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This is a "Boulders in Valleys" puzzle (previously called "Balls on Hills"). It's similar to Slitherlink and Corral; your goal is to draw a path along grid cell edges that partitions the grid into two regions. This can be a closed loop, or a path that starts and ends at the grid boundary. The clues are based on the intuition of a boulder rolling down a hill into a valley. There is a unique solution, and you should be able to use logic to find it.

The Puzzle

Penpa Link

Rules

Path Rules

  1. The solution is a single, non-self-intersecting path that partitions the grid into two zones (whether by looping or starting and ending at the grid boundary)

  2. The path travels along grid edges (like Slitherlink) within the grid boundary

  3. There is only one path that fits the constraints

Clue Rules

  1. There is a boulder at every number. The value of the number indicates the deepest depth the boulder could roll down to. It must roll down to this depth along at least one path, but it does not have to roll down to this depth along every path.

  2. Arrows indicate direction of gravity for the boulder.

  3. Grid boundary stops the boulder as if it were part of the curve.

Mechanics of Rolling Boulders

  1. A boulder can roll “down” any corner, and fall straight “down” if there is no edge below it (“down” defined with respect to gravity).

  2. A boulder cannot roll “up” (against gravity) a wall or along two “horizontal” (perpendicular to gravity) segments in a row (the hill would be too shallow!).

  3. Boulders do not interfere with each other (i.e. they can pass through each other, overlap each other, etc)

Examples of Boulder Rolling Mechanics

Examples

Green indicates the path that the deepest boulder took. The example in red is an impossible situation, as the deepest path (2) is deeper than the number on the boulder (1).

Extra

This is the third Boulders in Valleys puzzle. It was previously called Balls on Hills; thanks to Beastly Gerbil for the name suggestion. It was also very tempting to call this "Valleyball", but "Boulders in Valleys" sounded more relaxed.

In the future I'll make more of these, but I'll only post the ones that have an interesting gimmick (such as interesting clue placement/solution shape or a combination with another puzzle).

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2 Answers 2

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The solution:

solution


Method:

I have used red colour to indicate newly added borders, and shaded cells when necessary, in the order blue -> red -> yellow.

As usual, we can start with 0s, and the adjacent cells those are not consecutive numbers.

because cells with 0s should have a border right below them according to the gravity acting upon. For the second trick, consider two adjacent cells containing 2s (as an example): Unless there is a border between them, the ball in the cell below (according to the gravity) has to go two units, forcing the other ball to move three units, which is a contradiction.

Apart from that,

the number 1 in the top row cannot join the adjacent cells, therefore gravity pulls it directly down.
1

Then,

the 3s in blue (see the diagram below) must be separated from each other. And the 2 cannot go along the arrow because then it meets the 5 which is not possible. Thus we can add some lines.
2

Again,

the 3s marked below cannot follow the direction of arrows, because those make the 4 and 1 impossible.
3

Similarly,

the shaded 1s (be careful, not ones) cannot go along the arrows, since they are not allowed to meet 2 and 3.
4

The shaded

2s must be separated from a unique line. Then the 4 in the top allows to add a few lines.
5

The

blue shaded 2 and red shaded 1 have paths along the arrows. And the yellow shaded 1 also helps to add some lines. 6

Since

3 in blue cannot join the 2 in the bottom row, it must fall directly from the first bent. Then considering the yellow and green shaded cells, we can close the bottom half of the loop.
7

Then

we can reach this state.
8

Now,

the 3 in blue has a unique path. Then we can add the line between the must be separated 2s as below.
9

Then we can resume the loop at some points because they have become obvious, and

the 1 allows a few other lines of the loop. 10

The 5 in the first row

combined with the 1 in blue gives a unique path.
11

Then

the 5 shaded below cannot go along the direct arrow since it cannot meet the 1, so it must take a turn in between. Therefore considering the other neighbouring cells we can close that part.
13

And the final stroke

follows the 7.
14


Very fun to solve, but not easy at all :)

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    $\begingroup$ Awesome job! Yeah, this one is difficult. You could get a less complex deduction in some places (will show how when I get a chance) but I really liked the way you reasoned about the trajectories - that was quite clever and taught me some tricks. Glad you enjoyed it :) $\endgroup$
    – BaileyA
    Commented Jan 7, 2023 at 15:44
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This is meant as an addendum to ACB's answer. It's not a full answer, but rather I wanted to point out some tricks one could use in the "opening moves" of the puzzle that I couldn't easily add in a comment.

The key idea behind the opening moves is to look for recurring patterns in the clues. We can look at the possible local solutions to these patterns and see if they all have edges in common. If so, then we can copy-and-paste these to other areas on the board. When creating this puzzle I wanted it to be more Slitherlink-y, hence the emphasis on patterns and also, as you can see in ACB's answer, a more Slitherlink-shaped solution.

Observation 1:

The first pattern is based on adjacent zeros and ones in orthogonal directions: Elbow Pattern We already know how to handle the zero, but note that this means we can't have a line on the dashed edge or the one will fall too far! Thus whenever we see this pattern we can fill it in. This is what I've been calling the "elbow pattern". ACB did end up following this deduction path for one instance of the pattern, but it's so useful for this puzzle type that I wanted to explicitly point it out. Out of all patterns I'm aware of, this gives the most information.

Observation 2:

The second pattern is based on orthogonal ones pointing at the same empty square. Corner Pattern Thick dashed lines represent an edge that we're conditioning on. We can see that there are only two valid solutions (green background). The bottom right corner of this image shows what they have in common. The thin dashed diagonal line represents the fact that one side of the diagonal must have two edges complete. It can be useful to notice that for later in the puzzle. I've been calling this the "corner pattern".

Now that I've written it out like this, the corner pattern does seem quite complicated - sorry about that, this puzzle would probably have been a bit nicer if this pattern was more straightforward. The reason why I hadn't realized I was conditioning two layers deep is because from experience in these puzzles I know that zig-zag patterns of this form are a big no-no. From this orientation, once the zig-zag starts both ball needs to continue falling, forcing the other ball to continue falling, and so on - unless terminating at a grid boundary. If you're aware of that then you can instantly rule it out.

Next:

There are a few other patterns I put in the puzzle, but ACB noticed them so I won't detail them here ;). There is a bit of a red herring in the case where two adjacent ones directly point to each other. If you go through the work, no edge/cross is shared by all possible solutions due to a pesky mirror symmetry between two of the solutions. You could, of course, use this knowledge later on in the puzzle to simplify things - and I did make sure that you could solve it slightly quicker by taking advantage of it, as a reward for anyone who fell down the red herring trap.

Using these ideas, you can fill in a decent amount of the board. You can then use ACB's logic to solve the puzzle, just with some of the same deductions being easier to make, due to there being more lines set in stone, eliminating possible things to consider.

There's even a sneaky third example of the corner pattern - the rightmost 3 in the third row from the top must fall directly left two squares, as reasoned by ACB. Thus, the cell directly left by two squares is a like a "phantom 1" falling left.

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    $\begingroup$ Wow! Thanks for these nice tricks. Btw, may I know how you added numbers with arrows in the grid like that, in penpa? $\endgroup$
    – ACB
    Commented Jan 7, 2023 at 16:46
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    $\begingroup$ No problem! In the "numbers" mode in penpa, you can select the 'sub' "Arrow" (above where you can choose the number color). It's hard to actually figure out how to add them, but the trick is to swipe on the number in the direction you want the arrow to go. $\endgroup$
    – BaileyA
    Commented Jan 7, 2023 at 16:49

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