MHB Weight and distance of a binary code

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The discussion centers on the relationship between the weight and distance of a binary code, specifically examining a set of eight codewords. Participants explore the challenge of calculating the minimum distance without manually checking all possible pairs, which can be cumbersome. One proposed method involves using the triangle inequality to estimate the distance based on the weights of the codewords. Clarification is sought regarding the notation used in the proposed formula, specifically the meaning of Ci and Cj, which refer to different codewords. The conversation highlights the complexity of efficiently determining distances in binary codes.
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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.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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