4
$\begingroup$

Create a Nonogram / Picross which cannot be solved uniquely without giving at least one clue. That is, without prefilling at least one field in the puzzle there exists multiple valid solutions.

The above shouldn't be too difficult for those that have already solved many nonograms. For added difficulty consider the following - but only after you have solved the initial puzzle, since it also gives hints at a solution for the above.

Consider solutions with the following limitations:

* Non-symmetric - both mirroring and rotational.
* No empty or completely filled rows or columns.
* A generic solution working both for even and uneven number of rows and columns.

$\endgroup$
1
  • 1
    $\begingroup$ Great title. What next? "Give me a break!" for a dissection puzzle? Still, it continues to perplex me how an answer can have more ^votes than the puzzle that generated it. $\endgroup$
    – humn
    Commented Apr 9, 2020 at 19:39

4 Answers 4

8
$\begingroup$

A minimal example:

enter image description here

That is, of course, symmetrical. But one can incorporate the same idea in a non-symmetrical grid:

enter image description here

Solutions:

enter image description here
In both cases one can shade either the blue or the red cells to get a valid solution.

$\endgroup$
1
  • 1
    $\begingroup$ I see we had the same idea :-) I was editing in my generalisation just as I saw the "1 new answer" banner pop up. $\endgroup$ Commented Apr 9, 2020 at 9:54
6
$\begingroup$

Here’s a Nonogram/Picross that
    has multiple possible solutions if no cell is given as a clue,
    has a unique solution given any cell as a clue, filled or empty,
    has no self-symmetric solutions, mirrored or rotational,
    has no empty or full columns or rows,
and possibly has the fewest total cells for these conditions.

Note that a condition here is additional to the puzzle statement, that knowing any cell at all forces a unique solution. This is satisfied because these solutions have no cells in common, filled or empty.

Darrel Hoffman points out in a comment that the solutions above are symmetric to each other and wonders what minimal layout would have mutually asymmetric solutions. After all, solutions for the 4×4 layout in jafe’s answer are not mirror images. ­ To avoid redundancy with other answers, though, the present answer’s condition that a unique solution would result from clueing any cell whichsoever requires that no cells at all are deducible without without any clues. Otherwise the clueing of a deducible cell would still leave the layout’s original ambiguity. ­ The following 8×6 layout seems minimal for this answer’s parameters while having mutually asymmetric solutions.

Generalization to any larger even numbers of columns or rows can be straightforward by adding pairs of columns and rows, as demonstrated by extending a solution for the mirror-image-solutions layout presented first.

A layout must have even numbers of columns and rows to satisfy this answer’s additional condition that a unique solution is forced when any cell is given as a clue, filled or empty. This condition means that there are just two possible solutions, each a negative image of the other, because the filled-or-empty choice of any clued cell can only select between two solutions.
  As a layout fixes the number of filled cells for each column and row, and as empty cells of one solution are filled cells of the other solution, every column and row has an equal number of filled and empty cells. Only even numbers of columns and rows allow all rows and columns, respectively, to have half of their cells filled while half remain empty.

$\endgroup$
1
  • 1
    $\begingroup$ Your two solutions may be asymmetrical by themselves, but one is obviously a mirror of the other. I wonder what would be the minimal solution where this is not the case. Jafe's 4x4 answer proves it's possible - can it be done with a smaller grid? $\endgroup$ Commented Apr 9, 2020 at 20:28
4
$\begingroup$

A simple example, with no rows or columns empty or completely full, would be

an $n\times n$ square ($n>1$) where the number in every row and column is $1$. That just means we have exactly one filled square in each row and each column, which of course can be done in many different ways.

To make it asymmetric and more general, you could simply

make part of the grid like that, with $1$s in every row or column, while another part is solvable with any arbitrary (non-symmetric) configuration. Just make sure the two parts are distinguishable by both rows and columns. Like a block matrix, say:

  • top left corner is just 1 in each row and column there,

  • bottom right corner is anything you want (even something uniquely solvable is OK)

  • top right and bottom left corners are empty.

Here "corner" means a whole subgrid of the puzzle, not just one cell.

See also (shameless plug) the natural generalisation/extension of this problem:

How many possible starting positions are uniquely solvable for a nonogram puzzle?

$\endgroup$
0
$\begingroup$

l can't see any pictures from lmgur, so here is an ASCll Art version.

1 1 1 1 1 M . or 1 . M 1 . M 1 M .
M = FiIIed ceII

The answer to the grid isn't unique.
But if you fill any of the four cells, you can solve the rest easily.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.