Ball-avoiding randomness in an Hamming metric

This is an open ended question, I don't know the answer and would like your ideas on the topic.

Let me first give you a practical setting where this kind of problem arise: Banks need to issue account numbers for their customer. For example , let's say that $A, B$ are two customers of Tartaglia Bank and that they are issued the following 6-digits account numbers:

$A:$ 21-33-45
$B:$ 27-33-45

Now, it should be pretty evident that, by chance, the two account numbers happened to be pretty similar: they only differ in the second digit. This is not good! One digit is easily mistaken and A could possibly receive the money that someone inteded to send to B and viceversa.

It's clear that we would like any two random account numbers to be as different as possible, that is, in this case, to differ in all 6 digits.

This has now the following generalization, that is our actual question:

Let $V$ be a vocabulary set (a set of "letters", in our previous case it was the set of natural numbers) , let $n$ be a positive integer and $W = V^n$ the set of n-length words (in our previous example $n=6$ and $W$ was the set of all possible 6-digits account numbers). Define for $a,b \in W$ $H(a,b)$ to be the number of letters in which the two words differ (this is called an Hamming Distance and it can be shown to induce a Metric Space on W). Given any two words $a,b \in W$ , what is the chance that their distance is maxmised, i.e. that $H(a,b)=n$ ? Generalising further, given $d > 0$ what is the chance that $k \geq 2$ words $a_1,...,a_k \in W$ are such that $H(a_i,a_j)\geq d$ for all $1\leq a_i < a_j \leq k$ ?

I am fairly sure this problem or one of its variant should be fairly doable and I'll try to spend some time on it when I finish publishing these notes. It was just one of those real life inspired problems that I felt like to share on this platform as a way to test it (this is my first post here) and perhaps to interest some other people, so I am looking forward for your answers/ideas/generalisations/link to papers/critiques/suggestions.

Thanks for reading!

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
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Comments

Hey there! Nice job on your first post! Welcome to the community.

I do not know a lot about Hamming Codes. Could you explain the question to me in some informal way?

Agnishom Chattopadhyay - 4 years, 11 months ago

Thank you very much! To be honest with you, I don't know much about them myself, just happened to know the name and learnt about metric space and it's a pretty popular example. I was a bit worried that I might have added some useless complications , so I'll try to rexpress the final question informally: basically, take $k$ different $n-$ letters words . (Here the words "letter" and "word" are to be intended in a more general mathematical fashion: for example £$$$%&??£ is a valid 9-letters word where each letter belong to the set of "keyboard keys". This set is called in general a vocabulary set and we can assume it to be finite: so your answer is also going to contain the number of element of this set (the cardinality of the set) as one of the parameters!).

So now you have $k$ $n-$ letter words.Given any two of these, you can count in how many characters or letters they differ: for example:

% % £ & & % $ %
% $ £ & & ? ? %

Those two words above are both 8 letters long and they differ in the 2nd,6th and 7th postion. We say that their Hamming distance is three, that is , they differ in three places, very easy!

Now you fix a positive number $d$ and you ask what is the chance that taken any pair of the $k$ words is such that their distance is at least $d$ , i.e. each pair ar at least $d$ words "apart".

I hope this reformulation makes sense, let me know!

Ermes Trismegisto - 4 years, 11 months ago

With you, so far.

So, we are trying to find the probability that a set of $k$ $n$ length words over some alphabet has the property that no two strings are closer than $d$ . Right?

@Agnishom Chattopadhyay – Replace "closer" with "further apart" and that's precisely it. ( $H(a_i,a_j) \geq d$ ) so we want the distance to be at least d.

Ermes Trismegisto - 4 years, 10 months ago

Given any two words $a,b \in W$ , what is the chance that their distance is maxmised, i.e. that $H(a,b)=n$ ?

Isn't it $(V-1)^ n$

Lokesh Sharma - 4 years, 11 months ago

Yeah, I agree with you that's pretty easy: you choose the letters the first word $a$ and then when you choose the letters of $b$ you can't use one character for each letter, so that one is farily easy.

Generalising further, given $d > 0$ what is the chance that $k \geq 2$ words $a_1,...,a_k \in W$ are such that $H(a_i,a_j)\geq d$ ?

I assume both $i$ and $j$ can take any value from $1$ to $k$ except the case $i = j$ . Right?

@Lokesh Sharma – Oh shit, good point ,yes! I'll edit it now, thanks!

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$