IBM Research | Ponder This | October 2007 solutions

# Ponder This

## October 2007

<<September October November>>

Ponder This Solution:

• tetraheron -- 1/2
• cube -- 7/12
• octahedron -- 5/12
• dodecahedron -- 19/30
• icosahedron -- 11/30

They can be derived as follows. For any graph the resistance between two points, S and T, can be computed as follows. Set the potential of S to 1, T to 0. Solve a set of linear equations to find the potential of the remaining vertices. Compute the current flow out of S (which will necessarily equal the current flow into T). Invert the current flow to determine the resistance.

For our case for each graph let n be the number of vertices and d be the degree of each vertex. Specify adjacent vertices S and T. Setting the potential of S to 1 and T to 0 will determine the potential values on the other vertices. Consider rotating the solid around S (so T coincides with each of the neighbors of S). Average the rotated potential values at each vertex. The average at S is clearly 1 (since S remains fixed). The average potential value at each neighbor of S is clearly the same by symmetry. Let this value be x. We claim the average potential at every remaining vertex is also x. This follows because there is no current flow into or out of any of the remaining vertices in any of the rotated states. So what is x? By symmetry the average potential over all the vertices is .5. So since the potential at S is 1 we have x=(n/2 - 1)/(n-1). So the average current out S (which will be the same for each rotation) is d*(1-x)=d*(n/2)/(n-1). So the total resistance between S and T is the inverse of this or (2*n-2)/(d*n) = (2/d)*(1-1/n). Using the values of n and d for each solid we find the answers above.

Note this method will work for any sufficiently symmetric graph.

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