I have been working on this nonogram for some time and can't seen to figure out what to do next...
This is the link: https://nonograms-katana.com/play/#game
Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. It only takes a minute to sign up.
Sign up to join this communityI have been working on this nonogram for some time and can't seen to figure out what to do next...
This is the link: https://nonograms-katana.com/play/#game
Look at the seventh column:
With the X at the top of the column, you can completely fill out the rest of the column.
Row, 5 column 15 must be an X, which then yields additional deductions for column 15.
Here's a couple of different places you can go from there, I'll leave the rest for you to solve:
Red squares are black, and grey squares are crosses, using colours to highlight.
Quick explanation:
The top most red must be filled in, regardless whether the other black to the right is part of the 1 or the 3. This greys out the square below, and means the red 3 can be placed as it cannot fit to the right. Furthermore, the bottom grey on the right cannot be filled, as that would make a 4, and from that some of the 6 can be placed.
Good luck! If you need a bit more I can add some
In the fifth row,
the rightmost $3$ can't start from the right-hand edge, because then it would become $4$ with the black cell you've already filled. So the rightmost cell in this row is empty (cross).
Then in the rightmost column,
there's not enough space for the $6$ above the fifth row, so all those cells are empty. Now there's ten unknown cells to contain that $6$, so the middle two of them at least must be black. That enables you to cross off the rightmost $1$ in two rows.
In the fourth row,
if the lone black cell in the middle is the $1$, then there must be $3$ to the left of it with an empty cell in between. Then, reading down from the columns, the three cells below that $3$ must be empty, but then the fifth row won't have enough space for its $3,1,3$. Contradiction, so the fourth row can be completely filled in up to the middle (two empty cells then two black cells).