Convert image of graph to Graph [duplicate]

On my system, MorphologicalGraph with the default Method can't seem to find all the vertices of the graph. I ended up with the following to get an initial graph that at least has all the vertices:

g=MorphologicalGraph[ Thinning[Binarize @ ColorNegate @ image, 3, Method->"MedialAxis"] ] 

This version of the graph had even more vertices than yours:

VertexCount @ g (* 163 *) 

Rather than trying to adjust the Image so that MorphologicalGraph would produce fewer vertices, I decided to post-process the Graph to eliminate extra vertices. First, here are the initial vertices:

vertexRules = Thread @ Rule[ PropertyValue[g, VertexCoordinates], VertexList[g] ]; 

Now, using the fact that there should be 47 vertices, we can use FindClusters to group vertices that should be the same:

clusters = FindClusters[vertexRules, 47]; 

If I just use the FindClusters[vertexRules] I get 48 clusters, because one of the vertices at the top right gets displaced too far from the others. At any rate, using these clusters, we can come up with a list of rules to eliminate extra vertices:

substitutes = Flatten @ Replace[ clusters, { {_}->Sequence[], {a__,b_}:>Thread[Rule[{a},b]] }, {1} ] 

Using the above rules, we can replace the vertices in the original edge list with a representative from each cluster:

edges = DeleteDuplicates @ DeleteCases[ EdgeList[g] /. substitutes, x_<->x_ ] 

After replacing the vertices with the representatives, there are many self-loops and multiedges that need to be pruned, which explains the need for DeleteDuplicates and DeleteCases. Now, all that's left to do is to position the vertices in the right spots:

graph = Graph[ edges, VertexShapeFunction -> "Square", VertexSize -> Large, VertexCoordinates -> SortBy[vertexRules, Last][[VertexList[edges], 1]] ]