Special Binary Codes

Question

Find n binary codes of length n such that distance between each pair is n/2 , where n is a even number , if possible?How to generate all codes? for example n=4 we have 1110,1101,1011,0111 each pair have distance 2. Distance between a pair of code means number of different bits in both code words. For example 1110 and 1101 ,only last two bits are different so distance between this pair is 2.

@Kevin pair is not **that** abstract and distance is pretty clear from the question: a pair are two values like `(1110,1101)` and distance is the number of bits the two values differ by, like in the above pair: different bits are 0 and 1, thus distance=2 — , Sep 09 '15 at 16:56
Oh, I thought it might have been "difference in positioning of zeroes", which made me question the definition of "pair", because `1110` and `0111` have a "positional difference" of 3. — Kevin, Sep 09 '15 at 17:05

score 0 · Answer 1 · edited Jun 20 '20 at 09:12

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The main-difficulty with this question is simply that there are several solutions that produce the required solution. For example in your question:

1110 , 1101 , 1011 , 0111

1000 , 0100 , 0010 , 0001

1100 , 1001 , 0011 , 0110 , 1010 , 0101 , 0000 (*1 note concerning 0000 bellow)

Would be all valid solution (order ignored).

In general we can assume that in any solution, each value has the same number of 0-bits. Proof:
Let a solution S for n consist of {a , b , ...}. Let's assume each value in S has exactly x 0-bits, except for one value (c'), which has x + 1 0-bits. Now the distance to one value will still be n/2, but for the rest the distance is n / 2 + 1, due to the additional 1-bit.

a = a1  a2  a3  a4  a5  ...
b = b1  b2  b3  b4  b5 ...
c = c1  c2  c3  c4  c5  ...
c'= c1' c2' c3' c4' c5' ...

if for example c3 != c3' and cx == cx' for any x != 3, the distance to any other value v, where v3 != c3' will always be n / 2 + 1 or if v3 == c3' n / 2 - 1. Solution S would be valid, if c is element of S. If c' is an element of S instead, S isn't a valid solution anymore.

Next step: The solution with most elements is the one that holds max(binomial_coefficient(n , x)), where x is the number of 0s. Thus the optimal solution would be:

x = n/2, if 2 | n

This solution will be a set of binomial_coefficient(n , n/2) + 1 values.

So basically we search for all permutations of x 1s in a n-bit value. Won't post any code here though, since that question has been answered pretty often.

*1: Solutions with x = n / 2 allow 0000 aswell as solution for any n. This is not the case, if x != n / 2.

edited Jun 20 '20 at 09:12

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answered Sep 09 '15 at 17:35

can you give any example for n=6 or 8? or give any previous link? – subrat singh Sep 09 '15 at 17:53
@subratsingh this answer applies to any `n`. and `n = 8` produces a set of 70 different values, so... – Sep 09 '15 at 19:40
@subratsingh to simplify things: the relevant part of the answer is, that each value of the solution must have the same number of 0-bits and that the solution with the most bits will have `n/2` 0-bits per value and will produce `binomial_coefficient(n , n/2)` values – Sep 09 '15 at 19:58
@Paul, see my results. Maybe I interpreted the question wrong, but some of my solutions break your same number of zeros rule. – agentp Sep 09 '15 at 20:12
@agentp what's the language? linq? there **are** solutions to which the 0-rule doesn't apply. But if you're searching for the solution with the most values, you'll have to rely on the 0-rule. Though i've forgotten to mention one special rule: for `n=4` my largest solution is missing `{0,0,0,0}` as additional value. might be I've missunderstood ops question just aswell... – Sep 09 '15 at 20:19
@agentp ok. well, btw. what exactly is your code doing? if it'd simply calculate **all distinct** solutions it would produce by far more than 32 solutions. And these aren't the greatest solutions. My greatest solution has 7 values (after adding 0000, thanks btw. didn't check that special case). – Sep 09 '15 at 20:26
added some explanation. My approach specifically looks for length=n solutions. – agentp Sep 09 '15 at 20:59

agentp · Answer 2 · 2015-09-09T21:09:04.523

A totally brute force mathematica approach -- This unfortunately breaks my memory for n>4..

n = 4;
Select[ Subsets[Tuples[{0, 1}, {n}], {n}] ,
     And @@ Flatten@
      Outer[((# == n/2 || # == 0) &@Total@Abs@Subtract@##) & , # , #, 
          1 ] & ]

Some of the 32 solutions:

Edit, to explain a bit: Tuples[{0, 1}, {n}] gives all 2^n n-bit sequences. Subsets[ Tuples ,{n} ] gives all unique unordered length-n sets of the tuples ( 1820 for n=4 ) Select chooses from those the sets that match the criterea. Outer applies the test to every pair in the set ( note we have == n/2 or 0 because Outer tests each binary against itself with the self-test yielding zero )

Note each of the subsets can be reordered n! ways so for n=4 there should be 768 unique solutions if you consider order important.

Special Binary Codes

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