## Puzzle 189 (Indirect Yajilin)

This is a Thursday Indirect Yajilin puzzle.

Puzzle 190

There is an important rule about 0s on the edge that I recently clarified on the description page. Basically, the “invisible arrow” on a vanilla clue (as all of the 0s here are) can’t point to an adjacent edge.

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### 5 Responses to “Puzzle 189 (Indirect Yajilin)”

1. Jack Bross Says:

This was a very nice puzzle. If we were going on the Nikoli scale, I’d say this is on the easy end of “Hard” in difficulty. Not sure if that’s where you aim for Thursday, but it’s a substantial ramp up in difficulty from yesterday (which struck me as the easy end of medium). Again, the difficulty also depends on familiarity with Yajilin in general. Once you get going, the logical flow of the puzzle is steady and satisfying.

It’s very clear once you start solving the puzzle, but I’m not sure if it’s worth noting in the directions that the blocks and loop are uniquely determined, but that the direction arrows aren’t. In other words, there are quite a few clues with more than one “zero” direction, so we can’t be sure where the arrow is actually pointing.

2. MellowMelon Says:

Thanks for the responses about the difficulty. The arrow directions not being uniquely determined is on the description page, when it says “one (or more) of the four directions has that number of black squares”.

The easy end of “Hard” would optimally be Friday, but generally I don’t make seven puzzles that fit each day of the week perfectly and allow plenty of leeway. This week it just happened I made a bunch of puzzles and didn’t know what to do with any of them. I thought the Nurikabe was placed too early if anything, for instance.

3. Mark Fox Says:

Each new twist on a familiar puzzle adds a certain degree of difficulty the first time I tackle it. I don’t think this puzzle is intrinsically all that hard, it just took some extra pondering at the beginning.

4. Heidi Quist Says:

I’m very stuck on this one. The bottom right corner doesn’t seem to follow the rules. There has to be a black in the crux of the three zeroes–line can’t go in and end. Then, for each zero in that trio, the other (horizontal or vertical) must have zero blacks, right? This will put a right angle in the bottom left corner of that trio. Then, one row up from the bottom edge you have two zeroes. These must have a line underneath because neither can be a black. Now you have two lines pointing in to that bottom right corner. Now with the zero one in from the bottom and far right edge, you have to have a line going along that outside edge because none of those can be blacks, to prevent dead ends. But now you have three lines pointing into that corner, and only two of them can connect, forcing a dead end. What am I missing? If I could post a picture to show you, I would, but not sure how to do that.

• MellowMelon Says:

I think I was able to follow what you said. In that bottom right sector you had three loose ends going into, your logic obtaining the two loose ends along the bottom row is sound. That and your contradiction tells you that R8C15 (three left and two up from bottom right corner square) has to be a black square.

I am not quite sure what about the nearby 0s made you believe that was impossible. Remember: only one of the four directions needs to have no black squares for a 0 clue to be satisfied.