This is a different 'connect the dots' puzzle.

Your task is to connect the blue and yellow nodes in the map below with links such that the created network guides two signals emitted from the two red triangles to the green pentagram. In doing so, each node needs to be used by at least on of the two signals at least once.

Clarification: Your task is to provide the additional links. Once they are all set, signals start and move according to the rules until they either die or end in the green node. You can no longer influence this at this stage.

enter image description here

The following rules apply:


  • A node may be visited as often as one likes.
  • Each node needs to be visited at least once by at least one signal
  • The signal strength when reaching the final node (green pentagram) does not matter.
  • Node labels are only to allow easier writing of an answer.


  • You may create as many links as you wish.

  • Already drawn links are fixed and can not be removed.

  • A link has to connect two nodes in a straight line.

  • Links have to leave nodes radially.
    (If you prolong the line it has to go through the centre.)

  • Links must not cross each other at any point.

  • Links must not 'touch' any other node.

  • Links and nodes have the dimensions as in the original map.
    (They are not mathematical points and lines.)

  • Examples for valid and invalid links are:

    Valid and invalid links
    (valid; invalid because crossing; invalid because touching other node; invalid because not radial)


  • Each signal travels along a link from one node to the next node.

  • Each red triangle emits one signal of strength 5 towards the first connected node.

  • Signals do not affect each other.
    (You can think of them as being emitted in sequence, i.e. the second is emitted after the first reached it's goal.)

  • The moment a signal arrives at a node, the signal strength is either:

    • decreased by -1 (blue node)
    • increased by +2 (yellow node)
  • A signal of strength 0 dies and is removed.

  • Once a signal has arrived at the node it moves clockwise for as many outgoing links as its strength indicates. (It may do full circles multiple times.)
    i.e. for a strength of 3, the signal would leave the node via the 3rd link after the one it came from.

  • The following images illustrate the signal movement: enter image description here

    enter image description here


There exists at least one solution. It may not be unique. You can specify your answer either by an image (preferred) or by specifying all links (2 node-labels each) and the path each signal takes (by node-labels.)


2 Answers 2


See comment from OP: this doesn't work. I'm going to have a look and see if I can get it into shape.

Here’s my complete solution:

enter image description here

The green lines are the lines that I’ve added, and the thick lines indicate the path that the two signals follow.

A few notes:

  • Since the task says “blue and yellow nodes”, I assumed that I can’t add links to the red or green nodes.
  • This is an approximation of the original image. It was easier to start from scratch than edit the original. I don’t think any of the lines are close enough for it to make much of a difference.
  • I do have a version with numeric labels, but it just gets crowded.

I tackled a path at a time: first the path from the right-hand red node, then the left. It was about 50-50 logic and sensible guesses.

Here’s how I worked through the right-hand node:

  • With three lines at C, the signal is going to rebound to the red node. We need to add one extra edge, and I chose CG because it forces the signal away from the centre (so it’s less likely to conflict with the left-hand path.
  • At H the score is $4-1=3$, which leads to P. It doesn’t seem possible to get to the green node from here without the signal expiring or going via a yellow node, so P seems like a sensible choice. Don’t add any edges.
  • At P the score is $3+2=5$, which would rebound to H. While not technically illegal, it seems pointless. I initially looked at jumping directly to O, but this makes it difficult to get a path in from the left-hand side, so I went via M instead. (I also knew I would have one dangling edge to fill in coming from O, for the left-hand path.)

With that path in hand, I moved on to the left:

  • I know I want to end with a score of 2 entering O, so I worked backwards from there. Coming from a yellow node would mean a low score as I'd entered the yellow node, which seems unnecessarily circuitous, so I guessed that I'd come from N. That means I had score 3 when I entered N.
  • I need to add an odd number of edges to B so that I don't rebound. C is out because that'll break the right-hand path, and so I went with A just because that was the first next-node for which I saw the rest of the path.

It’s probably not a unique solution, although I haven’t gone back to find another yet. But unless I’ve misunderstood the rules, I’m fairly sure that is a valid solution.

  • $\begingroup$ +1 for taking this on, and your assumptions are correct. However, I think your solution isn't for 2 reasons: a) not all nodes are 'visited' by at least 1 signal and b) I think you signal-node interaction does not match the rules. As for b): Just look at the signal emitted from the right red node. It has strength 5. It hits node C and becomes strength 4. Now, the movement-action: Starting to count clockwise from the incoming link's clockwise neighbour, the 4th link is the one leading back to the red node. $\endgroup$
    – BmyGuest
    Mar 12, 2015 at 10:35
  • $\begingroup$ @BmyGuest Damn, I missed rule (a), and misread the signal emitting. I'll have another look at it later. :-) $\endgroup$
    – alexwlchan
    Mar 12, 2015 at 10:59
  • $\begingroup$ Please do! BTW, my solution (not necessarily unique) does involve signals visiting nodes multiple times. $\endgroup$
    – BmyGuest
    Mar 12, 2015 at 11:23

Assuming 'visit a node' means that a signal has to reach that node, not necessarily continue on from that particular node to the green node, here's one solution. Signal strengths at the start and end of each link are indicated after each link.

  • LRed-B 5-4
  • B-K 4-6
  • K-A 6-5
  • K-J 6-5, J-E 5-4, J-N 5-4
  • K-O 6-5
  • O-Green 5-4

  • RRed-C 5-4

  • C-F 4-3
  • F-L 3-5
  • L-O 5-4
  • O-Green 4-3

  • C-G 4-6 (originally from RRed-C)

  • C-H 4-3, H-M 3-2
  • H-P 3-5
  • P-I 5-4
  • I-D 4-3
  • $\begingroup$ I'll check that when I find a bit more time than at the moment. For clarity: Could you just list the links you've added first? I'm somewhat confused by your 4th point. At 3rd you have "K-A" but your fourth is "K-J" while the signal now is at A ? $\endgroup$
    – BmyGuest
    Mar 12, 2015 at 12:06
  • $\begingroup$ Clarification: You start out with 2 signals on a network with all links installed. These 2 signals process until they die or end in the green node - and you can no longer influence this. Your aim is to create a network where the two signals do not die, but end in the green node. Additionally, in doing so, all shown nodes need to be "move through = visited" by at least one signal at least once. $\endgroup$
    – BmyGuest
    Mar 12, 2015 at 12:09
  • $\begingroup$ Sorry, but I think your solution does not fit the requirements. Seems you're suggesting to add the following links to the picture: (BK, KO, LO, AK, CG, ID) With those links added, signals emitted from the two red nodes following the rules above will not end up in the green node. $\endgroup$
    – BmyGuest
    Mar 12, 2015 at 13:39
  • 1
    $\begingroup$ @BmyGuest Thanks for the clarification - I notice you've since edited your question. I wondered about the question wording, hence the assumption stated in my answer's preamble. Your signal diagram (3rd on the top row) seemed to indicate signals propagating at equal strength from multiple outbound links "for as many outgoing links as its strength indicates". I see now that you intended to have the signal propagating from exactly one outbound link. $\endgroup$
    – Lawrence
    Mar 12, 2015 at 14:26

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