Your situation is a made trivial by its symmetry. My comment and the other answers point this out. Were it less trivial, a direct tool (with limited application) called the delta-wye conversion could be applied here. And still more generally, nodal analysis (KCL) and mesh analysis (KVL) are commonly applied to problems still more complicated than this one. These (KCL and KVL) are very general/powerful tools, indeed. You need go no further to handle most of what you may find in electronics.
But there are still more powerful tools that not only handle KCL/KVL problems but far more. Everything from harmonics, probability, Dirichlet's principle, Thomson's principle, Tellegen's theorem, and even into string theory. (Electronics as only a minor bump along the road, so to speak.)
Spice uses a subset of the following method. So it's worth getting exposed even if you never directly use them.
A circuit is a directed graph. In your case, something like this:
In the above, I've removed the battery and simply show the battery current entering and leaving. (Rank-Nullity Theorem explains why. For another day.)
If the nodes are columns and the edges are rows then the connection matrix \$A\$ and conductance matrix \$C\$ look like this:
$$\begin{align*}
\begin{smallmatrix}
\fbox{A}&N_0&N_1&N_2&N_3\\
\hline
a:\vphantom{\frac1{R_b}}&\hfill -1&\hfill 0&\hfill 1&\hfill 0\\
b:\vphantom{\frac1{R_b}}&\hfill -1&\hfill 0&\hfill 0&\hfill 1\\
c:\vphantom{\frac1{R_b}}&\hfill 0&\hfill 0&-1&\hfill 1\\
d:\vphantom{\frac1{R_b}}&\hfill 0&\hfill 1&\hfill -1&\hfill 0\\
e:\vphantom{\frac1{R_b}}&\hfill 0&\hfill 1&\hfill 0&\hfill -1
\end{smallmatrix}
&&
\begin{smallmatrix}
\fbox{C}&a&b&c&d&e\\
\hline
a:& \frac1{R_a}& 0& 0& 0& 0\\
b:& 0&\frac1{R_b}&\hfill 0&0&0\\
c:&0& 0&\frac1{R_c}& 0&0\\
d:&0&0&0&\frac1{R_d}& 0\\
e:& 0& 0& 0&\hfill 0&\frac1{R_e}
\end{smallmatrix}
\end{align*}$$
For the connection matrix \$A\$ on the left, a -1 says that the arrow is pointing away from that node and a 1 says that the arrow is pointing into that node. An edge must point into and out of two nodes, so the sum on a row must be 0. (You don't have to use the same directions I did. You can choose them, arbitrarily.) For the conductance matrix \$C\$, all that is done is to list the edge conductances down the diagonal. The rest is all zero.
We can arbitrarily set \$V_0=1\:\text{V}\$ and \$V_1=0\:\text{V}\$ and try to find out what \$I\$ becomes. Then it's easy to work out the resistance.
From KCL (in the rowspace and right nullspace) and Ohm's Law (in the columspace and left nullspace) we know that \$A^T\,C\,A\,\left[\begin{smallmatrix}1 \\ 0\\v_2\\v_3\end{smallmatrix}\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\right]=\left[\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}\right]\$. We can break up the matrix product, \$A^T\,C\,A\$, into four quadrants that map out as \$\left[\begin{smallmatrix}P&Q^T\\Q&R\end{smallmatrix}\right]\$. From that, we have \$\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\begin{smallmatrix}P&& Q^T\\\\\\Q&& R\end{smallmatrix}\right]\,\left[\begin{smallmatrix}1 \\ 0\\v_2\\v_3\end{smallmatrix}\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\right]=\left[\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}\right]\$ and then we can apply the Schur complement concept to find \$\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}P-Q^T\,R^{-1}\,Q\right]\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\begin{smallmatrix}\hfill 1\\\\\\\\\hfill 0\end{smallmatrix}\right]=\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\begin{smallmatrix}\hfill I\\\\\\\\-I\end{smallmatrix}\right]\$.
(By using the Schur complement, the matrix work was reduced as we don't care about the voltages, \$V_2\$ and \$V_3\$, and so don't need to solve the entire system of KCL equations to find what we needed. We can focus on the bits of interest. This ability to narrow the focus is difficult to see when just applying KCL and/or KVL.)
That will solve for \$I\$ (and \$-I\$, which we don't need.) Since the voltage applied was \$1\:\text{V}\$, the resistance will be \$R=\frac{1\:\text{V}}{I}\$.
The matrix equation helps yield the following general solution for \$R\$:
$$R=\frac{R_{a} R_{b} R_{c} + R_{a} R_{b} R_{d} + R_{a} R_{b} R_{e} + R_{a} R_{c} R_{e} + R_{a} R_{d} R_{e} + R_{b} R_{c} R_{d} + R_{b} R_{d} R_{e} + R_{c} R_{d} R_{e}}{R_{a} R_{c} + R_{a} R_{d} + R_{a} R_{e} + R_{b} R_{c} + R_{b} R_{d} + R_{b} R_{e} + R_{c} R_{d} + R_{c} R_{e}}$$
But if you simplify things by saying that \$R_{a}=R_{b}\$ and \$R_{d}=R_{e}\$ then the above reduces to \$\frac12\left(R_a+R_d\right)\$ and you can see that \$R_c\$ is no longer any part of the solution answer. That means it doesn't matter what value it takes. It has no impact.
As already mentioned, the above is part of how Spice programs work.
A symbolic program like Sympy can also be used:
ra,rb,rc,rd,re = symbols('ra,rb,rc,rd,re',real=True)
A = Matrix([[-1,0,1,0],[-1,0,0,1],[0,0,-1,1],[0,1,-1,0],[0,1,0,-1]])
C = Matrix([[1/ra,0,0,0,0],[0,1/rb,0,0,0],[0,0,1/rc,0,0],[0,0,0,1/rd,0],[0,0,0,0,1/re]])
M = A.transpose()*C*A
P = M[[0,1],[0,1]]
Q = M[[2,3],[0,1]]
R = M[[2,3],[2,3]]
simplify(1/(((P-Q.transpose()*R.inv()*Q)*Matrix([[1],[0]]))[0]))
(rarbrc + rarbrd + rarbre + rarcre + rardre + rbrcrd + rbrdre + rcrdre)/(rarc + rard + rare + rbrc + rbrd + rbre + rcrd + rcre)
simplify(1/(((P-Q.transpose()R.inv()Q)*Matrix([[1],[0]]))[0]).subs({rb:ra,re:rd}))
ra/2 + rd/2
If they were swapped, the resistance going into each side of the 1Ω would be different, wouldn't that create difference in either current or voltage across the 1Ω?
– Gumpf Mar 09 '23 at 13:15