It’s strange that I have never posted any solving techniques for Picross-Hanjie-Griddlers-Nonograms puzzles. I’ve explained some of them in my picross books, but never here on the website.

So, that’s about to change and I’m immediately starting with an advanced solving technique, which I call “**bordering**“.

Consider this example (it’s a part of a griddlers puzzle that I’m working on right now).

Focus on the two bottom rows and on the corresponding clues. I claim that the cell marked with a red question mark cannot be black, it must remain white. Why? If it were black, the “5” clue, whichever of the two fives it might be, would stretch from this black either to the left or to the right, or a little bit to the left and a bit to the right. Now, see the clues on the top of the image. The clues highlighted in red are all greater than 1 and they are all last clues for the column they apply to. This means that the last patch of blacks in the corresponding columns consist of at least two black cells. In other words, if these patches of blacsk started in the bottom row, they would extend at least to the penultimate row of the puzzle. Get it? Now, if “?” were black, there would be a patch of 5 black cells which would all extend upwards for 2 or more cells, because of the clues on the top. This means that in the penultimate row there would be a patch of 5 cells, too, which must not happen, because the largest clue in the penultimate row is 2! Therefore, the “?” must be white! Get it?

But that’s not all! See if you can figure out how many other cells in the bottom row also can’t be black. I will reveal the answer at the bottom of this post.

Before that, let me tell you that I’ve published a new book with nonograms puzzles. There are 600 of them in this book! It’s the largest picross book out there. But more about that in a couple of days.

Ok, now, are you ready to see the answer? None of the cells marked with a red X cannot be black, they must be white!

The rule for the “bordering” hanjie solving technique can be generalized as follows (are you ready?):

If the smallest clue (we’ll call it X) in the bottom row is greater than the largest clue (we’ll call it Y) in the row above it, and if there is a string of at least 2*Y+1 adjacent columns in which the last clue is greater than 1, than the cell in the bottom row which belongs to the column in the middle of the string of Y columns cannot be black, it must be white!

Why this term 2*Y+1? Because the “X” clue could come either from the left or from the right, so you need at least twice as many clues plus 1 to get more than Y black adjacent cells in the penultimate row.

In my example, X=5, Y=2, and just by chance 2*Y+1=5 (but it doesn’t have to be the same as X). What’s important is that X>Y and that there are at least 2*Y+1 adjacent columns with bottom clues bigger than 1. In my example, there are actually 15 such adjacent columns, but be careful, you cannot put a certain “white” in all of them, only in the middle 15-2*Y. 🙂 Also, this partial puzzle shows that you can apply the same rule twice to the bottom row. There is another string, this time with precisely 5 adjacent columns and now you can mark only one cell as a certain white (5-2*Y). That’s the last red X in the image above.

Due to symmetry, the same principle applies to top rows and of course to first and last columns.

Study this example and think about it. It should all make sense. Let me know what you think!

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