# HAISU (Room Count): An original grid-logic challenge

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 this example puzzle clarifies the rules. Your actual challenge is this!

No guessing, no handwavy steps, just pure logic required to solve this puzzle!

Edit: A long awaited fix.

• Well this is hard... I'm stuck at this point: imgur.com/a/RDaNy Jan 16, 2017 at 2:37
• @greenturtle3141 about 99% the same as my current image lol. My only difference is that I have a line going up from the 2 in the bottom right corner. Jan 16, 2017 at 2:39
• Not sure I quite follow every deduction made on that link... for instance, picture 2 to 3, are you really sure those two blue lines must be different entries of the room? Jan 16, 2017 at 3:08
• ...It appears that I messed up. Jan 16, 2017 at 3:44

Here is what I came up with by adding lines 1 by 1 where I could prove it was 100%. Please forgive the atrocious paint display after all my redo and deletions.

bottom right : only way to hit the 2 on the second entrance while keeping an exit open to go up after.

bottom right : only way to hit every tiles without entering more than 3 times.

right : only way to hit the 1 on first entrance and the 2 on second entrance. Cannot loop any higher without blocking the final loop for the exit(takes quite a bit of thinking to see it; could have been left for later when it is more obvious)

Only way to hit the 1 on first entrance without dooming the tile to its right to eternal solitude.

Must touch the 1 on first time in and must only enter 2 times.

Follow the only possible path to allow the correct amount of entrances.

More logical deductions based on required number of entrances.

• Like, a step by step explanation, because clearly someone tried this process and failed miserably and now that person is very sad and is interested in the exact line of deduction used. Jan 16, 2017 at 4:54
• Can someone tell me why am I getting so many downvotes? it's my 4th one on this question. A little frustrating considering how many hours and hard work I put in this answer... Jan 16, 2017 at 4:57
• This is the correct solution, great job! Unfortunately I can't green tick until there is a step-ish by step-ish explanation of the solution, as per usual for grid-deduction puzzles. Your process sounds very good (my solution, too, begins in the bottom right and ends at the top left), but until it's explained sufficiently such that readers can follow your solve path, I can't accept this as a 'full' answer. :( Jan 16, 2017 at 5:22
• In your very first step, how did you eliminate this possibility, marked in pink? Jan 16, 2017 at 8:57
• @thecommexokid You have to go on the 2 on your second time, which mean you would have to do your loop to the right of the 2 first then go back to get the 2. But the loop to the right of the 2 put a "roof" over the 2 that can never be crossed again which makes it impossible to reach the exit.... Sorry if this is confusing... is it clear enough? Jan 16, 2017 at 9:25

EDIT: The puzzle has now been fixed to be unique. So here is my solution to the fixed puzzle. For posterity I'll keep my old post below.

Here is the full solution

Here is the logic I used to get there

First I want to start with explaining some common techniques. The number one thing we have to work with is the number of entrances/exits (from here on out I'll call them doors). This means in a room with a small 4 there must 8 doors. In general there are twice as many doors as the small number (unless the start or end is in the room but that doesn't happen in this puzzle). Furthermore, we know that alcoves (cells with three outward facing edges) must have a least one door.

We also know that every cell is in the path. This means that sometimes a cell will be "forced" if there are only two possible lines. An example is the corners in the beginning.

Finally, since the start and end are both along the border there is another technique we can use. We have to be careful not to trap the path away from the exit. For instance, in the given puzzle if we go from the start to the middle to the right edge and then down, our path is stuck in the lower right section of the puzzle making it impossible to reach the end.

Ok let's get started. We start with marking the corners. Note that we already know the direction that any path along the edge goes since going the other way would trap the line (see the third technique). So, whenever we can deduce the direction of a line I will mark it with an arrow. Looking at the room with the small one, we know one door is in the alcove. This means the second door is in the top right or bottom middle. At the bottom left, the 2x2 room with a small 2 must have 4 doors so we can mark those in. Now bottom right, the big 2 can't have 2 doors since we would have to path across the other cells first trapping the path. This means the top left cell of the room must have two doors. Now the red line isn't possible since we have to go through the top left path first.

Now focus on the bottom right room with a small 3. There are already 3 doors and at least 2 more have to come from the 2 alcoves. This leaves one free door. Let's try to put this along the right edge. The rest of the red lines are forced. This is where we run into a contradiction. We need the big 1 at the bottom to be first but after following that path to the end there is no way to enter that room again.

Now we know that the big path in the bottom right must go left. This forces the start of the small path to go left. From here we see that the last door must be in the bottom alcove. With all the doors found we can solve the rest of the room. Going back to the lower left, we have a forced corner. It can't continue right and only have 4 doors in the room, so it must go up. This makes another forced corner in the next room.

Let's make another guess at the bottom far right that the two paths connect without entering the room with a small two. This leaves 4 possible doors that we can mark in. There also are several forced cells at the bottom right. However, from here we see that there is no valid way to enter every cell in the room.

With the contradiction above, we know that the two paths in the bottom right must extend to the left. Also in order to reach the 1 first they must connect before the big 1. We can fill in the rest of the green lines but we are left with two solutions. Let's leave this for now and look towards the middle right. We are force to have the green lines in this area. If we continue to the left over the big two we run into a problem. If we follow the yellow lines to the red corner (which needs to be first before the big 2) we trap our path.

Instead the path must go up with the big 2 having 2 doors. This forces the rest of the green lines. The red line between the green lines is not possible since we have the same problem of being trapped to get the big 1 first.

Therefore, we have to loop the path around to get the big 1. This then forces the green lines in the top right. Now focusing on the bottom, we are forced to have this solution to the big room with a small two so that we don't enter the small 4 room as that would make the big 1 not be reached first. This leaves the rest of the borders being doors to reach 8 total. We can then finish off this room with forced cells.

After marking the stray green line, we try the right red line. However, this makes it impossible to enter all cells and reach the big 3 third.

Instead the arrow on the small two must go left and up to the big 3. To make this the third entry we need a door to the left and a door in the top alcove. Next looking at the big skinny section in the middle with a small 4, there are already 7 doors. The last door must be on the far left which gives the green lines in that room. This forces the green lines above as well.

Now focus on the top room with the small 4. We can force two corners and the backwards 'C'. From the small 4 cell we must go down and then left so we can reach the big 3 third. At the 2x2 room with a small 3 there are 7 slots for 6 doors. The lower left can't have two doors since it would form a closed loop. This forces the green lines in this room.

From the lower big 3 we are forced up. Following that same path we must go right. Continuing further we go left to the upper big 3. From the path connecting to the start we loop around to the room with the upper big 3 to make it the third entry. Then the top is just another forced cell.

Finally, we can solve the rest easily making sure that there are 6 doors to the top left room.

EDIT: Old post starts here

Unfortunately, I don't believe this puzzle is unique and therefore not solvable with just pure logic.

I could only solve the problem to here:

At this step @stackreader connected the big 3 towards the bottom to the square to the right. However this does not have to be true (as seen below).

From here I found an additional 5 solutions.

• Hiya! The unique version of this puzzle has a small '3' instead of a small '2' in the 2x2 square at the left. I'll edit it now that this post has come to the top of the PSE feed anyway (I didn't want to bump this puzzle after I found the fix.) Sep 14, 2018 at 11:48