forked from StudioInfinity/dyna3
Engine prototype (#13)
This PR adds code for a Julia-language prototype of a configuration solver, in the `engine-proto` folder. It uses Julia version 1.10.0. ### Approaches Development of this PR tried two broad approaches to the constraint geometry problem. Each one suggested various solution techniques. The Gram matrix approach, with the low-rank factorization technique, seems the most promising. - **Algebraic** *(In the `alg-test` subfolder).* Write the constraints as polynomials in the inversive coordinates of the elements, and use computational algebraic geometry techniques to solve the resulting system. We tried the following techniques. - **Gröbner bases** *(`Engine.Algebraic.jl`).* Symbolic. Find a Gröbner basis for the ideal generated by the constraint equations. Information about the solution variety, like its codimension, is then relatively easy to extract. - **Homotopy continuation** *(`Engine.Numerical.jl`).* Numerical. Cut the solution set along a random hyperplane to get a generic zero-dimensional slice, and then use a fancy homotopy technique to approximate the points in that slice. A few notes about our experiences can be found on the [engine prototype](wiki/Engine-prototype) wiki page. - **Gram matrix** *(in the `gram-test` subfolder).* A construction is described completely, up to conformal transformations, by the Gram matrix of the vectors representing its elements. Express the constraints as fixed entries of the Gram matrix, and use numerical linear algebra techniques to find a list of vectors whose Gram matrix fits the bill. We tried the following techniques. - **LDL decomposition** *(`gram-test.sage`, `gram-test.jl`, `overlap-test.jl`).* Find a cluster of up to five elements whose Gram matrix is completely filled in by the constraints. Use LDL decomposition to find a list of vectors with that Gram matrix. This technique can be made algebraic, as seen in `overlap-test.jl`. - **Low-rank factorization** *(source files listed in findings section).* Write down a quadratic loss function that says how far a set of vectors is from meeting the Gram matrix constraints. Use a smooth optimization technique like Newton's method or gradient descent to find a zero of the loss function. In addition to the polished prototype described in the results section, we have an early prototype using an off-the-shelf factorization package (`low-rank-test.jl`) and an visualization of the loss function landscape near global minima (`basin-shapes.jl`). The [Gram matrix parameterization](wiki/Gram-matrix-parameterization) wiki page contains detailed notes on this approach. ### Findings With the algebraic approach, we hit a performance wall pretty quickly as our constructions grew. It was often hard to find real solutions of the polynomial system, since the techniques we use work most naturally in the complex world. With the Gram matrix approach, on the other hand, we could solve interesting problems in acceptably short times using the low-rank factorization technique. We put the optimization routine in its own module (`Engine.jl`) and used it to solve five example problems: - `overlapping-pyramids.jl` - `circles-in-triangle.jl` - `sphere-in-tetrahedron.jl` - `tetrahedron-radius-ratio.jl` - `irisawa-hexlet.jl` We plan to use low-rank factorization of the Gram matrix in our first app prototype. ### Visualizations We used the visualizer in the `ganja-test` folder to visually check our low-rank factorization results. The visualizer runs [Ganja.js](https://enkimute.github.io/ganja.js/) in an Electron app, made with [Blink](https://github.com/JuliaGizmos/Blink.jl). Although Ganja.js makes beautiful pictures under most circumstances, we found two obstacles to using it in production. - It seems to have precision problems with low-curvature spheres. - We couldn't figure out how to customize its clipping and transparency settings, and the default settings often obscure construction details. Co-authored-by: Aaron Fenyes <aaron.fenyes@fareycircles.ooo> Co-authored-by: Glen Whitney <glen@studioinfinity.org> Reviewed-on: glen/dyna3#13 Co-authored-by: Vectornaut <vectornaut@nobody@nowhere.net> Co-committed-by: Vectornaut <vectornaut@nobody@nowhere.net>
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(proposed by Alex Kontorovich as a practical system for doing 3D geometric calculations)
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These coordinates are of form $I=(c, r, x, y, z)$ where we think of $c$ as the co-radius, $r$ as the radius, and $x, y, z$ as the "Euclidean" part, which we abbreviate $E_I$. There is an underlying basic quadratic form $Q(I_1,I_2) = (c_1r_2+c_2r_1)/2 - x_1x_2 -y_1y_2-z_1z_2$ which aids in calculation/verification of coordinates in this representation. We have:
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These coordinates are of form $I=(c, b, x, y, z)$ where we think of $c$ as the co-radius, $b$ as the "bend" (reciprocal radius), and $x, y, z$ as the "Euclidean" part, which we abbreviate $E_I$. There is an underlying basic quadratic form $Q(I_1,I_2) = (c_1b_2+c_2b_1)/2 - x_1x_2 -y_1y_2-z_1z_2$ which aids in calculation/verification of coordinates in this representation. We have:
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| Entity or Relationship | Representation | Comments/questions |
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| ------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
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| Sphere s with radius r>0 centered on P = (x,y,z) | $I_s = (1/c, 1/r, x/r, y/r, z/r)$ satisfying $Q(I_s,I_s) = -1$, i.e., $c = r/(\|P\|^2 - r^2)$. | Can also write $I_s = (\|P\|^2/r - r, 1/r, x/r. y/r, z/r)$ -- so there is no trouble if $\|E_{I_s}\| = r$, just get first coordinate to be 0. |
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| Plane p with unit normal (x,y,z), a distance s from origin | $I_p = (2s, 0, x, y, z)$ | Note $Q(I_p, I_p)$ is still -1. Also, there are two representations for each plane through the origin, namely $(0,0,x,y,z)$ and $(0,0,-x,-y,-z)$ |
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| Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. Because of this we might choose some other scaling of the inversive coordinates, say $(\||P\||,1/\||P\||,x/\||P\||,y/\||P\||,z/\||P\||)$ instead, but that fails at the origin, and likely won't have some of the other nice properties listed below. Note that scaling just the co-radius by $s$ and the radius by $1/s$ (which still preserves $Q=0$) dilates by a factor of $s$ about the origin, so that $(\|P\|, \|P\|, x, y, z)$, which might look symmetric, would actually have to represent the Euclidean point $(x/\||P\||, y/\||P\||, z/\||P\||)$ . |
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| ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by the above case. |
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| P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | |
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| Sphere/planes represented by I and J are tangent | $Q(I,J) = 1$ (??, see note at right) | Seems as though this must be $Q(I,J) = \pm1$ ? For example, the $xy$ plane represented by (0,0,0,0,1) is tangent to the unit circle centered at (0,0,1) rep'd by (0,1,0,0,1), but their Q-product is -1. And in general you can reflect any sphere tangent to any plane through the plane and it should flip the sign of $Q(I,J)$, if I am not mistaken. |
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| Sphere/planes represented by I and J intersect (respectively, don't intersect) | $\|Q(I,J)\| < (\text{resp. }>)\; 1$ | Follows from the angle formula, at least conceptually. |
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| P is center of sphere represented by I | Well, $Q(I_P, I)$ comes out to be $(\|P\|^2/r - r + \|P\|^2/r)/2 - \|P\|^2/r$ or just $-r/2$ . | Is it if and only if ? No this probably doesn't work because center is not conformal quantity. |
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| Distance between P and R is d | $Q(I_P, I_R) = d^2/2$ | |
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| Distance between P and sphere/plane rep by I | | In the very simple case of a plane $I$ rep'd by $(2s, 0, x, y, z)$ and a point $P$ that lies on its perpendicular through the origin, rep'd by $(r^2, 1, rx, ry, rz)$ we get $Q(I, I_p) = s-r$, which is indeed the signed distance between $I$ and $P$. Not sure if this generalizes to other combinations? |
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| Distance between sphere/planes rep by I and J | Note that for any two Euclidean-concentric spheres rep by $I$ and $J$ with radius $r$ and $s,$ $Q(I,J) = -\frac12\left(\frac rs + \frac sr\right)$ depends only on the ratio of $r$ and $s$. So this can't give something that determines the Euclidean distance between the two spheres, which presumably grows as the two spheres are blown up proportionally. For another example, for any two parallel planes, $Q(I,J) = \pm1$. | Alex had said: Q(I,J)=cosh^2 (d/2) maybe where d is distance in usual hyperbolic metric. Or maybe cosh d. That may be right depending on what's meant by the hyperbolic metric there, but it seems like it won't determine a reasonable Euclidean distance between planes, which should differ between different pairs of parallel planes. |
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| Sphere centered on P through R | | Probably just calculate distance etc. |
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| Plane rep'd by I goes through center of sphere rep'd by J | I think this is equivalent to the plane being perpendicular to the sphere, i.e.$Q(I,J) = 0$. | |
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| Dihedral angle between planes (or spheres?) rep by I and J | $\theta = \arccos(Q(I,J))$ | Aaron Fenyes points out: The angle between spheres in $S^3$ matches the angle between the planes they bound in $R^{(1,4)}$, which matches the angle between the spacelike vectors perpendicular to those planes. So we should have $Q(I,J) = \cos\theta$. Note that when the spheres do not intersect, we can interpret this as the "imaginary angle" between them, via $\cosh t = \cos it$. |
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| R, P, S are collinear | Maybe just cross product of two differences is 0. Or, $R,P,S,\infty$ lie on a circle, or equivalently, $I_R,I_P,I_S,I_\infty$ span a plane (rather than a three-space). | Not a conformal property, but $R,P,S,\infty$ lying on a circle _is_. |
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| Plane through noncollinear R, P, S | Should be, just solve Q(I, I_R) = 0 etc. | |
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| circle | Maybe concentric sphere and the containing plane? Note it is easy to constrain the relationship between those two: they must be perpendicular. | Defn: circle is intersection of two spheres. That does cover lines. But you lose the canonicalness |
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| line | Maybe two perpendicular containing planes? Maybe the plane perpendicular to the line and through origin, together with the point of the line on that plane? Or maybe just as a bag of collinear points? | The first is the limiting case of the possible circle rep, but it is not canonical. The second appears to be canonical, but I don't see a circle rep that corresponds to it. |
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| Entity or Relationship | Representation | Comments/questions |
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| ---------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
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| Sphere $s$ with radius $r>0$ centered on $P = (x,y,z)$ | $I_s = (\frac1{c}, \frac1{r}, \frac{x}{r}, \frac{y}{r}, \frac{z}{r})$ satisfying $Q(I_s,I_s) = -1,$ i.e., $c = r/(\|P\|^2 - r^2)$. | Note that $1/c = \|P\|^2/r - r$, so there is no trouble if $\|P\| = r$; we just get first coordinate to be 0. Using the point representation $I_P$ from below, let's orient the sphere so that its normals point into the "positive side," where $Q(I_P, I_s) > 0$. The vector $I_s$ then represents a sphere with outward normals, while $-I_s$ represents one with inward normals. |
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| Plane $p$ with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note that $Q(I_p, I_p)$ is still $-1$. We orient planes using the same convention we use for spheres. For example, $(-2, 0, -1/\sqrt3, -1/\sqrt3, -1/\sqrt3)$ and $(2, 0, 1/\sqrt3, 1/\sqrt3, 1/\sqrt3)$ represent planes that coincide in space, which have their normals pointing away from and toward the origin, respectively. Note that the ray from $(sx, sy, sz) \in p$ in direction $(-x, -y, -z)$ is the ray perpendicular to the plane through the origin; since $(-x, -y, -z)$ is a unit vector, $(sx, sy, sz)$ and hence $p$ is at distance $s$ from the origin. These coordinates are essentially the limit of a sphere's coordinates as its radius goes to infinity, or equivalently, as its bend goes to 0. |
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| Point $P$ with Euclidean coordinates $(x,y,z)$ | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$. This gives us the freedom to choose a different normalization. For example, we could scale the representation shown here by $(\|P\|^2+1)^{-1}$, putting it on the sphere where the light cone intersects the plane where the first two coordinates sum to $1$. |
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| ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by (some normalization of) the above case. |
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| Point $P$ lies on sphere or plane given by $I$ | $Q(I_P, I) = 0$ | Actually also works if $I$ is the coordinates of a point, in which case "lies on" simply means "coincides with". |
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| Sphere/planes represented by $I$ and $J$ are tangent | If $I$ and $J$ have the same orientation where they touch, $Q(I,J) = -1$. If they have opposing orientations, $Q(I,J) = 1$. | For example, the $xy$ plane with normal $-e_z$, represented by $(0,0,0,0,1)$, is tangent with matching orientation to the unit sphere centered at $(0,0,1)$ with outward normals, represented by $(0,1,0,0,1).$ Accordingly, their $Q$ - product is $-1$. |
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| Sphere/planes represented by $I$ and $J$ intersect (respectively, don't intersect) | $\lvert Q(I,J)\rvert \le (\text{resp. }>)\; 1$ | Follows from the angle formula and the tangency condition, at least conceptually. One subtlety: parallel planes have $Q$ - product $\pm 1$, because they intersect at infinity (and in fact, are "tangent" there)! |
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| $P$ is center of sphere rep'd by $I$ | $Q(I, I_P) = -r/2$, where $1/r = 2Q(I_\infty, I)$ is the signed bend of the sphere, and $I_P$ is normalized in the standard way, which is to set $Q(I_\infty, I_P) = 1/2$ | This relationship is equivalent to both of the following. (1) The point $P$ has signed distance $-r$ from the sphere. (2) Inversion across the sphere maps $\infty$ to $P$. |
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| Distance between points $P$ and $R$ is $d$ | $Q(I_P, I_R) = d^2/2$ | If $P$ and $R$ are represented by non-normalized vectors $V_P$ and $V_R$, the relation becomes $Q(V_P, V_R) = 2\,Q(V_P, I_\infty)\,Q(V_R, I_\infty)\,d^2$. This version of the relation makes it easier to see why $d$ goes to infinity as $P$ or $R$ approaches the point at infinity. |
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| Signed distance between point rep'd by $V$ and sphere/plane rep'd by $I$ is $d$ | In general, $\frac{Q(I, V)}{2Q(I_\infty, V)} = Q(I_\infty, I)\,d^2 + d$. When $V$ is normalized in the usual way, this simplifies to $Q(I, V) = d^2/r + d$ for a sphere of radius $r$, and to $Q(I, V) = d$ for a plane. | We can use a Euclidean motion, represented linearly by a Lorentz transformation that fixes $I_\infty$, to put the point on the $z$ axis and put the nearest point on the sphere/plane at the origin with its normal pointing in the positive $z$ direction. Then the sphere/plane is represented by $I = (0, 1/r, 0, 0, -1)$, and the point can be represented by any multiple of $I_P = (d^2, 1, 0, 0, d)$, giving $Q(I, I_P) = d^2/2r + d.$ We turn this into a general expression by writing it in terms of Lorentz-invariant quantities and making it independent of the normalization of $I_P$. |
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| Distance between sphere/planes rep by $I$ and $J$ | Note that for any two Euclidean-concentric spheres rep by $I$ and $J$ with radius $r$ and $s,$ $Q(I,J) = -\frac12\left(\frac rs + \frac sr\right)$ depends only on the ratio of $r$ and $s$. So this can't give something that determines the Euclidean distance between the two spheres, which presumably grows as the two spheres are blown up proportionally. For another example, for any two parallel planes, $Q(I,J) = \pm1$. | Alex had said: $Q(I,J)=\cosh(d/2)^2$ maybe where d is distance in usual hyperbolic metric. Or maybe $\cosh(d)$. That may be right depending on what's meant by the hyperbolic metric there, but it seems like it won't determine a reasonable Euclidean distance between planes, which should differ between different pairs of parallel planes. |
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| Sphere centered on point $P$ through point $R$ | | Probably just calculate distance etc. |
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| Plane rep'd by $I$ goes through center of sphere rep'd by $J$ | This is equivalent to the plane being perpendicular to the sphere: that is, $Q(I, J) = 0$. | |
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| Dihedral angle between planes or spheres rep by $I$ and $J$ | $\theta = \arccos(Q(I,J))$ | Aaron Fenyes points out: The angle between spheres in $S^3$ matches the angle between the planes they bound in $R^{(1,4)}$, which matches the angle between the spacelike vectors perpendicular to those planes. So we should have $Q(I,J) = \cos(\theta)$. Note that when the spheres do not intersect, we can interpret this as the "imaginary angle" between them, via $\cosh(t) = \cos(it)$. |
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| Points $R, P, S$ are collinear | Maybe just cross product of two differences is 0. Or, $R,P,S,\infty$ lie on a circle, or equivalently, $I_R,I_P,I_S,I_\infty$ span a plane (rather than a three-space). Or we can add two planes constrained to be perpendicular with one constrained to contain the origin, and all three points constrained to lie on both. But that's a lot of auxiliary entities and constraints... | $R,P,S$ lying on a line isn't a conformal property, but $R,P,S,\infty$ lying on a circle is. |
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| Plane through noncollinear $R, P, S$ | Should be, just solve $Q(I, I_R) = 0$ etc. | |
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| circle | Maybe concentric sphere and the containing plane? Note it is easy to constrain the relationship between those two: they must be perpendicular. | Defn: circle is intersection of two spheres. That does cover lines. But you lose the canonicalness |
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| line | Maybe two perpendicular containing planes? Maybe the plane perpendicular to the line and through origin, together with the point of the line on that plane? Or maybe just as a bag of collinear points? | The first is the limiting case of the possible circle rep, but it is not canonical. However, there is a distinguished "standard" choice we could make: always choose one plane to contain the origin and the line, and the other to be the perpendicular plane containing the line. That choice or Plücker coordinates might be the best we can do. If we use the standardized perpendicular planes choice, then adding a line would be equivalent to adding two planes and the two constraints that one contains the origin and the other is perpendicular to it. That doesn't seem so bad. The second convention (perpendicular plane through the origin and a point on it) appears to be canonical, but there doesn't seem to be a circle representation that tends to it in the limit. |
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| Inversion of entity represented by $v$ across sphere $s$, rep'd by $I_s$ | $v \mapsto v + 2Q(I_s, v)\,I_s$ | This is just an educated guess, but its behavior is consistent with inversion in at least two ways. (1) It fixes points on $s$ and spheres perpendicular to $s$. (2) It preserves dihedral angles with $s$. |
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The unification of spheres/planes is indeed attractive for a project like Dyna3. The relationship between this representation and Geometric Algebras is a bit murky; likely it somehow fits under the Geometric Algebra umbrella.
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