I am using the D3 force directed graph with collapsible nodes as described here
Is there a way to avoid the explosion that happens when the graph is initially loaded? Also, can we stop the movement/animation of nodes once the graph has loaded?
Related
When single node is dragged, other nodes and links don't follow along like in this graph example: https://observablehq.com/#d3/force-directed-graph
Here is the code: https://stackblitz.com/edit/js-xbmyr7?file=index.js
I have followed the 'template' solution typical of draggable graphs but am unable to understand why the rest of the graph does not move. All nodes and links are selected in the ticked function.
Is it possible to create a custom (link) force in a d3 force-directed graph, such that the link strength is asymmetric? What I mean is that I want two nodes A and B, and B is completely oblivious to node A, but node A is constantly trying to remain a certain distance (maybe 0?) from node B.
More generally, I'd like two graphs G1 and G2 where G1 is draggable and moves like a regular force-directed graph (oblivious to G2), and the nodes in G2 are constantly trying to re-position themselves relative to G1 in a certain way.
Is it possible to use D3 Trees and draw graph. Then let the user to rearange the nodes without changing their relationships?
I used d3 Tree functions to have the location of nodes and used my current forced directed graph. This way I could keep my current functionalities, and also having more than one parent for nodes, and user could move the nodes around. However, once user moves the node, I do not let d3 to rearrange the node when nodes is expanded. I want the node stays in the location that user moved and wants.
Although we can check a if a graph is bipartite using BFS and DFS (2 coloring ) on any given undirected graph, Same implementation may not work for the directed graph.
So for testing same on directed graph , Am building a new undirected graph G2 using my source graph G1, such that for every edge E[u -> v] am adding an edge [u,v] in G2.
So by applying a 2 coloring BFS I can now find if G2 is bipartite or not.
and same applies for the G1 since these two are structurally same. But this method is costly as am using extra space for graph. Though this will suffice my purpose as of now, I'd like know if there any better implementations for the same.
Thanks In advance.
You can execute the algorithm to find the 2-partition of an undirected graph on a directed graph as well, you just need a little twist. (BTW, in the algorithm below I assume that you will eventually find a 2-coloring. If not, then you will run into a node that is already colored and you find you need to color it to the other color. Then you just exit saying it's not bipartite.)
Start from any node and do the 2-coloring by traversing the edges. If you have traversed every edge and every node in the graph then you have your partition. If not, then you have a component that is 2-colored and there are no edges leaving the component. Pick any node not in the component and repeat. If you get into a situation when you have a few components that are all 2-colored, and there are no edges leaving any of them, and you encounter an edge that originates in a node in the component you are currently building and goes into a node in one of the previous components then you just merge the current component with the older one (and possibly need to flip the color of every node in one of the components -- flip it in the smaller component). After merging just continue. You can do the merge, because at the time of the merge you have scanned only one edge between the two components, so flipping the coloring of one of the components leaves you in a valid state.
The time complexity is still O(max(|N|,|E|)), and all you need is an extra field for every node indicating which component that node is in.
I want to use d3 force-directed layout for visualizing my graph. However, I want to show edge direction since it is a directed graph. Is it possible to show this using d3?