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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. Your actual challenge is this!

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1 Answer 1

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Here is my answer

Only good way to enter 5 times I could find. Also must hit the 1 on first entrance.

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Logical prolongation of lines that shouldn't touch each others.

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The path coming from the entrance will be forced to take this path eventually.

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Only way to hit the 3 on the 3rd entrance and respect the current flow of the lines.

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Further deductions based on straight forward paths.

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Some of my logical deductions were not so logical after all. Here is the result after some minor revisions.

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  • $\begingroup$ Nice! There is a purely logical solve path to the finish (which, interestingly, does things almost in reverse order to what you've done!), but you do present the correct answer :) $\endgroup$ Commented Feb 25, 2017 at 14:20
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    $\begingroup$ @TheGreatEscaper, could you explain how this answer works? It seems to me that this path enters the small-2 room 3 times. $\endgroup$
    – Passage
    Commented Feb 25, 2017 at 18:12
  • $\begingroup$ @Passage oops, I hadn't noticed that. The intended answer doesn't do that. The general topology of stack readers solution matched mine so I didn't look closely enough at the details. This is not the correct answer. $\endgroup$ Commented Feb 25, 2017 at 22:49
  • $\begingroup$ @TheGreatEscaper My bad, just fixed it... I think. $\endgroup$ Commented Feb 26, 2017 at 0:10
  • $\begingroup$ @passage Stackreader has just corrected his answer to the correct one. This solution no longer has a logical pathway leading to the answer, but I'll leave it green ticked anyway. $\endgroup$ Commented Feb 26, 2017 at 1:17

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