Hamming it up

The coordinates of the N=2dN=2^{d} vertices of an dd-dimensional cube can be written as the following strings of 00's and 11's: v1=(0,0,,0,0), v2=(1,0,,0,0),vd+1=(0,0,,0,1),vN=(1,1,,1,1).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 (00 or 11) are different. For example, the Hamming distance between v1v_{1} and v2v_{2} is 1 while the Hamming distance between the vertices v1 v_{1} and vNv_{N} is dd. 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 v1=(0,0), v2=(1,0), v3=(0,1),andv4=(1,1). 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 v1v_{1} to v2v_{2} and v3v_{3} and v4v_{4} to v2v_{2} and v3v_{3}. If you view the bits 00 and 11 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 ΩR=1~\Omega, determine the equivalent resistance in Ohms between v1 v_{1} and vNv_{N} in d=7d=7 dimensions?

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