Following relation is given between three binary codewords, u, v and w. Sum of the codewords u and w is given by 010110011, and sum of the codewords v and w is given by 100010010. What is the Hamming distance between u and v? *
Answers
Answer:
1. (Error probability): Consider a code of length six n = 6 defined as,
(a1, a2, a3, a2 + a3, a1 + a3, a1 + a2)
where ai ∈ {0, 1}. Here a1, a2, a3 are information bits and the remaining bits are redundancy
(parity) bits. Compute the probability that the decoder makes an incorrect decision if the
bit error probability is p = 0.001. The decoder computes the following entities
b1 + b3 + b4 = s1
b1 + b3 + b5 == s2
b1 + b2 + b6 = s3
where b = (b1, b2, . . . , b6) is a received vector.
We represent the error vector as e = (e1, e2, . . . , e6) and clearly if a vector (a1, a2, a3, a2+
a3, a1 + a3, a1 + a2) was transmitted then b = a + e, where ’+’ denotes bitwise
mod 2 addition.
The reason the decoder choose the above equations is the following,
b2 + b3 + b4 = a2 + e2 + a3 + e3 + a2 + a3 + e4 = e2 + e3 + e4 = s1
b1 + b3 + b5 == . . . = e1 + e3 + e5 = s2
b1 + b2 + b6 = e1 + e2 + e6 = s3
Therefore, based on the knowledge of b the receiver computes the linear combinations of error bits. Given s1, s2, s3 the decoder must choose the most likely
error pattern e which satisfies the three equations. Notice that there is a unique
solution for e for any (s1, s2, s3) 6= (1, 1, 1).
E.g. assume that s = (0, 0, 1) then obviously (e1, . . . , e6) = (0, 0, . . . , 1) satisfies
the above equations. Can there be some other e of weight one to get s = (0, 0, 1)
?
Let us try (e1, . . . , e6) = (1, 0, . . . , 0). Then s2 = 1 etc. Therefore, an error pattern
of weight 1 is decoded correctly.
Now if (s1, s2, s3) = (1, 1, 1) the decoder must choose one of the possibilities,
(1, 0, 0, 1, 0, 0)
(0, 1, 0, 0, 1, 0)
(0, 0, 1, 0, 0, 1)
4
The decoder need to select one of these patterns (usually in advance).
Remark: We will later see (when we introduce standard array and syndrome
decoding) that only if the coset leader that correspond to the syndrome appears
to be the error pattern then we make a correct decision. Among all other error
patterns there is one with two errors that is decoded correctly. Do not forget that
we only consider the solutions of minimum weight for e.
The probability of correct decoding is therefore,
PC = (1 − p)
6 + 6(1 − p)
5
p + (1 − p)
4
p
2 = 0.999986.
That is the decoder can correct if no error has occured in transmission (first term),
can correct all single error patterns (second term), and a single error pattern of
weight 2 (second term). Note that there are
6
2
error patterns of weight 2 but
only one can be corrected.
Explanation:
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