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I'm trying to find the gray coding for the 8-triangular signal constellation below, but no matter what I try, one of the neighbors always has 2 bits differing.

enter image description here

For example, two of the neighbors for 000 in the lower right quadrant is 101 and 100,, but 101 and 000 differ by 2 bits.

Is there a Gray coding for such a signal constellation?

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  • \$\begingroup\$ Are you sure that a solution is supposed to exist? The points at 110 and 000 in your example both have four neighbours (if only taking the diagonal ones), but there are only three bits per word. Can you put lines into your image which show where the constraint of only a single changed bit actually applies? \$\endgroup\$
    – Philipp Burch
    Commented Feb 21, 2016 at 8:04
  • \$\begingroup\$ I wasn't sure if there should be a solution. \$\endgroup\$
    – Rayne
    Commented Feb 21, 2016 at 8:21
  • \$\begingroup\$ Are you sure doesn't exist 2D gray? \$\endgroup\$
    – Marko Buršič
    Commented Feb 21, 2016 at 9:04

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There is no solution for that constellation.

Consider the point labeled 000, it has four (perhaps five) neighbours, for a three bit code they can't all differ from 000 by a single bit.

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  • \$\begingroup\$ How can I justify that my choice of bit mapping onto the signal points lead to the minimum Bit Error Probability (BEP)? Do I need to try out every combination and calculate the BEP? \$\endgroup\$
    – Rayne
    Commented Feb 21, 2016 at 13:35
  • \$\begingroup\$ I got my bit mapping above by looking at the bit mapping of a circular 8-PSK and giving the same code to points at the relative same location. Is there a systematic way of assigning bit mapping to the constellation? For example, if I pick a point to start, are there any rules to let me assign a code to its neighbors? How many different bit mappings in total is there for this constellation? \$\endgroup\$
    – Rayne
    Commented Feb 21, 2016 at 15:22
  • \$\begingroup\$ start by dividing the symbol plane into zones. 8psk has a central zone which is disallowed thus each symbol is in a zone which has only two neighbours. \$\endgroup\$
    – Jasen Слава Україні
    Commented Feb 22, 2016 at 0:59

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