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  (Bonus / bounty follow-up challenges have been moved to Semiïnfinite autobinomonorownonomicrogram)


  ☆   Be the  first  next to make your own nontrivial autobinomonorownonomicrogram   ☆

“Impossible!” you scoff?   Might indeed be, except that your autobinomonorownonomicrogram may be infinitely wide and include leading 0s, as in 01 or 0010. $ \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} $

Nontrivial  ?” you might ask.   Well, trivial autobinomonorownonomicrograms are just too common.

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

is a trivial one that solves to
  $\small\begin{array}{c|c|} \L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0& 1& 0& 1& \R \l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 \r \L &\0&\1&\0&\1&\0&\1&\1&\0&\1&\0&\1&\0&\1& \R \end{array}$

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.)   This example is called trivial as...

...nontrivial here means that multiple pairs of adjacent 1s occur among the cells. The example does not qualify because it has only one adjacent pair of 1s.

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. This already happens to be nontrivial as it has multiple sets of adjacent 1 cells.

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

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 ($\,\scriptsize\wedge\,$).

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

This nonetheless qualifies as nontrivial because two pairs of adjacent cells contain 1s.

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  • 6
    $\begingroup$ Can the solution to this just be a pronunciation of .. that word in the title? $\endgroup$ – Matt Jan 4 '17 at 14:22
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    $\begingroup$ It's 9°F here today. My car made this exact noise when I first went to start it. True story. $\endgroup$ – Rubio Jan 4 '17 at 14:26
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    $\begingroup$ Man I got tired just trying to pronounce that title. Which part of it was trivial, again? $\endgroup$ – Xenocacia Jan 4 '17 at 14:58
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    $\begingroup$ -1; unpronounceable title. $\endgroup$ – Mithrandir Jan 4 '17 at 16:03
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    $\begingroup$ I want this to reach HNQs just so that as many people as possible see that title. $\endgroup$ – Rand al'Thor Jan 4 '17 at 18:22
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This satisfies the requirements:

enter image description here

I've provided a decimal conversion row and highlighted the locations of 11s in the "clue".

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  • $\begingroup$ Looking good, Deusovi, good and quick! Thought I'd get to wait longer before adding a certain bonus/bounty challenge. You'll see. (This is also a feeler for a more complex nonogram puzzle in the works.) $\endgroup$ – humn Jan 4 '17 at 15:19
  • $\begingroup$ @humn: I see lots of ways this can be extended! Is it something to do with one of the 2s corresponding to itself in both interpretations, or maybe making it only extend infinitely in one direction? $\endgroup$ – Deusovi Jan 4 '17 at 15:32
  • $\begingroup$ Deusovi, did you know that your solution can be modified to extend in just one direction? It can, and was, duly credited. $\endgroup$ – humn Jan 6 '17 at 5:54
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(Extension of an answer to the original bonus question’s semi-infinite challenge.)

Here’s a “mid-infinite” nontrivial autobinomonorownonomicrogram, with neatly finite left and right margins in the spirit of an infinite dragon curve’s fixed endpoints.



(Derived from Deusovi’s solution.)

Lingering question:   Can a mid-infinite solution have infinitely many pairs of adjacent 1s (and thus infinitely many counts of 10). Only two pairs of adjacent cells here are 1s.

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