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How do I know my vertex has a connection?

By : user2175078
Date : October 16 2020, 08:10 AM
fixed the issue. Will look into that further Well, when you get the output [null, null, null, [null, null, null, null]], you know there is a reference to your connected node. So the second array is connected by the first, but not vice versa. Because your code don't connect the second node, but the first is connected.I think you should't print the List of nodes, this will lead to problems.
One case, the Nodes are connected together so you have Parent connected to Child and Child connected to Parent. When you now print this out, you will get a recursive execution.
code :
private Node north;
private Node south;
private Node west;
private Node east;
private Node[][] 

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For a struct vertex, what's the difference between map[int]vertex and map[int]*vertex?

Date : March 29 2020, 07:55 AM
I hope this helps you . The main difference is that map[int]vertex stores vertex values and map[int]*vertex stores vertex references (pointers). The output of the following program should help illustrate:
code :
package main

type vertex struct {
    x, y int

func main() {
    a := make(map[int]vertex)
    b := make(map[int]*vertex)

    v := &vertex{0, 0}
    a[0] = *v
    b[0] = v

    v.x, v.y = 4, 4
    println(a[0].x, a[0].y, b[0].x, b[0].y)
0 0 4 4

How do I get indexed vertex positions AND indexed vertex normals from wavefront .obj file into an OpenGL vertex shader?

By : Balagurunathan Mural
Date : March 29 2020, 07:55 AM
I wish this helpful for you One thing you must understand is, that a "vertex" is not synnonymous with "position". A vertex is the whole set of attributes; position, normal, texture coordinate, etc. OpenGL processes geometry one primitive at a time, where a primitive consists of 1 (point), 2 (line) or 3 vertices (triangle). You can not process position and normal separately, they always must go together.
Before you can process it in OpenGL you must expand the data in the .obj file. For each unique tuple of attributes introduce a new vertex with a new index and replace indexed attributes from the .obj with the indexed vertex OpenGL requires.

Vertex connection by Yes/No in R Language

By : shanik
Date : March 29 2020, 07:55 AM
should help you out This is the output from dput() as @Thomas suggested. This is a great way to share data with folks on stackoverflow, so that they can easily grab it and test code out on it.
code :
df <- structure(list(V1 = c("Yes", "Yes", "No"), V2 = c("Yes", "Yes", 
    "No"), V3 = c("Yes", "No", "Yes"), V4 = c("No", "No", "No"), 
    V5 = c("No", "No", "No"), V6 = c("Yes", "Yes", "No"), V7 = c("No", 
    "No", "No"), V8 = c("No", "No", "No"), V9 = c("No", "No", 
    "No"), V10 = c("Yes", "Yes", "No"), V11 = c("Yes", "No", 
    "No"), V12 = c("No", "No", "No"), V13 = c("Yes", "Yes", "No"
    ), V14 = c("Yes", "No", "No"), V15 = c("No", "No", "No"), 
    V16 = c("No", "No", "No"), V17 = c("Yes", "No", "No"), V18 = c("No", 
    "No", "No"), V19 = c("No", "No", "No"), V20 = c("No", "No", 
    "No"), V21 = c("No", "No", "No"), V22 = c("No", "No", "No"
    )), .Names = c("V1", "V2", "V3", "V4", "V5", "V6", "V7", 
    "V8", "V9", "V10", "V11", "V12", "V13", "V14", "V15", "V16", 
    "V17", "V18", "V19", "V20", "V21", "V22"), class = "data.frame", 
    row.names = c("Fish", "Squid", "Pigs"))
# define the vertices
vertices <- row.names(df)
L <- length(vertices)

# set up empty data frame for the edges
numedges <- choose(L, 2)
edges <- data.frame(v1=rep(NA, numedges), v2=NA, numrows=NA)

# cycle through all possible pairs of vertices
# total up the number of people answering "Yes" to both
k <- 0
for(i in 1:(L-1)) {
for(j in (i+1):L) {
    k <- k + 1
    edges$v1[k] <- vertices[i]
    edges$v2[k] <- vertices[j]
    edges$numrows[k] <- sum(df[vertices[i], ]=="Yes" & df[vertices[j], ]=="Yes")
     v1    v2 numrows
1  Fish Squid       5
2  Fish  Pigs       1
3 Squid  Pigs       0

Let G be a directed graph. There is a special vertex S such that there is a path to this vertex from every other vertex

By : Nitin Soni
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further It is not sufficient to show that v is in a sink SCC, even if we knew that the undirected form of the graph is connected. This is because there may be multiple sink SCCs and one sink cannot be reached from another.
If your question is: "given a vertex v, what is the best way to determine if v is reachable from every other vertex of the graph," then you should start at v and follow all of the edges backwards. If you can reach every vertex by following edges backwards, then that means v is reachable from every other vertex of the graph.

How to remove an edge connection to a vertex in OrientDB?

By : T.Kiarie
Date : March 29 2020, 07:55 AM
this will help It's not properly a GUI thing, but you can ignore it.
When you create an edge and connect it to a vertex, OrientDB creates a collection of links (a RIDBAG) as a property of the vertex. When you delete edges, the edge pointer is removed from the collection, but the collection itself is not removed.
code :
UPDATE User REMOVE in_HasAddress 

/* or out_HasAddress if you want to remove the outgoing edges collection */
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