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Discrete mathAnswered question
ymochelows ymochelows 2022-09-07

Is this coding exercise well thought?
I have to create a code C of 5 words with length n = 6, with the alphabet F 2 = { 0 , 1 } that corrects 1 mistake.
I am new to coding theory so I am having some troubles with this particular question...
If it corrects 1 mistakes, that means that d > 2 1 (I believe this is a well-known result [I prove it at my courses]), where d is the minimal distance of the code C defined as d = d ( C ) = min { d ( x , y ) | x , y C , x y }, and that distance in the definition is the Hamming distance defined as d ( x , y ) = # { i | 1 i n , x i y i }, being x = ( x 1 , x 2 , . . . , x n ) and y = ( y 1 , y 2 , . . . , y n ) elements in F q n (where F q is the alphabet of the code, and n denotes the length of the words).
So, in my particular case I want that the minimal distance of the code is greater than 2, i.e., I want that there are no two different words in the code that have Hamming distance equal or lesser than 2, i.e., no two words in the code that have more than 4 coordinates equal to each other (as If they have four equal to each other, there would be 6 4 = 2 different to each other, so the Hamming distance would be 2, and if there are 5 equal to each other, the Hamming distance would be 1 which is not valid). So, I begin to construct the code, by taking the first word:
( 1 , 1 , 1 , 1 , 1 , 1 ) F 2 6
Then, I could take another one like:
( 1 , 1 , 1 , 0 , 0 , 0 ) F 2 6
which satisfies the previous conditions. So, taking this kind of words, and taking on account the conditions that must hold, I could keep going with this ''constructive algorithm'' until I have 5 words. For example:
( 1 , 0 , 0 , 1 , 0 , 0 ) F 2 6
( 0 , 1 , 0 , 1 , 1 , 0 ) F 2 6
( 0 , 0 , 0 , 0 , 0 , 0 ) F 2 6
And this would conclude the exercise (by taking all this 5 words listed above). I am not sure if this is a great argument and I would really appreciate if someone could give me some feedback on it...

Discrete mathAnswered question
Terry Briggs Terry Briggs 2022-09-07

A 'non-numerical\analytic' proof that ( n k ) > ( n k 1 ) for large n N
The number of k-subsets of [n] is given by the formula ( n k ) or n C k . They famously occur in the expansion of ( 1 + x ) n and they are given by the formula
( n k ) = n ! ( n k ) ! k !
Using this formula, it is easy to prove the inequality that ( n k ) > ( n k 1 ) for large enough n N . What this inequality says is that the number of ways of choosing k 1-subsets is eventually smaller than the number of ways of choosing k-subsets of [n].
One more way of showing this is by observing that ( n k ) is a polynomial in n of degree k and then, we can see that it will outgrow ( n k 1 ) which is a lower degree polynomial in n.
Is there a more natural way of seeing this inequality without the use of calculations with the use of something more combinatorial-like? Possibly by exhibiting an injection between the k 1-subsets and k-subsets? Or by another interpretation of the numbers where it is easier to get such an injection? Or something else altogether?
Remark: My supervisor mentioned almost immediately that there was a way to see this using Symmetric Chain Decomposition. But I do not have the luxury of spending that much time on this. I apologise. I would be thankful if you could provide a proof based on the same.
Tl;dr: What I am looking for is something more along the lines of bijections or a different interpretation of the binomial coefficients that makes the inequality easier to see. I hope the approach doesn't rely heavily on calculations and at the same time, explains why the the inequality reverses for small n.

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