# 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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