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Unformatted text preview: Error Correcting Codes: Combinatorics, Algorithms and Applications (Fall 2007) Lecture 16: Plotkin Bound October 2, 2007 Lecturer: Atri Rudra Scribe: Nathan Russell & Atri Rudra In the last lecture we proved the GV bound, which states that for all with 1 1 q , there exists a q-ary code of distance and rate at least 1 H q ( ) , for every > . In fact we proved that with high probability, a random linear code C lies on the GV bound. We picked a generator matrix G at random and proved the latter result. At this point, we might ask what happens if G does not have full rank? There are two ways to deal with this. First, we can show that with high probability G does have full rank, so that | C | = q k . However, the proof from last lecture already showed that, with high probability, the distance is greater than zero, which implies that distinct messages will be mapped to distinct codewords and thus | C | = q k . Further, the proof required that 1 1 q because it is needed for the volume bound V ol q ( , n ) q H q ( ) n to hold. It is natural to wonder if the above is just an artifact of the proof or, for example, is it possible to get R > and > 1 1 q ? In todays lecture, we will show that this cannot be the case by proving the Plotkin bound. 1 Plotkin Bound We start by stating the Plotkin bound....
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