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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:

progress

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?

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5 Answers 5

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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 ...

new grid after the next deductions

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  • $\begingroup$ And yep, I've now solved it completely. facepalm $\endgroup$
    – Rand al'Thor
    Commented Jun 17, 2019 at 18:01
  • $\begingroup$ 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.. $\endgroup$
    – Nuclear Hoagie
    Commented Jun 17, 2019 at 18:05
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At least

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

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[Edit] 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).

Poorly editted image showing blocks that can be filled in

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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.

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  • $\begingroup$ OK, so the 5 must be in the upper block ... but then what? We don't know exactly where that 5 must be. $\endgroup$
    – Rand al'Thor
    Commented Jun 17, 2019 at 17:54
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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.

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    $\begingroup$ But why would it continue to the right necessarily? It could equally well continue to the left, as far as I can see. $\endgroup$
    – Rand al'Thor
    Commented Jun 17, 2019 at 17:53

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