Discussion
Consider the Euler variables of polyhedral objects namely, P1, P2, P3, and P4 as given in the table below.
Which one of the following options is NOT a topologically valid closed polyhedral object as per Euler's law?
| Polyhedral objects | Faces (F) | Edges (E) | Vertices (V) | Faces' inner loops (L) | Bodies (B) | Genus (G) |
| P1 | 6 | 12 | 8 | 0 | 1 | 0 |
| P2 | 5 | 8 | 5 | 0 | 1 | 0 |
| P3 | 5 | 12 | 8 | 0 | 1 | 0 |
| P4 | 10 | 24 | 16 | 0 | 1 | 0 |
Answer: A
The invalid object is P3.
The Euler-Poincare formula for solids is \(F - E + V - L = 2(B - G)\). Every object here has \(L = 0\), \(B = 1\) and \(G = 0\), so the right side is 2.
\[ \begin{aligned} P1 &: 6 - 12 + 8 = 2 \\ P2 &: 5 - 8 + 5 = 2 \\ P3 &: 5 - 12 + 8 = 1 \\ P4 &: 10 - 24 + 16 = 2 \end{aligned} \]
Only P3 fails. P1 is a cube, P2 a square pyramid, and P4 a valid object with 10 faces such as an octagonal prism, so they are all topologically valid.
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