HAISU with a twist

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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HAISU is a grid-deduction puzzle designed by this site's very own TheGreatEscaper. To quote them:




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.




For an example puzzle, see the original post: HAISU (Room Count): An original grid-logic challenge



This puzzle is a bit different, however: it is embedded on a Möbius strip. The top and bottom act like normal, but the left and right edges join up as shown by the letters. Good luck!



HAISU with a twist



Notes:



  • The letters aren't their own cells, they just tell you where to join the two sides up.

  • The rooms do not connect between the two sides of the puzzle (e.g. if an edge passes through d it will leave and enter the room again).

  • "No guessing, no handwavy steps, just pure logic required to solve this puzzle!" - TheGreatEscaper

  • Thanks Wen1now for testsolving!





grid-deduction







share|improve this question





















  • Great idea with the Möbius strip :)
    – npkllr
    8 hours ago






  • 2




    I'm having trouble understandig the big numbers. Aren't they basically the same as the small numbers?
    – Ian Fako
    7 hours ago










  • The small numbers refer to the amount of times you have to enter a room (room = thick lines). The large numbers refer to a cell (small square / thin lines). If a cell contains a large number (N), you have to pass over the cell the Nth time you enter the room that contains your cell.
    – npkllr
    6 hours ago














up vote
11
down vote

favorite
3












HAISU is a grid-deduction puzzle designed by this site's very own TheGreatEscaper. To quote them:




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.




For an example puzzle, see the original post: HAISU (Room Count): An original grid-logic challenge



This puzzle is a bit different, however: it is embedded on a Möbius strip. The top and bottom act like normal, but the left and right edges join up as shown by the letters. Good luck!



HAISU with a twist



Notes:



  • The letters aren't their own cells, they just tell you where to join the two sides up.

  • The rooms do not connect between the two sides of the puzzle (e.g. if an edge passes through d it will leave and enter the room again).

  • "No guessing, no handwavy steps, just pure logic required to solve this puzzle!" - TheGreatEscaper

  • Thanks Wen1now for testsolving!





grid-deduction







share|improve this question





















  • Great idea with the Möbius strip :)
    – npkllr
    8 hours ago






  • 2




    I'm having trouble understandig the big numbers. Aren't they basically the same as the small numbers?
    – Ian Fako
    7 hours ago










  • The small numbers refer to the amount of times you have to enter a room (room = thick lines). The large numbers refer to a cell (small square / thin lines). If a cell contains a large number (N), you have to pass over the cell the Nth time you enter the room that contains your cell.
    – npkllr
    6 hours ago












up vote
11
down vote

favorite
3









up vote
11
down vote

favorite
3






3





HAISU is a grid-deduction puzzle designed by this site's very own TheGreatEscaper. To quote them:




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.




For an example puzzle, see the original post: HAISU (Room Count): An original grid-logic challenge



This puzzle is a bit different, however: it is embedded on a Möbius strip. The top and bottom act like normal, but the left and right edges join up as shown by the letters. Good luck!



HAISU with a twist



Notes:



  • The letters aren't their own cells, they just tell you where to join the two sides up.

  • The rooms do not connect between the two sides of the puzzle (e.g. if an edge passes through d it will leave and enter the room again).

  • "No guessing, no handwavy steps, just pure logic required to solve this puzzle!" - TheGreatEscaper

  • Thanks Wen1now for testsolving!





grid-deduction







share|improve this question













HAISU is a grid-deduction puzzle designed by this site's very own TheGreatEscaper. To quote them:




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.




For an example puzzle, see the original post: HAISU (Room Count): An original grid-logic challenge



This puzzle is a bit different, however: it is embedded on a Möbius strip. The top and bottom act like normal, but the left and right edges join up as shown by the letters. Good luck!



HAISU with a twist



Notes:



  • The letters aren't their own cells, they just tell you where to join the two sides up.

  • The rooms do not connect between the two sides of the puzzle (e.g. if an edge passes through d it will leave and enter the room again).

  • "No guessing, no handwavy steps, just pure logic required to solve this puzzle!" - TheGreatEscaper

  • Thanks Wen1now for testsolving!





grid-deduction




grid-deduction






share|improve this question













share|improve this question











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asked 9 hours ago









boboquack

14.2k142110




14.2k142110











  • Great idea with the Möbius strip :)
    – npkllr
    8 hours ago






  • 2




    I'm having trouble understandig the big numbers. Aren't they basically the same as the small numbers?
    – Ian Fako
    7 hours ago










  • The small numbers refer to the amount of times you have to enter a room (room = thick lines). The large numbers refer to a cell (small square / thin lines). If a cell contains a large number (N), you have to pass over the cell the Nth time you enter the room that contains your cell.
    – npkllr
    6 hours ago
















  • Great idea with the Möbius strip :)
    – npkllr
    8 hours ago






  • 2




    I'm having trouble understandig the big numbers. Aren't they basically the same as the small numbers?
    – Ian Fako
    7 hours ago










  • The small numbers refer to the amount of times you have to enter a room (room = thick lines). The large numbers refer to a cell (small square / thin lines). If a cell contains a large number (N), you have to pass over the cell the Nth time you enter the room that contains your cell.
    – npkllr
    6 hours ago















Great idea with the Möbius strip :)
– npkllr
8 hours ago




Great idea with the Möbius strip :)
– npkllr
8 hours ago




2




2




I'm having trouble understandig the big numbers. Aren't they basically the same as the small numbers?
– Ian Fako
7 hours ago




I'm having trouble understandig the big numbers. Aren't they basically the same as the small numbers?
– Ian Fako
7 hours ago












The small numbers refer to the amount of times you have to enter a room (room = thick lines). The large numbers refer to a cell (small square / thin lines). If a cell contains a large number (N), you have to pass over the cell the Nth time you enter the room that contains your cell.
– npkllr
6 hours ago




The small numbers refer to the amount of times you have to enter a room (room = thick lines). The large numbers refer to a cell (small square / thin lines). If a cell contains a large number (N), you have to pass over the cell the Nth time you enter the room that contains your cell.
– npkllr
6 hours ago










1 Answer
1






active

oldest

votes

















up vote
6
down vote













So that should be the solution:




enter image description here




Step by step:




1. Top-Right: Cell with the 2 can't be connected to the O.

2. Below the O: Room has to be entered exactly once -> cells have to be connected to each other.

3. Bottom-Left: Only way to hit the 3 on the third entrance since it is next to the X.
enter image description here



1. Bottom-Left: Cell only has two free neighbour-cells left.

2. Bottom: Cell only has two free neighbour-cells left.

3. Top: Room with the 2 must be entered at least two times + the room below can only be entered once -> all cells have to be connected to each other.
enter image description here



1. Top-Right: Connect the Möbius strip.

2. Above the X: Room has to be entered two times and has two cells -> they can't be connected.

3. Next to it: Room has to be entered three times -> top-right cell can't be connected to any other cell in the room.
enter image description here



Now it gets a bit more complicated since we have to consider the direction of the path.

1. Bottom-Left: The path on the bottom-left is connected with the top-right. At the top-right the direction of the path is from left to right since we have to pass the 2 on our second entry. Therefore if we would connect the bottom-left path with the e or the f we would always end at the O which would lead to a direction-conflict.

2. Top-Right: Since we can't use e and f any more, this is the only possible path to connect the cells.
enter image description here







share|improve this answer










New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Nice job! Did you manage to get there with logic?
    – boboquack
    7 hours ago










  • Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
    – npkllr
    7 hours ago






  • 2




    Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
    – BmyGuest
    7 hours ago










  • Sure, I'm working on it ;)
    – npkllr
    7 hours ago










Your Answer




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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote













So that should be the solution:




enter image description here




Step by step:




1. Top-Right: Cell with the 2 can't be connected to the O.

2. Below the O: Room has to be entered exactly once -> cells have to be connected to each other.

3. Bottom-Left: Only way to hit the 3 on the third entrance since it is next to the X.
enter image description here



1. Bottom-Left: Cell only has two free neighbour-cells left.

2. Bottom: Cell only has two free neighbour-cells left.

3. Top: Room with the 2 must be entered at least two times + the room below can only be entered once -> all cells have to be connected to each other.
enter image description here



1. Top-Right: Connect the Möbius strip.

2. Above the X: Room has to be entered two times and has two cells -> they can't be connected.

3. Next to it: Room has to be entered three times -> top-right cell can't be connected to any other cell in the room.
enter image description here



Now it gets a bit more complicated since we have to consider the direction of the path.

1. Bottom-Left: The path on the bottom-left is connected with the top-right. At the top-right the direction of the path is from left to right since we have to pass the 2 on our second entry. Therefore if we would connect the bottom-left path with the e or the f we would always end at the O which would lead to a direction-conflict.

2. Top-Right: Since we can't use e and f any more, this is the only possible path to connect the cells.
enter image description here







share|improve this answer










New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Nice job! Did you manage to get there with logic?
    – boboquack
    7 hours ago










  • Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
    – npkllr
    7 hours ago






  • 2




    Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
    – BmyGuest
    7 hours ago










  • Sure, I'm working on it ;)
    – npkllr
    7 hours ago














up vote
6
down vote













So that should be the solution:




enter image description here




Step by step:




1. Top-Right: Cell with the 2 can't be connected to the O.

2. Below the O: Room has to be entered exactly once -> cells have to be connected to each other.

3. Bottom-Left: Only way to hit the 3 on the third entrance since it is next to the X.
enter image description here



1. Bottom-Left: Cell only has two free neighbour-cells left.

2. Bottom: Cell only has two free neighbour-cells left.

3. Top: Room with the 2 must be entered at least two times + the room below can only be entered once -> all cells have to be connected to each other.
enter image description here



1. Top-Right: Connect the Möbius strip.

2. Above the X: Room has to be entered two times and has two cells -> they can't be connected.

3. Next to it: Room has to be entered three times -> top-right cell can't be connected to any other cell in the room.
enter image description here



Now it gets a bit more complicated since we have to consider the direction of the path.

1. Bottom-Left: The path on the bottom-left is connected with the top-right. At the top-right the direction of the path is from left to right since we have to pass the 2 on our second entry. Therefore if we would connect the bottom-left path with the e or the f we would always end at the O which would lead to a direction-conflict.

2. Top-Right: Since we can't use e and f any more, this is the only possible path to connect the cells.
enter image description here







share|improve this answer










New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Nice job! Did you manage to get there with logic?
    – boboquack
    7 hours ago










  • Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
    – npkllr
    7 hours ago






  • 2




    Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
    – BmyGuest
    7 hours ago










  • Sure, I'm working on it ;)
    – npkllr
    7 hours ago












up vote
6
down vote










up vote
6
down vote









So that should be the solution:




enter image description here




Step by step:




1. Top-Right: Cell with the 2 can't be connected to the O.

2. Below the O: Room has to be entered exactly once -> cells have to be connected to each other.

3. Bottom-Left: Only way to hit the 3 on the third entrance since it is next to the X.
enter image description here



1. Bottom-Left: Cell only has two free neighbour-cells left.

2. Bottom: Cell only has two free neighbour-cells left.

3. Top: Room with the 2 must be entered at least two times + the room below can only be entered once -> all cells have to be connected to each other.
enter image description here



1. Top-Right: Connect the Möbius strip.

2. Above the X: Room has to be entered two times and has two cells -> they can't be connected.

3. Next to it: Room has to be entered three times -> top-right cell can't be connected to any other cell in the room.
enter image description here



Now it gets a bit more complicated since we have to consider the direction of the path.

1. Bottom-Left: The path on the bottom-left is connected with the top-right. At the top-right the direction of the path is from left to right since we have to pass the 2 on our second entry. Therefore if we would connect the bottom-left path with the e or the f we would always end at the O which would lead to a direction-conflict.

2. Top-Right: Since we can't use e and f any more, this is the only possible path to connect the cells.
enter image description here







share|improve this answer










New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









So that should be the solution:




enter image description here




Step by step:




1. Top-Right: Cell with the 2 can't be connected to the O.

2. Below the O: Room has to be entered exactly once -> cells have to be connected to each other.

3. Bottom-Left: Only way to hit the 3 on the third entrance since it is next to the X.
enter image description here



1. Bottom-Left: Cell only has two free neighbour-cells left.

2. Bottom: Cell only has two free neighbour-cells left.

3. Top: Room with the 2 must be entered at least two times + the room below can only be entered once -> all cells have to be connected to each other.
enter image description here



1. Top-Right: Connect the Möbius strip.

2. Above the X: Room has to be entered two times and has two cells -> they can't be connected.

3. Next to it: Room has to be entered three times -> top-right cell can't be connected to any other cell in the room.
enter image description here



Now it gets a bit more complicated since we have to consider the direction of the path.

1. Bottom-Left: The path on the bottom-left is connected with the top-right. At the top-right the direction of the path is from left to right since we have to pass the 2 on our second entry. Therefore if we would connect the bottom-left path with the e or the f we would always end at the O which would lead to a direction-conflict.

2. Top-Right: Since we can't use e and f any more, this is the only possible path to connect the cells.
enter image description here








share|improve this answer










New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this answer



share|improve this answer








edited 6 hours ago





















New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 8 hours ago









npkllr

3016




3016




New contributor




npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






npkllr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Nice job! Did you manage to get there with logic?
    – boboquack
    7 hours ago










  • Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
    – npkllr
    7 hours ago






  • 2




    Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
    – BmyGuest
    7 hours ago










  • Sure, I'm working on it ;)
    – npkllr
    7 hours ago
















  • Nice job! Did you manage to get there with logic?
    – boboquack
    7 hours ago










  • Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
    – npkllr
    7 hours ago






  • 2




    Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
    – BmyGuest
    7 hours ago










  • Sure, I'm working on it ;)
    – npkllr
    7 hours ago















Nice job! Did you manage to get there with logic?
– boboquack
7 hours ago




Nice job! Did you manage to get there with logic?
– boboquack
7 hours ago












Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
– npkllr
7 hours ago




Yes, of course. At first it was a bit confusing with the Möbius strip but it's a nice additional challange :)
– npkllr
7 hours ago




2




2




Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
– BmyGuest
7 hours ago




Nicely done. If you addtionally could describe (in spoiler tags) the individual logical deduction steps you've used, it would be worth addtional upvotes. :c)
– BmyGuest
7 hours ago












Sure, I'm working on it ;)
– npkllr
7 hours ago




Sure, I'm working on it ;)
– npkllr
7 hours ago

















 

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