How to Solve Perimeter: One Closed Loop from the Clues

Perimeter asks you to draw one continuous closed loop along grid edges. A number inside a cell tells you exactly how many of that cell's four edges belong to the loop. Tap an edge once to draw it and tap again to remove it. There is no separate cross or “empty edge” state, so your deductions need to live in the pattern of drawn and undrawn edges.

The cleanest way to reason about the puzzle is to keep two ledgers at once: a cell ledger for clue counts and a vertex ledger for the shape of the loop.

Ledger one: what does each clue still need?

If a 3 already has two drawn sides, exactly one of its remaining sides must be added. If a clue is already satisfied, do not add another side around it. If the number of available sides equals the number still needed, all of those available sides are forced.

Recount adjacent clues after every edge because one edge belongs to two cells. A single placement can satisfy one clue while reducing the choices for its neighbor.

Ledger two: every used vertex has degree two

A finished loop never branches and never ends. At every grid intersection touched by the loop, exactly two incident edges are used. If a vertex already has two drawn edges, no third edge may join it. If a partial route reaches a vertex with only one legal continuation, that continuation is forced.

This vertex rule finds deductions even in parts of the board with no nearby number.

Corner clues speak loudly

A clue near the outer border has fewer ways to distribute its edges while keeping the loop valid. A 3 in a corner, for example, tightly constrains the two border sides and the two interior sides. Test its possibilities against the degree-two rule at the corner vertices.

Do not apply memorized Slitherlink patterns blindly. Count the actual four cell edges and verify the loop can continue through both vertices.

Do not close the loop while work remains

If an edge would complete a small cycle while unsatisfied clues or disconnected line segments remain elsewhere, that edge cannot be correct. The final answer is one loop, not several.

This is one of the strongest late-game deductions. Two open ends may look eager to meet, but they must wait until every other required segment has joined the same circuit.

Use undrawn edges mentally

The interface does not store cross marks, but you can still reason with forbidden edges. An edge beside a satisfied clue is unavailable. An edge that would create a third branch at a vertex is unavailable. An edge that would close a premature loop is unavailable.

Once you mentally remove those choices, a clue or vertex often has only one possible continuation.

A reliable pass order

1. Count every numbered cell and draw edges that are immediately forced.
2. Inspect vertices around the new edges for forced continuations.
3. Revisit neighboring clues that share those edges.
4. Reject any move that branches or closes a separate loop.
5. Trace the nearly finished circuit and join its final ends only after all clues are satisfied.

Reading a clue and a vertex together

Consider a cell marked 3 with two sides already drawn. It needs one of the other two sides. If one candidate would enter a vertex that already has two loop edges, that candidate is forbidden, so the other side is forced. The clue ledger alone said “one of two”; the vertex ledger finished the deduction.

The reverse works too. A vertex with one drawn edge may have two possible continuations, but one may run beside a clue that is already satisfied. That leaves the other continuation. Most medium and hard Perimeter boards are solved by repeatedly letting the two ledgers break each other's ties.

Perimeter is won by alternating local arithmetic with global topology. The clues tell you how much line belongs nearby; the vertices and the single-loop rule tell you how those pieces are allowed to connect.

A new Perimeter puzzle is available every day. For more repetitions with the same rules, open Practice and choose Perimeter.