Case: Slide

Examples

Case 1

This is a 5x5 puzzle, so the lane peak is the $5$-skyscraper. It’s in the 3rd cell, which is the same number as the clue, so the case applies.

The lowest-case sequence we could have is $(125)$, and the highest-case is $(345)$. Hence the candidates for the head cell are $\{123\}$ and for the second cell $\{234\}$.

Case 2

Here the lane peak is 4 cells away from the $4$-clue, so the case applies.

The lowest sequence we could have is $(1235)$, and the highest $(2345)$. So for the head cell, we have $\{12\}$; then $\{23\}$ for the next, and so on.

Case 3

Now we have a 6x6 skyscraper so the lane peak is the $6$-skyscraper. Notice we can pinpoint a lane peak by Meet in the Middle since $3 + 4 = 6 + 1$.

This is 3 cells away from the $3$ clue, and 4 cells away from the 4 cells away from the $4$ clue, so the case applies on both sides.

On the $3$-clue side the lowest sequence is $(126)$ and the highest $(456)$, giving $\{1234\}$ and $\{2345\}$ for the candidates.

On the $4$-clue side we have $(1236)$ up to $(3456)$, giving $\{123\}$, $\{234\}$ and $\{345\}$.

Explanation

The Structure

When the lane peak is in the $c$-th cell of a half-lane, then the greatest possible number of visible skyscrapers is $c$. For instance, if it were in the tail cell of a 6x6 puzzle, i.e. the 5th cell…

…then we could have a maximum of 5 skyscrapers visible, by having a sequence of skyscrapers with none obscured:

Of course, this isn’t the only such sequence. We could very well have had $(12356)$, $(23456)$, etc. to achieve 5 visible skyscrapers.

If the clue were, say, $3$, then we’d need to choose which cells contain visible skyscrapers, and there are many different ways of doing that.¹

But if the clue were $5$, then the only way to have 5 skyscrapers visible is if every cell is visible.

The lane peak is guaranteed to be visible, so we need 4 more visible to make up the required 5. There are only 4 cells between the clue and lane peak – so they must all be visible!

When the lane peak is further away, we have a choice between the gaps. But when it’s in the $c$-th cell (as close as it can be), it leaves no wiggle room, so we have only 1 choice.

The Numbers

The structure that ensures all cells are visible is a dense sequence – a chain of skyscrapers where each is taller than the last. For instance, $\text{| 1 2 3 4 5}$ is a sequence, since $1 < 2 < 3 < 4 < 5$. All of these skyscrapers would be visible in the half-lane.

However, while we do have only 1 possible structure (all visible), there’s more than 1 sequence that could satisfy that structure. We can’t say for certain which one it is, so we can only pencilmark.

For instance, if we need 2 visible skyscrapers…

…then we could have $(15)$, $(25)$, $(35)$, or $(45)$. That’s a lot of options!

Let’s compare this to a sequence of 4 visible skyscrapers:

Well, one sequence we could choose is $(1235)$. But there’s nothing guaranteeing that – we could also have $(1245)$. Or $(1345)$. Or $(2345)$.

Focusing on the head cell, in all of the possible sequences it’s either $1$ or $2$. This makes sense because if it were $3$, the longest sequence we could have is $(345)$, which only has 3 skyscrapers, not 4 as the clue requires.

For the next cell, it’s $2$ and $3$. It needs to definitely be taller than $1$, and it can’t be $4$, since then we would only be able to see a maximum of 3 skyscrapers.

Continuing with the same logic, for the last cell we have $3$ and $4$.

In general, to find the candidates for each cell, we only need to consider the lower and upper bounding sequences. The lower bound is the sequence containing the shortest possible skyscrapers, and the upper bound contains the tallest possible skyscrapers. From there we know our candidates are all the skyscrapers between the lower and upper bound (inclusive).

For the lower bound, we start with the shortest possible skyscraper, $1$, and then keep picking the shortest available skyscraper. That gives $(1236)$.

For the upper bound, we can work backwards. Preceding the $6$, the tallest possible skyscraper is $5$, and then we keep picking the tallest available skyscraper. That gives $6, 5, 4, 3$, so we have the sequence $(3456)$.

Finally, we just need to add in all the intermediate values. For example, in first cell the range is $[1, 3]$, which gives $\{123\}$.

Solving

As we obtain information on the sequence, we can leverage its structure to actually solve some of these cells.

Forcing the Rest

The head cell acts as a ‘lower bound’ on the sequence, dictating where it starts from. If it starts from the tallest candidate, then every cell afterwards must also have their tallest candidates!

Suppose here we manage to find that the head cell is $2$, which is the tallest of $\{12\}$.

Then the only possible sequence collapses to $(2345)$, so we can solve the half-lane!

This doesn’t just apply for the head cell, though. At any point, if a cell is solved to be its tallest candidate, then all following cells must also contain their tallest candidate.

On the flip side, if a cell is solved to be its shortest candidate, then all preceding cells must also contain their shortest candidate!

All make sense? It’s important to not lose sight of it in larger puzzles when cells start to have more candidates. For instance:

Let’s say we find that the middle unsolved cell is $4$.

Purely relying on Sudoku-style eliminations, we’d leave the last unsolved cell as $[35]$.

But we can actually solve it, because there’s the sequence structure! Since every cell must be visible, we know it can’t be a $3$ since that’d be obscured. It must be a $5$.

This is why it’s important to remember these pencilmarks aren’t just plain pencilmarks, they also encode the sequence structure. When I look at them, I don’t actually think about the individual numbers, I just think “This is a sequence” – i.e. each skyscraper needs to be taller than the one before it.

Narrowing the Band