in reply to Re^8: encoding hdmi video byte
in thread encoding hdmi video byte

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;

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Re^10: encoding hdmi video byte
by BrowserUk (Patriarch) on Feb 25, 2010 at 10:37 UTC

    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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Re^10: encoding hdmi video byte
by hdmiguru (Initiate) on Feb 25, 2010 at 17:13 UTC
    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