I have a list (of lists) and like to identify elements that have the same "structure"! Here is a minimal example:
example = { {{m1,m2},{m2,m4}} , {{m1,m2},{m1,m4}} }
This list has two elements {{m1,m2},{m2,m4}} and {{m1,m2},{m1,m4}} (which both are lists again). Now, in the problem I am working on each element is associated to a product of Kronecker deltas according to
{{m1,m2},{m2,m4}} -> delta(m1,m2) delta(m2,m4)$\delta_{m1,m2} \delta_{m2,m4}$
with delta(i,j)$\delta_{i,j}$ the Kronecker delta. In that sense, {{m1,m2},{m2,m4}} and {{m1,m2},{m1,m4}} are mathematically identical (have the "same structure").
Question: How can I write a Mathematica code that identifies these elements as the same and returns me a new list, where each element with the same structure appears only once together with its "multiplicity" in the original list?
My (failed) solution: I thought I could work with graphs and associate graphs G1 and G2 to {{m1,m2},{m2,m4}} and {{m1,m2},{m1,m4}}, respectively, with corresponding edges as indicated by the pairs. Then, G1 and G2 literally look the same, yet when I try
Tally[{G1,G2}]
I do not get {G1,2}, but {{G1,1},{G2,1}}. So somehow Mathematica is unable to identify graphs with the "same structure" as the same.
How can I solve the problem? Any hint is greatly appreciated! Many thanks!
Edit in response to comments:
The example above is a real sample list: m1, m2, ... are indices that never take on any numerical value. I just want to order them according to the structure I mentioned. The full code I tried is
Test = {{{m1, m2}, {m2, m4}}, {{m1, m2}, {m1, m4}}};
graphs = {};
For[i = 1, i <= Length[Test], i++,
mvert = {m1, m2, m3, m4};
medges =
Table[Test[[i, j]][[1]] <-> Test[[i, j]][[2]], {j, 1,
Length[Test[[1]]]}];
graphs = Append[graphs, {Graph[mvert, medges]}];
]
Tally[graphs]
but it doesn't work. I think Dimensions and CountsBy also will not do the job, GroupBy might help but it seems complicated to find the right function according by which to group (that's why I used the graph method because I thought that's the best idea).
