How to Build Up the Betweenness Centrality Score

Question

Can the Betweenness Centrality score FOR EACH VERTEX in a graph be deconstructed as follows. For instance, in the graph below, the betweenness centrality score of vertex D is 3.5. Can this calculated value of 3.5 be broken out in terms of the following columns in a table.

Here is the code for the graph and the overall betweenness centrality measure.

Clear[g]
edges = {A \[UndirectedEdge] B, A \[UndirectedEdge] C, 
   B \[UndirectedEdge] D, C \[UndirectedEdge] D, 
   D \[UndirectedEdge] F};
g = Graph[edges, VertexLabels -> "Name", VertexLabels -> Automatic, 
  VertexSize -> 0.01, VertexLabelStyle -> 14]

BetweennessCentrality[g];
SortBy[{VertexList[g], BetweennessCentrality[g]}\[Transpose], N@*Last]

 {{F,0.}, {A,0.5}, {B,1.}, {C,1.}, {D,3.5}}

Suppose the graph is directed, such as

edges = {TX -> R1, R1 -> R3, R3 -> R5, R5 -> RX, TX -> R2, R2 -> R4, 
   R4 -> R6, R6 -> RX, R1 -> R2, R2 -> R3, R3 -> R4, R4 -> R5, 
   R5 -> R6, TX -> R7, TX -> R8, R7 -> R9, R9 -> R6, R9 -> RX, 
   R8 -> R6};
g = Graph[edges, VertexLabels -> Automatic]

In this case, the Betweenness of R5=4 which is not the same value from the code provide for the undirected graph case at the top of this note.

Answer 1

ClearAll[spathsallpairs, spathsthrough, numberofshortestpaths, numberofspthrough]

spathsallpairs = Module[{g = #}, 
    FindPath[g, ##, {GraphDistance[g, ##]}, All] & @@@ Subsets[VertexList@g, {2}]] &;

spathsthrough[g_, v_] := DeleteDuplicates[
  Join @@ (Cases[{a_, ___, v, ___, b_} :> {a, b}] /@ spathsallpairs[g])]

numberofshortestpaths = Length @ FindPath[#, ##2, {GraphDistance[#, ##2]}, All] &;

numberofspthrough[v_] := Count[{_, ___, v, ___, _}]@
    FindPath[#, ##2, {GraphDistance[#, ##2]}, All] &;

Example:

edges = {"A" \[UndirectedEdge] "B", "A" \[UndirectedEdge] "C", 
   "B" \[UndirectedEdge] "D", "C" \[UndirectedEdge] "D", "D" \[UndirectedEdge] "F"};
g = Graph[edges, VertexLabels -> "Name", VertexLabels -> Automatic, 
  VertexSize -> 0.01, VertexLabelStyle -> 14]

 spathsthrough[g, "D"]

{{"A", "F"}, {"B", "C"}, {"B", "F"}, {"C", "F"}}

ndst = numberofspthrough["D"][g, ##] & @@@ spathsthrough[g, "D"]

{2, 1, 1, 1}

nst = numberofshortestpaths[g, ##] & @@@ spathsthrough[g, "D"]

{2, 2, 1, 1}

ratios = ndst/nst;

Grid[Join[{{"{v1,v2}", "number of sp thru D", "number of sp", "ratio"}}, 
  Transpose[{spathsthrough[g, "D"], ndst, nst, ratios}], 
  {{"", "", "Total: ", N@Total@ratios}}], 
 Dividers -> All,  Alignment -> Center]