Tangled Cord Puzzles: How to Remove Every Crossing

To solve a tangled cord puzzle, find the dots that sit in the most crossings, start with the one that has the fewest cords, and move it into the gap between the dots it is joined to. Repeat one dot at a time, counting crossings after every move, until none are left or you can be sure that one crossing must stay.

This guide explains what counts as a crossing, how to count crossings quickly, a step-by-step method, a full worked example and a practice puzzle with its answer. It also explains why a few networks can never be laid flat, however patiently you move the dots.

What is a tangled cord puzzle?

A tangled cord puzzle gives you a handful of dots, or pegs, joined by straight cords. The drawing starts in a mess, with cords crossing each other all over the page. Your job is to move the dots so that no two cords cross.

The cords stay attached. When you move a dot, every cord tied to it stretches and follows. You cannot cut a cord, add one, or swap which dots it joins. Only the positions of the dots change.

Some versions let a dot go anywhere on the page. Others give you a fixed set of spots, like holes in a pegboard, and a dot must sit in one of them. The fixed-spot version is easier to check by hand, and it is the version used in the examples below.

What counts as a crossing?

A crossing is a point where two cords pass through each other somewhere along their length. A paper published on 25 October 2012 sets out the rules for a drawing of a network: each line joins its two end dots and does not run through any other dot ^[1]. The same paper adds that two lines meeting only at a shared end dot are not counted as crossing ^[1].

So a dot with five cords leaving it is perfectly tidy, even though the cords touch there. What you must avoid is two cords that cut across each other in open space, or a cord that runs straight through a dot it does not belong to.

That paper also defines the key measure for this whole kind of puzzle: the crossing number of a network is the smallest count of crossings you can reach across every possible way of drawing it ^[1]. A network you can lay perfectly flat has a crossing number of zero. Your goal in any tangled cord puzzle is to reach that smallest count.

How do you count crossings quickly?

Counting crossings by eye is easy to get wrong, so use a system. Go through the cords one by one, and for each cord, check every cord after it on your list. Write each crossing as a pair, such as AB with DE. Pairs that share a dot can usually be skipped, because two straight cords from one dot meet only at that dot. The exception is two cords leaving the dot in the same direction along one line, which overlap and are not allowed at all.

There is a fast shortcut when cords run between two rows of pegs. Suppose pegs A, B, C, D and E sit in a top row, and each one is tied to a peg in a bottom row. Two cords cross exactly when their bottom ends come in the opposite order to their top ends.

Here the pairs out of order are A with B, A with D, C with D and C with E. That makes four crossings, and you never had to look at the drawing. To clear them all, rearrange the bottom row so the cords leave in the same order they arrive.

What is the step-by-step method?

Work through these steps in order. They turn a messy picture into a short list of decisions.

1. List the cords. Write every cord as a pair of letters. A written list stops you inventing or losing cords as the picture changes.
2. Count the crossings. Use the pair method above and write down each crossing pair. This number is your score, and every good move lowers it.
3. Find the busiest dots. Look for the letters that appear in most of your crossing pairs. Often two or more dots tie, because every crossing involves four dots. Keep every tied dot on a short list of candidates.
4. Check their cords. A dot with only two or three cords is easy to move, because it only has to sit close to a few neighbours. A dot with many cords is better left where it is, with others moved around it. Try the candidate with the fewest cords first, but do not trust this rule blindly.
5. Find a triangle or short loop. Three dots that are all joined to each other form a triangle. Lay out one triangle or small loop cleanly first, and use it as a frame. Every other dot must then go either inside it or outside it.
6. Test each candidate with a recount. Move one dot at a time into the gap between the dots it is joined to, and recount. If no spot for that dot reaches zero, put it back and test the next candidate on your list.
7. Place loose ends last. A dot with a single cord can almost always be tucked in beside its partner at the end.
8. Suspect a hard core. If the score will not drop below one or two whatever you try, look for the special networks described later. You may already be at the best possible drawing.

How does the method work on a real puzzle?

Here is a worked example on a pegboard of twelve spots, three rows by four columns. Seven dots, A to G, sit on seven of the spots.

The cords are AB, AD, AE, AF, AG, BD, BG, CD, CF and DE. That is ten cords in all. The rule for this puzzle is strict: you may move exactly one dot to one empty spot, and afterwards no two cords may cross.

Step one, count. Checking every pair gives three crossings: AB crosses DE, AE crosses BG, and BG crosses DE again.

Step two, find the busiest dots. Two dots tie. Dot E appears in all three crossing pairs, once through its cord to A and twice through its cord to D. Dot B also appears in all three, once through its cord to A and twice through its cord to G. Both go on the list of candidates.

Step three, test B. B has three cords, to A, D and G, and E has only two, so the cord rule already favours E. Testing B anyway shows why the recount matters. Its best spots, the top-right corner and the middle row in the second or fourth column, each still leave one crossing. The top-left corner leaves three, and the middle row, third column, would put B on the cord from A to E. No single move of B clears the board, so B goes back where it was.

Step four, check E. E has only two cords, one to A and one to D. It only needs a spot that is near both A and D without stepping over anything.

Step five, find a frame. A, D and E are all joined to each other, so they form a triangle. Wherever E goes, that triangle must not trap a cord that crosses its sides.

Step six, try the spots for E. E has five empty spots to choose from. The top-left corner leaves one crossing. The top-right corner and the right end of the middle row each leave three. The middle row, third column, is not allowed at all, because E would sit right on the cord from B to G.

That leaves the middle row, second column. With E there, the cord from E to D is a short step to the left, and the cord from E to A is a short diagonal down to the right. Neither of them touches another cord.

Step seven, recount. Every pair of cords now either shares a dot or stays apart, so the crossing count is zero. A search through every possible single move confirms that this is the only one that clears the board.

Why can some networks never be laid flat?

Some networks keep at least one crossing however you arrange them. A small but stubborn case is the three-and-three network, where each of three dots is tied by a cord to each of three other dots. The paper from 25 October 2012 relies on the fact that its crossing number is exactly one, so a single crossing is reachable but zero is not ^[1].

A slightly larger case is a well-known network of ten dots and fifteen lines. The same paper shows that it can be drawn with two crossings and that no drawing has fewer ^[1]. If a tangled cord puzzle hides this network inside it, you will need at least two crossings, and more if the rest of the network adds its own.

There is no upper limit either: the paper proves that for every whole number k there are networks whose crossing number is exactly k ^[1]. That measure allows lines to curve, and straight cords on pegs can only do as well or worse, so a stubborn tangle can force as many crossings as its maker likes.

This is why step eight matters. When a puzzle asks for zero crossings and you are stuck at one, try to spot three dots that are each joined to the same other three. If you find them, the puzzle as set cannot be finished, and the honest answer is the smallest count you can reach.

What happens when the dots sit on a circle?

A common variant fixes the dots at the corners of a round or many-sided frame and lets you choose which cords to draw. A paper published on 1 January 2012 calls a network drawn on the corners of a convex polygon non-crossing when its straight lines meet only at corners ^[2].

The same paper describes the fullest version, a triangulation, which holds as many lines as possible without a crossing and has a triangle for every region inside it ^[2]. When a puzzle asks you to add as many cords as you can without any crossing, you are building one of these, and you know you have finished when every inside region has exactly three sides.

That paper also works out what such drawings look like when one is picked at random from all the possible ones ^[2]. For drawings picked at random from all non-crossing ones, from those in one piece, or from those with no weak point where removing one corner splits them, the busiest corner has only a small number of lines, growing in step with the logarithm of the number of corners ^[2]. For drawings that are all in one piece, the longest shortest route between two corners grows roughly like the square root of the number of corners ^[2].

Our own rule of thumb follows from that picture, and it is our advice only: if a corner in your attempt has a great many cords, check whether some of them could start from another corner instead.

What mistakes should you avoid?

The table lists the slips that cost the most time.

Can you solve this practice puzzle?

Here is a new board of the same size. Move exactly one dot to one empty spot so that no two cords cross.

The cords are AB, AD, AF, BF, BG, CD, CE, CG, DE and EG.

Start by counting the crossings, list every dot that appears in all of them, and test each one with a recount. Try it before reading on.

Answer. The starting board has three crossings: AD crosses CE, AD crosses EG, and CD crosses EG. Two dots tie, because D and E each appear in all three pairs, and each has three cords. Test E first: its only legal spots are the top-left corner, which leaves six crossings, and the bottom row in the first or third column, which still leave three. So E cannot fix the board. Now test D. Move D to the bottom row, first column, where its cords to A, C and E fan out to the right without touching anything else. Every crossing disappears, and a search through every possible single move confirms that this is the only one that works.

Where can you practise next?

Tangled cords also turn up in picture riddles, where you trace which cord leads where instead of moving the dots. This one asks you to follow a tangle from switch to lantern.

Trace each cord with a fingertip, keep track of which cord you are on at every crossing, and the answer follows.