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Alastair Hewitt
Harvard Extension School
Cambridge, MA

Registered: Feb 2004
Posts: 35

Demonstrating ECA rules in 2,5 Turing Machines

There looks to be a general scheme for generating all 256 elementary CAs in a 2,5 Turing Machine. Some of the "simpler" ones can be generated by a 2,4 machine, but this is not the case for rule 30. The jury is still out on getting 110 in a 2,4 machine although I doubt it. And of course, the 2,3 machine can even do a version of rule 60.

I couldn't find the written rules for 110 on a 2,5 Turing Machine anywhere, but this is what I derived:
{1, 0} -> {2, 2, -1}
{2, 0} -> {3, 2, -1}
{1, 1} -> {0, 1, 1}
{2, 1} -> {4, 1, 1}
{1, 2} -> {4, 2, 1}
{2, 2} -> {4, 2, 1}
{1, 3} -> {0, 1, 1}
{2, 3} -> {0, 1, 1}
{1, 4} -> {1, 1, -1}
{2, 4} -> {2, 1, -1}

These rules should generate rule 30:
{1, 0} -> {1, 2, -1}
{2, 0} -> {3, 2, -1}
{1, 1} -> {0, 2, 1}
{2, 1} -> {4, 2, 1}
{1, 2} -> {0, 2, 1}
{2, 2} -> {4, 2, 1}
{1, 3} -> {4, 1, 1}
{2, 3} -> {0, 1, 1}
{1, 4} -> {1, 1, -1}
{2, 4} -> {2, 1, -1}

BTW, I haven't tested these, but a comparison of the 110 rules with the graphic in the book confirms the right structure.

I was thinking about defining the general scheme and then doing a demonstration project: You would select the ECA rule and then see the 2,5 Turing Machine evolution for that rule.

Of course, someone may have already done this, so if that is the case please let me know so I don't reinvent the wheel!!

Thanks, Alastair

Last edited by Alastair Hewitt on 07-24-2007 at 01:24 AM

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