How to replace hidden nodes by "virtual" edges? - filter

Take a simple network A=B=C, where node B has the attribute minor. When I just filter out that attribute, I'll get the graph A C. However, I want a visualisation of the fact that A and C are indirectly connected, i.e. A---C. How can this be achieved? Are filters even the proper way?

I could be wrong but I don't think this is something you can achieve without processing your data first, looping through all the edges that involve B, creating an array of B's neighbours and then creating new edges linking them.
The exact way to do this depends on the format of your data and the programming language.
You could also add edges manually in Gephi, but that is of course only realistic for small graphs.
I don't think filters are the right tool for this because they simply exclude nodes (and edges) but don't create new ones. https://github.com/gephi/gephi/wiki/Filter

Related

How do you compute a Djikstra Heat Map with multiple sources?

Currently, so far I am using a level-wise BFS which traverses throughout the grid and the values(calulcated by looking at the level) essentially is the distance its from the source. How, do I implement this so I have multiple heat sources? Does this mean I will have to compute multiple level-wise BFS, after iterating one level for the BFS for source 1, I move on to source 2 and etc
Nevermind, I can do to iteratively for one. And for the next few sources I have to have a couple of conditional statements which prevent me from overwriting too much.

JGraphT create a random graph from pre existing vertices

I need to create a random graph from a few pre existing vertices.
I didn't find a way to do this yet.
Each tutorial is using the "VertexFactory" (which btw cannot be resolved even though importing org.jgraph.generate.*) to make up vertices while creating the graph.
But I rather would like to generate the graph from already existing vertices. I'm especially interested in the WattsStragotz algorithm but I don't know yet.
Is this possible with using the already written RandomGeneration classes of Jgrapht?
Thanks a lot
Yes you can do this, but now with vertices that have already been added to the graph. If, for whatever reason, you want to use pre-existing objects as your vertices, you could do something along the following lines:
Put all your pre-existing vertex objects into a list
Implement a custom Vertex supplier (the old VertexFactory has been taken out of commission in favor of the Supplier paradigm in the newer java versions). This vertex supplier iterates over your list of pre-defined vertices, and returns a new vertex each time the get() method of the supplier is invoked.
Create a new graph with your custom vertex supplier. Have a look at this example on our wiki page.
Invoke any of the random graph generators on the graph.

Standard names for "stacked" versus "hanging" layered graph drawing algorithms?

Here are two different ways of drawing the same hierarchy. Notice that in the "stacked" layout, nodes are always one layer higher than their highest "child" node. (Important: See edit at bottom of question for another example)
Do these two types of layered drawing methods have specific names? I'm trying to find existing algorithms for the "stacked" one, but can't seem to surface any info because I don't know what it's called.
If they don't have names to distinguish them because they rely on the same algorithm, are there well known sets of parameters for attaining the "stacked" version of the graph with existing algorithms? Thanks!
Edit: Although the above graphs are strict "trees", the algorithm I'm looking for should be able to handle cases where nodes have more than one parent, and cases where there is more than one path from root to leaf. Here's an example, and here's another.
Edit2: In case it's useful to anyone, a hacky (and slow) force-directed approach with pre-computed node layers (y-axis contraints) seems to work all right. Here's what it looks like. That example uses cytoscape.js and cola.js, and it's upside down. It's not at all a solution to this question so I'm just putting this here as an edit.
(SO wouldn't let me submit the JSBin link without a code block...)
I don't know of any specific names for the above. It looks like the layering algorithm in both cases is the longest path algorithm that minimizes height but essentially ignores width. If you layer the graph from the bottom-up and the graph has many sinks (vertices with zero out-degree) then you will get a wide bottom layer (a "stacked" layout?). If you layer the graph from the top-down and it has many sources (vertices with zero in-degree) then you will get a wide top layer (a "hanging" layout?).

How to find the neighbors of a graph effiiciently

I have a program that create graphs as shown below
The algorithm starts at the green color node and traverses the graph. Assume that a node (Linked list type node with 4 references Left, Right, Up and Down) has been added to the graph depicted by the red dot in the image. Inorder to integrate the newly created node with it neighbors I need to find the four objects and link it so the graph connectivity will be preserved.
Following is what I need to clarify
Assume that all yellow colored nodes are null and I do not keep a another data structure to map nodes what is the most efficient way to find the existence of the neighbors of the newly created node. I know the basic graph search algorithms like DFS, BFS etc and shortest path algorithms but I do not think any of these are efficient enough because the graph can have about 10000 nodes and doing graph search algorithms (starting from the green node) to find the neighbors when a new node is added seems computationally expensive to me.
If the graph search is not avoidable what is the best alternative structure. I thought of a large multi-dimensional array. However, this has memory wastage and also has the issue of not having negative indexes. Since the graph in the image can grow in any directions. My solution to this is to write a separate class that consists of a array based data structure to portray negative indexes. However, before taking this option I would like to know if I could still solve the problem without resolving to a new structure and save a lot of rework.
Thank you for any feedback and reading this question.
I'm not sure if I understand you correctly. Do you want to
Check that there is a path from (0,0) to (x1,y1)
or
Check if any of the neighbors of (x1,y1) are in the graph? (even if there is no path from (0,0) to any of this neighbors).
I assume that you are looking for a path (otherwise you won't use a linked-list), which implies that you can't store points which have no path to (0,0).
Also, you mentioned that you don't want to use any other data structure beside / instead of your 2D linked-list.
You can't avoid full graph search. BFS and DFS are the classic algorithms. I don't think that you care about the shortest path - any path would do.
Another approaches you may consider is A* (simple explanation here) or one of its variants (look here).
An alternative data structure would be a set of nodes (each node is a pair < x,y > of course). You can easily run 4 checks to see if any of its neighbors are already in the set. It would take O(n) space and O(logn) time for both check and add. If your programming language does not support pairs as nodes of a set, you can use a single integer (x*(Ymax+1) + Y) instead.
Your data structure can be made to work, but probably not efficiently. And it will be a lot of work.
With your current data structure you can use an A* search (see https://en.wikipedia.org/wiki/A*_search_algorithm for a basic description) to find a path to the point, which necessarily finds a neighbor. Then pretend that you've got a little guy at that point, put his right hand on the wall, then have him find his way clockwise around the point. When he gets back, he'll have found the rest.
What do I mean by find his way clockwise? For example suppose that you go Down from the neighbor to get to his point. Then your guy should be faced the first of Right, Up, and Left which he has a neighbor. If he can go Right, he will, then he will try the directions Down, Right, Up, and Left. (Just imagine trying to walk through the maze yourself with your right hand on the wall.)
This way lies insanity.
Here are two alternative data structures that are much easier to work with.
You can use a quadtree. See http://en.wikipedia.org/wiki/Quadtree for a description. With this inserting a node is logarithmic in time. Finding neighbors is also logarithmic. And you're only using space for the data you have, so even if your graph is very spread out this is memory efficient.
Alternately you can create a class for a type of array that takes both positive and negative indices. Then one that builds on that to be 2-d class that takes both positive and negative indices. Under the hood that class would be implemented as a regular array and an offset. So an array that can start at some number, positive or negative. If ever you try to insert a piece of data that is before the offset, you create a new offset that is below that piece by a fixed fraction of the length of the array, create a new array, and copy data from the old to the new. Now insert/finding neighbors are usually O(1) but it can be very wasteful of memory.
You can use a spatial index like a quad tree or a r-tree.

Circular representation of a tree structure

I have some data in a tree structure, and I want to represent them in a graphical way, with the root node in the middle of the stage, his children displaced in a circle around him, and so on for every children, around their parent.
I don't want overlapping nodes, so the question is how to arrange space in an optimal way.
Something less or more like (found via google)
What algorhythms I have to search to realize something like this?
If you don't care about how it's done, but just that you are visualizing the data, then take a look at graphviz's radial layout. Although the example doesn't look exactly what you want, it is the layout you'd need. It'll also give you some ideas on how it's done too with the loads of research papers in there. Good luck!
You could also see how easy it is to extend this paper into a circular structure.
You can do it in an emergent way by setting up a system in which each tree node tries to keep as much distance from all other nodes (except parent) as possible, but as short a distance as possible from the parent (down to some minimum distance which it must maintain). If you run that algorithm for each node repeatedly until it stabilizes, you'll have an arrangement like the one you describe. I'm sure there are a lot of optimizations you can do to it, but I'm pretty sure this is going to be the simplest approach. Trying to calculate all the layout up front would be very complex...
You are trying to draw a planar representation of a graph.
Find some buzzwords and perhaps a resource here
And in wikipedia
Ah and I forgot: You can do this the newtonian way with forces.
Simply give all nodes a repelling potential, like make them all Protons which push each other away. Give the edges the properties of newtonian springs, exerting forces that pull them together and you are all set.
Could even create nice animations that way.
This is also an official way of graph drawing, but I don't know the name.
If you want to draw the tree with a minimum of wasted space and short connections, then you're in for a computationally expensive solution. It will be hard to get this to be realtime on a tree of decent size, not to mention that making small changes to the tree might result in a radically different equilibrium.
Another approach would be to abandon the physical simulation and just build it iteratively. I've done something similar last week, but my trees are probably a lot less involved than yours.
For this tree-layout, each node object has to store an angle and an offset. These two numbers control where on the graphics surface they end up.
Here is my basic algorithm:
1) recurse over your entire tree-data and find all the Leaf nodes.
2) while you're doing this, be sure to measure the length of each branching structure, so you know which is the longest.
3) once you have all your leaf nodes, distribute them equally over a concentric circle. You can either use the entire circle, or only some part of the angle domain.
4) once all Leaf nodes have been solved, you recurse again over the tree, going from the outside in. Each node you encounter that is not a leaf node is in need of layout. Essentially, every node from here on has an angle which is the average of all it's child nodes, and the offset is the graph_radius * (depth_of_node / maximum_depth)
I found this gives me a very decent and humanly readable distribution, albeit not a very efficient one in terms of screen usage. I uploaded an animation of my tree-display here: GIF anim

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