Understanding degrees of freedom of a twisted cubic

geometrymatricesprojection-matricesprojective-geometry

I am trying to understand twisted cubics as explained in Multiple View Geometry in Computer Vision (Page 76). The twisted cubic is given by the equation:

$\begin{bmatrix}X1\\X2\\X3\\X4\end{bmatrix} =A\begin{bmatrix}1\\\theta\\\theta^2\\\theta^3\end{bmatrix} =\begin{bmatrix}a_{11} + a_{12}\theta+a
_{13}\theta^2+a_{14}\theta^3\\a_{21} + a_{22}\theta+a
_{23}\theta^2+a_{24}\theta^3\\a_{31} + a_{32}\theta+a
_{33}\theta^2+a_{34}\theta^3\\a_{41} + a_{42}\theta+a
_{43}\theta^2+a_{44}\theta^3\end{bmatrix}$

The text says that twisted cubic has 12 DoF. Now A is a 4 x 4 matrix and it has a scaling factor. So, the DoF becomes 16 – 1 = 15. Now, the text further says,

…less 3 for a 1D projectivity on the parameterization $\theta$

This will make DoF = 15 – 3= = 12.

Question 1: I am trying to understand the 1D projectivity on $\theta$. I have a theory for this. I assume that this parameter represents angle just like as in $(r, \theta)$ representation of a circle. Now, the angles lie in the range $0 – 2\pi$ rad or 0 – 360 deg. Any angle beyond this range can be mapped back to a value in this range ((n mod 360 + 360) mod 360 assuming degrees). Consider these angles represented as Homogeneous Coordinates. Is it possible to represent this mapping by a 2 x 2 matrix with 3 DoF? Are these the 3 parameters subtracted from the DoF of A? If that's not the case, how to explain the 1D projectivity of $\theta$?

Further the book also says that a unique twisted cubic can be defined by 6 points in general position. The explanation is that the above equation provides 2 constraints from a single point, once $\theta$ is eliminated. Since one of the values on the RHS is a scale factor, there are already only 3 ratios. I am trying to understand how one more constraint is reduced/ elimination of $\theta$.

Question 2: What does '$\theta$ is eliminated' mean in this context?

Edit: This video explains a twisted cubic as an intersection of parabolic cylinder and cubic cylinder and explains elimination of a parameter without H.C. It gives a sense of what's going on here. But feel free to improve upon this.

Best Answer

It's not apparent what angle is being discussed (after all, the twisted cubic lives in $\Bbb P^3$), or what is intended by "consider these angles represented as homogeneous coordinates," for example). Projective geometry has no invariant notion of angle (projective transformations change angles between curves, just as they change lengths${}^*$), so I'd suggest jettisoning that line of thinking as likely to mislead your intuition anyway.

I'd say that the description

...less $3$ for a $1$D projectivity on the parameterization $\theta$...

from the text is imprecise. What's going on here is the following:

(1) By way of analogy, recall that we can parameterize any (say, connected, piecewise $C^1$) curve $\gamma$ in Euclidean space by arc length---because these parameterizations respect the Euclidean structure (i.e., the notion of length) induced on the curve, they have special status among all parameterizations of $\gamma$. On the other hand, arc length parameterizations are not unique: If $\gamma(s)$ is one such parameterization, so is $\gamma(s + A)$ for any constant $A$. On the other hand (if $\gamma$ is oriented) all (oriented) parameterizations arise this way, so our parameterization has exactly one degree of freedom.

There is an analogous description for curves $\gamma$ in projective geometry. This time, projective space induces a projective structure on the curve itself, and while $\gamma$ has many parameterizations, only some of them respect this structure. In this case, if $\gamma(\theta)$ is one such parameterization, the others are precisely the parameterizations $$\theta \mapsto \gamma\left(\frac{A \theta + B}{C \theta + D}\right),$$ where $A D - B C \neq 0$---we call these projective parameterizations or projective changes of variable (or often outside the context of projective geometry, linear fractional transformations). But $(A, B, C, D)$ and $(\lambda A, \lambda B, \lambda C, \lambda D)$ define the same parameterizations, so projective reparameterizations only entail $3$ degrees of freedom.

We can of course make this explicit for our twisted cubic: If our original parameterization is $${\bf X}(\theta) = [X^i(\theta)] = [a_{i1} + a_{i2} \theta + a_{i3} \theta^2 + a_{i4} \theta^3] ,$$ then applying a projective transformation gives the new parameterization \begin{align} {\bf X}\left(\frac{A \theta + B}{C \theta + D}\right) = &\left[X^i\left(\frac{A \theta + B}{C \theta + D}\right)\right] \\ = &\left[a_{i1} + a_{i2} \left(\frac{A \theta + B}{C \theta + D}\right) + a_{i3} \left(\frac{A \theta + B}{C \theta + D}\right)^2 + a_{i4} \left(\frac{A \theta + B}{C \theta + D}\right)^3\right] \\ = &\left[a_{i1} (C \theta + D)^3 + a_{i2} (A \theta + B) (C \theta + D)^2\right. \\ &\qquad \left. + a_{i3} (A \theta + B)^2 (C \theta + D) + a_{i4} (A \theta + B)^3\right] \end{align} We can collect like terms in $\theta$ to produce a new parameterization of the same twisted cubic of the form $${\bf X} = {\bf B} [\theta^i]$$ for some matrix $\bf B$ whose entries are linear in the entries $a_{ij}$.

Since $3$ of our $15$ degrees of freedom only change the parameterization and not the choice of twisted cubic, we expect that it takes $15 - 3 = 12$ real parameters to specify a twisted cubic. Any point specifies $2$ real parameters, so generically we expect that it takes $\frac{12}{2} = 6$ points to specify a twisted cubic. (You might already be familiar with the fact that it takes only $5$ points to specify a conic in $\Bbb P^2$.)

(2) Here, "eliminating" $\theta$ just means eliminating the parameter $\theta$ in a system of equations. Analogously, with some algebra we can describe an affine line in $\Bbb R^2$ given by a parameterization $t \mapsto (v t + x, w t + y)$ with an equation of the form $a x + b y + c = 0$, which in particular does not involve $t$. Notice that do not actually need to solve for the parameter explicitly to carry out the relevant counting.

Remark It's natural to ask why the linear fractional transformations are exactly the reparameterizations that respect the projective structure. We can, for example, view any projective parameterization of the twisted cubic as an embedding $\theta : \Bbb P \to \Bbb P^3$. In particular, if we apply a projective transformation $T$ to $\Bbb P^3$ that fixes a particular cubic, then $\theta^{-1} \circ T \circ \theta : \Bbb P \to \Bbb P$ is a projective transformation of $\Bbb P$, but we know what these transformations are: They are given in homogeneous coordinates exactly by the action of $PGL(2, \Bbb R)$, which in any affine chart on $\Bbb P$ acts by linear fractional transformations.

${}^*$ In fact there is a projectively invariant notion of arc length, but it is rather different in flavor from the familiar Euclidean one---which is emphatically not preserved by projective transformations. If you're interested in computational aspects of projective invariants like this one, I highly recommend Clelland's From Frenet to Cartan: The Method of Moving Frames, though you should have some elementary differential geometry under your belt (curves and surfaces in Euclidean space, vector fields, $1$-forms, exterior derivative) before taking it up.

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[Math] Java Tetris – Using rotation matrix math to rotate piece

The operations you are wanting to perform can be easily implemented without the need of rotation matrices since the transforms you desire are limited to right angle rotations and mirroring.

Assuming $B$ is a two dimensional array representing a block and $B[i,j]$ is the $i^{\text{th}}$ row and $j^{\text{th}}$ column of the block taking on values of 1 for a single block segment and 0 for an empty space.

The first operation you need is a transpose which is effectively a 90 degree clockwise rotation:

B T(B b)
    B b' <- new B[b.columns, b.rows]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns -1
            b'[j,i] <- b[i, j]

    return b'

The second operation is mirroring a block about the y-axis:

B Y(B b)
    B b' <- new B[b.rows, B.columns]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns - 1
            b'[i,j] <- b[i, b.columns - 1 - j]

    return b'

And finally, mirroring a block about the x-axis:

B X(B b)
    B b' <- new B[b.rows, B.columns]

    for i = 0 to b.rows - 1
        for j = 0 to b.columns - 1
            b'[i,j] <- b[b.rows - 1 - i, j]

    return b'

Given a block $B$ in a neutral position, a 90 degree clockwise rotation is $T(B)$, 180 is $T(X(T(B)))$, 270 is $Y(T(B))$ and 360 is simply $B$. counter clockwise rotations are just 360 minus the complementing clockwise rotations.

Also, as @StevenStadnicki pointed out in the comments, you can hard code much of this and you will find that if you are replicating the original Tetris, you'll encounter more that requires a hard coded solution than not.

[Math] Angle of the diagonal of a convex quadrilateral.

I found a solution. I would like to know if there is a better/simpler solution.

Here is a picture in which I added names in some line segments.

Suppose we know $\alpha, \beta, \gamma, \delta$ and $\phi$, and we want to find $\theta$.

Using the law of sines we have: $$\frac{\sin\phi}{X}=\frac{\sin\beta}{N}, \quad\frac{\sin\theta}{Y}=\frac{\sin\gamma}{M},\quad\frac{\sin(\pi-\theta-\gamma)}{X}=\frac{\sin\gamma}{M},\quad\frac{\sin(\alpha-\phi)}{Y}=\frac{\sin\delta}{N}$$

Eliminating the lengths we end up with the following equation: $$\frac{\sin(\theta+\gamma)}{\sin\theta}=\frac{\sin\delta\sin\phi}{\sin(\alpha-\phi)\sin\beta}$$

We use the angle addition formula for the left side of the equation: $$\frac{\sin(\theta+\gamma)}{\sin\theta}=\frac{\sin\theta\cos\gamma+\cos\theta\sin\gamma}{\sin\theta}=\cos\gamma+\cot\theta\sin\gamma$$

We can now solve for $\theta$: $$\theta=\cot^{-1}\left(\frac{\sin\delta\sin\phi}{\sin(\alpha-\phi)\sin\beta\sin\gamma}-\cot\gamma \right)$$

Best Answer

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[Math] Angle of the diagonal of a convex quadrilateral.

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