What if “no self loops” assumption is not taken into account? Whether the graphs can be uniquely determined in this case?
W1 =
2
664
0 1 4 0
1 0 5 2
4 5 0 1
0 2 1 0
3
775
W2 =
2
664
0 2 3 0
1 0 3 0
1 5 0 4
0 2 1 0
3
775
:
If \self loops" assumption is taken into account, we need to have the edge weight of the self
loop for obtaining a unique answer. Without the edge weight of the self loop, we will not be
able to obtain a unique answer. The answer below assumes a unit weight for self loops for
each node.
W1 =
2
664
0 1 4 0
1 0 5 2
4 5 0 1
0 2 1 0
3
775
W2 =
2
664
0 2 3 0
1 0 3 0
1 5 0 4
0 2 1 0
3
775
:
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