# Semiïnfinite autobinomonorownonomicrogram

(Scroll down for a bonus / bounty follow-up question)

You too can be the very first to make your very own
☆   semiïnfinitely significant autobinomonorownonomicrogram   ☆

“But I already have my very own mid-infinite autobinomonorownonomicrogram,” you pule?   Of course you have —yawn — by now, who doesn’t?

As contrast, here is a run-of-the-mill biïnfinite (bi-infinite) autobinomonorownonomicrogram. $\require{begingroup}\begingroup \def \l { \kern-.3em\cdots~ } \def \L { & ~\cdots\kern -.1em } \def \r { ~\cdots } \def \R { \kern-.2em\cdots~\\\hline } \def \p { \phantom{ \Rule {2.5ex}{2.0ex}{0.5ex}} } \def \X {\kern-.5em \Rule{2.5ex}{2.0ex}{0.5ex} \kern-.5em} \def \b {\kern-.5em \p \kern-.5em} \def \1 {\kern-.5em\rlap {\normalsize \bf \kern .2em 1 } \p \kern-.5em} \def \0 {\kern-.5em \rlap{ \scriptsize \kern.3em 0 } \p \kern-.5em}$

$\small\begin{array}{c|c|} \sf\scriptsize Consecutive~counts~(in~binary) \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R \l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r \L &\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b&\b& \R \end{array}$

And its solution.
$\small\begin{array}{c|c|} \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R \l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r \L &\0&\1&\0&\1&\0&\1&\1&\0&\1&\0&\1&\0&\1& \R \end{array}$

Note how the same infinite digit sequence ... 0 1 0 1 0 11 0 1 0 1 0 1... constitutes both the margins’ counts and the interior’s cells. (These counts are binary, so 10 = 2.)   The present puzzle, however, seeks:

A semiïnfinite (semi-infinite) grid, which extends infinitely only leftward or rightward but not both.   (The example above is biïnfinite as it extends in both directions.)

A significant grid, meaning that it contains infinitely many row counts of 10 and thus infinitely many 1 1 adjacencies among cells.   (The example has only one row count of 10 and only one corresponding 1 1 cell adjacency.)

Here is a schematic of a right-extended semiïnfinitely significant autobinomonorownonomicrogram.

$$\small \def \s #1#2{{\scriptsize\raise1.3ex\matrix{ \sf #1 \\[-1ex] \sf #2 }}} \begin{array}{c|c|} &\s {some}{counts} & 1& 1&\cdots & 1& 1&\cdots & 1& 1&\R \s{some}{counts} ~ 10 ~ \cdots ~ 10 ~ \cdots ~ 10 \r & \s{some}{cells} &\1&\1& \cdots&\1&\1& \cdots&\1&\1& \R \end{array}$$

Autobinomonorownonomicrogram?”   It’s short for auto-bino-monorow-nono[micro]gram.
auto:   Self-descriptive — cells’ contents match the margin counts’ actual digits.
bino:   Binary numbers.
monorow:   Exactly one row tall.
nonogram:   This type of grid puzzle.
micro:   No numbers greater than 2, which shows as binary 10 (or 010, 0010, ...).
(Thus 3 consecutive cells cannot be all 1s.)

Evolutionary path of autobinomonorownonomicrograms. Begin with a familiar nonogram such as this 3×8, where numbers at its left and top margins are length counts of consecutive filled cells in their respective rows and columns.

$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & 2 & 1 & 1 & 1 & 1 & 1 & 2 \kern.05em \\ \hline 2 ~~ 3 & \b & \b & \b & \b & \b & \b & \b & \b \kern.05em \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \kern.05em \\ \hline 3 ~~ 2 & \b & \b & \b & \b & \b & \b & \b & \b \kern.05em \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & 2 & 1 & 1 & 1 & 1 & 1 & 2 \kern.05em \\ \hline 2 ~~ 3 & \b & \b & \X & \X & \b & \X & \X & \X \kern.05em \\ \hline 1 ~~ 1 ~~ 1 & \b & \X & \b & \b & \X & \b & \b & \X \kern.05em \\ \hline 3 ~~ 2 & \b & \X & \X & \X & \b & \X & \X & \b \kern.05em \\ \hline \end{array}$$

Turn that into binary, where 0s and 1s indicate empty and full cells while binary numbers are used for counts.

$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & \! 10 \! & 1 & 1 & 1 & 1 & 1 & \! 10 \\ \hline 10 ~~ 11 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 11 ~~ 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 & \! 10 \! & 1 & 1 & 1 & 1 & 1 & \! 10 \\ \hline 10 ~~ 11 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \1 \\ \hline 1 ~~ 1 ~~ 1 & \0 & \1 & \0 & \0 & \1 & \0 & \0 & \1 \\ \hline 11 ~~ 10 & \0 & \1 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline \end{array}$$

Empty a couple of corner-cell 1s to attain the micro quality, having counts only of 0, 1 and 10.

$$\small\begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 &\bf 1 & 1 & 1 & 1 & 1 & 1 &\bf 1 \\ \hline 10~~\bf 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline 1 ~~ 1 ~~ 1 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline{\bf 10}~~ 10 & \b & \b & \b & \b & \b & \b & \b & \b \\ \hline \end {array} \qquad \begin{array}{r|c|} & & & 1 & 1 & & 1 & 1 & \\[-1ex] & 0 &\bf 1 & 1 & 1 & 1 & 1 & 1 &\bf 1 \\ \hline 10~~\bf 10 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline 1 ~~ 1 ~~ 1 & \0 & \1 & \0 & \0 & \1 & \0 & \0 & \1 \\ \hline{\bf 10}~~ 10 & \0 & \0 & \1 & \1 & \0 & \1 & \1 & \0 \\ \hline \end{array}$$

Almost there, pare down to just one row. The following 1×$\kern1mu\raise1mu\infty$ nonogram would be an autobinomonorownonomicrogram if only its digits were exactly matched. But its cells contain a 1 (circled) where the counts’ digits do not ($\,\scriptsize\wedge\,$).

$$\small\begin{array}{r|c|} \L & 0& 1& 1& 0& 1& 1& 0& 1& 0& 1& \R \l ~0 1~~1 0~~1 \rlap{\kern-.25em\scriptsize\raise-1.5ex\wedge} 0~~1~~0 1 \r \L &\0&\1&\1&\0&\1&\1 \rlap{\kern-.65em\Large\raise-.1ex\bigcirc} &\0&\1&\0&\1& \R \end{array}$$

Bonus / bounty question:   Can a biïnfinite significant autobinomonorownonomicrogram infinitely repeat an unchanging pattern?   (Doubtful at pose time, but also a hint for the present puzzle.)

$$\small \def \s #1#2{{\scriptsize\raise1.3ex\matrix{ \sf #1 \\[-1ex] \sf #2 }}} \begin{array}{c|c|}\L & 1& 1& \cdots & 1& 1& \cdots & 1& 1& \cdots & 1& 1&\R \l 10 ~ \s{some}{counts} ~ 10 ~ \s {same}{counts} ~ 10 \r \L &\1&\1&\s{some}{cells}&\1&\1& \s {same}{cells}&\1&\1& \s{same}{cells}&\1&\1& \R \end{array}$$

$\endgroup$

• '“But I already have my very own mid-infinitely autobinomonorownonomicrogram,” you pule?' - what does pule mean? – boboquack Feb 4 '17 at 1:27
• Don't take it wrong, @boboquack, I'm not calling you a pule, that would be silly! Actually, that would be puler. (Pule is a verb: to whimper, whine, cry weakly.) – humn Feb 4 '17 at 3:28

This one I can confirm is possible!

Mine's left-infinite - ...110101010101101010101101010110101101 with ever-decreasing-length segments (from L to R) of 01010...10s between the 11s

This can be broken up like so:

...1 1 01 01 01 010 1 1 01 01 010 1 1 01 010 1 1 010 1 10 1 where the first 1 10 1 is not part of any pattern but before that we have 010s with 2 1s and the correct number of 01s to pad which is also decreasing (from L to R) at the correct rate.

Picture from @humn:

• I knew you'd be ready for this one, boboquack. Alternate bounty for another semiïnfinitely significant autobinomonorownonomicrogram that isn't a superset of this, or a could-be-less-than-rigorous proof of none such possible. (I'm already very curious to know how you got this particular semiïnfinitely significant autobinomonorownonomicrogram so quickly.) – humn Feb 4 '17 at 3:52
• @humn a bit of a case bash with some common-sense did it, was this your solution? – boboquack Feb 4 '17 at 5:05
• Yes, same, boboquack, reached with a fair amount of analysis after the previous puzzle got solved surprisingly quickly. I doubt anyone else would have solved this, with a cold start, even half as quickly as you, and am surprised again anyway. This is essentially a cellular-automaton situation in disguise. – humn Feb 4 '17 at 5:15