Question: I am trying to get the output of my jQuery in google when displaying.
However my jQuery does not work anymore and it keep displaying this debugger:///VM###. 

This is my code:

```
 $(document).ready(function () {
        var getJsonData = $.getJSON("json/travel.json")
            .done(function (data) {
                console.log("hello");
            });

```

Method I have tried:

1. clear console
2. restarting computer
3. restarting app
4. refreshing my cache

Did it save to the server? Any solution or advice? Thanks!

[Image](https://i.stack.imgur.com/Abdnd.png)

Answer: 
Since you want 29,281 resulting graphs, labelling must be important for you (IOW, you're not modding out by isomorphism.) Using a brute-force approach in networkx:

```
from itertools import combinations, product
import networkx as nx

def gen_dag(num_nodes):
    all_edges = list(combinations(range(num_nodes), 2))
    for p in product([None, 1, -1], repeat=len(all_edges)):
        G = nx.DiGraph()
        G.add_nodes_from(range(num_nodes))
        edges = [edge[::edge_dir] for edge, edge_dir in zip(all_edges, p) if edge_dir]
        G.add_edges_from(edges)
        if nx.is_directed_acyclic_graph(G):
            yield G

```

which gives

```
In [75]: graphs = list(gen_dag(5))

In [76]: len(graphs)
Out[76]: 29281

```

and (for example)

```
In [79]: graphs[1234].edges()
Out[79]: OutEdgeView([(3, 1), (3, 4), (4, 0), (4, 2)])

In [80]: nx.adjacency_matrix(graphs[1234]).todense()
Out[80]: 
matrix([[0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0],
        [0, 1, 0, 0, 1],
        [1, 0, 1, 0, 0]], dtype=int64)

```
It may be easier to think of your graphs simply as a connection of edges (the nodes can be kept constant).

Start from numEdges = 0 (unconnected graph) and go to numEdges = numNodes - 1 (fully-connected graph). For each numEdges, simply place edges in every possible permutation.

Set your possible edges using the fully-connected graph. For five nodes for example:

```
AB
AC
AD
AE
BC
BD
BE
CD
CE
DE

```

Note the pattern: A (B, C, D, E), B (C, D, E), C(D, E), D(E)

Once you prepare your list of possible edges, it should be straightforward to place down edges according to the permutations.