Case: Middle Ground

Examples

Case 1

The $N-2$ skyscraper is the $3$-skyscraper, so the 2nd cell can contain $[124]$ but not $3$.

Case 2

The $N-2$ skyscraper is the $4$-skyscraper. In the left column, the 2nd cell can contain $[1235]$ but not $4$. In the right column, the case doesn’t apply since the lane peak ($6$) isn’t in the tail cell.

Case 3

The $N-2$ skyscraper is the $5$-skyscraper, so the 2nd cell can contain $[12346]$ but not $5$.

Then, we can note that $1$ has already been used in the column, and by Ascendant we know the $6$ cannot go in the 2nd cell of the $2$-clue column. This lets us eliminate a few more candidates.

Explanation

The most significant skyscrapers in a Skyscrapers puzzle are the $1$ and $N$ skyscrapers. After those, we can often make meaningful deductions about the $N-1$ skyscrapers. But the others are just sort of… ‘uninteresting’, and exist only to fill in the blanks.

That being said, in a $3$-clue lane exclusively, there’s a very unique constraint around the $N-2$ skyscraper!¹

Consider the following lane:

The $N-1$ skyscraper here is the $4$-skyscraper. Think about which cells we could place it in.

It could very well go in the head cell, leaving the $5$ to go somewhere in the gap:

And you might think the $4$ could go in any of the other cells too. Suppose we put it in this one here:

Now, by the rules of Skyscrapers we know the $5$ can’t go between the $4$ and $6$:

If this were the case, then we’d see 4 or more skyscrapers, because there’s guaranteed to be at least 1 unobscured skyscraper before the $4$.

In this situation, we know the $5$ must come before the $4$:

This way, it obscures the $4$, giving us 2 guaranteed peaks, with the last one coming someplace before the $5$. Keep this constraint in mind!

Now… notice we couldn’t place the $5$ in the head cell, because this would allow only 2 visible skyscrapers, not 3:

This is pretty obvious, but here’s the trick – combine it with our previous constraint.

We said that (1) $5$ must come before $4$, and (2) $5$ can’t go in the head cell. What happens, then, if $4$ is in the 2nd cell?

We need to place the $5$ before it, but we can’t, because the cell before it is the head cell. There’s nowhere valid to place the $5$!

We’ve reached a contradiction, so we can conclude $4$ cannot go in the 2nd cell.

Pretty crazy, huh? Unexpected that the $4$ is randomly gone.

So $1$, $2$, $3$ can all go in any cell in the lane, but specifically $4$, the $N-2$ skyscraper, cannot go in the 2nd cell.

This deduction works for any NxN puzzle – but only in a $3$-clue lane, because that’s what facilitates this interaction of constraints.

Challenges

Puzzle 1

Solution

Explanation

The $1$-skyscraper in front of the $2$-clue means that the next skyscraper must be the lane peak, since this is the only way to have 2 visible skyscrapers.

Now we have the lane peak in the tail cell of the $3$-clue lane, so the case applies. The $N-2$ skyscraper is $4$, so we exclude that candidate.

Puzzle 2

Solution

Explanation

Applying the case, we have all candidates except $4$.

However, since the two $[12]$ cells form a couple, between them they consume $\{12\}$ so we eliminate those as candidates.

Then, by Ascendant we know the $5$ must be past-peak in the $2$-clue row, which leaves $3$ as the solution.

Now, to have 3 visible skyscrapers, we know the sequence must be $(456)$.

Puzzle 3

Solution

Explanation

In the lowermost row, the $3$-clue creates a dense sequence.

Notice this means the $5$ cannot go in the tail cell of the $3$-clue lane, so the case effectively applies.

The $2$ and $3$ have already been used in the column and row, respectively, so we can eliminate those candidates.