Thee Hamming distance $H(S_1, S_2)$ between two binary strings $S_1, S_2$ of length $n$ is the number of positions on which the two strings disagree. It is straightforward to show that if $S_1, S_2$ are two random strings of length $n$, the expected Hamming distance $\mathbb{E}[H(S_1, S_2)]$ is $n/2$.

Now, say I'd like to have a set of $k$ strings $S =\{S_1, \ldots, S_k\}$ such that for every $S_i \in S$, the average distance between $S_i$ and the other strings of $S$ is at least $n/2$. In other words, for every $S_i \in S$, $\sum_{j \neq i} H(S_i, S_j) \geq (k - 1)n/2$.

Actually, I'm not really interested in constructing $S$. Rather, I'm interested in whether the following argumentation, which shows that $S$ exists, is valid - for it seems to me that there is a hole to fill. This came up during an informal discussion on math stuff, and I couldn't refute the argument on the spot. Here's the argument.

We show that such an $S$ exists by picking $S_1, \ldots, S_k$ randomly (uniformly and independently). Then for any $S_i \in S$, the exected sum of distances $\mathbb{E}[\sum_{i \neq j} H(S_i, S_j)] = (k - 1) \mathbb{E}[H(S_i, S_j)] = (k - 1)n/2$, by linearity of expectation. Thus there is a set of strings $S$ such that $\sum_{i \neq j}H(S_i, S_j) \geq (k - 1)n/k$. Since this holds for every $S_i \in S$, we deduce that there exists a set of strings $S$ such that every string $S_i \in S$ has $\sum_{i \neq j}H(S_i, S_j) \geq (k - 1)n/2$

The part that's bugging me is the end, which basically says "since this holds for every $S_i$, then there is a $S$ in which every $S_i$ has the desired sum". The argument shows that for any $S_i$, there is a $S$ having the desired sum. But does this show that there is a single $S$ satisfying every $S_i$ simultaneously?

  • $\begingroup$ Note that there's another error, apart from the one you're rightly pointing out: You wanted a set of strings, and the argument applies to ordered tuples of strings, with repetitions allowed. This isn't a problem, though, since disallowing repetitions increases the expected Hamming distance. $\endgroup$ – joriki Jun 28 '16 at 9:30

You're right, it doesn't. The argument merely shows that there are $k$ possibly different sets that each have the desired property for one of the $S_i$.

  • $\begingroup$ OK, since we share the same point of view, I trust you, thank you! $\endgroup$ – Manuel Lafond Jun 28 '16 at 11:57
  • $\begingroup$ @ManuelLafond: This is perhaps easier to see (so you don't have to take it on trust ;-) in the case of uniformly randomly picking an ordered $k$-tuple $a$ of different integers from $0$ to $n$. Then each component of $a$ has expected value $n/2$, so for each component $a_i$ there must be a tuple with $a_i\ge n/2$, but obviously this doesn't imply that there's a tuple in which all components are $\ge n/2$ (and in fact there isn't for sufficiently large $k$). $\endgroup$ – joriki Jun 28 '16 at 12:15
  • $\begingroup$ Yes, that's a great counter-argument! Exactly what I need :) Thanks again $\endgroup$ – Manuel Lafond Jun 28 '16 at 12:37

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