I need to solve a problem and I'm really stuck with it, so I want to summon the power of the community to see if anyone has an idea on how to handle it.
I need to create a porous material from a given surface . So, I have a point cloud representing the surface of a cylinder, like the one in the figure, and I need to generate a graph inside it from the points on the surface to be filled with volume. It is mandatory that all the points of the surface are used (some more can be added if necessary), the length of the edges must be an input parameter of the function and the angle between two nodes must always be greater than 45º with respect to the horizontal plane.
My initial idea was to make a while loop that in each iteration creates a random point cloud inside the cylinder between the current z (the current z is the maximum z value of the previous iteration) and the current z + the given edge length. Once this point cloud is created, it joins the surface nodes of the last iteration with the points of this point cloud that satisfy the condition of edge length and angularity. And it continues until the current z is greater than the maximum z of the cylinder surface.
The problem with this idea is that it is not consistent and the results are disastrous. So I would like to ask if anyone has a better idea or if anyone knows if any python libraries could help me. I am currently using networkx and numpy-stl but those are not meant to do what I want. ChatGPT is unable to understant me too :(.
Thank you so much for you time!!
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Given (N+1) points. All the N points lie on x- axis. Remaining one point (HEAD point) lies anywhere in the coordinate plane.
Given a START point on x- axis.
Find the shortest distance to cover all the points starting from START point.We can traverse a point multiple times.
Example N+1=4
points on x axis
(0,1),(0,2),(0,3)
HEAD Point
(1,1) //only head point can lie anywhere //Rest all on x axis
START Point
(0,1)
I am looking for a method as of how to approach this problem.
Whether we should visit HEAD point first or HEAD point in between.
I tried to find a way using Graph Theory to simplify this problem and reduce the paths that need to be considered. If there is an elegant way to represent this problem using graphs to identify a solution, I was not able to find it. This approach becomes very inefficient as the n increases - the time and memory is O(2^n).
Looking at this as a tree graph, the root node would be the START point, then each of its child nodes would be the points it is connected to.
Since the START point and the rest of the points aside from the HEAD all lay on the x-axis, all non-HEAD points only need to be connected to adjacent points on the x-axis. This is because the distance of the path between any two points is the sum of the distances between any adjacent points along the path between those two points (the subset of nodes representing points on the x-axis does not need to form a complete graph). This reduces the brute force approach some.
Here's a simple example:
The upper left shows the original problem: points on the x-axis along with the START and HEAD points.
In the upper right, this has been transformed into a graph with each node representing a point from the original problem. The edges represent the paths that can be taken between points. This assumes that the START point only represents the first point in the path. Unlike the other nodes, it is only included in the path once. If that is not the case and the path can return to the START point, this would approximately double the possible paths, but the same approach can be followed.
In the bottom left, the START point, a, is the root of a tree graph, and each node connected to the START point is a child node. This process is repeated for each child node until either:
A path that is obviously not optimal is identified, in which case that node can just be excluded from the graph. See the nodes in red boxes; going back and forth between the same nodes is unnecessary.
All points are included when traversing the tree from the root to that node, producing a potential solution.
Note that when creating the tree graph, each time a node is repeated, its "potential" child nodes are the same as the first time the node was included. By "potential", I mean cases above still need to be checked, because the result might include a nonsensical path, in which case that node would not be included. It is also possible a potential solution results from the path after its child nodes are included.
The last step is to add up the distances for each of the potential solutions to determine which path is shortest.
This requires a careful examination of the different cases.
Assume for now START (S) is on the far left, and HEAD (H) is somewhere in the middle the path maybe something like
H
/ \
S ---- * ----*----* * --- * ----*
Or it might be shorter to from H to and from the one of the other node
H
//
S ---- * --- * -- *----------*---*
If S is not at one end you might have something like
H
/ \
* ---- * ----*----* * --- * ----*
--------S
Or even going direct from S to H on the first step
H
/ |
* ---- * ----*----* |
S
A full analysis of cases would be quite extensive.
Actually solving the problem, might depend on the number of nodes you have. If the number is small < 10, then compete enumeration might be possible. Just work out every possible path, eliminate the ones which are illegal, and choose the smallest. The number of paths is I think in the order of n!, so its computable for small n.
For large n you can break the problem into small segments. I think its enough just to consider a small patch with nodes either side of H and a small patch with nodes either side of S.
This is not really a solution, but a possible way to think about tackling the problem.
(To be pedantic stackoverflow.com is not the right site for this question in the stack exchange network. Computational Science : algorithms might be a better place.
This is a fun problem. First, lets try to find a brute force solution, as Poosh did.
Observations about the Shortest Path
No repeated points
You are in an Euclidean geometry, thus the triangle inequality holds: For all points a,b,c, the distance d(a,b) + d(b,c) <= d(a,c). Thus, whenever you have an optimal path that contains a point that occurs more than once, you can remove one of them, which means it is not an optimal path, which leads to a contradiction and proves that your optimal path contains each point exactly once.
Permutations
Our problem is thus to find the permutation, lets call it M_i, of the numbers 1...n for points P1...Pn (where P0 is the fixed start point and Pn the head point, P1...Pn-1 are ordered by increasing x value) that minimizes the sum of |(P_M_i)-(P_M_(i-1))| for i from 1 to n, || being the vector length sqrt(v_x²+v_y²).
The number of permutations of a set of size n is n!. In this case we have n+1 points, so a brute force approach testing all permutations would have complexity (n+1)!, which is higher than even 2^n and definitely not practical, so we need further observations to improve this.
Next Steps
My next step would now be to see if there are any other sequences that can be proven to be not optimal, leading to a reduction in the number of candidates to be tested.
Paths of non-head points
Lets look at all paths (sequences of indices of points that don't contain a head point and that are parts of the optimal path. If we don't change the start and end point of a path, then any other transpositions have no effect on the outside environment and we can perform purely local optimizations. We can prove that those sequences must have monotonic (increasing or decreasing) x coordinate values and thus monotonic indices (as they are ordered by ascending x coordinate between indices 0 and n-1):
We are in a purely one dimensional subspace and the total distance of the path is thus equal to the sum of the absolute values of the differences in x coordinates between one such point and the next. It is clear that this sum is minimized by ordering by x coordinate in either ascending or descending order and thus ordering the indices in the same way. Note that this is true for maximal such paths as well as for all continuous "subpaths" of them.
Wrapping it up
The only choices we have left are:
where do we place the head node in the optimal path?
which way do we order the two paths to the left and right?
This means we have n values for the index of the head node (1...n, 0 is fixed as the start node) and 2x2 values for sort order. So we have 4n choices which we can all calculate and pick the shortest one. One of the sort orders probably determines the other but I leave that to you.
Anyways, the complexity of this algorithm is O(4n) = O(n). Because reading in the input of the problems is in O(n) and writing the output is as well, I believe that is an algorithm of optimal complexity. However, if we could reformulate the problem somewhat, so that we could read and write the input and output in some compressed form, as in only the parameters that we actually need to solve the problem, then it is possible that we could do better.
P.S.: I'm not a mathematician so I probably used wrong words for some concepts and missed the usual notation for the variables and functions. I would be glad for some expert to check this for any obvious errors.
I am trying to understand how to manually generate objects.
I have a mesh, part of which I delete and create a new geometry in its place. I have information about the normals of deleted vertices. On the basis of which I have to build new faces (in a different size and quantity) looking in the same direction.
But I don’t understand how to choose the correct winding. It sounds easy when the lessons talk about CCW winding in screen space. But what if I have a bunch of almost chaotic points in the model space? How then to determine this CCW, which axis is used for this? I suggest that the nearest old normals might help. But what is the cheapest method to determine the correct order?
It turned out to be easier than I thought. It is necessary to find the cross product of the first two vectors from the vertices of a triangle, then find the dot of the resulting vector and the normal vector, if the result is negative, then during generation it is necessary to change the order of vertices.
I have the coordinates of a bunch of car crashes and I reasoned that the place where they are the densest is where crashes are the most likely to occur. The solution I found when looking into this problem is to use point density, which is simply the number of points within a given radius divided by the area of a circle with that radius. I have two problems with this method:
1) Changing the size of the radius can significantly change your answer. For example, if all points are between 1 and 2 miles of point P, then P will be the most dense if r > 2 and P will be the least dense if r < 1.
2) If you want to find the point with the second highest density, then the result will always be a point immediately next to the point of the highest density.
What is a better method to find density?
P.S. I apologize if this is a duplicate of someone else's question. Whenever I searched for something related to this problem, things about point density always came up.
I'm working on a PyMEL script that allows the user to duplicate a selected object multiple times, using a CV curve and its points coordinates to transform & rotate each copy to a certain point in space.
In order to achieve this, Im using the adjacent 2 points of each CV (control vertex) to determine the rotation for the object.
I have managed to retrieve the coordinates of the curve's CVs
#Add all points of the curve to the cvDict dictionary
int=0
cvDict={}
while int<selSize:
pointName='point%s' % int
coords= pointPosition ('%s.cv[%s]' % (obj,int), w=1)
#Setup the key for the current point
cvDict[pointName]={}
#add coords to x,y,z subkeys to dict
cvDict[pointName]['x']= coords[0]
cvDict[pointName]['y']= coords[1]
cvDict[pointName]['z']= coords[2]
int += 1
Now the problem I'm having is figuring out how to get the angle for each CV.
I stumbled upon the angleBetween() function:
http://download.autodesk.com/us/maya/2010help/CommandsPython/angleBetween.html
In theory, this should be my solution, since I could find the "middle vector" (not sure if that's the mathematical term) of each of the curve's CVs (using the adjacent CVs' coordinates to find a fourth point) and use the above mentioned function to determine how much I'd have to rotate the object using a reference vector, for example on the z axis.
At least theoretically - the issue is that the function only takes 1 set of coords for each vector and I have absolutely no Idea how to convert my point coords to that format (since I always have at least 2 sets of coordinates, one for each point).
Thanks.
If you wanna go the long way and not grab the world transforms of the curve, definitely make use of pymel's datatypes module. It has everything that python's native math module does and a few others that are Maya specific. Also the math you would require to do this based on CVs can be found here.
Hope that puts you in the right direction.
If you're going to skip the math, maybe you should just create a locator, path-animate it along the curve, and then sample the result. That would allow you to get completely continuous orientations along the curve. The midpoint-constraint method you've outlined above is limited to 1 valid sample per curve segment -- if you wanted 1/4 of the way or 3/4 of the way between two cv's your orientation would be off. Plus you don't have to reinvent all of the manu different options for deciding on the secondary axis of rotation, reading curves with funky parameterization, and so forth.
I am trying to animate an object, let's say its a car. I want it go from point
x1,y1,z1
to point x2,y2,z2 . It moves to those points, but it appears to be drifting rather than pointing in the direction of motion. So my question is: how can I solve this issue in my updateframe() event? Could you point me in the direction of some good resources?
Thanks.
First off how do you represent the road?
I recently done exactly this thing and I used Catmull-Rom splines for the road. To orient an object and make it follow the spline path you need to interpolate the current x,y,z position from a t that walks along the spline, then orient it along the Frenet Coordinates System or Frenet Frame for that particular position.
Basically for each point you need 3 vectors: the Tangent, the Normal, and the Binormal. The Tangent will be the actual direction you will like your object (car) to point at.
I choose Catmull-Rom because they are easy to deduct the tangents at any point - just make the (vector) difference between 2 other near points to the current one. (Say you are at t, pick t-epsilon and t+epsilon - with epsilon being a small enough constant).
For the other 2 vectors, you can use this iterative method - that is you start with a known set of vectors on one end, and you work a new set based on the previous one each updateframe() ).
You need to work out the initial orientation of the car, and the final orientation of the car at its destination, then interpolate between them to determine the orientation in between for the current timestep.
This article describes the mathematics behind doing the interpolation, as well as some other things to do with rotating objects that may be of use to you. gamasutra.com in general is an excellent resource for this sort of thing.
I think interpolating is giving the drift you are seeing.
You need to model the way steering works .. your update function should 1) move the car always in the direction of where it is pointing and 2) turn the car toward the current target .. one should not affect the other so that the turning will happen and complete more rapidly than the arriving.
In general terms, the direction the car is pointing is along its velocity vector, which is the first derivative of its position vector.
For example, if the car is going in a circle (of radius r) around the origin every n seconds then the x component of the car's position is given by:
x = r.sin(2πt/n)
and the x component of its velocity vector will be:
vx = dx/dt = r.(2π/n)cos(2πt/n)
Do this for all of the x, y and z components, normalize the resulting vector and you have your direction.
Always pointing the car toward the destination point is simple and cheap, but it won't work if the car is following a curved path. In which case you need to point the car along the tangent line at its current location (see other answers, above).
going from one position to another gives an object a velocity, a velocity is a vector, and normalising that vector will give you the direction vector of the motion that you can plug into a "look at" matrix, do the cross of the up with this vector to get the side and hey presto you have a full matrix for the direction control of the object in motion.