Oriental HAISU - An unoriginal grid-deduction challenge

Question

TheGreatEscaper comes up with several inventive (and original) grid-deduction genres, so I thought I'd try my hand at combining two of them.

1. HAISU

   The rules are simple - draw a path from the O to the X, passing through every cell in the grid exactly once. The grid is divided into several rooms. When your path passes over a cell with the big number N, it must be the Nth time you have entered the room. If a room has a small number m in the top left corner, you must enter that room a total of m times.
   See HAISU (Room Count): An original grid-logic challenge for an example puzzle.

2. Oriental House

   The rules are as follows: Draw a path from S to F O to X, passing through every square exactly once.
   The areas defined by bolded lines are 'rooms'. You may pass through rooms more than once. When the path passes over an arrow, the path segment inside that room over that arrow must either have entered or exited in that direction.
   See Oriental House: An original grid-deduction challenge for examples of valid and invalid configurations.

Enjoy! The solution should (hopefully) be unique, so in the words of TheGreatEscaper, "No guessing, no handwavy steps, just pure logic required to solve this puzzle!"

ERRATUM: an 8 was added in the S box. This shouldn't adversely affect the solution path until the end.

Answer 1

Answer

Full Step-by-step solution

First, the easy steps:

We start with the 3 box here:

We enter this room 3 times, so each entry we stay exactly one square's worth. Therefore we can conclude this:

And also due to the arrow beneath, that entry must enter from there:

But look at the right arrow next to it. Clearly it cannot enter from that area, or else we would have a conflict in that square (as the path cannot split up.) Moreover, that region does not have a right border that can result in a right exit, so it must be a right entrance from a few squares away:

So we can complete the path as such:

COMMERCIAL BREAK: A BRIEF EXCURSION IN TOPOLOGY
Let's ignore the grid entirely for now. Let's suppose, for a moment, that you have two path parts that both touch the edge of the grid, one which goes clockwise and the other which goes counterclockwise.

WLOG, the clockwise part comes before the counterclockwise part. Thus they must connect eventually as follows:

Where could we start? Where could we end? Somewhere in these regions, specifically:

Importantly, however, notice that these two regions are separated by part of the path.

Unfortunately that can't happen here. Therefore we conclude that all edge path pieces must orient the same direction.
END INTERMISSION

Thanks to our intermission we can conclude the orientations of the free bits.
But wait, there's more! You need to connect these two, or else that side would be counterclockwise.

So, more free stuff.

You can't go left due to this, as that would result in that region having at most two entries.

So it goes right.

COMMERCIAL BREAK: MORE EXCURSIONS IN TOPOLOGY
Last time on A Brief Excursion in Topology... We showed that edge pieces must orient the same way!

Today, we'll look at in what order you can touch these pieces.
Really, it is quite simple. You need to go in order.
What happens if we don't go in order?

You get a plural loop. Our paths are highly trained. Please do not try this at home.
Therefore, today we concluded that: you must travel these edge pieces in order!
END INTERMISSION
First, some more things. The down arrow on R4C1 can't be an entry arrow, it must be an exit arrow. So we get this:

Likewise, the down arrow on R2C9 is also not an entry arrow, or else it would collide with the other edge piece we've already drawn. So it must exit somehow:

This gets us R3C10... wait a moment:
COMMERCIAL BREAK: TOPOLOGY STRIKES BACK
Our fascinating question of the day: is this legal? (supposing that the arrows on the edge travel clockwise)

Well, this results in:

If the top edge comes first, we need this:

If the bottom edge comes first, we need this:

Either way, we've once again isolated the start and endpoints by a path in between. Not allowed.
END INTERMISSION
Wait, so hold up. Are you saying that that arrow on R2C9 is actually an entry arrow? Yes. But not there. It would have to be here at the nearest:

Thanks, Topology Man!
And in fact, it must enter there, for if it entered any later, R3C8 would end up being inaccessible:

So yeah, we do the rest of the free deductions:

COMMERCIAL BREAK: TOPOLOGY 4: ELECTRIC BOOGALOO
Actually, let's just cut to the chase and say the fundamental fact.
Suppose we actually have a path from O to X.

Well, we can change that path to a closed loop:

This loop has winding number 1! In particular, though, one corollary is that you can assign an interior and an exterior to this loop, and that the orientation of the loop will necessarily be consistent with respect to this. Furthermore, this means that for any two parts of the loop that are next to each other with no intervening parts, the loop's orientation will be traveling in opposite directions.
END INTERMISSION
So ok. We can orient this as a result:

Consider the left arrow in R5C9. It must go left a few squares, at least, in order to exit left. (We know the orientation of the line due to our topology exercise.)

That gives us some more information in R4C7:

And conveniently tells us where to exit:

Likewise, the arrow in R7C9 must exit there, since exiting in R6C10 would be traveling counterclockwise on that edge.
Also, much like the earlier deduction involving R2C9, the top left region resolves similarly, as if R4C1 resolved at R2C3 that would close off R2C2. So it must resolve at R1C2.

R1C4 cannot go down, as then it would have to exit up, and the only way to do that is to go up at R3C3, creating a plural loop:

So it goes straight instead. That region needs to be entered three times, so it exits immediately.

Similar to earlier, R2C5 resolves going left...

And then it must exit up, and the only way to do so is at R3C3. So it does.

This lets us conclude a few things, first around the left center:

And also R3C5 and the top border.

R4C6 has nowhere else to go...

Some more easy deductions...

Consider the end at R1C8. It has two outlets it can go to. (since R2C7 makes an easy loop). What if it went to R3C8?

Let's consider the 8 condition. Can this still be satisfied?

The answer is no. So R1C8 connects to R1C9 and thus this resolves as such:

Oh yeah, and if the trick works twice, it's bound to work again. R10C6's right arrow resolves at R9C10, not R8C9, or else R9C9 is isolated.
Now, R6C9 has three potential exits - but only one potential entrance, from the left.

We consider R5C9/R5C10/R6C9/R6C10/R7C9/R8C9 as a whole. If R8C9 didn't connect to R7C9, then R7C9 must connect to R6C9, and we are left with three ends to pair up - not good!

So instead this must happen:

Now consider R8C8. There are three possibilities for it; two of them are easily ruled out. If the path goes up/left, then the arrow is violated. If the path goes left/down, then there is a quick plural loop due to R7C9.

So it goes up/down and we get some more progress:

We now use the 3 entry condition in R8C6.

More progress...

More progress...

More progress...

R7C3 can't immediately go right, since that would violate R6C9 (that would be the 2nd segment traversed).
COMMERCIAL BREAK: TOPOLOGY: THE FINALE
Consider this completely unrelated scenario, nope this doesn't look familiar at all.

Congrats, you managed to trap the exit. Now you can't get out. I hope you're proud of yourself.
Really, this is just a corollary of the "complete the loop thing" + the "adjacent edges should run against each other" thing.
END INTERMISSION
So I heard that you aren't supposed to go up from R7C4.

Let's not have a plural loop:

And again,

R6C4 is forced, so we've got one endpoint down. Yay!

It's a lot of forcing, really.

More forcing...

Finally, the ambiguity can be resolved by using the final Haisu 3 clue.

And we're done!