## Does an exact reconstruction of the Rule 30 cone contain a route to answering the challenges?
I have not used Wolfram products to explore the Rule 30 challenges. I don’t have a formal background for this work but I enjoy impossible problems. In the pursuit, I developed a formula capable of reproducing the possibly complete Rule 30 cone.
I think I may have come close to the solution, become confused or perhaps I have it and do not know it. In any case, I was happy to see the invitation to ask the human community what you think.
The formula works by flattening the cone row by row, reading each row from left to right.
Starting with (F(0)=1), for each index (n\geq1), let
𝑡=𝑙𝑓𝑙𝑜𝑜𝑟𝑠𝑞𝑟𝑡𝑛𝑟𝑓𝑙𝑜𝑜𝑟,𝑞𝑞𝑢𝑎𝑑𝑟=𝑛−𝑡2.
Here, (t) is the row number and (r) is the position within that row, both counted from zero. Perfect squares mark the beginnings of rows.
The three parent values are
Misplaced &
Misplaced &
and
Misplaced &
These conditions set any parent outside the preceding row to zero. The value at (n) is then
𝐹(𝑛)=𝐿𝑛𝑜𝑝𝑙𝑢𝑠(𝐶𝑛𝑙𝑜𝑟𝑅𝑛),
where (\oplus) means exclusive OR and (\lor) means OR.
The center column is selected at the pronic positions—the products of consecutive integers:
𝑐𝑡=𝐹𝑏𝑖𝑔𝑙(𝑡(𝑡+1)𝑏𝑖𝑔𝑟).
Im not familiar with the Wolfram software so I wrote a standalone Python program that compares this formula with conventional Rule 30 evolution. It checks the 65,536 cells of the first 256 complete rows and also checks the first 4,096 center states separately, all show exact agreement.
The program uses only the Python standard library, I posted it to github here:
r30_test
These are finite checks, so I only assume that the agreement continues. I was satisfied that I had broken the problem out of its single origin enclosure so I could approach the problem from an alternate origin. Which may have been the challenge itself. Then I became confused and went to bed.
My question is:
Can this formula help prove that the center column never becomes eventually periodic, has equal limiting frequencies of zeros and ones, and requires at least linear time to compute its (n)th bit? Or have I simply expressed Rule 30 differently?