Hamming it up

Let $x$ denote the no. of vertices from which current enters a junction and $y$ denote the no. of vertices to which current leaves from the junction.Let us consider the vertices:

$V_{0}:$ $(0,0,0,0,0,0,0)$ $\rightarrow x = 0 , y = 7$

$V_{1}:$ $0,0,0,0,0,0,1)$ $\rightarrow x = 1 , y = 6$

$V_{2}:$ $(0,0,0,0,0,1,1)$ $\rightarrow x = 2 , y = 5$

$V_{3}:$ $(0,0,0,0,1,1,1)$ $\rightarrow x = 3 , y = 4$

$V_{4}:$ $(0,0,0,1,1,1,1)$ $\rightarrow x = 4 , y = 3$

$V_{5}:$ $(0,0,1,1,1,1,1)$ $\rightarrow x = 5 , y = 2$

$V_{6}:$ $(0,1,1,1,1,1,1)$ $\rightarrow x = 6 , y = 1$

$V_{7}:$ $1,1,1,1,1,1,1)$ $\rightarrow x = 7 , y = 0$

Now , due to symmetry , the current coming to a junction from each wire would be same, and also the current going out to each wire would be same.

Let us introduce $I_{0}$ current at $V_{0}$ .

At $V_{r}$ , $x = r , y = 7-r$ .

Let $I_{r}$ denote the current incoming from 1 wire at $V_{r}$

Clearly , using kirchoff's current law,

$xI_{r} = yI_{r + 1}$

$\Rightarrow rI_{r} = (7 - r)I_{r + 1} \Rightarrow \boxed{I_{r + 1} = \frac{rI_{r}}{7 - r}}$

Now , $I_{1} = \frac{I_{0}}{7}$ , we can evaluate all of $I_{1} , I_{2} , \dots$ ,and $I_{7}$ .

Now, $\Delta V_{V_{0} \rightarrow V_{7}} = I_{1}R + I_{2}R + \dots + I_{7}R = I_{0}R_{eff}$ .

Hence $R_{eff} = \frac{I_{1} + I_{2} + \dots + I_{7}}{I_{0}} R = \boxed{0.36 \Omega}$

Suppose when we put a voltage difference between $v_1$ and $v_N$ , the vertices whose coordinates have the same number of bit 1s will have the same voltage.

Therefore, we can group points whose voltages are the same as one point, $2^d$ points will become $d+1$ points. We can name these points by the number of bit 1s in their coordinates, so they are $0,1,2,...,d$ .

It is obvious that point $k$ only connect directly with points $k-1$ and $k+1$ but we need to find out how many wires with resistance are between them.

From the given condition, we know that there are $d \choose k$ vertices which have k bit 1s in coordinate and each of these vertices will connect directly to $(d-k)$ vertices which have $k+1$ bit 1s in coordinate. Therefore, there are $(d-k) {d \choose k}$ wires with resistance between point $k$ and $k+1$ .

Therefore, the equivalent resistance between $v_1$ and $v_N$ is $r=\displaystyle \sum_{k=0}^{d-1} \frac{R}{(d-k) {d \choose k}}$ .

With $d=7$ and $R=1 \Omega$ , we obtain that $r=0.36 \Omega$

Work with the $n$ -dimensional hypercube, so that $N = 2^n$ . By symmetry, all the vertices a given Hamming distance $k$ from $v_1$ are at the same potential $V_k$ . A vertex at Hamming distance $k$ from $v_1$ is connected to $k$ vertices at Hamming distance $k-1$ from $v_1$ , and connected to $n-k$ vertices at Hamming distance $k+1$ from $v_1$ . Kirchoff's Law tells us that $k\big(V_{k-1} - V_k\big) + (n-k)\big(V_{k+1} - V_k\big) \; = \; 0 \qquad 1 \le k \le n-1$ from which we get ${n-1 \choose k}\big(V_{k+1} - V_k\big) \; = \; {n-1 \choose k-1}\big(V_k - V_{k-1}\big) \qquad 1 \le k \le n-1$ Thus we deduce that there exists a constant $\alpha$ such that ${n-1 \choose k}\big(V_{k+1} - V_k\big) \; = \; \alpha \qquad 0 \le k \le n-1$ The potential difference between $v_1$ and $v_N$ is $V \; = \; V_n - V_0 \; = \; \alpha\sum_{k=0}^{n-1} {n-1 \choose k}^{-1}$ while the current in the circuit (considering Kirchoff's Law at $v_1$ ) is $I \; = \; n(V_1 - V_0) \; = \; n\alpha$ and hence the effective resistance of the circuit is $R \; = \; \frac{1}{n}\sum_{k=0}^{n-1}{n-1 \choose k}^{-1}$ Putting $n=7$ gives $R = 0.3595$ .