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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.

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Hopefully this example puzzle clarifies the rules.

My previous Haisus have been over-complicated and/or difficult, so here is a puzzle which is MUCH more moderate in difficulty!

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Here is the unique solution:

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and here's one way to get there, all steps being forced so that the outcome is known to be unique. Let's begin with the 3-visit region near the top left. There's only one way to arrange that.

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And now the regions to its left are straightforward.

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Let's add the obvious things in the corners, which we should really have done first on autopilot.

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Now we can fill in the centre-left region:

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The path can't cross the north and west boundaries of the central region for fear of giving too many visits to the L-shape above it.

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Thinking about that region at top centre, we have to do this for fear of too many visits to adjacent regions:

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Count entrances and exits for the 2-visit L-shape near the middle:

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If we join the upper two bits of path in that region near the bottom left, they'll end up being part of a loop that doesn't visit every square. So don't do that:

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We can't enter the single-visit region near top right from the left, because that would make the top-right region impossible, so:

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There's very little flexibility at top right and we fairly quickly find we have to do this:

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Counting entrances and exits of the bottom area, we must have this:

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leading quickly here:

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There's only one way to handle the cell to the right of the start, and then we can count entrances and exits of the 3-visit L-shaped region:

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Now there's only one way to fill in that narrow corridor:

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and all that remains is the earliest portion of the path:

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  • $\begingroup$ You little ninja... oh well. $\endgroup$ – stack reader Feb 25 '17 at 13:28
  • $\begingroup$ @stackreader You've still got a chance with the Rose puzzle! $\endgroup$ – TheGreatEscaper Feb 25 '17 at 13:37
  • $\begingroup$ Sorry! If I'd known someone else was working on it I'd have been happy to leave it alone. $\endgroup$ – Gareth McCaughan Feb 25 '17 at 16:15
  • $\begingroup$ Don't feel bad, you won fare and square. $\endgroup$ – stack reader Feb 25 '17 at 16:43
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Here is my solution

Only way to get 2 entrances in a 2X2 square.

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The 2 lines must connect so that only 1 entrance is done in the room bellow.

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Once again, Only way to get 2 entrances in a 2X2 square.

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Logical deductions to only allow 2 entrances in the big room.

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The 2 red dots must connect somewhere or else the room to the right will be impossible to solve in just 1 entrance. From there, we can do further logical deductions.

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Only possible way to hit the 2 on the second entrance without forcing more then 1 entry on the room to the left.

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Only way to enter 3 times in the lower left room.

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The rest just comes together on its own.

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