全文链接：https://tecdat.cn/?p=33183
PROBLEM 1) Creating Random Adjacency Matrices
Script Name: adjMatrix Input: n... The number of vertices in the graph p... Probablity two vertices are connected plot... whether or not the matrix should be plotted as a graph Output: The nxn matrix of zero and ones Error Checking: The dimension is postive (else return NULL)
Description: The matrix is related to a simple, undirected graph of n vertices. In the graph is Vertex i and Vectex j are joined by an edge, then in the matrix A[i,j] = 1, if no edge exists then A[i,j]=0. There are differenct ways of handling the diagonal. We will require the diagonal elements to be NA.
The matrix must be symmetric. That is A[i,j]=A[j,1]
Whether or not an entry in the matrix is 0 or 1, is determined by drawing numbers from the binomial distribution. This distribution will yield a 1 with probability p and a o with probability (1-p). In R it will work like this, in the ith column (vertex i) , we require A[i,i] to be NA. This leaves n-1 entries to be determined. We use the command rbinom(n-1,1,p) to general these entries all at once, or rbinom(1,1,p) to generate them one at a time.
Plot: Using igraph create a simple plot. Use a color easy on the eyes such as something found in the Brewer color palette (discussed in class)
代码语言：javascript复制requires(igraph)
adjMatrix<-function(n,p,plot=FALSE)
{m=matrix(1,n,n)
for(i in 1:n){
for(j in 1:n){
if(i<j){
m[i,j]=ifelse(i==j,NA,rbinom(1,1,p) )
m[j,i]=m[i,j]
}
}
}
colpal<-brewer.pal(8,"Dark2")
gA <- graph.adjacency(m, mode="undirected",diag=FALSE )
V(gA)$color<-colpal[1] You choose 1 to 8... try some others
plot(gA, layout=layout.auto,main="Random Graph using Brewer Color Palette")
s
}PROBLEM 2) Person with the most friends in the network
Script Name: vowelMax Input: A... An adjacency matrix plot... Plot the graph with the "friends" network highlighted Output: The number of the vertex with the most edge connections Error Checking: None
Description: The number of "friends" a person has in a network is the number of edges connected to that vertex. The ith person in the network, will be connected to the jth person if A[i,j]=1. So the number of 1's in the ith column reports how many edges connect person i to others In mathematical terms, this is called the "degree" of vertex i.
So you must determine the number of ones in each column and then find which column has the most ones.
Hint: There is a quick trick to use here. Since a column has only 0 or 1 in it, the sum of all of the numbers in the column is the same as the number of 1's in the column. Prove that to yourself.
Plot: Use igraph to plot the graph. Use three colors for the vertices: one for the most friendly person, another for his/her immediate friends, and another for the remaining vertices. Color red the edges from the most friendly person to his/her friends. All remaining edges are black.
代码语言：javascript复制mostFriends<-function(A,plot=FALSE)
{
A[is.na(A)]=0
s=which(apply(A,2,sum)==max(apply(A,2,sum)))
n=which(A[,s]==1)
pastel<-brewer.pal(8,"Pastel2") Try a different palette
gA <- graph.adjacency(A, mode="undirected",diag=FALSE )
V(gA)[n]$color<-pastel[1]
V(gA)[s]$color<-pastel[2] changes the 5 vertex to the 4th color of the palette
V(gA)[n]$color<-pastel[3]
E(gA)[s%--%n]$color<-"red"
plot(gA, layout=layout.auto,main="Random Graph With A Range of Vertices Colored")
s
}PROBLEM 3) Find a lonely person
Script Name: noFriends Input: A... An adjacency matrix Output: The number of the vertex with the no edge connections Error Checking: None
Description: The number of "friends" a person has in a network is the number of edges connected to that vertex. The ith person in the network, will be connected to the jth person if A[i,j]=1. So the number of 1's in the ith column reports how many edges connect person i to others In mathematical terms, this is called the "degree" of vertex i.
So you must determine the number of ones in each column and then find which column has the most ones.
Hint: There is a quick trick to use here. Since a column has only 0 or 1 in it, the sum of all of the numbers in the column is the same as the number of 1's in the column. Prove that to yourself.
Plot: Create a plot of the graph using igraph. Make the "lonely" person/people a different color
代码语言：javascript复制noFriends<-function(A,plot=FALSE)
{
s=which(apply(A,2,sum)==0))PROBLEM 4) Number of friendship relationships in the network
Script Name: numFriendships Input: A... An adjacency matrix Output: The number of uniqe edges in the graph
Description: The number of friendships in the network is the number of edges in the graph. Basically, count the ones in the adjacency matrix, but be careful. (Remember, an edge from i to j is the same as an edge from j to i)
代码语言：javascript复制numFriendships<-function(A)
{
num
}PROBLEM 5) Add a person to the friendship network
Script Name: addPerson Input: A... An adjacency matrix p... Probability a relationship forms plot... Show the new person and relationships Output: New adjacency matrix
Description: As in creating an adjacency matrix, a new column is added and the binomail distribution is used to determine the 1 entries.
Plot: Plot the graph using igraph. Use three colors for the vertices: one for the newly added person, another for his/her immediate friends, and another for the remaining vertices. Color red the edges from the new person to his/her friends. All remaining edges are black.
代码语言：javascript复制addPerson<-function(A,p,plot)
{
n=nrow(A)
new=rep(1,n)
for(i in 1:n){
new[i]=rbinom(1,1,p)
}
A=cbind(A,new)PROBLEM 6) Add several people to the friendship network
Script Name: growNetwork Input: A... An adjacency matrix p... Probability a relationship forms n... the number of people to add plot... Show the new person and relationships Output: New adjacency matrix
Plot: Plot the graph using igraph. Use three colors for the vertices: one for the newly added person, another for his/her immediate friends, and another for the remaining vertices. Color red the edges from the new person to his/her friends. All remaining edges are black.
代码语言：javascript复制growNetwork<-function(A,n,p,plot=FALSE)
{row=nrow(A)
for(ii in 1:n){
nn=nrow(A)
new=rep(1,nn)
for(i in 1:nn){
new[i]=rbinom(1,1,p)
}
