Case: High-Rise

$N$ pinned between $2$ clues

When a lane has $2$ -clues on both sides, one head cell must contain the $N-1$ skyscraper.


2						2


2	*N-1*				*N-1*	2

Examples

Case 1




2						2

The 2 possibilities are:




2	4	2




2	4	2

Case 2


	2

2		2

As in case 1, we note the $4$ must go in one of the head cells of the row:


	2

2	4	4	2

However, notice in the left column, the $4$ cannot go in the 2nd cell, since then there would be no way to have 2 skyscrapers visible in that column.


	2

2	4	4	2

We can’t place the lane peak before the $4$ , since then we see only $(5)$ , but if we put it after, we see $(x45)$ .

That eliminates the $4$ from that cell, leaving it with just the other head cell to go in.


	2

2		4	2

Explanation

From Ascendant we know in a half-lane with a $2$ -clue, the $N-1$ skyscraper must either be past-peak or in the head cell.

When we have two $2$ -clues on either side, though, we get an interesting crossfire of these constraints.


2			5			2

Let’s start with the left half-lane. The $N-1$ skyscraper is $4$ . It could go in the head cell, in which case happy days; but suppose it doesn’t. What then?


2	3		5			2

Let’s take the head cell to not be $4$ , but a $3$ , for example.

Well, it must be past-peak.


2	3		5	4	4	2

But wait, we’ve got the restrictions from the $2$ -clue on the other side too!

From the perspective of the other half-lane, the $4$ is pre-peak, which means it must go in the head cell.


2	3		5		4	2

The same logic applies for the other side, since it’s all symmetric. And that’s it!

Delightfully simple, but very powerful.