Random graphs with fixed numbers of neighbours

In connection with Le Monde puzzle #46, I eventually managed to write an R program that generates graphs with a given number n of nodes and a given number k of edges leaving each of those nodes. (My early attempt was simply too myopic to achieve any level of success when n was larger than 10!) Here is the core of the R code:

A=42 #number of nodes
L=13 #number of edges
ApL=A+L

if ((A*L)%%2==1){
 print("impossible graph")
}else{
 con=matrix(0,A,A)
 diag(con)=A      #eliminate self-connection
 suma=apply(con,1,sum)-A

 while (min(suma)<L){

 if (sum(suma<L)==1){ #bad news: no correspondence!
                      #go back:
   con=aclrtr(con,L)
   diag(con)=A
   suma=apply(con,1,sum)-A

 }else{

   j=sample((1:A)[suma<L],1)
   slots=(1:A)[con[j,]==0]  #remaining connections
   if (length(slots)==1){
     vali=slots
     if (sum(con[vali,]>ApL-1)) vali=NULL
   }else{
     vali=slots[apply(con[slots,],1,sum)<ApL]
   }

   if (length(vali)==0){

     con=aclrtr(con,L)
     diag(con)=A
     suma=apply(con,1,sum)-A

   }else{

    if (length(vali)==1){
      k=vali[1]
    }else{
      k=sample(slots[apply(con[slots,],1,sum)<ApL],1)
    }
    con[k,j]=con[j,k]=1
    suma=apply(con,1,sum)-A
}}}}

and it uses a sort of annealed backward step to avoid simulating a complete new collection of neighbours when reaching culs-de-sac….

aclrtr=function(con,L){
#removes a random number of links among the nodes with L links

A=dim(con)[1]
ApL=A+L
while (max(apply(con,1,sum))==ApL){

 don=sample(1:(L-1),1)
 if (sum(apply(con,1,sum)==ApL)==1){
 i=(1:A)[apply(con,1,sum)==ApL]
 }else{
 i=sample((1:A)[apply(con,1,sum)==ApL],1)
 }
 off=sample((1:A)[con[i,]==1],don)
 con[i,off]=0
 con[off,i]=0
 }
con
}

There is nothing fancy or optimised about this code so I figure there are much better versions to be found elsewhere…

Ps-As noticed before, sample does not work on a set of length one, which is a bug in my opinion…. Instead, sample(4.5,1) returns a random permutation of (1,2,3,4).

> sample(4.5)
[1] 4 3 1 2