Talk:Jagged lines

Analysis

Dean Hickerson's page's RLE's comment says

#C Jagged lines of gliders, formed by a drifting collision of two LWSS
#C streams, crash to form an approximately vertical jagged line of pairs
#C of blocks. I don't know if the line stays within a bounded distance
#C of the center line, or extends infinitely far to the left, or to the
#C right, or both.
#C Dean Hickerson, 5/3/2005

(I'm not sure whether this ought to be quoted in the page.)

Anyway, I have simulated it to a billion iterations in Golly (HashLife seems to struggle with it, I suppose due to its aperiodicity), and it crosses the zero line infinitely many times, and indeed does extend infinitely far in both directions.

Where the bi-blocks' horizontal displacements are divided by 40 and they are arranged in a 0-indexed list, where the first one (which lies on the line of symmetry of the initial pattern) has displacement 0, the index of the list's nth occurrence of 0 is given by the nth term of the generating function

λ x: x*(7+5*x+6*x²+10*x³-27*x⁴+75*x⁵-23*x⁶+311*x⁷-375*x⁹-9*x¹⁰+89*x¹¹-16*x¹²-105*x¹³+30*x¹⁴-361*x¹⁵+340*x¹⁷-40*x¹⁸+84*x¹⁹)/((1-x)²*(1+x)*(1+x²)*(1-2*x²)*(1+2*x²)*(1+x⁴))

And its number of cells of displacement from the first is

λ x: 80*x*(5+9*x+14*x²+21*x³+2*x⁴+57*x⁵+38*x⁶+273*x⁷+273*x⁸-8*x⁹-16*x¹⁰+52*x¹¹+36*x¹²-40*x¹³-16*x¹⁴-288*x¹⁵-288*x¹⁶-32*x¹⁷-64*x¹⁸)/((1-x)²*(1+x)*(1+x²)*(1-2*x²)*(1+2*x²)*(1+x⁴))

After the initial period of chaos, the zeroes between large gaps come in quartets, the three after the first have index delays and displacements of (1,10,11) and (80,560,720) respectively, alternating between forwards and backwards. If these "trivial zeroes" (as it were :-) after the first in each quartet are excluded, the behaviour (upon regularity) is somewhat nicer to describe, zeroes occur at indexes given by g.f.

λ x: (1525-116*x+1408*x²)/((1-x)*(1+x)*(1-4*x))

or explicitly

λ n: (22*4ⁿ⁺⁶-30*n-2097*(-1)ⁿ-2335)/180

And at displacements with g.f.

λ x: 80*(357+73*x-372*x²)/((1-x)*(1+x)*(1-4*x))

or explicitly

λ n: 16*(22*4ⁿ⁺³+132*(-1)ⁿ-145)/3

Note that this begins to apply, for n=0, at the block with a displacement of 7440 cells from the first, despite the fact that it has only a single trivial successor, not three like the rest of them, and that since the bi-blocks are created at a speed of c/6 upwards, there is thus the simpler statement that a bi-block pair is produced on generation 11*4ⁿ+4224*(-1)ⁿ-4187, for all n>=6.

The absolute maxima are not so easily describable as the zeroes, they seem to asymptotically double with each sign change (making its absolute version's running maximum grow as O(sqrt(n))) but seem not to be so precisely describable, the sequence of the maximum between each (and its index) goes ((0,0),(94,-7),(373,12),(1656,-25),(6575,48),(26240,-98),(105054,186),(418889,-370)). The sequence of indexes of minima between maxima between sign changes (and their first occurrences) goes ((52,-1),(213,2),(864,-6),(3449,12),(13821,-28),(55345,59),(221624,-118),(866737,235)).

Does there exist a recursive definition that allows the nth term to be computed in O(log(n)) operations?

Preblockless symmetrical variant

I have spoken with Magma about the variant without the initial pre-block (mentioned in Game of Life News), and the sequence of 0s and 1s generated by the sequence. The GoLN article mentions its equivalent substitution system, I asked whether the fact that this causes the length to multiply by 5 per iteration implies that it cannot be given by the first differences of a Beatty sequence (of the form a(n)=⌊a*n⌋ for a quadratic irrational a), and he pointed out that there is the simpler fact that it contains 5 of the 8 possible length-3 substrings, whereas Sturmian words (a superset of Beatty sequences, I think) have each length-n substring occur exactly n+1 times.

It appends elements while reading existing ones, so is equivalent to the concatenation of the subsequences A035263[:4ⁿ] for each n>=0. 0s are substituted for two 0s and two 1s, while 1s are substituted for one 0 and three 1s, so it is given by the matrix multiplication

(2,2,^. (1, ((4ⁿ+2)/3,
.1,3)ⁿ* 0)= (4ⁿ-1)*2/3)

So after it has concatenated together the first n iterations of this, and produced (4ⁿ-1)/3th bits, it will have (4ⁿ+6*n-1)/9 0s and 2*(4ⁿ-3*n-1)/9 1s, and of course asymptotic density 2/3.

DroneBetter (talk) 13:39, 3 August 2023 (UTC)