Join adjacent numbers in the below grid to reveal the hidden content.

The rules are:

1: Join all pairs of vertically and horizontally adjacent numbers that are the same.

2: Join all pairs of vertically, horizontally, and diagonally adjacent numbers that are next to each other in the circular sequence.

The sequence consists of the ten digits wrapping round so there is no beginning or end. Work out the order of those digits from the following information:

  • When all the lines have been drawn in, there's a single shape - if the lines were a boardwalk in a swamp you could walk from anywhere on the boardwalk to anywhere else on it without getting your feet wet.

  • There's exactly one diagonal line segment (a line segment is the line between two numbers).

(I'm having problems deciding on the appropriate tag(s) for this! Please feel free to change!)

What Goes Round Begriddled puzzle grid


@AxiomaticSystem has already found the correct solution. However, as this question is tagged as I thought it merited a second answer which demonstrates that this solution is entirely logically deducible. Here goes!

Each step of this solution will be illustrated using the diagram provided (thick blue lines are part of the solution; thin red lines are logically impossible) and a table alongside which shows the possibilities for numbers that can be consecutive in the cycle (a cross means not possible, 'Y' means 'Yes').

Step 1:

First, carry out step 1 of the instructions and join all horizontal and vertical pairs of identical numbers. Furthermore, rule out all pairs of numbers which would result in more than 1 diagonal line when carrying out the OP's instruction step 2:
enter image description here

Step 2:

Note next that 1 shares a diagonal with all of 3, 4, 6, 7, 8 and 9 - if there is only one diagonal max, then 1&5 must be consecutive. Furthermore, the one diagonal MUST involve the number 1 - we can rule out all other number pairs where a diagonal does not involve 1:
enter image description here

Step 3:

Let's rule out a few more potential pairings...
- The 5-6 in the rightmost column is now isolated and cannot be connected to the boardwalk - rule out 5&6;
- Similarly, the 4-8 in the leftmost column;
- Also, we cannot have 1&3, or this would isolate the top left corner (since 1 cannot be connected to 4 or 9 as well as 3, as it is already connected to 5);
- Neither can we have 1-9 at the base of the 5th column, as to connect that to the boardwalk we would need to connect 9 both to 7 and to 5 in addition to 1 (and a number can only be connected to 2 different numbers).
enter image description here

Step 4:

Note now that if 5 connects to 0, there is no longer any way to link the left and right parts of the boardwalk. In fact, to link the two sides we will need to link 5 to 9. We now know both of 5's neighbours:
enter image description here

Step 5:

9 has one more potential connection, and its only remaining options are 2 or 7. If it is 7 then there is again no way to link the left and right parts of the boardwalk, as without being able to use the 9-2-2 connection in the top-right, the only way to link the 2-2 to the rest of the boardwalk would involve linking through 1-8 AND 1-7 - but 1 has only one potential partner left. We must therefore link 9&2 and cross out all 9-7's. enter image description here

Step 6:

2 must now link to 8, or there is no way to link 8-8 to the boardwalk. Also, 1&7 can be ruled out, as the centre-bottom 7 requires both the 1-8 and 1-7 links, which is impossible (as previously ruled out in Step 5). enter image description here

Step 7:

Now, only one link exists between the 3rd and 4th columns: 1-4. This must be the remaining connection with 1, and provides our one diagonal, in the top left corner. enter image description here

Step 8:

The boardwalk is now complete, as no more connections are possible. So how to fit 3, 6, 7 and 0 into the cycle? Note that 3 cannot be adjacent to 6, or this would create a second unconnected boardwalk in the bottom left corner. This means 3 must be adjacent in the cycle to 0 and 7.

Also, 7 must connect to 6, as there are no other options available for 7. Thus we need to fit 0-3-7-6 into the cycle. Since 6 cannot be adjacent to 8 (ruled out in Step 2), it must be adjacent to 4 instead, giving us the final configuration. enter image description here

Now that all is said and done (and logically deduced!) we can see the final answer is really very appropriate indeed:

The blue lines in the grid depict the Greek letter PI, the mathematical constant whose value is 3.141592... Now hold on a moment - where have we seen those digits before?! I suggest you look at the first line of numbers in the grid again...!

EDIT: In fact, as the even more astute (like @LannyStrack in comments) might notice, this puzzle actually contains the first twenty digits of pi, wrapped around the edge of the grid, starting top left! 3.1415926535897932384...
enter image description here

| improve this answer | |
  • 1
    $\begingroup$ Well your answer came in second but so well done that it has to be the accepted one :) $\endgroup$ – Nick Rice Apr 19 at 22:29
  • 2
    $\begingroup$ rot13(Gur qvtvgf bs cv ner abg whfg va gur svefg yvar bs ahzoref, ohg jenccrq nebhaq gur ragver obeqre!) $\endgroup$ – Lanny Strack Apr 19 at 22:57
  • $\begingroup$ It seems to me that your statement rot13(Abgr arkg gung 1 funerf n qvntbany jvgu nyy bs 3, 4, 6, 7, 8 naq 9 ) should read rot13(Abgr arkg gung 1 funerf n qvntbany jvgu nyy bs 0, 2, 3, 4, 6, 7, 8 naq 9 ). $\endgroup$ – Lanny Strack Apr 19 at 23:05
  • $\begingroup$ @LannyStrack Re your second comment: we've already ruled out 0 and 2 in the previous step (multiple diagonals). Re your first: Didn't spot that at 11pm last night! Will modify, thanks :) $\endgroup$ – Stiv Apr 20 at 6:11
  • $\begingroup$ I see. Thanks.. $\endgroup$ – Lanny Strack Apr 20 at 6:27

From the three initial pairs, the top row suggests an image like

this one from 415928, which has exactly one diagonal link.
From there, the only placement of the remaining four numbers that creates no diagonal links is 0376415928.

| improve this answer | |
  • $\begingroup$ On reflection (previous comment since deleted) I now follow your argument but it's still difficult to do so without more explanation. Could you provide some more detail about your logical deductions? Particularly the steps that led to your ordering of the final 4 numbers? That would make it easier to process mentally for the rest of us. Thanks :) $\endgroup$ – Stiv Apr 19 at 20:38
  • $\begingroup$ Upvoted because right and first! But I had to accept @Stiv's answer, as I'm sure you understand :) $\endgroup$ – Nick Rice Apr 20 at 8:28

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