The lines in this shape need to be numbered in a way following the rules below.


  • Each line can only have one number
  • The Red line must have a number 1
  • Numbers may be reused on other lines
  • No lines with the same number may intersect at the dots
  • A line cannot have a difference greater than 1 increment/unit from any other line its intersecting
  • If the line intersects with at least one line with a difference of 1, the rule above is void
  • There are no negative numbers
  • Only whole numbers are allowed

Feel free to comment any changes I should use to make the question easier to understand, or questions regarding rules or correct answers.

  • $\begingroup$ @OP You've already accepted an answer here which was correct for the problem as originally posed. Rather than change the problem to something different and look for updated responses, why not leave the original here and post the new version as a new question? $\endgroup$
    – Rubio
    Nov 14, 2016 at 22:11
  • $\begingroup$ @Rubio I'd rather suggest a part two to the question instead of a new question entirely; it would likely be flagged as duplicate $\endgroup$
    – Areeb
    Nov 14, 2016 at 22:34

2 Answers 2


If only integers are allowed, then the puzzle is impossible: since lines connecting at the same dot must be numbered with distinct integers, any dot with three or more lines would be connecting numeric differences of 2 at least.

If rationals are allowed, things are much easier though.

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  • $\begingroup$ Now that I know that it is impossible given the updated rules, could you help me find a configuration in which only the lowest point intersects with a difference greater than 1? $\endgroup$ Nov 13, 2016 at 8:25
  • 1
    $\begingroup$ @Tommy if I understand the problem correctly, that would still be impossible, because there are points above the lowest one with three and four lines. $\endgroup$
    – GOTO 0
    Nov 13, 2016 at 8:31
  • $\begingroup$ You can use as many or as little numbers as you need, and you can use the same number multiple times in the puzzle $\endgroup$ Nov 13, 2016 at 8:35
  • 1
    $\begingroup$ @TommyWoldt: As long as you have a dot with more than two lines connecting to it, it's impossible. Since the number cannot be the same, then if one line's number is n, one other line must be n-1 or n+1. Suppose it's n-1. The third line doesn't have any available option, since it's connected to a line with number n, so it should be n-1 or n+1, but n-1 has already been taken by the second line, so it should be n+1, but n+1 differs by 2 to n-1, the number of the second line. $\endgroup$
    – justhalf
    Nov 14, 2016 at 2:40
  • $\begingroup$ @justhalf Though I may not fully understand your explaination, ill try to explain my own rules. What i mean is that one could have a line labeled 1, another labeled 2, and so forth for 3 and 4. These lines may intersect because 4 intersects point 3, which has a difference if 1, so its happy. 3 has 4 and 2 so its also happy, same with 2 and following the same logic pleases 1. $\endgroup$ Nov 14, 2016 at 3:17

Label the points, starting at the bottom of the red line, going clockwise around the outside: A, B, C, D, E, F. The internal two points going up are G,H.

The red line is AB, with value 1. Set values on CH, DE, FG, each to value 1. These lines do not intersect, so they can all have that value.

Let BC, DH, EG all have value 2.

Let BG, CD, EF all have value 3.

The final line, GH, has value 4.

No two lines with the same value intersect. Every line intersects another line with a value that differs by exactly 1, so the rule about no difference greater than one never applies. All the numbers are positive integers.

All the rules seem to apply.


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