Cryptography Examples: Hamming Code in Practice

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Task:

Using the Hamming code, find the error in the message: 1001 0001 1101 1110 0000 000.

Solution:

The message length is 27 digits; from them 22 are informational and 5 are control.

Control digits are b1 = 1, b2 = 1, b4 = 1, b8 = 1, b16=0.

Let’s calculate the number of J for error detection.

For convenience, let’s introduce the following sets:

V1 = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27… – all numbers whose first digit is equal to 1;

V2 = 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27… – all numbers whose second digit is equal to 1;

V3 = 4, 5, 6, 7, 12, 13, 14, 15, 20, 21, 22, 23, 28… – all numbers whose third digit is equal to 1;

V4 = 8, 9, 10, 11, 12, 13, 14, 15, 24, 25, 26, 27, 28… – all numbers whose fourth digit is equal to 1;

V5 = 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28 … – all numbers whose fifth digit is equal to 1.

Digits numbered J are defined as follows:

j1 = b1 + b3+b5+b7+b9+b11+b13+b15+b17+b19+b21+b23+b25+b27 = 1;

j2 = b2+ b3+b6+b7+b10+b11+b14+ b15+ b18+ b19+ b22+ b23+ b26+ b27= 0;

j3 = b4+ b5+b6+b7 +b12+b13+ b14+ b15+ b20 +b21+b22+b23 = 0;

j4 = b9+b10+b11+b12+b13+b14+b15+b24+b25+b26+b27 = 0;

j5 = b16+ b17+b18+b19+b20+b21+b22+b23+b24+b25+b26+b27 = 1,

so the number J = 10001 2   =  1710 .

Thus, an error occurred in the seventeenth digit of the transferred number and 1 should be replaced to 0. Then we obtain 1111 1011 0010 1100 0101 1100 110.

Now let’s remove the control digits. We get 1101 0010 1100 1011 1001 10 – the transmitted number.

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