Hey R.E.
Nobody really picked this up, but I experimented for a little and produced this
code to produce the Wolfram 30 puzzle, thought it might be useful for anyone
wanting to explore.
Turns out to be a nice use case for 3 u;._3 to extract length 3 subsets from a
list as in a tessellation …
The following can be pasted into a script (careful of quotes copying in here):
=====================================================================
load 'viewmat'
NB. Wolfram’s 8 match cases for the existing row
NB. NB. Convert each match case in base 2 (eg 1 1 1) to an integer (eg 7) as
this is then faster to look up in addrow
cases=: #.8 3$1 1 1, 1 1 0, 1 0 1, 1 0 0, 0 1 1, 0 1 0, 0 0 1, 0 0 0
value=: 0 0 0 1 1 1 1 0
addrow=: 3 : 0
lastrow=. ,>_1 {. y
NB. Calculate the #. of length 3 subarrays from the lastrow, then look up in
cases and return corresponding value to create newrow
newrow=. value {~ cases i. 3 #.;._3] 0 0,lastrow,0 0
y,<newrow
)
view=:3 : 0
(|.-i.1{.$ mat)|."0 1 mat=:>y
)
diagonal =:3 : 0
(<0 1)|: y NB. Return diagonal of matrix y
)
smoutput 'To run for example ...'
smoutput ' viewmat view addrow^:50 <1'
=====================================================================
So the sample line above:
viewmat view addrow^:50 <1
reproduces the graphic triangle shown in the first link you gave below …
I did not spend much time on analysing the centre column (which in my case as I
maintain each row as a list, may be found using diagonal as here;
diagonal >addrow^:6 <1
1 1 0 1 1 1 0
Hope others find this useful.
@Cliff Reiter: maybe even scope to consider as another chaotic instance as
Woflram suggest…/Rob
> On 2 Oct 2019, at 6:49 pm, R.E. Boss <[email protected]> wrote:
>
> This was submitted in math-fun
>
> https://rule30prize.org
> https://writings.stephenwolfram.com/2019/10/announcing-the-rule-30-prizes
>
>
> R.E. Boss
> ----------------------------------------------------------------------
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