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     ☆   Be the first to make your own semiminibinononohohohologram   ☆

“  !  ” you interject?   Might not be quite as supercalifragilisticexpialidocious as it sounds, though. What could look like an infinitely satisfactory solution can easily hide mistakes in plain view or around an infinite corner.   The goal is for margin counts to be pixel-per-cell copies of the edgeless filled-in image, where numbers are 3 cells (pixels) tall and column counts are 5 cells (pixels) wide.

The circled 11 demonstrates how two 1s of a 11 (binary 3) row count may belong to two separate 11 column counts.   The black cells and digits highlight one way to include a 0 (as part of 01 this time).

Mistakes in this example are column counts that disagree with columns’ cells. Almost all column counts show infinitely many 11s (threes) whereas every fifth column is empty or almost so.   Notice how the center cell column of the large-image 0 has accurate counts above it, with 01 and 1 for the only two places where filled-in cells do not belong to 3-tall triples.

So, “semiminibinononohohohologram?”   It’s short for semi-mini-bino-nono[hoho­ho-holo]gram.
      mini:   No numbers greater than 3. (Thus 4 consecutive cells cannot be all filled.)
      semi:   Half of all available numbers typically occur: 1 and 3, rarely 0, never 2.
      bino:   Binary numbers, leading zeros allowed: 1, 01, 11 (for 3) and 011 (also 3).
    nonogram:   This type of grid puzzle.
     hohoho:   For fun.
      holo:   The cells’ image is a pixel-per-cell magnification of the margin counts.
        (Large pixels at that: 3 × 3 pixels make 0,  1× 3 pixels make 1.)


Click to expand this menagerie of valid and invalid cell configurations. No need to memorize them; they just reflect what is stated elsewhere. Faint shading indicates cells that may be occupied by neighboring numbers without touching.   Notes:

  • Digits do not touch each other, even at corners.

  • In row counts, digits separated by 2 or more blank cells belong to separate numbers.

  • In column counts, all digits within a column’s 5-cell (5-pixel) span are considered together as a single number, regardless of how many blank cells separate any digits.

  • There are no 10 (as in 2) counts because pixels in 0 and 1 only line up singly or in triples.

  • Each column or row either has a single 0 count, if empty, or any combination of 1, 01, 11 and 011 counts.

  • Mid-infinite intervals are cheered, as long as they have consistent ends. For example, 1 1 1 ... 1 1 1 0 1 0 1 0 ... 1 0 1 0 1  is fine but  1 0 1 0 1 ... 11 0 11 0 11  would be inconsistent.   Instances of debatable consistency are welcome on grounds of general interest.

Got this far? Congratulations!   No complete detailed solution comes straight to mind?   A visit to the autobinomonorownonomicrogram exhibit might help familiarize some underlying concepts.   Incidentally, this came out looking tougher, but solving easier, than intended.

                               Text-reduced elaboration of the example

                               |   :     :     :     |     :     :     :     :     :     :
                               |   | 3 3 | 3 3 | 3 3 | 3 3 | 3 3 | 3 3 | 3 3 | 3 3 | 3 3 |
                               |   |  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|
                               |   |3  3 |3  3 |3  3 |3  3 |3  3 |3  3 |3  3 |3  3 |3  3 |
                               |   | 3  3| 3  3| 3  3| 3   | 1  3| 3  3| 3  3| 3  3| 3  3|
                               |   |3 3  |3 3  |3 3  |3 3  |3    |3 3  |3 3  |3 3  |3 3  |
                               |   | 3 3 | 3 3 | 3 3 | 3 3 | 1 3 | 3 3 | 3 3 | 3 3 | 3 3 |
                               |   |  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|  3 3|
                               |   :     :     :     :     :     :     :     :     :     :
------------------------------ |------------------------------------------------------------
                               |   :     :     :     :     :     :     :     :     :     :
       1 1         1 1         |   :     : O  O:     |     :O  O :     : O  O:     |     :
          1 1         1 1      |   :     : O  O:     |     :O  O :     : O  O:     |     :
 1 1         1 1         1 1   |   :     : O  O:     |     :O  O :     : O  O:     |     :
    1 1         1 1            |   :O  O :     :     |     :     :O  O :     : O  O|     :
       1 1         1 1         |   :O  O :     :     |     :     :O  O :     : O  O|     :
          1 1         1 1      |   :O  O :     :     |     :     :O  O :     : O  O|     :
 1 1         1 1         1 1   |   :     :     :O  @ |@  O :     :     :O  O :     | O  O:
    1 1         1 1            |   :     :     :O  @ |@  O :     :     :O  O :     | O  O:
                   1 1         |   :     :     :O  @ |@  O :     :     :O  O :     | O  O:
          1 11 1      1 1      |   :     :     :     |     :@@@ O:     :     :O  O |     :
 1 1             01      1 1   |   :     :     :     |     :O O O:     :     :O  O |     :
    1 1      11                |   :     :     :     |     :@@@ O:     :     :O  O |     :
       1 1       1             |   : O  O:     :     |@ @  :     :     :     :     |O  O :
          1 1      1 1         |   : O  O:     :     |@ @  :     :     :     :     |O  O :
             1 1      1 1      |   : O  O:     :     |@ @  :     :     :     :     |O  O :
 1 1            1 1      1 1   |   :     : O  O:     |     : @   :     : O  O:     |     :
    1 1            1 1         |   :     : O  O:     |     : @   :     : O  O:     |     :
       1 1            1 1      |   :     : O  O:     |     : @   :     : O  O:     |     :
          1 1            1 1   |   :O  O :     : O  O|     :     :O  O :     : O  O|     :
             1 1               |   :O  O :     : O  O|     :     :O  O :     : O  O|     :
 1 1            1 1            |   :O  O :     : O  O|     :     :O  O :     : O  O|     :
                               |   :     :     :     :     :     :     :     :     :     :
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  • $\begingroup$ Examples are guaranteed 95%+ error-free. Doubts/corrections welcome. $\endgroup$ – humn Jan 17 '17 at 7:27
  • 2
    $\begingroup$ If I have to say the name then I don't think I'll want to make it :P $\endgroup$ – Beastly Gerbil Jan 17 '17 at 7:48
  • $\begingroup$ Are 001 and 0011 allowed? $\endgroup$ – boboquack Jan 30 '17 at 22:47
  • $\begingroup$ Row counts of 001 and 0011 are fair game, @boboquack, with the warning that they will create some column counts of 0, which would be the only counts for their columns, and those columns would have all blank cells. $\endgroup$ – humn Jan 30 '17 at 23:12
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@humn, can you please check this, but I contend that there is no valid semiminibinononohohohologram.


First, we will prove that there is no row count of 0.

If there existed such a row count, consider the group of 3 rows that the row labelled by it is in. Since both 0's and 1's take up all three rows, there cannot exist any 0's or 1's in the group of 3 rows, and so a row goes unlabelled - contradiction.

So then there must be a 1 or a 01 after every 0.

0| Can't be ... or . -> 0|  -> (Nothing)| <Wrong
?|          . .    .    0|
?|          ...    .    0|

Let a self-referential column group be one which provides a column count for one of its 5 sub-columns.

Second, we will prove that there is a 0 in a self-referential column group.

If there was no 0 in the self-referential column group, there would be no 1 in the self-referential column group.

But then all columns would have 3's in it, because parts of the 3's in the 4 sub-columns that aren't self-referring are 3-high, so the column labels would also have to be threes, and by induction (can be formalised) all column counts have 3's in them, so no 0's exist anyway (because 011 can't fit in a column group). Contradiction.

          (Same number of 3s as 1s)     3     3  3    3   3
 -----              -----               - <- ----- <- ----- <- etc. 
No ... ->           .               ->        .  .    .   .
   . .              .                         .  .    .   .
   ...              . (3-high)                .  .    .   .

Third, we will prove that there is a 0 in a non-self-referential column group.

If there was no such 0, all such column groups would consist of 1's. Consider a column group that does not refer to a self-referential column. If there is a lone 1, we are done (because there is a 0 in the row it refers to), otherwise, everything must be in 3's.

Now at least one of the 3's must also be 3's as row counts, otherwise there would be a middle column gap, giving a 0 in the column group referring to it. So we have a 0, which must be in the self-referential column.

Since there are infinitely many such column groups, there are infinitely many 0's in a self-referential column group. However, this means that the column has infinitely many 01's (because if 1 was just a 0, there wouldn't be anything else in the column).

(Couldn't find the rest of this proof, so still trying to remember. TBC...)


Fourth, we will prove that there exists a row with at least 2 1's.

Take a row group with a 0 in it. Then take the middle row. Since the 0 has two 1-wide parts, in the row that referred to it, there must be 2 1's.


Fifth, we will prove that there is no 01 in a self-referential column group.

If there was, since there can't be a 0 without a 1 in the same column as other numbers, the only 1's for the column count would be in the second column.

But since the column group has 01, it must refer to the second column, because it is self-referential and the second column is the only one with a 1-tall number (the 0).

But then all other column groups would have 3's in it, because parts of the 01's, 1's or 3's in the 4 sub-columns that aren't self-referring are 3-high, so the column labels would also have to be threes, and by induction (can be formalised) all other column counts have 3's in them, so there is only 1 column that can have 1's in it. Contradiction (of the 3rd point).

So a self-referential column consists of a single 0.


Sixth, we will prove that there can't be 3 empty columns adjacent to each other.

This would mean that 3 adjacent column counts each have a single 0.

Because there can't exist a lone 0, the 3 0's must be adjacent to each other. It is easy to see that this means that the rightmost of the 3 columns has the 0 on the left.

But then 1 cannot be to the right of the 0, because there are two empty columns in the same column group as the right 0. Contradiction.


Seventh, we will prove that there cannot be two empty columns, a 1, followed by another two empty columns.

Again, consider the column counts. The two 0's must be to the right and in the middle of their column groups respectively.

Then the first one of the 3 column count must be on the left, because otherwise there would be a 00 row count.

But then because of the sixth point, the second 1 must be in the fourth column of that group.

So the first 0 must be in the centre, to avoid a 10 and to avoid a 3-column gap.

The other 0 must be next to it to avoid a 0 row count, i.e. on the right, but then since a 1 can't go in the column group, for he same reasons as before, we have a 00 row count. Contradiction.


Now with all the proving out of the way, we can case bash (prove by exhaustion) over a self-referential column group.

First, if the 0 is on left, there must be a 1 in the fifth column of the group to avoid a 0 row count. However, there can't be a 01. Contradiction.

If the 0 is on the right, it can either refer to the first or second column in the group. If it refers to the first column, the next column also needs a 0 in the centre, and the next four after that have 3, two 1's, 3 and 0 respectively (by the column counts). Let this be called a 031130 position (note that the first 0 is in the centre of its column) - these positions will be disproved later. If it refers to the second column, the column group before it must also contain a 0, but there is no way to place that 0 adjacent to the 0 in the self-referential column group, so there is a row count with a 0. Contradiction.

If the 0 is in the centre, it can either refer to the first or fifth column in the group. If it refers to the first column, the next four columns are 3, two 1's, 3 and 0, forming another 031130 position (again by the column counts). If it refers to the fifth column, consider the column group to the left of it. It must also be self-referential, this time to the first column of the group.

Now consider a 031130 position. We will make an assumption tree that represents all possibilities:

  • The first 1 of the left 3 column group is in the first column, because otherwise the 0 doesn't have a 1 after it and we have a 0 row count.
    • The second 1 of the left 3 column group is in the third column. Then the first 1 of the 11 column group is in the first column.
      • The second 1 of the 11 column group is in the first or second column. Then the first 1 of the right 3 column group would be too far away (3 empty columns between).
      • The second 1 of the 11 column group is in the third column. Then the first 1 of the right 3 column group is in the first column. Then the second 1 must be in the third column. So the right 0 is on the left of its group, but then a one must be in the fifth column of its group.
      • The second 1 of the 11 column group is in the fourth column. But this is too far away, the first 1 has two empty columns on each side.
    • The second 1 of the left 3 column group is in the fourth column. Then the first 1 of the 11 column group is in the first column because otherwise the second 1 of the left 3 column group has 2 empty columns on each side.
      • The second 1 of the 11 column group is in the first or second column. Then the first 1 of the right 3 column group would be too far away (3 empty columns between).
      • The second 1 of the 11 column group is in the third column. Then the first 1 of the right 3 column group is in the first column. Then the second 1 must be in the third column. So the right 0 is on the left of its group, but then a one must be in the fifth column of its group.
      • The second 1 of the 11 column group is in the fourth column.
        • The first 1 of the right 3 column group is in the first column, to avoid two empty columns around a 1.
          • The second 1 is in the third column. So the right 0 is on the left of its group, but then a one must be in the fifth column of its group.
          • The second 1 is in the fourth column. So the right 0 is on the left of its group, to avoid two empty columns around a 1, but then a one must be in the fifth column of its group.

If someone can pick a hole in the proof, it might just give you the leverage you need to construct a semiminibinononohohohologram!


Diagrams coming soon...

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  • $\begingroup$ After a better check, boboquack, I'd like to cast doubt earlier than before, on "Third, ... there is a 0 in a non-self-referential column group." My test solution, with regained confidence, agrees with your reasoning in having a self-referential column and "the column has infinitely many 01's." Yet the solution has no other 0s, leaving us just where part of your proof went missing. Please @humn me with a contradiction to look for in the test solution. Regardless, your considerations are gratifyingly similar to what went into this puzzle and a bounty is on its way for this much already. $\endgroup$ – humn Feb 2 '17 at 17:29
  • $\begingroup$ @humn I had to get my computer repaired, so I lost some of my recent work. Anyways, I'm trying to use the gap in my proof to find one. $\endgroup$ – boboquack Feb 2 '17 at 20:28

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