quadtree | Includes remove | Learning library

 by   asim Go Version: v0.1.0 License: MIT

kandi X-RAY | quadtree Summary

kandi X-RAY | quadtree Summary

quadtree is a Go library typically used in Tutorial, Learning, Example Codes applications. quadtree has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Golang implementation of the quadtree algorithm. Includes removal, update and knearest search.
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              quadtree has a low active ecosystem.
              It has 24 star(s) with 7 fork(s). There are 3 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 1 open issues and 3 have been closed. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of quadtree is v0.1.0

            kandi-Quality Quality

              quadtree has 0 bugs and 0 code smells.

            kandi-Security Security

              quadtree has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              quadtree code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              quadtree is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              quadtree releases are available to install and integrate.
              Installation instructions are not available. Examples and code snippets are available.
              It has 307 lines of code, 21 functions and 1 files.
              It has high code complexity. Code complexity directly impacts maintainability of the code.

            Top functions reviewed by kandi - BETA

            kandi has reviewed quadtree and discovered the below as its top functions. This is intended to give you an instant insight into quadtree implemented functionality, and help decide if they suit your requirements.
            • kNearestRoot finds the nearest neighbor of i that intersect with i .
            • Bound the point at the given coordinates .
            • earthRadius returns the radius of the earth
            • New creates a new quad tree
            • Create a new AABB .
            • NewPoint returns a new Point
            • rad2Deg converts radians radians to radians
            • deleToRad converts a given deg to radians .
            Get all kandi verified functions for this library.

            quadtree Key Features

            No Key Features are available at this moment for quadtree.

            quadtree Examples and Code Snippets

            No Code Snippets are available at this moment for quadtree.

            Community Discussions

            QUESTION

            How to free all memory blocks of a picture using quadree structure with different cases?
            Asked 2022-Jan-15 at 16:18

            Few weeks ago, I was trying to implement a function to display a quadtree. My current problem is concerning the same work, so I pass you this link for all the context: How to display composed quadtrees using C language?
            (I'm using a few features that come from this post)

            the quadtree structure:

            ...

            ANSWER

            Answered 2022-Jan-15 at 16:18

            A robust way to free the memory from the quadtree is to make sure you feedback that a pointer no longer is pointing to valid memory. Since your current freeMemory only takes an block_image pointer, the function cannot convey this information back to the caller.

            Better would be to change its interface to provide this facitily with another level of indirection.

            Source https://stackoverflow.com/questions/70722684

            QUESTION

            Rendering imageData to new canvas
            Asked 2022-Jan-12 at 07:53

            I'm following a tutorial by George Francis in the tutorial after some initial examples he shows how to use image data to create random layouts.

            I'm trying to work out how to get the image data from a canvas created using paper.js, as I need to get the rgb values from each individual pixel on the canvas

            Link to codepen

            Unknowns:

            • Do I need to use the rasterize() method on the shape I've created?

            Currently I am attempting the following:

            ...

            ANSWER

            Answered 2022-Jan-12 at 07:53

            I think that your issue was that at the time where you are getting the image data, your scene is not yet drawn to the canvas.
            In order to make sure it's drawn, you just need to call view.update().

            Here's a simple sketch demonstrating how it could be used.
            Note that you don't need to rasterize your scene if you are using the Canvas API directly to manipulate the image data. But you could also rasterize it and take advantage of Paper.js helper methods like raster.getPixel().

            Source https://stackoverflow.com/questions/70625826

            QUESTION

            How to display composed quadtrees using C language?
            Asked 2021-Dec-31 at 20:13

            Currently I’m trying to implement different functions to realize operations on quadtrees using C language.

            Context:

            A quadtree (black and white picture) is:

            • White
            • Black
            • Is broken down into 4 sub-images: top-left,top-right,bottom-left,bottom-right

            Pictures are represented by the next struct:

            ...

            ANSWER

            Answered 2021-Dec-31 at 18:43

            how I could display a composed image with a postfix notation by explaining how to modify my function so that it can consider a maximum of cases?

            That's really trivial. Just display it for this cell or display for the childs.

            Source https://stackoverflow.com/questions/70543148

            QUESTION

            How can I use d3 force to space out overlapping points on a map
            Asked 2021-Dec-05 at 06:03

            I have a map that has some overlapping points. I'm using a quadtree and turf.js to determine, when I click on a point, how many other points are within a 30 mile radius.

            What I'd like to do (if there's more than one point in that radius) is use d3.forceSimulation to equally spread out the overlapping points.

            This is a pretty close example of what I want to do, but uses d3v3 and google maps: http://bl.ocks.org/cdmahoney/raw/9876525/?raw=true

            I have included d3.forceSimulation and when I click on a place with more than one point within the 30 mile radius, the points do take on the force-- but they move up to the left hand corner of the page.

            How can I get the points to push out in even distances from where I click on the map, like this:

            Help very much appreciated!!

            ...

            ANSWER

            Answered 2021-Dec-05 at 06:03

            While I'd be tempted to not use a force layout for this, I'll work with the code you have here (though the question of the lines connecting the circles to their original location is not addressed here) and quickly address why the circles do not behave as you expect.

            A force layout will create the appropriate properties on a node if they don't exist. For position of a node, these properties are d.x and d.y. Your data does not have x or y properties, so when you create the force, the nodes are initialized with values around the origin, [0,0], which is why they migrate to the top left corner. This problem can be solved by creating x and y properties:

            Source https://stackoverflow.com/questions/70230656

            QUESTION

            How to index nearby 3D points on the fly?
            Asked 2021-Nov-30 at 02:43

            In physics simulations (for example n-body systems) it is sometimes necessary to keep track of which particles (points in 3D space) are close enough to interact (within some cutoff distance d) in some kind of index. However, particles can move around, so it is necessary to update the index, ideally on the fly without recomputing it entirely. Also, for efficiency in calculating interactions it is necessary to keep the list of interacting particles in the form of tiles: a tile is a fixed size array (eg 32x32) where the rows and columns are particles, and almost every row-particle is close enough to interact with almost every column particle (and the array keeps track of which ones actually do interact).

            What algorithms may be used to do this?

            Here is a more detailed description of the problem:

            1. Initial construction: Given a list of points in 3D space (on the order of a few thousand to a few million, stored as array of floats), produce a list of tiles of a fixed size (NxN), where each tile has two lists of points (N row points and N column points), and a boolean array NxN which describes whether the interaction between each row and column particle should be calculated, and for which:

              a. every pair of points p1,p2 for which distance(p1,p2) < d is found in at least one tile and marked as being calculated (no missing interactions), and

              b. if any pair of points is in more than one tile, it is only marked as being calculated in the boolean array in at most one tile (no duplicates),

              and also the number of tiles is relatively small if possible (but this is less important than being able to update the tiles efficiently)

            2. Update step: If the positions of the points change slightly (by much less than d), update the list of tiles in the fastest way possible so that they still meet the same conditions a and b (this step is repeated many times)

            It is okay to keep any necessary data structures that help with this, for example the bounding boxes of each tile, or a spatial index like a quadtree. It is probably too slow to calculate all particle pairwise distances for every update step (and in any case we only care about particles which are close, so we can skip most possible pairs of distances just by sorting along a single dimension for example). Also it is probably too slow to keep a full (quadtree or similar) index of all particle positions. On the other hand is perfectly fine to construct the tiles on a regular grid of some kind. The density of particles per unit volume in 3D space is roughly constant, so the tiles can probably be built from (essentially) fixed size bounding boxes.

            To give an example of the typical scale/properties of this kind of problem, suppose there is 1 million particles, which are arranged as a random packing of spheres of diameter 1 unit into a cube with of size roughly 100x100x100. Suppose the cutoff distance is 5 units, so typically each particle would be interacting with (2*5)**3 or ~1000 other particles or so. The tile size is 32x32. There are roughly 1e+9 interacting pairs of particles, so the minimum possible number of tiles is ~1e+6. Now assume each time the positions change, the particles move a distance around 0.0001 unit in a random direction, but always in a way such that they are at least 1 unit away from any other particle and the typical density of particles per unit volume stays the same. There would typically be many millions of position update steps like that. The number of newly created pairs of interactions per step due to the movement is (back of the envelope) (10**2 * 6 * 0.0001 / 10**3) * 1e+9 = 60000, so one update step can be handled in principle by marking 60000 particles as non-interacting in their original tiles, and adding at most 60000 new tiles (mostly empty - one per pair of newly interacting particles). This would rapidly get to a point where most tiles are empty, so it is definitely necessary to combine/merge tiles somehow pretty often - but how to do it without a full rebuild of the tile list?

            P.S. It is probably useful to describe how this differs from the typical spatial index (eg octrees) scenario: a. we only care about grouping close by points together into tiles, not looking up which points are in an arbitrary bounding box or which points are closest to a query point - a bit closer to clustering that querying and b. the density of points in space is pretty constant and c. the index has to be updated very often, but most moves are tiny

            ...

            ANSWER

            Answered 2021-Nov-27 at 13:46

            I suggest the following algorithm. E.g we have cube 1x1x1 and the cutoff distance is 0.001

            1. Let's choose three base anchor points: (0,0,0) (0,1,0) (1,0,0)
            2. Associate array of size 1000 ( 1 / 0.001) with each anchor point
            3. Add three numbers into each regular point. We will store the distance between the given point and each anchor point inside these fields
            4. At the same time this distance will be used as an index in an array inside the anchor point. E.g. 0.4324 means index 432.
            5. Let's store the set of points inside of each three arrays
            6. Calculate distance between the regular point and each anchor point every time when update point
            7. Move point between sets in arrays during the update

            The given structures will give you an easy way to find all closer points: it is the intersection between three sets. And we choose these sets based on the distance between point and anchor points.

            In short, it is the intersection between three spheres. Maybe you need to apply additional filtering for the result if you want to erase the corners of this intersection.

            Source https://stackoverflow.com/questions/70026017

            QUESTION

            d3 svg shadow difference
            Asked 2021-Oct-29 at 22:26

            I make a copy of crypto bubbles where I want the coins to float around like bubbles now I run into the problem that a squares are completely filled and some are not this is because of the shadow? How do I solve this?

            ...

            ANSWER

            Answered 2021-Oct-29 at 13:30

            Use a smaller stdDeviation or enlarge your filter region. Something like this might work:

            Source https://stackoverflow.com/questions/69768303

            QUESTION

            Why a quadtree sometimes need a max number to hold in a node?
            Asked 2021-Sep-11 at 17:37

            I am doing a compuational geometric issue which uses TriangularMeshQuadtree from a C# library, and some of its constructors is written as follows (from metadata, so I cannot see the details of implementations):

            constructor 1:

            ...

            ANSWER

            Answered 2021-Sep-11 at 17:37

            Your understanding "... until every point is unique in one section" is not quite correct. It describes a very special kind of quadtree that is usually used to in example to explain the concept.

            In general, a quadtree can hold many more items per node. This is often done to reduce the number of nodes (if we have more entries per node than we need fewer nodes). The benefit of reducing node count is:

            • Reduced memory usage (every node adds some memory overhead)
            • Usually faster search (every nodes adds an indirection which is slow), i.e. a very 'deep' tree is slow to traverse.

            The maxItem should not be too large, because inside a node the points are usually stored in a linear list. A linear list obviously requires linear search, so that slows things down if the list is too large. In my experience sensible values for maxItem are between 10 and 100.

            Another parameter that is often given is maxDepth. This parameter limits the depth of the tree, which is equal to the number of parents of a given node. The idea is that a bad dataset can result in a tree that is very 'deep', which makes it expensive to traverse. Instead, is a node is at depth=maxDepth, it is prevented from splitting, even if it exceeds maxItem entries.

            Having said all the above, there are useful real-world quadtree-type structures that allow at most one entry per quadrant. One example is the PH-Tree (disclaimer: self-advertisement). It uses other techniques to limit the depth to 32 or 64. It takes a while to explain and it wasn't part of the question, so I just reference the documentation here.

            Source https://stackoverflow.com/questions/69116013

            QUESTION

            2D overlapping rectangles occlusion
            Asked 2021-Sep-04 at 19:31

            I'm looking for an algorithm that can find intersecting rectangles that overlap each other.

            The catch is that my data structure is similar to a quadtree with bounding boxes instead of points. I'm doing a basic rectangle intersection check but the problem is as I zoom into the tree the child nodes get detected and the parents, I would like to exclude the parent if its fully occluded by a child for the given camera rectangle.

            zoom animation

            As you can see from the image above as the camera rectangle (black box) sits inside the green node the purple node is still highlighted (filled), consequently as I zoom more and more the parents are always highlighted, even though the camera rectangle can be fully filled with child nodes only.

            This makes sense since the camera rectangle is still inside the parent but I have searched and thought about the problem for a while and can't seem to figure out an elegant solution. There seems to be several ways of doing this for 3D spaces but I can't find anything simple for 2D AABB rectangles.

            A few solutions that I thought of:

            • Subtract the child nodes from the parents resulting in concave polygons and then perform polygon intersection.
            • Use the fill color to check which rectangles are visible, therefore occluding the ones behind.
            • Perform raycasting or sub-division and check which is the smallest node for a given section.

            Is there a better way to do this? Thank you

            Update 1

            I have solved the problem by subdividing the camera into smaller sections and for each section the smallest intersecting node is found. This works but there must be a more efficient and cleaner way of doing this.

            Update 2

            Thank you Trentium for your answer. I can clearly see that an algorithm like that would be a lot more performant than what I'm currently doing.

            Eventually I will implement it as splitting a rectangle into smaller rectangles and not polygons sounds like a fun challenge.

            Also, I did some very non scientific benchmarks on the current approach and it takes 0.5ms-1ms to both filter and draw everything, so for now performance is still not a concern.

            ...

            ANSWER

            Answered 2021-Sep-03 at 22:42

            Suggest considering a variation of your first solution proposed "Subtract the child nodes from the parents resulting in concave polygons and then perform polygon intersection."

            Specifically, if the immediate children are wholly contained within the parent, then suggest that for each rectangle, that an associated array of visible residual rectangles also be maintained. And then use this array of residual rectangles as the means of determining if the camera / viewport rectangle includes the parent or not.

            For example, let's say the parent rectangle is (0,0) - (100,100) and there is an initial child rectangle added at (75,75)-(100,100). The data structure will appear as...

            • parent.rectangle = {0,0,100,100}
            • parent.visible = [ {0,0,100,75}, {0,75,75,100} ]
            • child1.rectangle = {75,75,100,100}
            • child1.visible = [ {75,75,100,100} ]

            ...and then if a second child comes along, say at {50,50,75,90}, then this new child is checked against each residual rectangle in the parent.visible array, subdividing the parent visible rectangles further as necessary...

            • parent.rectangle = {0,0,100,100}
            • parent.visible = [ {0,0,100,50}, {0,50,50,75}, {75,50,100,75}, {0,75,50,100}, {50,90,75,100} ]
            • child1.rectangle = {75,75,100,100}
            • child1.visible = [ {75,75,100,100} ]
            • child2.rectangle = {50,50,75,90}
            • child2.visible = [ {50,50,75,90} ]

            This method will add a bit of work up front adjusting the immediate parent's visible rectangles as children are added, but should greatly reduce the amount of rectangle intersection tests relative to the current algorithm that involves subdivision of the camera / viewport. Plus this proposed algorithm only makes use of rectangle-to-rectangle intersection tests (both when adding a child to a parent, and when testing the camera / viewport intersections!) rather than the suggested rectangle-to-polygon tests...

            Source https://stackoverflow.com/questions/68993199

            QUESTION

            Convert d3v6 Observable example using zoom.center
            Asked 2021-Jul-28 at 13:58

            I have a project where I'm trying to zoom to specific points. In d3v3, I could do that with this function

            ...

            ANSWER

            Answered 2021-Jul-27 at 03:28

            There is no zoom.center in any d3-zoom release, from the first one to the current (v3.0.0 at the time of writing).

            That zoom.center method was created by that Observable's author, which actively contributes to D3 (see here). You can see the proposal here: https://github.com/d3/d3-zoom/pull/212

            However, if you inspect the code the author is using, you'll see that adding your own center method is not complicated. All you need is:

            Source https://stackoverflow.com/questions/68530758

            QUESTION

            Why this Quadtree Query often returns nothing?
            Asked 2021-May-12 at 06:13

            Any "W" refers to half-width and any "H" refers to half-height. Any "X" and "Y" refers to the center of a rectangle aligned with the co-ordinate axes.

            I am attempting to write some general purpose quadtree code for use in physics simulations and game design. At first it was returning ALL points in the contents of a quadtree that was completely enclosed by the search region. After adding a bunch of print() statements to illuminate the process I managed to solve that problem but now I have a new one I can't seem to figure out. For some reason when I query the quadtree for the "Placeables" within a rectangular region I either get a totally correct result, but more often than not I get nothing returned at all. Placeables are defined as follows:

            ...

            ANSWER

            Answered 2021-May-12 at 06:13

            Wow, this was a difficult problem to find (at least, I hope I solved it).

            The problem is with your definition of rW and rH in setup():

            Source https://stackoverflow.com/questions/67421797

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