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I want to decode the following bit stream using Hamming code: 01110110

Once using even parity and once odd parity. I will show my work. For even parity:

$p_{1} \; p_{2} \; 0 \; p_{3} \; 1 \; 1 \; 1 \; p_{4} \; 0 \; 1 \; \; 1 \; 0$

Now, we check:

$p_{1}: p_{1} \; 0 \; 1 \; 1 \; 0 \; 1 $, since there is an odd number of 1's, $p_{1}=1$

$p_{2}: p_{2} \; 0 \; 1 \; 1 \; 1 \; 1 $, since there is an even number of 1's, $p_{2}=0$

$p_{4}: p_{3} \; 1 \; 1 \; 1 \;0 $, since there is an odd number of 1's, $p_{3}=1$

$p_{8}: p_{4} \; 0 \; 1 \; 1 \;0 $, since there is an even number of 1's, $p_{4}=0$

so the following hamming code is $$1 \; 0 \; 0 \; 1\; 1 \; 1 \; 1 \; 0 \; 0 \; 1 \; \; 1 \; 0$$

As for odd, the parities will just be the opposite and the code will be

$$0\; 1 \; 0 \; 0 \; 1 \; 1 \; 1 \; 1 \; 0 \; 1 \; \; 1 \; 0$$

Can someone tell me if my work is correct? Because I am a bit confused and would appreciate the help.

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    $\begingroup$ Some background is missing here. What does it mean to use "even parity" or "odd parity"? What is the Hamming code for you? $\endgroup$ – Yuval Filmus Nov 4 '19 at 11:39
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There are many equivalent Hamming codes depending on the ordering of parity and data bits. Assuming that:

  • You are using the specific bit ordering described here: https://en.wikipedia.org/wiki/Hamming_code#General_algorithm
  • By "even parity" you mean that the parity bits are computed as the sum (modulo 2) of the corresponding groups of bits
  • By "odd parity" you mean that the parity bits are computed as 1 minus the sum (modulo 2) of the corresponding groups of bits,

then yes, your simulation of the algorithm is correct.

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I'm not really understanding some parts of your question, The Hamming algorithm is used to count the number of 1 bits of a binary number. The idea is to create a mask = 1 and check for every of the 32 bits if (mask & n) == 1, then shift the mask to the left by one. If true is returned, it means that the binary representation of n has a 1 at that position, so increase the bit number. Here is an example:

int hamming(int n) {
        int bits = 0, mask = 1;
        for (int i = 0; i < 32; i++) {
            if ((n & mask) != 0) bits++;
            mask <<= 1; //shift left by one
        }
        return bits;
    }
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