For picking the root, is there any rule to follow?
Let say if I pick E as the root, do I need to split E as G is crossing E?
I can understand the example on wiki example, but this one is more complicated (More polygons cross another)
If I don't need to split it, then what is the next polygon that I should pick as subset
Thanks in advance, this would help my study during Christmas
Related
I'm very new to blender and started testing out geometry nodes following a tutorial.
(Donuts and Gummies)
I am now attempting a small project to polish my skills.
I want to have only one instance of the traverse on the long shaft every n times. Currently, I have two instances in the middle and 4 at the ends. (see image below)
My objectives are:
Get one instance every n times on the shaft
Remove the end pieces completely.
I managed to solve my first issue by merging the edges like so
Still need assistance on the 2nd issue, please.
The solution to my second issue was fixed using the Trim Curve node.
I had to do some calculations to plugin the values, but that was the easy part.
Hope someone finds this useful.
I know how to implement union find in general, but I was thinking of whether there would be a way to utilize the set structure in python to achieve the same result.
For example, we can union sets pretty easily. But I'm not sure how to determine if two elements are in the same set using just sets.
So, I am wondering if there is a data structure in python that would support such operation, other than the usual implementation?
You could always solve this problem by visualizing it as a tree and its nodes connecting to each other via the root, and then looking up the tree if you want to know if two nodes are connected. If the two nodes you are comparing has the same root (they are in the same tree), than they are connected.
To connect two nodes, just go to the root of each tree they are in, and make one root become the parent of the other.
This video will give you a great intuition about it:
https://www.youtube.com/watch?v=YIFWCpquoS8&list=PLUX6FBiUa2g4YWs6HkkCpXL6ru02i7y3Q&index=1
The connection between the tree nodes can be made via pointers in a language which supports it, but if your language dont (python), than you can create your own pointers by storing positions and links via an array.
The array would be such that its positions would represent your nodes, and the values inside it represents the connection of the specific node to its root. On the beginning, the position in the array is filled with the node number because the nodes has initially no parent, but as you connect nodes, the roots changes, and the array has to represent this. Actually, the value stored there is the identificator of the root.
But try visualizing the problem visually first instead of thinking of arrays and too much mathematical artificats. Visually dealing with it makes the solution sound banal, and can be a good guidance while writing code.
I say this because I have watched the video from Robert Sedgewick I just posted, with a graphical simulation of the solution, and implemented myself without paying too much attention to the code on his book. The intuition the video gave me is much more valuable than any mathematics.
It will help you to encapsulate the nodes into a class, with the following methods:
climbTreeFromNodeUpToRoot
setNewParentToThisNodeAndUpdateHeights
The first method, as the name says, takes you from a node and goes up the tree until finding the root of it, which is then returned.
If you compare two nodes with this method (actually, the roots returned by it), you know easily if they are connected by just comparing their roots.
Once you want to connected them, you go up the trees of both nodes, and ask one root to take the other one as its parent.
The trees can grow very big in height (sorry I dont use the official nomeclature, but this is the one that makes sense to me), so this simple approach will get very slow when you have to climb the tree at a later time.
To prevent trees from becoming to high, dont just set one root as the parent to another without criterium, but attach the smallest tree (in terms of height, not quantity of elements) to the highest one.
For this, you need to know the heights of each tree, and this information you can store on their respective root (via an extra array in your case, or an extra pointer from each node in other languages). This information should be updated everytime another tree connects to it.
It is not possible for a tree to know that she just got a new tree attached to it, so its important that every tree attaching to a second one informs the second as to update its height.
This information can be sent to the root of the second tree, and later used to judge (as writen before) which tree is the smallest. Remember, attaching a small tree to a big one instead of the opposite will save you incredible amounts of time.
Do you want something like this?
myset = ...
all(elt in myset for elt in (a,b))
Normally Intersection of two planes A and B (not parallel) will return a line L. I know how to implement this, but if now given a plane A and the line of intersection L to find plane B. Is there a solution? Thanks in advance!
No, it is not possible to find (or "recover") the plane B, because an infinite number of planes (Bs) can intersect plane A exactly at the line L but still are allowed to "hinge" (or rotate) about it (within certain limits of course so as to not be parallel as you mention).
You need a little bit more information to define one single plane (three points, a point and a line, a point and a normal vector, for more information please see here). Also, Paul Bourke's website contains really a wealth of information if you are working in computer graphics.
Perhaps there is a way to get this little bit of information from your problem (?)
(By the way, i am not sure that this a question for Stackoverflow, perhaps it would fit better to the Mathematics part)
I have a tile based map where several tiles are walls and others are walkable. the walkable tiles make up a graph I would like to use in path planning. My question is are their any good algorithms for finding a path which visits every node in the graph, minimising repeat visits?
For example:
map example http://img220.imageshack.us/img220/3488/mapq.png
If the bottom yellow tile is the starting point, the best path to visit all tiles with least repeats is:
path example http://img222.imageshack.us/img222/7773/mapd.png
There are two repeat visits in this path. A worse path would be to take a left at the first junction, then backtrack over three already visited tiles.
I don't care about the end node but the start node is important.
Thanks.
Edit:
I added pictures to my question but cannot see them when viewing it. here they are:
http://img220.imageshack.us/img220/3488/mapq.png
http://img222.imageshack.us/img222/7773/mapd.png
Additionally, in the graphs I need this for there will never be a situation where min repeats = 0. That is, to step on every tile in the map the player must cross his own path at least once.
Your wording is bad -- it allows a reduction to an NP-complete problem. If you could minimize repeat visits, then could you push them to 0 and then you would have a Hamiltonian Cycle. Which is solvable, but hard.
This sounds like it could be mapped onto the traveling salesman problem ... and so likely ends up being NP complete and no efficient deterministic algorithm is known.
Finding a path is fairly straight forward -- find a (or the minimum) spanning subtree and then do a depth/breadth-first traversal. Finding the optimal route is the really difficult bit.
You could use one of the dynamic optimization techniques to try and converge on a fairly good solution.
Unless there is some attribute of the minimum spanning subtree that could be used to generate the best path ... but I don't remember enough graph theory for that.
For instance:
An approach to compute efficiently the first intersection between a viewing ray and a set of three objects: one sphere, one cone and one cylinder (other 3D primitives).
What you're looking for is a spatial partitioning scheme. There are a lot of options for dealing with this, and lots of research spent in this area as well. A good read would be Christer Ericsson's Real-Time Collision Detection.
One easy approach covered in that book would be to define a grid, assign all objects to all cells it intersects, and walk along the grid cells intersecting the line, front to back, intersecting with each object associated with that grid cell. Keep in mind that an object might be associated with more grid-cells, so the intersection point computed might actually not be in the current cell, but actually later on.
The next question would be how you define that grid. Unfortunately, there's no one good answer, and you need to consider what approach might fit your scenario best.
Other partitioning schemes of interest are different tree structures, such as kd-, Oct- and BSP-trees. You could even consider using trees combined with a grid.
EDIT
As pointed out, if your set is actually these three objects, you're definately better of just intersecting each one, and just pick the earliest one. If you're looking for ray-sphere, ray-cylinder, etc, intersection tests, these are not really hard and a quick google should supply all the math you might possibly need. :)
"computationally efficient" depends on how large the set is.
For a trivial set of three, just test each of them in turn, it's really not worth trying to optimise.
For larger sets, look at data structures which divide space (e.g. KD-Trees). Whole chapters (and indeed whole books) are dedicated to this problem. My favourite reference book is An Introduction to Ray Tracing (ed. Andrew. S. Glassner)
Alternatively, if I've misread your question and you're actually asking for algorithms for ray-object intersections for specific types of object, see the same book!
Well, it depends on what you're really trying to do. If you'd like to produce a solution that is correct for almost every pixel in a simple scene, an extremely quick method is to pre-calculate "what's in front" for each pixel by pre-rendering all of the objects with a unique identifying color into a background item buffer using scan conversion (aka the z-buffer). This is sometimes referred to as an item buffer.
Using that pre-computation, you then know what will be visible for almost all rays that you'll be shooting into the scene. As a result, your ray-environment intersection problem is greatly simplified: each ray hits one specific object.
When I was doing this many years ago, I was producing real-time raytraced images of admittedly simple scenes. I haven't revisited that code in quite a while but I suspect that with modern compilers and graphics hardware, performance would be orders of magnitude better than I was seeing then.
PS: I first read about the item buffer idea when I was doing my literature search in the early 90s. I originally found it mentioned in (I believe) an ACM paper from the late 70s. Sadly, I don't have the source reference available but, in short, it's a very old idea and one that works really well on scan conversion hardware.
I assume you have a ray d = (dx,dy,dz), starting at o = (ox,oy,oz) and you are finding the parameter t such that the point of intersection p = o+d*t. (Like this page, which describes ray-plane intersection using P2-P1 for d, P1 for o and u for t)
The first question I would ask is "Do these objects intersect"?
If not then you can cheat a little and check for ray collisions in order. Since you have three objects that may or may not move per frame it pays to pre-calculate their distance from the camera (e.g. from their centre points). Test against each object in turn, by distance from the camera, from smallest to largest. Although the empty space is the most expensive part of the render now, this is more effective than just testing against all three and taking a minimum value. If your image is high res then this is especially efficient since you amortise the cost across the number of pixels.
Otherwise, test against all three and take a minimum value...
In other situations you may want to make a hybrid of the two methods. If you can test two of the objects in order then do so (e.g. a sphere and a cube moving down a cylindrical tunnel), but test the third and take a minimum value to find the final object.