Dot-to-Dot Path Puzzles: How to Join Every Dot Exactly Once

A dot-to-dot path puzzle asks you to join every dot in a single unbroken path that visits each dot exactly once, and the surest way to solve one is to start with the dots that have the fewest possible partners. Fix those links first, check the start and finish against any hop or number rules, and the rest of the path usually falls into place.

This guide covers the three common kinds of path puzzle, the partner-counting habit, how hop-length rules work, and why one-way arrows are kinder than they look. The last sections solve one puzzle in full and finish with a grid for you to try.

What is a dot-to-dot path puzzle?

The classic picture puzzle numbers its dots, so you simply join them in order. Path puzzles take the numbers away, or most of them, and leave you to find the order yourself. The rule is always the same: one path, every dot once, no dot used twice.

Three common kinds are:

• Grid paths with a few numbers. Dots sit in a grid, some carry numbers, and you fill in the rest so that each number touches the next.
• Drawn networks. Dots are joined by printed lines, and your path may only travel along those lines.
• Hop-length puzzles. Dots stand in a row or around a ring, and every hop must have an allowed length, such as two or three places.

Some versions also fix the start, the finish or both. A fixed end is a gift, because it removes most of the guesswork at the edges.

Why should you count each dot's partners?

A dot's partners are the dots it is allowed to link to. Counting them is the single most useful habit in this kind of puzzle.

• A dot with one partner can only be an end of the path. If the puzzle already names a different start and finish, there is no answer.
• A dot with two partners that is not an end must use both of them. Draw both links at once.
• A dot with two drawn links is full. Cross out its other partners, which may leave neighbouring dots with fewer choices.

Corners of a grid and dots at the ends of a row are where this starts, because they have the fewest partners. Work inward from them, and keep updating the counts as links get drawn and crossed out.

How do hop-length rules work?

Picture dots standing in a row and numbered in order. A hop from one dot to another has a length equal to the gap in their numbers, so a hop from dot 4 to dot 7 has length three. The puzzle tells you which lengths you may use, or exactly how many hops of each length.

A paper published on 15 March 2010 studied exactly these rows of dots, with every hop one, two or three places long ^[1]. Several of its results make handy sanity checks before you start.

• One length only. If every hop has the same length, whether one, two or three, a row path through all the dots is possible only when that length is one ^[1]. Hops of two alone, for example, leave half the dots untouched.
• Ones and twos. Any list with at least one hop of length one and the rest of length two can be laid out along a row ^[1].
• No ones at all. Twos and threes alone can still work: any list with at least four twos and at least three threes can be laid out along a row ^[1].
• First dot to last dot. A row path that begins on the first dot and ends on the last, using hops of one, two and three, always uses an even number of twos ^[1]. This applies only when the path ends on the last dot, so it says nothing about the worked example below.

The paper sorted every possible list of ones, twos and threes into those that can be laid out along a row and those that cannot ^[1]. You will not need the full list, but it shows that a hop puzzle can be impossible however cleverly you search, so check the counts first.

What changes when the dots form a ring?

When the dots sit around a ring, a hop is measured the shorter way round. A ring has no end dots squeezed against an edge, so rings are often easier to plan.

The same paper showed that around a ring, any list that uses at least one hop of each length, one, two and three, and has at least six hops in total, can always be laid out ^[1]. So a list with all three lengths present and six or more hops is safe.

A single length can still fail. On a ring with an even number of dots, a path made only of hops of two is impossible ^[1]. The reason is easy to see: hops of two only ever land on every second dot, so half the ring is never reached. On a ring with an odd count of five or more dots, the hops of two eventually wrap around and reach every dot.

Why are one-way arrows easier than they look?

Some puzzles draw arrows instead of plain lines, so each link can only be used in one direction. That sounds harder, but one kind of arrow puzzle always has an answer.

Suppose every pair of dots is joined by exactly one arrow, pointing one way or the other. Mathematicians call this a tournament. Every tournament has a path that follows the arrows and visits every dot once ^[2].

You can build that path by hand, one dot at a time:

1. Start with any single dot as your path.
2. Take a new dot. If it points to the first dot of the path, put it at the front.
3. Otherwise walk along the path and look for two neighbours where the earlier one points to the new dot and the new dot points to the later one. Slip the new dot between them.
4. If no such gap exists, the last dot of the path points to the new dot, so add it at the end.

Our own computer check ran this method on all 1,024 ways of pointing the arrows between five dots, and it built a complete path every time.

How do the puzzle types compare?

How do you solve a path puzzle, step by step?

1. Read the rules. Note which links are allowed, and whether the start and finish are fixed.
2. List partners. For each dot near an edge or corner, write how many partners it has.
3. Run the quick check. Use the table above to rule out an impossible puzzle early.
4. Draw forced links. Every dot with two partners that is not an end uses both.
5. Cross out full dots. A dot with two links takes no more, so update its neighbours.
6. Join the chains. Forced links form short chains. Ask how the start can reach the nearest chain without stranding a dot.
7. Check the finished path. Every dot once, every link allowed, the right start and finish.

Worked example: can you hop along this row?

Ten dots stand in a row, numbered 1 to 10. Every hop must have length two or three. Start on dot 1 and finish on dot 3, visiting every dot exactly once.

By our own computer count, without the fixed finish this row has 11 different paths from dot 1, so the fixed finish is what makes the answer unique. Here is how to find it.

Step 1: list partners at the ends. Dot 10 can only link to dots 8 and 7. Dot 9 can only link to dots 7 and 6. Neither is an end, so both links are forced, which gives a chain: 8, 10, 7, 9, 6.

Step 2: the second dot. Dot 2 can only link to dots 4 and 5. It is not an end, so the chain 4, 2, 5 is forced too.

Step 3: the start. Dot 1 can hop to dot 3 or dot 4. Hopping to dot 3 would reach the finish at once, with eight dots unvisited. So the path opens 1, 4, and then the forced chain carries it on to 2 and 5.

Step 4: leave dot 5. Dot 5 has partners 2, 3, 7 and 8. Dot 2 is used. Dot 3 is the finish, too early. Dot 7 already has two links in the chain from step 1, so it is full. That leaves dot 8.

Step 5: run the chain. From dot 8 the forced chain runs to 10, 7, 9 and 6. Dot 6 is three places from dot 3, so the path ends exactly where it should.

The full path is 1, 4, 2, 5, 8, 10, 7, 9, 6, 3. It uses three hops of two and six hops of three, and an exhaustive computer search finds no other path from dot 1 to dot 3.

Which slips break a dot path?

• Joining a forced chain in the middle. A dot inside a chain already has both its links.
• Racing to the finish. Reaching the last dot early leaves others stranded, as in step 3 above.
• Ignoring short-way-round on a ring. On a ring, a hop of length eight among ten dots is really a hop of two.
• Guessing before counting. One minute of partner counting saves many wrong paths.

Can you solve this grid path?

Sixteen dots sit in a 4 by 4 grid. Number them 1 to 16 so that each number touches the next, sideways, up, down or on a slant. Five numbers are given.

Hint: number 15 must touch both 14 and 16, and the corner dot 11 has only three partners.

Answer: There is exactly one way to fill the grid.

From 1 the path steps right to 2, then runs down the right-hand column to 5, along the bottom row through 6, 7 and 8, up the left-hand column to 11, then through 12 and 13 to the centre, finishing 14, 15, 16.

Now try a classic numbered dot puzzle and see what picture the right order reveals: