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# We (almost) have a ternary Gray code solution

Clive Maxfield
10/9/2008 02:00 PM EDT

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re: We (almost) have a ternary Gray code solution
1/30/2012 7:16:54 PM
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hey clive, i realize its a bit late, but here's an idea for converting from a ternary gray code to binary. first, here's my ternary gray code. 0+ ++ +- +0 00 -0 -+ -- 0- note it's reflective about the center with + and - being switched. then to convert to binary, let + = 1, 0 = 0, and - be determined by the trit to the left. if +, = 0; if 0, = 1; if -, then again look at the left trit. then the above becomes... 01 11 10 10 00 10 11 11 01 what do you think?

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re: We (almost) have a ternary Gray code solution
9/26/2009 11:38:52 PM
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Ternary reflective Gray code is covered in the Wikipedia article. http://en.wikipedia.org/wiki/Gray_code

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re: We (almost) have a ternary Gray code solution
1/6/2009 6:54:30 AM
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What if we turn the problem on its head and use transition signalling to avoid the problem of two transitions at the same time?... We could have 3 binary signal wires for which only one signal changes state from 0 to 1 or from 1 to 0. Since we'd have 3 wires, each could represent one of the ternary states from the time there has been a transition on that particular wire. If you think of the transitions 0->1, 1->2, 2->0 as cyclic, we could have just 2 binary signal wires. One to indicate a transition in the clockwise direction, and the other to indicate a transition in the counterclockwise direction. This solves the problem of there being a transition across two threshold levels between two states...but the state would not be obvious to an outside observer monitoring the wires unless they knew the starting state. Would that make circuits implemented in this manner more secure, I wonder? Another related question. Is there a way to have K-maps for ternary, where it is possible to make loops around ternary variables so that they can be grouped into larger blocks?

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re: We (almost) have a ternary Gray code solution
10/13/2008 2:52:09 AM
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In an email to Max I expounded on the following statements that make the proposition impossible: 1. There is no reason to consider more than 2 bits in the expansion of the ternary digits. 2. If you choose any three of the possible 2-bit patterns (00, 01, 10, 11), there will always be one transition with two bits changing. 3. The ternary gray code includes all three transition possibilities: 0<->1, 1<->2, 0<->2. So there is no way to map the ternary digits onto binary patterns without changing two bits in at least one of the transitions. It seems that gray codes with unit steps will always allow a binary expansion that is itself a gray code, while those (like the ternary gray code) that require non-unit steps, don't. The reason is, with a unit step you could avoid the transition (0->2, for example) that maps onto the forbidden binary transition.