By Paul P. Martin

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**Sample text**

In f2 , however, not every binary code of length 3 is the image of an element of T , so if 101 was corrupted to 001, say, we would know at least that there had been an error in transmission. Indeed with this encoding every single element transmission error would show up. However double errors could still appear to be OK. 3. 5 above and f3 (N) = 00000, f3 (S) = 01101, f3 (E) = 10110, f3 (W ) = 11011. 6. Exercise. Verify that if any two errors occur then the received message is not the image of any sent message, signaling an error.

By the singleton bound A2 (3, 2) ≤ 23−(2−1) = 22 = 4 But our example (2) is a 2-ary (3,4,2)-code, so A2 (3, 2) ≥ 4. Hence A2 (3, 2) = 4. A much better upper bound is generally given by the ‘ball packing argument’. 17. Lemma. If x ∈ S n then t |Bt (x)| = r=0 n (q − 1)r r Proof: |Sr (x)| is the number of strings in S n differing from x in precisely r places. 18. (Ball packing bound) Let C be a q-ary (n, M, d)-code with d ≥ 2t + 1. Then t n (q − 1)r ≤ q n M r r=0 Proof: Since d ≥ 2t + 1, the t-balls centred on codewords are all disjoint.

For √ example π, e and 2. Such abstractions are presently among the features distinguishing humans from computers... but that is another story. 1. CODING example it is a sound, so the decoder is an ear/brain system; but obviously the sound heard by B at the appropriate point in time will have only a negligible amount of correlation with the original encoding). An attempt is made to decode this version. 5. Some approximation to the original message arrives for use at the target. 2. A code C for a source alphabet T is a function f : T → S + to sequences in code alphabet S.

### Bottom up Computing and Discrete Mathematics by Paul P. Martin

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