Weight and distance of a binary code

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SUMMARY

The discussion centers on the relationship between the weight and distance of a binary code, specifically analyzing a set of codewords. The codewords provided include eight binary strings, and the weight can be calculated manually. However, the participants seek a more efficient method to determine the minimum distance, referencing the triangle inequality with the formula d(C) ≤ min(w(ci) + w(cj)). The brute force method for calculating the minimum distance in this context requires checking 28 distances, highlighting the need for a more streamlined approach.

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  • Understanding of binary code and codewords
  • Knowledge of Hamming weight and distance
  • Familiarity with the triangle inequality in coding theory
  • Basic combinatorial mathematics for distance calculations
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Researchers, coding theorists, and computer scientists interested in error-correcting codes and their properties will benefit from this discussion.

annie122
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Is there a relationship between the distance and weight of a binary code?
I want to find the weight and distance of the code consisting of the codewords:

0000 0000
0010 1110
0101 1100
1011 1010
1110 0101
1100 1011
0111 0011
1001 0111

(spaces inserted for readability)

The weight can be checked easily by hand, but I don't want to check 21 distances.
Is there a quicker way?

One idea I had is d(C) \leq min(w(ci) + w(cj)), from the triangle inequality.
 
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Re: weight and distance of a binary code

Yuuki said:
Is there a relationship between the distance and weight of a binary code?
I want to find the weight and distance of the code consisting of the codewords:

0000 0000
0010 1110
0101 1100
1011 1010
1110 0101
1100 1011
0111 0011
1001 0111

(spaces inserted for readability)

The weight can be checked easily by hand, but I don't want to check 21 distances.
Is there a quicker way?

One idea I had is d(C) \leq min(w(ci) + w(cj)), from the triangle inequality.

The 'brute force procedure' for find the minimum distance in a code of size M=8 requires to check 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 distances... ... just a little question : in Your formula what are $C_{i}$ and $C_{j}$?... Kind regards $\chi$ $\sigma$
 
The 'brute force procedure' for find the minimum distance in a code of size M=8 requires to check 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 distances...
I said 21 because I already know the distance from the code 0000 0000 from the calculation of the weights.

... just a little question : in Your formula what are Ci and Cj?
They are two different codewords.
I should have written this there, sorry.
 

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