Haisu: MathGrant

Question

HAISU is a portmanteau of three Japanese words - 'hairu', to enter, 'su', number, and 'hausu', an English borrow word meaning house, of course.
Together, we get a meaning of 'enter number house', which I have roughly translated to English as 'Room Count'.

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. An example Haisu puzzle and its unique solution are shown below.

Hopefully that example puzzle clarifies the rules.

I made the puzzle below based on a fellow puzzler's logo. The shading doesn't mean anything, it's just part of the original design.
It turned out to be quite a difficult and interesting puzzle, so I've decided to post it here anyway! Enjoy :)

EDIT: The puzzle has now been changed fairly drastically past the opening steps. Apologies to those who spent extended periods of time on this question. Stackreader's and Trenin's progress still stand! It's just the later steps of the puzzle which have changed. Once again, super sorry, and this version has DEFINITELY been checked many times to make sure that it DEFINITELY works!!!

Answer 1

Starting over since the puzzle changed.

The main things from before are the same - the corners are all known, and the 4 pentomino still has the same logic. Here is what we have.

There are two choices for the order of the remainining cells of the 4 pentomino. We can either hit the top one first and then loop around the origin to hit the other, then come all the way back around to the 3. We need to ensure there is enough room to get to the X.

Alternatively, we can go to the bottom one first and then make one path around to the top one, then into the 3 and around to the X.

Lets assume the second scenario. Thus:

Lets look at the S pentomino with both a 1 and a 2 in the top right hand corner. Clearly, we need to hit the 1 cell immediately when we enter, leave, re-enter and then hit the 2 cell. If we do so on the red path, we will need to enter the pentomino to the left of the 1, go through the 1 and then exit somewhere. Re-enter at the 2 and the follow the corner around. The problem with this is that we will then enter the top pentomino with a 3 in the 3 cell for the first time. Perhaps if our path took us through this pentomino a couple of times first?

The following is one attempt.

The problem with this is obviously, there is no path to the finish from the 4, neither the 5 nor 2 are satisfied. Attempts to fix this on the left side are impossible because we need to keep the path that connects to the 4 pentomino around the perimeter.

Similar failures await if we try to reconcile this with the blue path.

Thus, we are forced to try the other scenario.

Clearly, the red, green, and orange paths are fairly decided on the right hand side in certain spots. Also, because we are entering the top of the 4 pentomino first, the path through that cell is also decided. We also can see that we can extend other paths as a result.

The orange path cannot touch the 1, since this is not the first time in that pentomino. By going far enough to avoid it other lines are extended as well.

In order to enter-reenter the two pentominos on the bottom, we need to move the orange line up and make room.

Now we can see that there aren't very many entrances into the large block with the 5 yet, so the orange line must go through it. Also, we must enter the top one twice before we return through it on the green line. Lastly, the bllue line must traverse the 1 first since there are no other ways to connect it.

It is a simple matter to finish it off in the only way possible.

Answer 2

Here is step 1, I will add more as I go.
Odds are I will run out of time and can't finish it though.

There is only 1 way to do the 4 at the bottom left.
We must enter the room 4 times before stepping on the middle square.
We need 1 side to enter and 1 side to exit. So we must find a way to enter the room 3 times by only blocking 2 sides of the "4" square. So we must enter at least one time in the room, without blocking 1 side, which is only doable by the little loop to the bottom right.

Answer 3

Pah - Just read that there's an error in the question. I've been struggling to crack it for a while. Here's my best effort so far:

I've not added the steps I followed as it's not complete yet. Hopefully it's close to the intended answer and will only take a little modification when the update is posted.