# 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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