Yajilin · Solving guide
Draw one closed loop through the empty cells and shade the cells it skips, guided by the numbered arrows. Every board we make is solvable by pure logic, never a guess. These are the techniques that get you there, in the order the puzzles ask for them: the everyday moves first, the loop-wide reasoning last.
The everyday moves. Every puzzle uses these, and easy boards need nothing else.
Technique 1
No two shaded cells may share a side. So every cell next to a shaded one has to be part of the loop, and the loop can never enter a shaded cell.
The cell in the bottom-left corner is shaded, so its neighbour to the right can't be shaded too and joins the loop. The border between them is ruled out. (The next two techniques keep solving this same board.)
Technique 2
A loop cell has the line passing straight through it, using exactly two of its four sides. So when only two sides are still open, the loop has to take both of them.
The cell we just added to the loop can't go left (the shaded cell) or down (the grid edge), so its only two open sides are up and right, and the line turns through both.
Technique 3
An arrow counts the shaded cells in the direction it points. Once that many are placed (or the arrow is a 0) every other cell along its line must be on the loop.
The down arrow at the top of the left column needs one shaded cell below it, and the shaded cell at the bottom already provides it. So the two empty cells in between must be on the loop.
Technique 4
When an arrow's line has only just enough room for its shaded cells, their places are fixed. (Since shaded cells can't touch, a line of a given length only holds so many.)
This arrow needs two shaded cells among the three to its left. Since shaded cells can't touch, the only way to fit two is at the two ends, leaving the middle on the loop.
Looking a step further: spotting cells the loop can no longer reach.
Technique 5
The loop is one continuous line, so any cell on it needs at least two free neighbours to come and go by. A cell that can reach fewer than two can never be on the loop, so it is shaded.
Both highlighted cells are shaded for this reason. One is boxed in on all four sides by arrows and the grid edge, with no way out at all; the other has just one open side, and a single way on is not enough for a line that must both enter and leave.
Technique 6
Every empty cell has to end up on the loop, and the loop needs two open sides to pass through a cell. So if a cell already has only two open sides, shading either neighbour would leave it with a single way on, which no loop can use. Both neighbours are forced onto the loop.
The corner cell (dashed) has only two neighbours. Shading either one would leave the corner with a single way onto the loop, which is impossible, so both neighbours are forced onto the loop.
Playing two clues against each other, and reasoning about the single loop as a whole.
Technique 7
When two arrows sit on the same row or column, one arrow's stretch lies inside the other's. What the shorter arrow accounts for can be subtracted from the longer one, pinning down the cells in between.
The two arrows point toward each other and share the cells in the middle. One arrow's single shaded cell has to sit in that shared stretch, which also completes the other arrow, so the one extra cell it reaches must stay empty and join the loop.
Technique 8
The finished puzzle is a single loop through every empty cell. Joining two loose ends into a closed loop is only allowed when nothing is left outside it, or those outside cells could never join.
Closing the line here would seal it into a loop while cells elsewhere are still left outside it, so this border is ruled out.
The deepest move: a parity argument spanning the whole grid.
Technique 9
Pick any straight line that splits the grid in two (the dashed line). The loop crosses it an even number of times, because every step to one side is matched by a step back. When only one border on that line is still open, it is fixed by whichever choice keeps the count even. Another way to see it: each region a cut carves off must have an even number of loop-ends poking out of it, or the line inside could never pair them all up into one closed loop.
The dashed cut is already crossed an even number of times, so the last open border on it can't carry the line.