nput V = (111100). We ended up at the decoder receiving (100) after multiplying by H. If we could tell the decoder that, if after multiplying by H, if you see (100), assume the error is e = (100000), then we can correct for this error. In other words, if we can match what we get from V H to errors e, then we can correct the errors. Let's follow engineering convention and call what we get from vH the vector S (i.e., vH = S), which is called syndrome. So, you've received V, and you multiplied it by H and you got S. Now look at this: H H e H H S. (6.14) (We know V H = e H from Equation (6.13).) So, for every error e there is a syn- drome S. In fact, it's easy to show that there are more e's than there are S's, SO more than one e will share the same S. For example, in the 3-bit set to 6-bit set case of Figure 6.10, there are 8 S's while there are 63 e's. Here's what the channel decoder must do (an example follows the explanation): 1) For each possible value of S, determine which error e you think has occurred. Do this as follows, using our (6,3) code as an example: a) Realize the value ofS= (000) = o indicates no error. That means there are 8-1=7 = possible S values that we can consider. b) Start with the most common errors, which are the one-bit errors, i.e., the errors represented by the vectors el (100000), e2 = (010000), e3=(001000) to e6 = (000001). For each of these errors e, figure out the S using e H = S. So you'll get S1, S2, S3, up to S6, one for each error vector. That accounts for a total of 6 of the 7 remaining S values. c) There is oneS value left unused. Start to consider the e values correspond- ing to two-bit errors (e.g., e= (101000)), and find an e value such that eH leads to the remaining S. Chapter Six 000000 0 0 0 ) Step 1.a. 000001 000010 Step 1.b. 000100 001000 010000 100000 010001 111 Step 1.c. Figure 6.11 Mapping errors e to syndromes S 2) Now, when a vector V arrives, create vH=eH, and you'll get S. From the results of part (1), decide which error e this S corresponds to, and correct the er