Very tricky nonogram - where to go next?

I've become somewhat addicted to Simon Tatham's "Pattern" (nonogram) puzzles recently. I thought I was becoming fairly adept with them, but this unusually difficult one has me stumped. I've got as far as this by using the usual tricks:

But now I can't figure out how to make any further progress. What am I missing?

How can I make the next step to solve this puzzle?

5 Answers

Duh, I got it.

Bottom row: the lone square must be part of the 6-block, but there's not enough space for it to go all the way to the right: it must extend at least two more cells to the left.

Then

edge cells are always useful because we can start from there to fill in whole blocks: in this case, the 5 and 3 at the bottom of the fourth and fifth columns.

I'm guessing the deductions will fall like dominoes from there ...

• And yep, I've now solved it completely. facepalm – Rand al'Thor Jun 17 '19 at 18:01
• You can infer even more about the 6. The center column is (7 2), flanked on either side by a 1 in the bottom position. Nothing can be filled in in the bottom row center column, otherwise you'll have an isolated square in the second-to-bottom row that violates (1 4 4). You can fill in two more squares to the left of the partial 6 block you show in the image, white out the bottom row center column and directly to the right of it, and place the 2 of (6 2) in the hole on the bottom right.. – Nuclear Wang Jun 17 '19 at 18:05

At least

on the bottom row, we know the 4th and 5th cell have to be part of the 6.

 Darn, you got it as I was typing this.

There are the ones I saw immediately that can be filled in (the red letters, the blue ones are the blocks that show that the red ones need to be filled in).

• Thanks @Rubio for fixing my spoiler formatting – Ted Jun 19 '19 at 22:00

You can use the

1,5,1 column (number 9), because the bottom row block cannot house the 5, which limits it at the top of the column.

Then:

the five block must start in at least row 3 and at most row 5, and this means rows 5,6,7 of column 9 can be filled.

• OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be. – Rand al'Thor Jun 17 '19 at 17:54

I’m thinking

Based on your 1 4 4 row (second from bottom), that you have a loner black square on the left side. If you continue that to the right, that should be your second set of 4.

• But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see. – Rand al'Thor Jun 17 '19 at 17:53