From f829f81a27ac426d2664aac7808d85c83d6be215 Mon Sep 17 00:00:00 2001 From: Glen Whitney Date: Thu, 8 Feb 2024 22:54:09 +0000 Subject: [PATCH] Update Engine prototype --- Engine-prototype.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/Engine-prototype.md b/Engine-prototype.md index c8675c9..8999393 100644 --- a/Engine-prototype.md +++ b/Engine-prototype.md @@ -33,7 +33,7 @@ Some of these are obvious, but we may as well write them down. Note we may want ## Overall solver approaches * Find the Gröbner basis of the constraints. This definitely tells us a lot about the possible configurations. For example, the basis will be 1 if and only if there are no solutions (correct?). And we should be able to tell if the system is rigid, i.e., has only finitely many solutions (correct?). In this case, can we find all of the algebraic number solutions? (I am worried that is in general a high-complexity problem that may become intractable in practice for realistic constructions??) And hopefully in the case of nonzero dimension components, we will have ways to find some rational points if they exist; we could certainly hit a variety with no rational points, correct?. We're of course also fine with finding say quadratic or other low-degree algebraic numbers that satisfy the constraints. I assume for any fixed degree, it could have only finitely many algebraic points of that degree over Q. I think that's true, but according to https://math.stackexchange.com/questions/21979/points-of-bounded-degree-on-varieties if it has infinitely many points, then there is a least degree for which it has infinitely many points. Are we interested in that degree? Can we find it from the Gröbner basis? If we could generate points of a given degree well, then we would know what degree to go to in order to allow a "smooth drag" among exact solutions. E.g, for a circle/sphere of quadratic algebraic radius there are already infinitely many rational points, if I am not mistaken, see https://mathoverflow.net/questions/125224/rational-points-on-a-sphere-in-mathbbrd. Except for the occasional "holes" around particularly low-denominator solutions, this would allow for a smooth drag just among rational points. ) -* Use Newton's method to minimize the sum of the squares of the constraints. If we do find an approximate zero this way, is there a way to find a nearby algebraic zero? Sadly, if we don't find a zero this way, we may have just gotten stuck at a local minimum, so that's no proof the constraints are inconsistent. +* Use Newton's method to minimize the sum of the squares of the constraints. If we do find an approximate zero this way, is there a way to find a nearby algebraic zero? Sadly, if we don't find a zero this way, we may have just gotten stuck at a local minimum, so that's no proof the constraints are inconsistent. Note that if we suspect that we are near a quadratic algebraic solution, then the approach described at https://mathoverflow.net/questions/2861/how-should-i-approximate-real-numbers-by-algebraic-ones could well be fruitful, if we take the approximation to high accuracy. * Use homotopy continuation (see https://www.juliahomotopycontinuation.org/) to get numerical information about the solution variety (and generate numerical approximations to points on it, correct?) If I understand correctly, an advantage of this method is that we get an almost-sure proof that the system is inconsistent (if it is), correct? And become almost sure whether it is rigid (has only isolated solutions), correct? As with Newton's method, we still have the question of whether having approximate numerical solutions helps us find exact algebraic ones. Also, if I understand correctly, this approach comes with a natural way to update solutions as one element is dragged, is that right? * Since generally speaking each constraint is quadratic, there _might_ be some useful information here: https://math.stackexchange.com/questions/2119007/how-do-you-solve-a-system-of-quadratic-equations