Is it possible to compute a valid Laplacian matrix from an effective resistance matrix? I am wondering whether it is possible to retrieve a node-admittance matrix $G$ (also called Laplacian matrix) in a purely resistive network composed of nets $\{1, \dots, i, \dots, j, \dots, n\}$, from the effective resistance matrix $R$ of the network.
Generally, $R$ is calculated from the pseudo-inverse $G^+$ of $G$ such that $R_{ij}$, the element $(i,j)$ in $R$, which corresponds to the effective resistance between nets $i$ and $j$, is:
$R_{ij} = G_{ii}^+ + G_{jj}^+ - 2\cdot G_{ij}^+$.
What I would like to do is the contrary, i.e. to deduce the Laplacian matrix $G$ from the matrix $R$. Let's consider an example involving a resistor network made of $4$ nets. Here, I am supposed to only know the matrix $R$. It is symmetric and its diagonal elements are zero:
$R= \begin{bmatrix}
0 & R_{12} & R_{13} & R_{14} \\
R_{12} & 0 & R_{23} & R_{24} \\
R_{13} & R_{23} & 0 & R_{34} \\
R_{14} & R_{24} & R_{34} & 0
\end{bmatrix}$.
The matrix $G$ to infer can be written as follows:
$G = \begin{bmatrix}
G_{11} & G_{12} & G_{13} & G_{14} \\
G_{12} & G_{22} & G_{23} & G_{24} \\
G_{13} & G_{23} & G_{33} & G_{34} \\
G_{14} & G_{24} & G_{34} & G_{44}
\end{bmatrix}$.
As $G$ is symmetric, I have only $10$ values to find in this example.
Moreover, let be its pseudo-inverse $G^+$ such that:
$G^+ = \begin{bmatrix}
G_{11}^+ & G_{12}^+ & G_{13}^+ & G_{14}^+ \\
G_{12}^+ & G_{22}^+ & G_{23}^+ & G_{24}^+ \\
G_{13}^+ & G_{23}^+ & G_{33}^+ & G_{34}^+ \\
G_{14}^+ & G_{24}^+ & G_{34}^+ & G_{44}^+
\end{bmatrix}$.
The approach I followed is to write a relation between the non-zero upper triangular blocks of $R$ and $G^+$ in the form below:
$\begin{bmatrix}
R_{12} \\
R_{13} \\
R_{14} \\
R_{23} \\
R_{24} \\
R_{34}
\end{bmatrix}
=\begin{bmatrix}
1 & -2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & -2 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
1 & 0 & 0 & -2 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & -2 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & -2 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -2 & 1
\end{bmatrix} \cdot
\begin{bmatrix}
G_{11}^+ \\
G_{12}^+ \\
G_{13}^+ \\
G_{14}^+ \\
G_{22}^+ \\
G_{23}^+ \\
G_{24}^+ \\
G_{33}^+ \\
G_{34}^+ \\
G_{44}^+ \\
\end{bmatrix}$.
Let's use a simpler expression:
$R_{vec} = A \cdot G_{vec}^+$.
I then computed the pseudo-inverse $A^+$ of $A$ in order to find $G_{vec}^+ = A^+ \cdot R_{vec}$.
Afterward I reconstructed the matrix $G^+$ from $G_{vec}^+$ and computed the pseudo-inverse of $G^+$, $G^{++}$, hoping that I would obtain a valid Laplacian matrix. Well... in the example I used, $G^{++}$ is symmetric, has positive diagonal elements and negative off-diagonal elements, but the absolute value of each diagonal element is not equal the the sum of absolute values of off-diagonal elements in the same row, which is an essential property of the Laplacian matrix.
Please could you help me to figure out how I can find a valid Laplacian matrix $G$, if possible?
I thank you in advance.
 A: Yes, it is possible to recover the Laplacian matrix $G$ from the resistance matrix $R$.
The Laplacian matrix is essentially the inverse of the resistance matrix, up to a scaling constant and adding a rank-one symmetric matrix ''correction term''.
Let $\mu$ be the vector in $\mathbb R^n$ defined by
$$\mu_i = 2 - \sum_{j \in N(i)} R_{ij},$$
where $N(i)$ denotes the neighbor indices of node $i$.
Then
$$ G =  - 2 R^{-1} + \frac{2}{\mu^T R \mu} \mu\mu^T .$$
The above formula is due to Bapat, ''Resistance matrix of a weighted graph'', Theorem 3.
See also Bapat and Sivasubramanian, ''Identities for minors of the Laplacian, resistance and distance matrices'', Theorem 5.

As an example, consider the ''tadpole graph'' consisting of an edge attached at one end to a triangle.
Then the resistance matrix and Laplacian matrix are
$$R = \begin{bmatrix}
0 & 2/3 & 2/3 & 5/3 \\
2/3 & 0 & 2/3 & 5/3 \\
2/3 & 2/3 & 0 & 1 \\
5/3 & 5/3 & 1 & 0
\end{bmatrix}
\qquad\text{and}\qquad
G = \begin{bmatrix}
2 & -1 & -1 & 0 \\
-1 & 2 & -1 & 0 \\
-1 & -1 & 3 & -1 \\
0 & 0 & -1 & 1
\end{bmatrix}.
$$
In terms of the above formula we have
$$
\mu = \begin{bmatrix}2/3 \\ 2/3 \\ -1/3 \\ 1 \end{bmatrix},
\qquad 
\mu \mu^T = \frac1{9}\begin{bmatrix}
4 & 4 & -2 & 6 \\
4 & 4 & -2 & 6 \\
-2 & -2 & 1 & -3 \\
6 & 6 & -3 & 9
\end{bmatrix},
\qquad
\frac{2}{\mu^T R \mu} = \frac{9}{17},
$$
and
$$
-2 R^{-1} = \frac1{17}\begin{bmatrix}
30 & -21 & -15 & -6 \\
-21 & 30 & -15 & -6 \\
-15 & -15 & 50 & -14 \\
-6 & -6 & -14 & 8
\end{bmatrix}.
$$
