Solution: Hyperthetical

When one insane hypothetical deduction solves the entire puzzle.


	2	3	1
						3

							5

4			3

		4		3	3

Opening

The opening is fairly long and painful. We’ll start by skylining as always.


	2	3	1
			6			3

							5

4			3

		4		3	3

By Meet in the Middle we can find another lane peak.


	2	3	1
			6			3

		6					5

4			3

		4		3	3

By firing range we can find another.


	2	3	1
			6			3

		6					5

4			3			6

		4		3	3

This also pinpoints one more.


	2	3	1
			6			3

		6					5

4			3			6
	6
		4		3	3

Unfortunately, the last 2 lane peaks will come …much later.

By Blockade we can also solve the $2$ -clue half-lane on the left.


	2	3	1
	5		6			3

		6					5

4			3			6
	6
		4		3	3

Finally, we’ll fill in the candidates for the $5$ -clue half-lane.


	2	3	1
	5		6			3

		6	45	34	23	12	5

4			3			6
	6
		4		3	3

Oh wait, notice in that lane there’s only one place for the $5$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4			3			6
	6
		4		3	3

Chasing on from there, this also puts the $5$ in the lower-right for the lower row. Neat!


	2	3	1
	5		6			3

		6	5	34	23	12	5

4			3			6
	6					5
		4		3	3

This wraps up the opening, and now we look for the non-obvious.

The Long Road Ahead

The most fruitful lane here will be the $4$ -clue row. First, we know the head cell can only contain $[123]$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	*123*		3			6
	6					5
		4		3	3

However, $3$ is already taken in the lane, so we’re left with $[12]$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	12		3			6
	6					5
		4		3	3

Now consider the second cell. Continuing the sequence, the candidates are $[1234]$ . Again, $3$ is taken, so we’re left with $[124]$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	12	*124*	3			6
	6					5
		4		3	3

We’re now going to make our first hypothetical deductions, to try eliminate some of these candidates. Let’s take $4$ first.


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	12	4	3			6
	6					5
		4		3	3

What can we deduce if we place a $4$ in the second cell?

Could this work? Yeah, we could just put $(x456)$ in each of the $4$ -clue half-lanes (with $x < 4$ ).


	2	3	1
	5		6			3

		6	5	34	23	12	5
		5
4	*low*	4	3	5		6
	6	*low*
		4		3	3

So $4$ is possible. Let’s consider $2$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	12	2	3			6
	6					5
		4		3	3

This would force the head cell to be $1$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	1	2	3			6
	6					5
		4		3	3

What can we deduce if we place a $2$ in the second cell?

Ah. But now we have a problem. This $4$ -clue row is going to see more than 4 skyscrapers, since either the $4$ or $5$ -skyscraper (or both) will be a peak between the $3$ and $6$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	1	2	3	45		6
	6					5
		4		3	3

Hence we can deduce $2$ is invalid. Finally, let’s consider $1$ .


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	12	1	3			6
	6					5
		4		3	3

What can we deduce if we place a $1$ in the second cell?

Now an issue arises with the $4$ -clue column. This is a dense sequence, so having a $1$ here is definitely not allowed. No matter what we do, we can’t have 4 skyscrapers visible.


	2	3	1
	5		6			3

		6	5	34	23	12	5
		3
4	12	1	3			6
	6	2				5
		4		3	3

$1$ cannot be a peak, so we can only attain a maximum of 3 peaks, not 4 as required.

So it can’t be $1$ either.

We checked $4, 2, 1$ and found that $2$ and $1$ would lead to unsolvable situations. Hence we can eliminate $1$ and $2$ as candidates, and conclude the second cell must contain a $4$ . Nice!


	2	3	1
	5		6			3

		6	5	34	23	12	5

4	12	4	3			6
	6					5
		4		3	3

In order to fulfil the sequence, we also now know this cell must be a $5$ , since it’s the only skyscraper between $4$ and $6$ .


	2	3	1
	5		6			3

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

Now for some pencilmarking. In the $3$ -clue half-lane we’ll need a sequence.


	2	3	1
	5	12	6			3
		23
		6	5	34	23	12	5
		5
4	12	4	3	*125*	*125*	6
	6	*123*				5
		4		3	3

At this point, none of the half-lanes we’ve looked at so far can really offer us anything. There’s nothing further to help us narrow down what they could be.

We can try more pencilmarking, but it doesn’t really lead anywhere here.

So, after much deliberation, the only sensible choice is to look at the two $3$ -clue upwards columns.

Absolute Madness, Part 1

Well well, what’s going on here? Whatever it is, there is a lot of structure here, because of the constraints already placed on some of the cells.


	2	3	1
	5	12	6			3
		23
		6	5	34	23	12	5
		5
4	12	4	3	125	125	6
	6	123				5
		4		3	3

The first thing to notice is that we only have 2 lane peaks left unsolved.


	2	3	1
	5	12	6			3
		23
		6	5	34	23	12	5
		5
4	12	4	3	125	125	6
	6	123				5
		4		3	3

So far, we’ve found 4/6 of the lane peaks.

This means the final 2 form a diagonal matrix.


	2	3	1
	5	12	6			3
		23		6	6
		6	5	34	23	12	5
		5		6	6
4	12	4	3	125	125	6
	6	123				5
		4		3	3

There’s more. The same also applies to the $5$ -skyscrapers.


	2	3	1
	5	12	6			3
		23		6	6
		6	5	34	23	12	5
		5		6	6
4	12	4	3	125	125	6
	6	123				5
		4		3	3

This means the remaining $5$ -skyscrapers also form a diagonal matrix.


	2	3	1
	5	12	6			3
		23		56	56
		6	5	34	23	12	5
		5		6	6
4	12	4	3	*125*	*125*	6
	6	123				5
		4		3	3

Crucially, the matrices of $6$ and $5$ overlap. That is super interesting. There’s definitely something to be said here.

Time to pull out the hypothetical deductions again. WLOG, suppose we have this permutation:


	2	3	1
	5	12	6			3
		23			6
		6	5	34	23	12	5
		5		6
4	12	4	3	125	125	6
	6	123				5
		4		3	3

Although the two lanes differ in the candidates $[34]$ and $[23]$ , these are effectively irrelevant in our current considerations, so we can safely treat the lanes as indistinguishable.

In the right lane, this pinpoints the $5$ .


	2	3	1
	5	12	6			3
		23			6
		6	5	34	23	12	5
		5		6
4	12	4	3	125	5	6
	6	123				5
		4		3	3

This, in turn, pinpoints the $5$ in the left lane.


	2	3	1
	5	12	6			3
		23		5	6
		6	5	34	23	12	5
		5		6
4	12	4	3	12	5	6
	6	123				5
		4		3	3

Take note of the structure here. We have something that looks like this:



	5	6

	6
		5
	3	3

And importantly, even if we had the opposite permutation of $6$ -skyscrapers, it would be the same configuration reflected:



	6	5

		6
	5
	3	3

This allows us to make some powerful deductions.

Firstly, notice in both configurations those upper two cells use $5$ and $6$ between them.



	5	6

	6
		5
	3	3



	6	5

		6
	5
	3	3

In other words, these upper 2 cells must form a couple of $[56]$ .


	2	3	1
	5	12	6			3
		23		56	56
		6	5	34	23	12	5
		5		6	6
4	12	4	3	125	125	6
	6	123				5
		4		3	3

Secondly, going back to our hypothetical scenario, focus on the $3$ -clue half-lane with the closer lane peak:


	2	3	1
	5	12	6			3
		23		5	6
		6	5	34	23	12	5
		5		6
4	12	4	3	12	5	6
	6	123				5
		4		3	3

The second cell here can’t be $1$ , because then we couldn’t have $3$ skyscrapers visible.


	2	3	1
	5	12	6			3
		23		5	6
		6	5	34	23	12	5
		5		6
4	12	4	3	1	5	6
	6	123				5
		4		3	3

In other words, the only possible way to fulfil the $3$ -clue is for the half-lane to be $\text{3 | 1 2 6}$ .

Again, since this is WLOG we can say the same for the other permutation. Overall, we deduce that the second cell can’t contain a $1$ -skyscraper, in both lanes.


	2	3	1
	5	12	6			3
		23		56	56
		6	5	34	23	12	5
		5		6	6
4	12	4	3	1	1	6
	6	123				5
		4		3	3

Well, now that means there’s only one place left for the $1$ to go in this row. Hence we can solve the head cell of that $4$ -clue half-lane!


	2	3	1
	5	12	6			3
		23		56	56
		6	5	34	23	12	5
		5		6	6
4	1	4	3	25	25	6
	6	123				5
		4		3	3

That’s pretty wild, isn’t it?

This $1$ is critical. It’s going to open a whole lot more deductions for us!

The Road in Sight

Let’s start by pencilmarking some more. Nothing too crazy here, just making explicit which skyscrapers haven’t been taken in each lane.


	2	3	1
	5	12	6	*124*	*124*	3
	*234*	23	*124*	56	56	*124*
	*234*	6	5	34	23	12	5
	*234*	5	*124*	6	6	*124*
4	1	4	3	25	25	6
	6	123	*124*			5
		4		3	3

While doing this, the consideration going through my head is “Could any of these have candidates eliminated to collapse to a solution?”

Looks messy I know, but do not fear the pencilmarks. Once we start eliminating they’ll make chasing effortless.

Now that we’ve done this, it’s quite apparent we can pinpoint the $1$ in the 3rd row.


	2	3	1
	5	12	6	124	124	3
	234	23	124	56	56	124
	234	6	5	34	23	1	5
	234	5	124	6	6	124
4	1	4	3	25	25	6
	6	123	124			5
		4		3	3

This eliminates the $1$ from the other unsolved cells in the rightmost lane.


	2	3	1
	5	12	6	124	124	3
	234	23	124	56	56	24
	234	6	5	34	23	1	5
	234	5	124	6	6	24
4	1	4	3	25	25	6
	6	123	124			5
		4		3	3

Now this, in turn, pinpoints the $1$ in the 2nd row.


	2	3	1
	5	12	6	124	124	3
	234	23	1	56	56	24
	234	6	5	34	23	1	5
	234	5	124	6	6	24
4	1	4	3	25	25	6
	6	123	124			5
		4		3	3

Which now eliminates $1$ from its column.


	2	3	1
	5	12	6	124	124	3
	234	23	1	56	56	24
	234	6	5	34	23	1	5
	234	5	24	6	6	24
4	1	4	3	25	25	6
	6	123	24			5
		4		3	3

This is great, but even better, all this chasing has led to us discovering a couple! – notice the two $[24]$ cells in the 4th row.


	2	3	1
	5	12	6	124	124	3
	234	23	1	56	56	24
	234	6	5	34	23	1	5
	234	5	24	6	6	24
4	1	4	3	25	25	6
	6	123	24			5
		4		3	3

These eliminate the $2$ and $4$ from the leftmost cell, making it a $3$ .


	2	3	1
	5	12	6	124	124	3
	234	23	1	56	56	24
	234	6	5	34	23	1	5
	3	5	24	6	6	24
4	1	4	3	25	25	6
	6	123	24			5
		4		3	3

Eliminating…


	2	3	1
	5	12	6	124	124	3
	24	23	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	6	6	24
4	1	4	3	25	25	6
	6	123	24			5
		4		3	3

What’d’y’know, another couple!


	2	3	1
	5	12	6	124	124	3
	24	23	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	6	6	24
4	1	4	3	25	25	6
	6	123	24			5
		4		3	3

These now eliminate the $2$ from the second cell in the row, making it a $3$ .


	2	3	1
	5	12	6	124	124	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	6	6	24
4	1	4	3	25	25	6
	6	123	24			5
		4		3	3

Eliminating…


	2	3	1
	5	12	6	124	124	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	6	6	24
4	1	4	3	25	25	6
	6	12	24			5
		4		3	3

The great thing about having so many couples, is that they act as a catalyst for even more couples to form. Now we can fill out the rest of those sparse $[6]$ cells’ candidates.


	2	3	1
	5	12	6	124	124	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	16	16	24
4	1	4	3	25	25	6
	6	12	24			5
		4		3	3

Whew, wasn’t that satisfying! That hypothetical deduction really paid off, didn’t it?

Absolute Madness, Part 2

Well guess what, we weren’t done with that hypothetical deduction. There’s one more deduction we could draw from it that we haven’t addressed yet.

We could’ve included this earlier, but that would be no fun. Then the chasing would be really mindless. Also, this is how I solved the puzzle myself (I only noticed later on), so it’s probably a more natural way of navigating it ;P

So again, WLOG take one of the possible permutations for the two lanes.


	2	3	1
	5	12	6	124	124	3
	24	3	1	5	6	24
	24	6	5	34	23	1	5
	3	5	24	6	1	24
4	1	4	3	2	5	6
	6	12	24			5
		4		3	3

This time, we can also solve the $1$ and $2$ cells that formed couples with the $6$ and $5$ , respectively. We won’t chase further since it won’t be necessary (or helpful, really).

I actually dropped this earlier in a note. For the left half-lane, the only possible solution is $\text{3 | 1 2 6}$ , since $1$ is the only skyscraper shorter than $2$ .


	2	3	1
	5	12	6	124	124	3
	24	3	1	5	6	24
	24	6	5	34	23	1	5
	3	5	24	6	1	24
4	1	4	3	2	5	6
	6	12	24	1		5
		4		3	3

Again, applying this symetrically to both lanes, we find that one of them must contain a $1$ .


	2	3	1
	5	12	6	124	124	3
	24	3	1	5	6	24
	24	6	5	34	23	1	5
	3	5	24	6	1	24
4	1	4	3	2	5	6
	6	12	24	1	1	5
		4		3	3

Now notice that means the $1$ in this row cannot go over on the left:


	2	3	1
	5	12	6	124	124	3
	24	3	1	5	6	24
	24	6	5	34	23	1	5
	3	5	24	6	1	24
4	1	4	3	2	5	6
	6	1	24			5
		4		3	3

If the $1$ went here, it wouldn’t be available for one of the two $3$ -clue half-lanes.

Hence we can eliminate $1$ as a candidate there!


	2	3	1
	5	12	6	124	124	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	16	16	24
4	1	4	3	25	25	6
	6	2	24			5
		4		3	3

And now we’re on the home straight. From here, the puzzle just collapses.

Endgame


	2	3	1
	5	1	6	124	124	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	24	16	16	24
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	24	24	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	2	16	16	24
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	24	24	3
	24	3	1	56	56	24
	24	6	5	34	23	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	24	24	3
	24	3	1	56	56	2
	24	6	5	34	23	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	24	24	3
	4	3	1	56	56	2
	24	6	5	34	23	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	24	24	3
	4	3	1	56	56	2
	2	6	5	34	23	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3

Almost there…


	2	3	1
	5	1	6	24	24	3
	4	3	1	56	56	2
	2	6	5	34	3	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	24	24	3
	4	3	1	56	56	2
	2	6	5	4	3	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	2	24	3
	4	3	1	56	56	2
	2	6	5	4	3	1	5
	3	5	2	16	16	4
4	1	4	3	25	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	2	4	3
	4	3	1	56	56	2
	2	6	5	4	3	1	5
	3	5	2	16	16	4
4	1	4	3	5	25	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	2	4	3
	4	3	1	6	56	2
	2	6	5	4	3	1	5
	3	5	2	16	16	4
4	1	4	3	5	2	6
	6	2	4			5
		4		3	3


	2	3	1
	5	1	6	2	4	3
	4	3	1	6	5	2
	2	6	5	4	3	1	5
	3	5	2	1	16	4
4	1	4	3	5	2	6
	6	2	4			5
		4		3	3


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


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

A very satisfying solve :]


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