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Hard nonogram

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

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    $\begingroup$ how come there are 2 pictures? $\endgroup$
    – Aaron
    Feb 26 at 19:03
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Look at the seventh column:

With the X at the top of the column, you can completely fill out the rest of the column.

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Row, 5 column 15 must be an X, which then yields additional deductions for column 15.

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Here's a couple of different places you can go from there, I'll leave the rest for you to solve:

enter image description here

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

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  1. 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.

  2. 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).

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