Hamming it up

The coordinates of the \(N=2^{d}\) vertices of an \(d\)-dimensional cube can be written as the following strings of \(0\)'s and \(1\)'s: \[v_{1}=(0,0,\ldots,0,0), \ v_{2}=(1,0,\ldots,0,0),\\ v_{d+1}=\ldots (0,0,\ldots,0,1), v_{N}=\ldots (1,1,\ldots, 1,1).\] The Hamming distance between two strings of equal length is defined as the number of positions at which the corresponding bits (\(0\) or \(1\)) are different. For example, the Hamming distance between \(v_{1}\) and \(v_{2}\) is 1 while the Hamming distance between the vertices \( v_{1}\) and \(v_{N}\) is \(d\). The edges of hypercube connect the vertices which are one unit-distance apart (in the Hamming distance sense) are connected. For example, in two dimensions we have 4 vertices \[ v_{1}=(0,0),\ v_{2}=(1,0),\ v_{3}=(0,1),\quad \text{and}\quad v_{4}=(1,1).\] Then, as discussed above, you should connect \(v_{1}\) to \(v_{2}\) and \(v_{3}\) and \(v_{4}\) to \(v_{2}\) and \(v_{3}\). If you view the bits \(0\) and \(1\) as the coordinates of the 4 vertices what you obtain is just a square. Clearly, in 3 dimensions, you will get a cube. If each edge of the hypercube is a wire with resistance \(R=1~\Omega\), determine the equivalent resistance in Ohms between \( v_{1}\) and \(v_{N}\) in \(d=7\) dimensions?

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