Case: Higher-Rise

Explanation

Via Meet in the Middle, we obtain 2 dense sequences in each of the half-lanes:

Every skyscraper must be visible. For the $1$-skyscraper, that means it can only go in one of the edge cells.

If one of them becomes taken, then the $1$ is forced to go in the other.

If that other edge cell was taken by a $2$, then we’re now left with 2 unsolved cells containing the candidates $[34]$.

If it was instead taken by a $3$, then we can solve the entire lane! That half-lane must have the sequence $(345)$, and on the other side we must have $(125)$.

Notice we can’t say anything much about the converse – if one terminal cell is taken up by a $1$, the following cell could still be any of $\{234\}$, and on the other side we just get $\text{| [23] [34]}$.

It’s a very nice symmetry, and only exists in odd-size puzzles. The next time you encounter it is 7x7, but there it gets unwieldy; with a 5x5, it’s at the perfect size to have some nice structure and tight deductions.

Challenges

Solve as much as you can (including pencilmarks) with the available context!

Puzzle 1

Solution

Explanation

First, we can pinpoint the lane peak by Meet in the Middle.

Then, in the left column we need a $3$ or $4$ between the $2$ and $5$.

Now we can apply Higher-Rise! Since the $1$ isn’t in the left edge cell, it must be in the right edge cell.

Also, notice the left edge cell can’t actually contain a $4$, since then we wouldn’t be able to see 3 skyscrapers. That leaves $3$ as the only option.

So we must then have $\text{| 3 4 5}$ in the left half-lane, which then makes the other half-lane $\text{| 1 2 5}$.

Puzzle 2

Solution

Explanation

As usual, we can pinpoint the lane peak by Meet in the Middle.

From here, it might not be clear where to proceed.

The key lies in considering the $4$-skyscraper. There are only 2 cells it can go in:

However, notice the $2$-clue on the right, and consider that upper cell carefully.

By Ascendant we know that in a $2$-clue half-lane, the $4$ can go anywhere except the second cell. That means $4$ can’t go in that cell, which pinpoints it to the lower cell!

Now let’s start pencilmarking. Normally, the top edge cell could be $[123]$.

But the $2$ has already been used in the top row, so we can eliminate it as a candidate. We also can’t have $3$ – we’d need $(345)$, but the $4$ has been used on the other side of the lane peak.

Hence we’re left with just $1$, solving the cell.

This is as far as we can get, and we don’t have enough information to solve any more cells. To finish, we’ll pencilmark in the remaining candidates to set us up for future deductions.