Very strong puzzle. Enjoyed it greatly. Thanks for the creation.
With that said, after I had gotten 60% through the puzzle, I hit a wall and it became too deep a trial and error to truly be called a logic puzzle. More akin to an extreme sudoku where you have to run out too much of the solution to see the failure mode.
Love your blog. Appreciate the time and effort you put in for our benefit.
Of all the puzzles on this blog, this is the one that least deserves the label of a logic puzzle, you’re right. That’s why I posted the note below. There’s a certain pattern in this puzzle that’s repeated several times, and the “intuition” I spoke of is assuming this pattern generalizes (something you can trust me to have verified). The puzzle can be completed without trial and error if you employ that assumption.
If you have no idea what I’m talking about, I can email you the intended solution.
How taboo would it be for us to post our own solving techniques (spoilers)? I got through this one without any trial and error, but I did have to think deeper than usual and recognize some higher-level patterns. Once you ask the right questions, the answers aren’t too hard.
OK then… for me, the puzzle was at least half done before I got to what I perceived to be the hard part. There are 2 clues near the right edge, a 5-with-leftward-pointing-arrow above a 6-with-leftward-pointing-arrow. To the left of those 2 clues, I had a big empty 9×2 area. The left half of the puzzle was basically done, and it contributed 2 black squares to the 5-left clue and 3 black squares to the 6-left clue, so each of those rows needed 3 more black squares. That’s a total of 6 black squares that needed to be in the 9×2 empty area. The key question: What’s the maximum number of black squares you can fit into a 3×2 empty area? The answer is 2. (Try to visualize 3 black squares in a 3×2 area and you will find it causes trouble. It can only work if there’s a number/arrow clue square in there somewhere.) So 6 black squares in a 9×2 empty area is as crowded as it can be. Obviously each of the 9 columns contains either 1 black square or 0 black squares. 6 of the columns must have a black square, and 3 must not, and no 3 consecutive columns can contain a black square. That plus the surrounding context made it possible to proceed. I learned/invented a new rule (logically derived from the base rules) and applied it to a tough situation… exactly what I hope to do in a logic puzzle.
That’s similar to my intent, although not exactly there.
This image shows the pattern I was getting at in this puzzle. The case where there is a 3×2 rectangle with clues giving one square in each of the two rows (or columns) appears four times in this puzzle and is solved the same way every time.
There is one instance where there is a 6×2 rectangle with 2s pointing in, at the bottom a bit to the left. By the time you reach that point in the puzzle you should have already done the 1 case two or three times, so the hope was that you would generalize the pattern you found.
Then at the right part of the puzzle, you get the 9×2 rectangle with 3s (or what are effectively 3s) pointing into it. At this point it’s very difficult to verify that the black cell configuration shown in the image (or some symmetry) is the only possible one, although trial and error can do it. My intent was that you would generalize the previous appearances of this pattern and intuitively realize that’s what it had to be.
Sounds like it wasn’t as obvious as I was hoping. But I’m glad you still enjoyed solving it.
It’s not hard to see that in that 9×2, the black columns must split as 1+2+2+1, though, which forced some white cells which allowed solving the situation gradually for me. This may be what Alan described.
I got the pattern you were intending, but also agree with the rest about that 9×2 region, if you stare at it a few minutes you get that just one arrangement is possible. I liked that the 1 arrow pointing upwards wasn’t pointing at a shaded region in the 9×2 region. Not really sure why I liked it though 😛