An 84-cell ring of Rule 30 that comes back to itself every six steps. The beads are the ring now. The strip inside is the same ring cut open at the bottom and laid flat, sites −41 to 42 across, with time running down for six periods. Four beads, spread round the ring, are traced: each dotted line runs from the bead to the cell that holds its present value, and slides down its column as time runs. The centre cell ticks white, black, white, black, and the cell beside it, read at even times, says 1 0 0, 1 0 0 for ever: the short loop S, and nothing else.
The period-two problem asks whether a finite start can keep the centre ticking 0 1 0 1 for ever. Read at even times, the column beside it is a string of gap words; the record's two are S = 100 and L = 10000. Question 4 of the portfolio asks how freely real Rule 30 rows can string S and L together. This ring is the first actual Rule 30 row that plays S for ever, so arbitrarily long all-S stretches are physically possible.
Every all-S trace has to look like this one near the centre. Its first five rows repeat with period six, whatever lies further out.
Each step of Rule 30 turns this ring by 14 of its 84 beads, so a fixed comb over the 14 beads at the top right reads the whole ring once every six steps. Sound, off until you press it, plays it that way: one beat per step, each beat a sweep along the comb in which every black bead plucks a note of A minor pentatonic, with a low A under it when site 0, the clock, is black and an E when site 1 is. The ring's whole infinite history is one bar of six beats, playing for ever. The music keeps its own time and goes on while the page is out of sight, until you pause it.
Listen for the alternation Gareth heard: arpeggio, staccato, arpeggio, staccato. The even beats run up the scale in arpeggios (25 notes, runs averaging 2.3 notes), the odd beats are staccato (18 notes, mostly alone, runs averaging 1.4), and the clock's low A falls on the staccato beats. It is Rule 30 breathing. Inside a black run every cell but the leftmost turns white at the next step, and a lone black cell becomes three, so each beat's runs break up into scattered notes and the scattered notes grow back into runs.
The ring repeats along the whole line, so it has infinitely many black cells. The prize asks about finite starts, so this is not a counterexample.
It is also rigid. In its search domain, the live states form a single 84-cycle and only one of 20 entrance paths survives, so there are no free choices and the family it gives has zero entropy. The route through this fixed six-step domain is closed; general all-S behaviour stays open.
84 bits, site 0 at bit 0. Repeat this word along the whole line and run Rule 30; this page does exactly that when it loads.