could reconstruct the right order of digits of the original numbers, but you would lose positive/negative and decimal point placement. Also you have to define the range of the function before you can qualify it as a bijection.

I did skip over decimals and negatives because it adds tedium without benefiting the general question. We may as well take a look. What I had in mind involved creating a little number-language...

Ignoring decimals and negatives, convert reals $a$ and $b$ to binary. Interpolate these digits, creating $c$. Let $d$ be an indicator for if $a$ is positive or negative, $e$ an indicator for $b$, using 2="Positive", 3="Negative", 4="Zero". In $c$, place a 5 after the corresponding $a$-digit where $a$'s decimal point would go; place a 6 in the place for $b$. Then string all these together: "$d$"+"$e$"+"$c$", and to make it a real number (and not just an infinite string of digits), add a decimal point after $e$. This is a real number that you could extract $a$ and $b$ out of. It contains all the information you need to uniquely identify any two reals. That makes it injective.

The range we're mapping in to is $\mathcal{D}=\{r\in\mathcal{R}| floor(r)\in\{22,23,24,32,33,34,42,43,44\}$, and decimal digits are 1's and 0's except for exactly one 5 and one 6 $\}$.

I think this puts $\mathcal{R}^2$ into a bijection with $B$ with $B\subset\mathcal{R}$. This is more of an algorithm than a function, but it works.
 
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rashtastic

New member
Jan 3, 2013
11
There is a natural bijective correspondence $\mathcal{C}\leftrightarrow\mathbb{R}^2 \times\mathbb{R}$ between the set of all circles in the plane and the space $\mathbb{R}^2 \times\mathbb{R}$ (given by taking the point in $\mathbb{R}^2$ to be the centre of the circle, and the point in $\mathbb{R}$ to be its radius).
Original hw friend had a nice comment:

I think one aspect of the "naturalness" of the "point + radius" parametrization is that it is continuous - changing any of the parameters transforms shapes smoothly.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
There is a natural bijective correspondence $\mathcal{C}\leftrightarrow\mathbb{R}^2 \times\mathbb{R}$ between the set of all circles in the plane and the space $\mathbb{R}^2 \times\mathbb{R}$ (given by taking the point in $\mathbb{R}^2$ to be the centre of the circle, and the point in $\mathbb{R}$ to be its radius).
Of course I should have said "... a natural bijective correspondence $\mathcal{C}\leftrightarrow\mathbb{R}^2 \times\mathbb{R}^+$ ...", where $\mathbb{R}^+$ means the positive real numbers. The radius has to be positive.