This preview shows page 1. Sign up to view the full content.
Unformatted text preview: n extend our approach by producing an embedding with more space
between valid codewords! Suppose we limit our selection of messages in S even further,
as follows:
HD(wi , wj ) ≥ 3 for all wi , wj ∈ S where i = j
(6.4)
How does it help to increase the minimum Hamming distance to 3? Let’s deﬁne one
more piece of notation: let Ewi be the set of messages resulting from corrupting wi with a
single error. For example, Eo00 = {001, 010, 100}. Note that HD(wi , an element of Ewi ) = 1.
With a minimum Hamming distance of 3 between the valid code words, observe that
there is no intersection between Ewi and Ewj when i = j . Why is that? Suppose there was a
message wk that was in both Ewi and Ewj . We know that HD(wi , wk ) = 1 and HD(wj , wk ) =
1, which implies that wi and wj differ in at most two bits and consequently HD(wi , wj ) ≤ 2.
That contradicts our speciﬁcation that their minimum Hamming distance be 3. So the Ewi
don’t intersect.
Now we can correct single bit errors as well: the received m...
View Full
Document
 Fall '13
 HariBalakrishnan

Click to edit the document details