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Unformatted text preview: wo bits, then either (1) the differing bits are in the same
row, in which case the row parity calculation is unchanged but two column parity
calculations will differ, (2) the differing bits are in the same column, in which case the
column parity calculation is unchanged but two row parity calculations will differ,
or (3) the differing bits are in different rows and columns, in which case there will be
two row and two column parity calculations that differ. So in this case HD(wi , wj ) ≥ SECTION 6.4. LINEAR BLOCK CODES AND PARITY CALCULATIONS 11 4.
• If Mi and Mj differ by three or more bits, then in this case HD(wi , wj ) ≥ 3 because wi
and wj contain Mi and Mj respectively.
Hence we can conclude that HD(wi , wj ) ≥ 3 and our simple “rectangular” code will be
able to correct all single-bit errors.
Decoding the rectangular code: How can the receiver’s decoder correctly deduce M
from the received w, which may or may not have a single bit error? (If w has more than
one error, then the decoder does not have to produce a correct answe...
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This document was uploaded on 02/26/2014 for the course CS 6.02 at MIT.
- Fall '13