Talk:STXTRM

While the [ command would seem to make STXTRM easier to program than MSM, I suspect the fact that non-instructions are not copied from the bottom to the top any longer will make : hard to use without losing the rest of the stack data, at least without combining it cleverly with |. Also, I think there is no way to get the equivalent of Underload's a command, not that it is needed for Turing completeness. --Ørjan (talk) 01:02, 12 August 2013 (UTC)

Oh, I think I was wrong about this. I just realized you can use [[]]:, and [[]]:/, to get hold of raw [ and ] characters even if in a literally balanced subprogram. From this it's not so hard to implement the analogy of Underload's a command:

[[]]:,.[[]]:/,/.

This followed by : gives you a command to preserve a (balanced) string for one iteration, and you can "quine" this to preserve indefinitely, thus getting around the non-copying of non-commands part:

[ ; [[]]:,.[[]]:/,/. ./ [[]]:,.[[]]:/,/. .: ] ; [[]]:,.[[]]:/,/. ./ [[]]:,.[[]]:/,/. .:

This makes me convinced that STXTRM is Turing-complete even without |. You still cannot use : directly for Underload ^, though, because it messes up the order of evaluation.

With the help of | I think you can simulate most Underload commands more directly, with a slightly more complicated encoding of Underload ():

[ |: [ A ] ] :|

will run subprogram A immediately. --Ørjan (talk) 19:11, 27 August 2013 (UTC)

Oh hm that is wrong, I forgot that just reversing [ A ] makes it hit a ] first. Should be fixable though. --Ørjan (talk) 23:50, 28 August 2013 (UTC)

[ [|:] , A [] ] :,|

should work. --Ørjan (talk) 23:56, 28 August 2013 (UTC)

Argh no, that of course breaks if A contains any [] characters. Looking awkward now. :( --Ørjan (talk) 23:58, 28 August 2013 (UTC)