How to Solve Miti Puzzles

Miti, created by Nishiyama Yukari, requires that you create a single closed loop in the grid, exactly one cell wide, that passes through all spaces in the grid. When complete, it will resemble a maze with no dead ends. To accomplish this, you will be drawing line segments on the lines of the grid.

At intersections with a dot, exactly 3 lines must meet. At every other intersection in the grid, there can be no more than two lines that meet.

At the beginning, the simplest step is to begin with dots on the border, because each of them already has two lines.

As a result, you can simply draw a line segment from that dot pointing into the rest of the grid.

Next, we’re going to look at this dot, because it is only one cell away from the border.

Notice that the purple line leading away from it would touch the border. Because there is no dot in that position, we cannot use that line, which means the remaining three lines in blue must be the correct ones.

In addition, if you were to leave that purple line in place, you would find that no matter which one of the other three sections you removed, you would be forming a dead end in the upper right corner.

One useful thing to remember is that the eventual path must be only one cell wide.

This means that cells next to the border will usually have a wall on their side that is opposite to the border. I’ve highlighted some spaces which cannot have a line at a right angle to the border, because there are no dots on the border wall itself. Note that I’m not connecting them with our known lines, except on the right side of the grid. This is because the path may snake around these existing lines.

These highlighted cells must have a wall on the side that faces opposite the border in order to meet the one-cell-wide path rule.

Now we can move on to our other dots, to see how three lines can intersect without breaking any rules.

Starting with this dot, which of the four possible lines can be removed?

If we take away the blue line, we would be creating two possible paths toward the upper right corner of the puzzle. This is not allowed, and we also know we couldn’t block off one of them with a line on the bottom of the highlighted cell, because that would create a dead end. Therefore, the blue line MUST remain.

This lets us quickly see that the green line is the correct one to remove, because otherwise, it forms a dead end with the blue line.

So we know we keep blue, red, and orange.

This final dot is a little more tricky.

What we can see immediately is that the line to remove must be either the red or blue one, because if you keep both, you will be creating a dead end.

If you remove the blue, you will need to extend the line from the wall to maintain the one-cell-wide path. If you remove the red one, it looks like you might be creating two possible paths, which isn’t allowed.

This might be taken care of, depending on what we do with the large open area to the left of center. For now, let’s just mark in the green and orange lines, and see what we can do with the open area.

At the highlighted yellow cell, we can see it needs an extra wall, because otherwise, the path is too wide through here. Since it can’t have one perpendicular to the wall without a dot, that means the new line must be on the side opposite the wall.

Meanwhile, this blue cell has too many possible adjacent spaces leading into it, so it also needs another wall. We can’t use the blue or red lines, because they would form dead ends, so the orange line must be the correct one to add.

Now, we’re going to examine a couple of possible paths through the bottom of the grid as it currently stands, in order to maintain the one-cell-wide path rule.

If we draw walls to create the green path, it would create a dead end at the highlighted cell, as well as a 3-way intersection just below the currently highlighted dot. Because that grid crossing doesn’t have a dot, it may not have more than two lines meeting there. Even if the green line turned upward instead, the dead end at the highlighted cell would be created.

This leaves us only with the red path being valid. so we will draw two lines to account for that turn at the bottom.

Once again, we have a cell with too many paths – when we’re finished, every cell will have an entry side and an exit side, with the other two sides walled off.

With this cell, we can see that if we place the second wall on the bottom or the right, it will create a dead end.

This means that our third wall at the green dot will be the top side of this space.