A Curious Crossways

24 May 2025

This was a 5x5 which I thought I could speedrun, but I ended up running into a hitch which took some careful consideration to overcome.


	3

3			1
4
3
			2
		2

Original puzzle is from brainbashers.com^↗.

Opening

The start is straightforward, pretty speedrunnable.


	3

3			5	1
4
3
				2
		2

This pinpoints the $5$ skyscraper for the $4$ clue…


	3

3			5	1
4		5
3
				2
		2

Which pinpoints that for the lowermost $3$ clue…


	3

3				5	1
4			5
3		5
					2
			2

And that pinpoints the $5$ for the vertical $3$ clue, putting the final $5$ in the top-left.


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

The only place for $4$ in the left column is in the last row, since $3$ and $4$ clues can’t have a $4$ in front of them.


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

This means the $3$ in the last row must go in front of the leftwards $2$ clue.


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

From the upwards $2$ clue we can deduce it has to be $[2, 1]$ , and this fills the final cell in the last row with the $1$ skyscraper.


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

Marking

Now we start making marks, except… nothing really pops up?


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

With nothing to really go off, I just marked up some more of the grid. The interesting “crossways” we’ve got here is the two $3$ clues we’ve got in the upper-left, both opposite terminal $5$ skyscrapers. So by Middle Ground we know the second cell in each of those lanes can’t be a $3$ .


		3
	5	*123*		34
3	*123*	*124*		34	5	1
4	12	23	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

Crux

At this point I was stuck. The lower $4$ and $3$ clues don’t really give anything, so we know the secret must lie in the interaction of the two $3$ clues.

It turned out to be exactly that! Let’s think about the vertical $3$ clue. If the first cell were $1$ , we could have $[1, 4, 2, 3, 5]$ , which is chill.


		3
	5	1		34
3	123	4		34	5	1
4	12	23	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

But if we put a $2$ there, $[2, 4, -, -, 5]$ wouldn’t leave any place for $1$ . So this forces the $1$ into the second cell…


		3
	5	2		34
3	123	1		34	5	1
4	12	23	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

But then our only possible order is $[2, 1, 3, 4, 5]$ , which would be 4 visible skyscrapers, not 3.

Hence there aren’t any valid configurations for the column if we put the $2$ in the first cell. So we eliminate it from the options.


		3
	5	13		34
3	123	124		34	5	1
4	12	23	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

Now do the same for the $2$ in the second cell. If we start the column with $1$ , we have $[1, 2, -, -, 5]$ , which will exceed 3 visible skyscrapers no matter what. If we start with $3$ , we have $[3, 2, -, -, 5]$ . But notice the third cell can only be $3$ or $2$ , so we can’t use both of them in the first two cells.


		3
	5	3		34
3	123	2		34	5	1
4	12	23	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

Like before, we can conclude the second cell can’t be a $2$ .


		3
	5	13		34
3	123	14		34	5	1
4	12	23	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

Now there’s only one place for the $2$ in that column!


		3
	5	13		34
3	123	14		34	5	1
4	12	2	34	5	124
3	23	34	5	1	24
	4	5	1	2	3	2
				2

After that fiddly deduction it’s plain sailing, and we clear the rest with Sudoku deductions.

Solve


		3
	5	13		34
3	23	14		34	5	1
4	1	2	34	5	14
3	23	34	5	1	24
	4	5	1	2	3	2
				2


		3
	5	13		34
3	23	14		34	5	1
4	1	2	34	5	4
3	23	34	5	1	2
	4	5	1	2	3	2
				2


		3
	5	13		34	1
3	23	14		34	5	1
4	1	2	3	5	4
3	3	34	5	1	2
	4	5	1	2	3	2
				2


		3
	5	13		34	1
3	2	14		34	5	1
4	1	2	3	5	4
3	3	4	5	1	2
	4	5	1	2	3	2
				2


		3
	5	3		34	1
3	2	1		34	5	1
4	1	2	3	5	4
3	3	4	5	1	2
	4	5	1	2	3	2
				2


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


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