How to Solve a Slitherlink

One real 6×6 with sixteen numbers, from an empty lattice to a finished loop. Every move below is the move the solver made, in the order it made it.

The order to sweep in

Start with the zeros, every time. A zero settles four segments for free and each of those four belongs to a neighbouring square as well, so a single zero routinely sets off a chain three or four squares long. If a board has zeros and you have not used all of them, you have not started yet.

Then sweep the dots. Anywhere a line already arrives, check whether all but one of the ways out have been crossed off, and if so, continue the line. Anywhere nothing is drawn, check whether all but one route has been ruled out, and if so cross that one off too. Both are instant to see and both create work for the numbers.

Only when neither of those moves anything is it worth reaching for the patterns. The corners come first because a corner dot has just two segments and that settles the number sitting in it immediately, and the threes come after, because reading two numbers against each other is genuinely more effort than reading one. The rhythm that finishes boards is: zeros, dots, numbers, dots again — and back to the dots after every single thing you draw.

Five positions from one board

The 6×6 below carries sixteen numbers and is graded expert — chosen not because a beginner should start there but because its opening happens to use all five arguments within twenty moves, which no easier board does. The five positions shown are the five moments where something new is being used; the moves in between are more of the same.

Read each pair of boards left to right. On the left is the position as it stood, with the numbers and dots the argument read marked and the segments it decides outlined. On the right is the same position a moment later. A small cross is a segment ruled out — that is a real result and is very often the entire content of a move — and a hairline is a segment nobody has decided yet.

1. The 0 at row 1, column 4 is the cheapest thing on the board and therefore the first move. A zero uses none of its four sides, so all four are crossed out before anything else has been read.

   Counting a cell

2. Look at the dot at row 1, column 4 of the lattice. Two of the three segments touching it were crossed out by that zero, nothing is drawn, and a line cannot reach a dot and stop — so the third is crossed out too. Nothing appears, and two squares just lost a side.

   Following the line

3. Now the free moves run out and the corners are worth reading. The 1 in the top-left corner sits under a dot with only two segments, which must both be used or neither — and both would give a 1 two sides. So both are crossed out.

   The corner cases

4. The 3 at row 4, column 6 and the 3 at row 5, column 5 touch only at a corner. On each of them the two sides facing away from that shared dot are drawn: four segments, in two places, from two numbers read against each other.

   Threes together

5. And here is the argument that makes this board expert. Two loose ends of the same run of line sit one segment apart near the right-hand edge. Joining them would close a small ring, and a closed ring can never grow — so it would have to be the whole answer, which the numbers elsewhere flatly refuse.

   No early closing

what the reasoning read what it decides what it wrote

Fifty-eight more moves of the same five ideas, and the two ends meet. Here is the finished loop.

When nothing at all will move

Go back to the last thing you drew and work outwards from it. By far the commonest cause of a stalled board is a dot next to a segment you have just settled where the line now has only one way out and nobody has noticed. Every segment you decide changes two dots and two squares, and checking those four is free.

If that fails, count the squares rather than eyeballing them. Take every number in turn and compare what it already has drawn and crossed against what it needs; a square that is one short of its number and has exactly one side left looks identical to a comfortable one until you actually do the sum. This is slow and it is worth doing exhaustively rather than where you suspect.

And if a full sweep and a full count both come up empty, the board wants the closing argument: look for two loose ends of the same piece of line that are one segment apart, and ask what would happen if you joined them. What the board never wants is a guess — every one published here was checked to be finishable without one.

Common questions

More Slitherlink pages

• SlitherlinkUnlimited boards
• Daily slitherlinkA new loop every day
• Slitherlink rulesThree rules, in full
• Loop techniquesThe whole ladder
• Squares and dotsTechnique — easy
• The corner casesTechnique — medium
• Threes togetherTechnique — hard
• Closing too earlyTechnique — expert
• Guessing at loopsTechnique — refused, here too
• Printable slitherlinkFor paper
• Slitherlink archiveEvery past loop