# Extending the Delta-Wye/∆-Y Transformation to higher polygons

I was looking at the derivation of the ∆-Y Transformation and decided to replicate the process with a square and its diagonals, i.e. I tried to create a Square-Cross/□-X transformation. But the equations so formed do not seem to have a solution. There are 4 variables and 6 equations, and to check whether they have a solution I used the Rouché-Capelli theorem. The rank of the coefficient matrix is 4 while the rank of the augmented matrix is 5, which implies that no solutions exist. Am I doing something wrong or is extending the Delta-Wye transformation to higher polygons just not possible? If it's the latter, why not? • Food for thought: try triangulating the polygon (giving the added edges an admittance of 0, or a resistance of ∞Ω) and then applying the delta/wye transform wherever you want — so long as it doesn't require computing 0/0 or ∞/∞. That might lead to interesting equivalent forms for some polygons. Jul 9 at 16:52

The biggest problem is that the cross has more symmetry than the square does.

In the cross, none of the resistors have any special relationship to each other; you could exchange any two of the resistors, and re-label the corresponding terminals, and the result would be identical to what you started with. In the square, on the other hand, each resistor has two adjacent resistors and one opposite resistor. If you exchange two adjacent resistors in the square, the result will be substantially different.

The cross has a special property related to that symmetry. Suppose you measure the resistance between two terminals, and then you measure the resistance between the other two terminals, and you add the two readings. All three ways of doing this will yield the same result (which is simply the sum of the values of the four resistors).

The square doesn't have the same property. Suppose all the resistors are 4 ohms. If you measure the resistance between two diagonally opposite terminals, you'll get 4 ohms (8 in parallel with 8), but if you measure the resistance between two adjacent terminals, you'll get 3 ohms (4 in parallel with 12). So a square with four equal resistors can't be converted to a cross. I suspect that the same goes for every square where all the resistor values are positive.

I think that, instead of converting a cross to a square, it should be possible to convert any cross to a tetrahedron, where there's a resistor between every terminal and every other terminal. However, not every tetrahedron can be converted back to a cross. (That follows from the fact that every square can be converted to a tetrahedron, but not every square can be converted to a cross.)

Cassie has the right of it. A generalized star-triangle conversion would be something like this:

• $$\Y\to \Delta\$$: Suppose $$\n\$$ vertices $$\\{v_1,v_2,\dots,v_n\}\$$ where $$\v_n\$$ is a special central vertex, with edges $$\\{e_1,e_2,\dots,e_{n-1}\}\$$ connecting them in a star. Then this transformation removes all these edges and the vertex $$\v_n\$$ and then adds $$\m=\frac12 \left(n-1\right) \left(n-2\right)\$$ edges fully connecting each remaining vertex to all other remaining vertices.
• $$\\Delta\to Y\$$: Suppose $$\n\$$ vertices $$\\{v_1,v_2,\dots,v_n\}\$$, with $$\m=\frac12 n \left(n-1\right)\$$ fully connecting edges $$\\{e_1,e_2,\dots,e_m\}\$$ forming triangle meshes. Then this transformation removes all these edges, adds the vertex $$\v_{n+1}\$$, and then adds $$\n\$$ edges connecting vertices $$\\{v_1,v_2,\dots,v_n\}\$$ to vertex $$\v_{n+1}\$$.

So, in your example case, you'd need six resistors to create the necessary triangle meshes, not four. You could definitely look at it as a tetrahedron as Cassie mentioned, already.

By the way, here's code I'd use for the usual $$\\Delta-Y\$$ conversion:

ra,rb,rc,rac,rbc,rab,va1,vb1,va2,vb2,vx1,vx2,i0=symbols('ra,rb,rc,rac,rbc,rab,va1,vb1,va2,vb2,vx1,vx2,i0',real=True,positive=True)
e1=Eq(va1/rac+va1/rab,i0+vb1/rab)
e2=Eq(vb1/rbc+vb1/rab,va1/rab)
e3=Eq(va2/rac+va2/rab,vb2/rab)
e4=Eq(vb2/rbc+vb2/rab,i0+va2/rab)
e5=Eq(va1/ra,i0+vx1/ra)
e6=Eq(vb1/rb,vx1/rb)
e7=Eq(va2/ra,vx2/ra)
e8=Eq(vb2/rb,i0+vx2/rb)
e9=Eq(vx1/ra+vx1/rb+vx1/rc,va1/ra+vb1/rb)
e10=Eq(vx2/ra+vx2/rb+vx2/rc,va2/ra+vb2/rb)
solve([e1,e2,e3,e4,e5,e6,e7,e8,e9,e10],[va1,va2,vb1,vb2,vx1,vx2,ra,rb,rc])
[(i0*rac*(rab + rbc)/(rab + rac + rbc),
i0*rac*rbc/(rab + rac + rbc),
i0*rac*rbc/(rab + rac + rbc),
i0*rbc*(rab + rac)/(rab + rac + rbc),
i0*rac*rbc/(rab + rac + rbc),
i0*rac*rbc/(rab + rac + rbc),
rab*rac/(rab + rac + rbc),
rab*rbc/(rab + rac + rbc),
rac*rbc/(rab + rac + rbc))]
sage: solve([e1,e2,e3,e4,e5,e6,e7,e8,e9,e10],[va1,va2,vb1,vb2,vx1,vx2,rac,rbc,rab])
[(i0*(ra + rc),
i0*rc,
i0*rc,
i0*(rb + rc),
i0*rc,
i0*rc,
ra + ra*rc/rb + rc,
rb + rc + rb*rc/ra,
ra*rb/rc + ra + rb)]


The same set of equations work to solve in either direction. (Note that the arbitrarily injected current, $$\i_0\$$, drops out when all is said and done. Also, node $$\C\$$ is treated as ground so star-resistor $$\R_{_\text{C}}\$$ has one end at ground.)

The idea can be expanded, as I indicated, and then solved just as above (except with more equations, obviously, and more variables to solve.)

You may wish to read "A degree preserving delta wye transformation with applications to 6-regular graphs and Feynman periods" by Shannon Jeffries and Karen Yeats, July 29, 2022. Physicists need this stuff for quantum field theory. So that's the place to go to see some serious applied mathematics used on problems like this one.

I think the problem might lie in what shape the square of resistors converts to. For instance consider your original square and add two imaginary resistors like this: - Then if you did two lots of delta to star conversion you'd get this: - And, even though the two imaginary resistors I added will make some values simplified in the conversion, I doubt that you'd get an equivalent shape to your original X-shape: - I suspect this is impossible. Look at this example. Zero Ohms could really be 1Ω or any small value, and ∞Ω any very large value. The two top corners, the ones A attaches to, are basically shorted, and likewise the two bottom corners. But the top left and bottom left don't communicate at all, no current flow. Likewise on the right side.

In the squared-up wye network, or star network if we can call it that, how can we have current flow between the two top terminals, and also between the bottom two (pink arrows), yet not allow any current flow between the two left, or the two right, terminals (green X'd off arrows)? There is one central node that all branches attach to, marked !? that makes this expanded delta-wye transform impossible.

If there is some sort of topology transform possible between a ring of elements (polygon) and something else, that something else will need more than one internal node.

• A useful brain exercise is to explain why (pun intended) things work fine for the usual three terminal Delta-Wye transform. Jul 11 at 0:10