@Heeka Note that there is a one-to-one correspondence between the Eulerian circuits in the original graph and in my enlarged graph.

@Markiyan Hirnyk My first answer was to a question that you deleted---a question about the examples on the help page ?LinearAlgebra,Modular,LinearSolve . I have since posted an answer to your second question.

@Markiyan Hirnyk My first answer was to a question that you deleted---a question about the examples on the help page ?LinearAlgebra,Modular,LinearSolve . I have since posted an answer to your second question.

@Markiyan Hirnyk There are two problems with your command

B:= LinearAolve(2, A, 2);

The first is that the return value of LinearSolve with the default inplace is NULL. So even if the command worked, B would just be NULL. 

The second is that the second 2 in your command says that the rightmost two columns of A are "augmented". In other words, the matrix A represents a linear system C.X = D where C is 3x2 and D is 3x2, the leftmost two columns of A being the C and the rightmost two columns being the D. This system is clearly inconsistent, its bottom row being < 0 0 | 1 1 >. If LinearSolve chooses to call that "singular" rather than "inconsistent", I am willing to forgive it. If it is to be considered a bug, it is only because of the choice of words in the error message.

@Markiyan Hirnyk There are two problems with your command

B:= LinearAolve(2, A, 2);

The first is that the return value of LinearSolve with the default inplace is NULL. So even if the command worked, B would just be NULL. 

The second is that the second 2 in your command says that the rightmost two columns of A are "augmented". In other words, the matrix A represents a linear system C.X = D where C is 3x2 and D is 3x2, the leftmost two columns of A being the C and the rightmost two columns being the D. This system is clearly inconsistent, its bottom row being < 0 0 | 1 1 >. If LinearSolve chooses to call that "singular" rather than "inconsistent", I am willing to forgive it. If it is to be considered a bug, it is only because of the choice of words in the error message.

@Markiyan Hirnyk The examples all seem consistent with what I said. With option inplace, the default, the result is obtained by re-examining the input matrix rather than by using the direct output of the command, which is NULL.

@Markiyan Hirnyk The examples all seem consistent with what I said. With option inplace, the default, the result is obtained by re-examining the input matrix rather than by using the direct output of the command, which is NULL.

@Markiyan Hirnyk Yes, of course, what I have created does not represent the original graph internally in Maple. This is a workaround required to create an approximate visualization of the Asker's original, which seemed to be what she wanted.

@Markiyan Hirnyk Yes, of course, what I have created does not represent the original graph internally in Maple. This is a workaround required to create an approximate visualization of the Asker's original, which seemed to be what she wanted.

@erik10 What is the maximum number of dintinct entries in A? It can't be 10000 because you wouldn't be able to visualize 10000 distinct vertical lines on one plot.

I multiplied by 1000 because I thought that that would cover any perceivable visual differences.

I don't understand what you're saying about no repetitions in the A vector. My code does not rely on there being repetitions.

If only the frequencies the given, then A could default to [$1..nops(B)] (i.e. [1, 2, ..., |B|]).

@erik10 What is the maximum number of dintinct entries in A? It can't be 10000 because you wouldn't be able to visualize 10000 distinct vertical lines on one plot.

I multiplied by 1000 because I thought that that would cover any perceivable visual differences.

I don't understand what you're saying about no repetitions in the A vector. My code does not rely on there being repetitions.

If only the frequencies the given, then A could default to [$1..nops(B)] (i.e. [1, 2, ..., |B|]).

Do you mean the package Gravitation written by John Fredsted?

Good solution. I like it better than my own.

Good solution. I like it better than my own.

@itsme Rather than copying and pasting, use lprint(%) and you'll get

hypergeom([2, 2, (-I*beta*epsilon+Pi)/Pi, (I*beta*epsilon+Pi)/Pi], [1, (-I*beta*epsilon+2*Pi)/Pi, (I*beta*epsilon+2*Pi)/Pi], 1/exp(4))/((beta^2*epsilon^2+Pi^2)*exp(4))