Help is needed. I have been stuck on this puzzle for some time. Any help or tips would be appreciated.

Nonogram 15 x 15 size puzzle

  • $\begingroup$ I've found an "in" for you, but it's a bit difficult to explain from the picture you've provided. Is there an interactive version you can link to? Where did you find this puzzle? $\endgroup$ Jun 16 '20 at 12:23
  • $\begingroup$ This is taken from Nonograms Katana app. Unfortunately, I cannot find a link which you are looking for. This puzzle is called Arrow and the author is eil5026 sent by users. $\endgroup$ Jun 16 '20 at 13:30
  • $\begingroup$ OK, no worries. I wrote up an answer just in words - hopefully it's clear enough to follow the logic. $\endgroup$ Jun 16 '20 at 13:33

Important realisation (which you've probably noticed already, but I put it here because I'm assuming knowledge of it in the answer below):

the clues are completely symmetrical about the leading NW-SE diagonal of the board. That means the solution must also be symmetrical. So whenever you make a deduction in one row, you can make a corresponding deduction in the corresponding column, or vice versa.

Now a few steps to make headway on this puzzle.

  • Look at the first column. It contains $3,3$, and there's a gap of size 4 at the bottom. Assume that gap contains the lower $3$.

    Then all $1$s for the second and fourth rows from the bottom are done, so we can blank out the rest of those rows. Then the $2,3,2$ column can't be filled in the bottom four rows, so we have the $3$ and the lower $2$ already. Then the $2,4,1$ row has the $1$ and the $4$ must be made by joining the two already-filled cells, with a blank to the left of that. But now the $6$ in the $1,3,6$ column doesn't have enough space.

    So the gap at the bottom of the first column is empty. By symmetry, the same for the gap at the end of the first row.

  • Look at the $1,3,4$ row and column. The $4$ is done in each case, and the $3$ is either joining the two filled cells (in which case the $1,1,1,1$ row and column are done) or going down/right to meet at a corner (in which case the $1,3,4$ are all filled and we can put blanks up/left to the end, leaving a gap of size only 9 for the $3,3$ in the first row/column, which means the last $1$ in the $1,1,1,1$ row/column must be the left/top cell).

    In either case, you can put three extra blanks in the $1,1,1,1$ row and in the $1,1,1,1$ column, since there's only two possibilities for where the last $1$ might be.

  • Then examine the $5,1$ row and column. You'll see that the position for the $5$ is now almost completely determined, and you can fill in the top left chunk of the board from that.

I haven't done a complete solution for you, but these starting steps should be enough to get you there.

  • $\begingroup$ Thanks for that. However when you say “by examining the 3,3 in the first row/column, we see that the last 1 in the 1,1,1,1 row/column must be the left/top cell.” How did you come to that conclusion? $\endgroup$ Jun 16 '20 at 14:46
  • $\begingroup$ @Clisleno1 Sorry, that was a pretty big jump over several steps of reasoning. Edited to hopefully make it clearer. $\endgroup$ Jun 16 '20 at 17:33
  • $\begingroup$ Ah right, yes I understand now. Thanks for your help. $\endgroup$ Jun 16 '20 at 17:41
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    $\begingroup$ Slight nitpick: the clues being symmetric already implies that the deductions must be symmetric. It is not necessary to make use of the solution being symmetric (this requires the additional assumption that the solution is unique, which is a standard convention but more of a meta-deduction that oughtn’t be required in a properly designed nonogram). $\endgroup$
    – Erick Wong
    Jun 16 '20 at 18:15

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