Autobinomonorownonomicrogram

  (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
0
s, as in 01 or 0010. $ require{begingroup}begingroup
def l { kern-.3emcdots~ } def L { & ~cdotskern -.1em } def
r { ~cdots } def R { kern-.2emcdots~\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-.5emrlap {normalsize bf kern .2em 1
} p kern-.5em} def {kern-.5em rlap{ scriptsize kern.3em 0
} p kern-.5em} $

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

  $smallbegin{array}{c|c|} sfscriptsize
Consecutive~counts~(in~binary) L & 0& 1& 0& 1& 0& 1& 1& 0& 1& 0&
1& 0& 1& R l 0 1~~0 1~~0 1~~1 0~~1~~0 1~~0 1 r L
&b&b&b&b&b&b&b&b&b&b&b&b&b&
R end{array}$

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

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.

$$smallbegin{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.

$$smallbegin{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 & & & 1 & 1 & & 1 & 1 & 1
\ hline 1 ~~ 1 ~~ 1 & & 1 & & & 1 & & & 1 \
hline 11 ~~ 10 & & 1 & 1 & 1 & & 1 & 1 & \ hline
end{array}$$

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

$$smallbegin{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 & & & 1
& 1 & & 1 & 1 & \ hline 1 ~~ 1 ~~ 1 & & 1 & &
& 1 & & & 1 \ hline{bf 10}~~ 10 & & & 1 & 1 &
& 1 & 1 & \ hline end{array}$$

Almost there, pare down to just one row. The
following 1×$kern1muraise1muinfty$ 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 ($,scriptsizewedge,$).

$$smallbegin{array}{r|c|} L & 0& 1& 1& 0& 1& 1& 0& 1& 0& 1&
R l ~0 1~~1 0~~1 rlap{kern-.25emscriptsizeraise-1.5exwedge}
0~~1~~0 1 r L &&1&1&&1&1
rlap{kern-.65emLargeraise-.1exbigcirc}
&&1&&1& R end{array}$$

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

$ endgroup $

ベストアンサー

これは以下の要件を満たします:

enter image description here

十進数の変換行を用意し、 11 の位置を「手がかり」にハイライト表示しました。

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