Just to clarify, is the proper interpretation of your OP that of Re^7: encoding hdmi video byte, or is it more like

Data byte D(0:7) encode this to make 9bit data q_m(0:8)
new      previous
bit of   bit of
q_m(0) =            D(0);
q_m(1) = q_m(0) XOR D(1);
q_m(2) = q_m(1) XOR D(2);
q_m(3) = q_m(2) XOR D(3);
q_m(4) = q_m(3) XOR D(4);
q_m(5) = q_m(4) XOR D(5);
q_m(6) = q_m(5) XOR D(6);
q_m(7) = q_m(6) XOR D(7);
q_m(8) =            1;
[download]

For the record. Once the translation tables are correct, whatever the proper encoding is, tr/// will process 1GB in 17 seconds, including the read and write time,

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Yes what you have said is correct. The encoding starts with taking the bit 0 (D[0]) of input data and that becomes the bit 0 of result (Q[0]). Then XOR the bit 0 of the result with bit 1 of input data. That becomes bit 1 of the result (Q1). Then XOR the bit 1 of the result with the bit 2 of input data. That becomes bit 2 of the result (Q2). Then XOR the bit 2 of the result with bit 3 of input data. That becomes bit 3 of the result (Q3). Then XOR the bit 3 of the result with the bit 4 of input data. That becomes bit 4 of the result (Q4). Then XOR the bit 4 of the result with bit 5 of input data. That becomes bit 5 of the result (Q5). Then XOR the bit 5 of the result with the bit 6 of input data. That becomes bit 6 of the result (Q6). Then XOR the bit 6 of the result with bit 7 of input data. That becomes bit 7 of the result (Q7). Th bit 8 of the result is always 1 (which indicates that we used XOR to the receiver). example input data 1010 0101 output data 1 0110 0011