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

vocabularyset (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-lengthwords(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 anHamming Distanceand 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!

## Comments

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TopNewestIsn't it \((V-1)^ n \) – Lokesh Sharma · 9 months, 1 week ago

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– Ermes Trismegisto · 9 months, 1 week 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.Log in to reply

I assume both \(i\) and \(j\) can take any value from \(1\) to \(k\) except the case \( i = j \). Right? – Lokesh Sharma · 9 months, 1 week ago

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– Ermes Trismegisto · 9 months, 1 week ago

Oh shit, good point ,yes! I'll edit it now, thanks!Log in to reply

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 · 9 months, 1 week ago

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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 distanceis 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 · 9 months, 1 week ago

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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 · 9 months, 1 week ago

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– Ermes Trismegisto · 9 months, 1 week ago

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.Log in to reply