Solution: The Power of Sudoku

An unusually high proportition of steps in solving this Skyscrapers are Sudoku-style deductions.


			3		5
						3
				2		4

		3
3	2
						3
		2

Opening

We start by skylining as usual, though we only manage to find one lane peak as usual, though we only manage to find one [lane peak).


			3		5
						3
				2		4

		3
3	2				6
						3
		2

With some firing range we can fill in some of the $5$ -clue half-lane too.


			3		5
						3
				2		4

		3			5
3	2				6
						3
		2

The rest becomes a sequence.


			3		5
					12	3
				2	23	4
					34
		3			5
3	2				6
						3
		2

And we have our first Sudoku-style deductions!


			3		5
					12	3
				2	3	4
					34
		3			5
3	2				6
						3
		2


			3		5
					12	3
				2	3	4
					4
		3			5
3	2				6
						3
		2

For completeness we’ll also fill out the tail cell’s candidates.


			3		5
					12	3
				2	3	4
					4
		3			5
3	2				6
					12	3
		2

Midgame

Next I want to turn our attention to this $3$ -clue lane. It has a very nice structure, with the lane peak in the tail cell and a low skyscraper in the head cell.


			3		5
					12	3
				2	3	4
					4
		3			5
3	2				6
					12	3
		2

With the lane peak in the tail cell, we only have two possibilities to consider:

$\text{3 | 2 5 \_ \_ \_ \_ 6}$
$\text{3 | 2 1 5 \_ \_ \_ 6}$

There can only be 1 more peak after $2$ , so that peak must be the $5$ -skyscraper.

Let’s consider the former case.


			3		5
					12	3
				2	3	4
					4
		3			5
3	2	5			6
					12	3
		2

This is invalid by Dangerzone, since now the $2$ -clue half-lane will have at least 3 skyscrapers visible.

Hence it must be the latter case!


				3		5
						12	3
					2	3	4
						4
		3				5
3	2	1	5			6
						12	3
		2

Rarely a bad idea to keep up with pencilmarking cells with only 2 candidates ;)


				3		5
						12	3
					2	3	4
						4
		3				5
3	2	1	5	34	34	6
		45				12	3
		2

Next up, let’s focus on this $4$ -clue lane.


				3		5
						12	3
					2	3	4
						4
		3				5
3	2	1	5	34	34	6
		45				12	3
		2

Since $3$ is in the head cell, the peaks must go $(3456)$ .

Let’s consider the best case. We could place the $4$ , but a $5$ after would raise issues.


				3		5
						12	3
			5	4	2	3	4
						4
		3				5
3	2	1	5	34	34	6
		45				12	3
		2

By Sudoku rules, this $5$ conflicts with the $5$ lower down in the column.

That means we can only push the $5$ back to the next cell, which forces the $6$ to be in the tail cell too.


				3		5
						12	3
	6	5			2	3	4
						4
		3				5
3	2	1	5	34	34	6
		45				12	3
		2

The $4$ was only a hypothetical deduction, so we don’t keep it.

Pencilmarking:


				3		5
						12	3
	6	5	14	14	2	3	4
						4
		3				5
3	2	1	5	34	34	6
		45				12	3
		2

What’d’y’know, more Sudoku-style elimination!


				3		5
						12	3
	6	5	14	14	2	3	4
						4
		3				5
3	2	1	5	34	34	6
		4				12	3
		2

Notice this also forces the $6$ to come before the $5$ in that column. Neat!


				3		5
						12	3
	6	5	14	14	2	3	4
		6				4
		3				5
3	2	1	5	34	34	6
		4				12	3
		2

And that leaves the $2$ in the tail cell too.


				3		5
		2				12	3
	6	5	14	14	2	3	4
		6				4
		3				5
3	2	1	5	34	34	6
		4				12	3
		2

…and that eliminates the $2$ from the upper-right cell, which also solves the lower-right cell. So many chained deductions!


				3		5
		2				1	3
	6	5	14	14	2	3	4
		6				4
		3				5
3	2	1	5	34	34	6
		4				2	3
		2

We can now pinpoint the lane peak in the uppermost row.


				3		5
		2	6			1	3
	6	5	14	14	2	3	4
		6				4
		3				5
3	2	1	5	34	34	6
		4				2	3
		2

This then pinpoints the lane peak in the lowermost row.


				3		5
		2	6			1	3
	6	5	14	14	2	3	4
		6				4
		3				5
3	2	1	5	34	34	6
		4		6		2	3
		2

And that pinpoints the final lane peak.


				3		5
		2	6			1	3
	6	5	14	14	2	3	4
		6				4
		3			6	5
3	2	1	5	34	34	6
		4		6		2	3
		2

Endgame

In the $3$ -clue half-lane, we need a high-low structure.


				3		5
		2	6	34	45	1	3
	6	5	14	14	2	3	4
		6				4
		3			6	5
3	2	1	5	34	34	6
		4		6		2	3
		2

These $[34]$ unsolved cells now form a couple, eliminating the $4$ from the third wheel.


				3		5
		2	6	34	45	1	3
	6	5	14	1	2	3	4
		6				4
		3			6	5
3	2	1	5	34	34	6
		4		6		2	3
		2


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
		6				4
		3			6	5
3	2	1	5	34	34	6
		4		6		2	3
		2

We can now pinpoint the $5$ in this column, which also pinpoints the $2$ .


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
		6		5		4
		3			6	5
3	2	1	5	34	34	6
		4		6		2	3
		2


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
		6		5		4
		3		2	6	5
3	2	1	5	34	34	6
		4		6		2	3
		2

And in the adjacent column, we can’t have $1$ in the $3$ -clue half-lane:


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
		6		5		4
		3		2	6	5
3	2	1	5	34	34	6
		4		6	1	2	3
		2

If the $1$ goes here, the lowermost row does not have 3 skyscrapers visible.

So it can only go here:


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
		6		5	1	4
		3		2	6	5
3	2	1	5	34	34	6
		4		6		2	3
		2

Pencilmarking:


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
	23	6	23	5	1	4
	14	3	14	2	6	5
3	2	1	5	34	34	6
		4		6	35	2	3
		2

Which leads to a whole slew of eliminations ;)


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
	3	6	23	5	1	4
	14	3	1	2	6	5
3	2	1	5	34	34	6
		4		6	35	2	3
		2


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	34	34	6
		4		6	35	2	3
		2


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	34	34	6
		4	3	6	35	2	3
		2

We’ve reached critical mass now, and the rest of puzzle collapses via Sudoku-style deductions.


				3		5
		2	6	34	45	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	34	34	6
		4	3	6	5	2	3
		2


				3		5
		2	6	34	4	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	34	34	6
		4	3	6	5	2	3
		2


				3		5
		2	6	3	4	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	34	34	6
		4	3	6	5	2	3
		2


				3		5
	5	2	6	3	4	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	4	34	6
		4	3	6	5	2	3
		2


				3		5
	5	2	6	3	4	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	14	3	1	2	6	5
3	2	1	5	4	3	6
	1	4	3	6	5	2	3
		2


				3		5
	5	2	6	3	4	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	4	3	1	2	6	5
3	2	1	5	4	3	6
	1	4	3	6	5	2	3
		2


				3		5
	5	2	6	3	4	1	3
	6	5	4	1	2	3	4
	3	6	2	5	1	4
	4	3	1	2	6	5
3	2	1	5	4	3	6
	1	4	3	6	5	2	3
		2

$\blacksquare$