Fibonacci DubTrip
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Motivation:
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I was looking for something line-based that could wiggle through the grid
with some kind of ambiguity, like German Whispers, so I chose to connect
the members of the fibonacci sequence. This is more like Dutch Whispers,
as you can connect 1 to 1, and change the "parity" of the sequence.

I didn't want to contrive a particular pattern with the fib connections, as
I wanted the grid to be organic. I hate patterns appearing.

I also wanted colouring to be useful, so I didn't want disambiguation between
the high fibs and low fibs to happen too early. Similarly, the non-fibs, 4, 6,
7, and 9 I wanted to remain ambiguous for as long as possible.

Walk-through Solve:
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Preparatory hints as to what markup should be used:
1) Use of colouring to show the distinction between [12358] and [4679] is not
as useful during the very early solve as you might think, but becomes much more
useful later on, so is worth doing.
2) Colouring to distinguish between the high and low values in [18] and [25]
cells can be useful if you start getting more than a handful of such cells.
3) Colouring just to know if a cell can be a [25] or a [138]/[18] is only
marginally useful early on, and can be skipped.

Let's Get Cracking!

Firstly, some truisms about the fibonacci links:
In every box the middle of the cell can have 0, 1, or 2 lines coming out of it. As it has all 8 possible neighbours: 0 links means it's [4679], 1 link means it's [18], and 2 links means it's [235].
One member of every link that doesn't diagonally cross a box boundary, must be one of [25].
Therefore, except when there's a 1-1 link, the lines will alternate between [25] and [138].
If there are 3 cells on a line that can all see each other, one of them must be a 3.

In box 9, the existence of the diagonal link forces the 3 into the ends of the dogleg, and [25] into the bend.
The existence of two vertical segments in column 8 forces r9c8 to be from [138], placing a second [25] in box 9.
The 3 in the box is pushed into r8c7.
The DubTrip kropke means r7c8 is 8 not 1, and all of 12358 resolve.

Looking at box 6, the 3 can no longer be in col 6.
r4c8 cannot be the 3 on that line, as r5c6-r5c8 would all have to be from [12] or [58].
=> r5c8 is 3, and the 2 and 5 can be placed.
=> r4c9 and r3c9 are 8 and 5
=> r5c6 and r6c6 are 8 and 5
=> box 3's 3 is in col 9, and can't abut the 5 - start singing!

Considering the two lines that enter box 5, if r5c4 was 1, then r6c4 and r5c2 would both need to be from [25], which r6c6 forbids. Therefore it's 2.
=> The line can be marked with [13], 2, [13], [25], [138].
The 3 in box 5 can only be placed in r6c4.
The 1 in box 5 can only be placed in r4c6
=> box 5 can be [12358] vs. [4679] coloured.
=> col 5 can be similarly coloured.

r8c4 can only be 1 or 8 and must orthogonally connect to something that's not 2.
=> r8c4 is 8, r9c is 5
=> the 2 and 3 can be placed on the line in row 9
Box 8 can be [12358] vs. [4679] coloured, and the 1 placed.

The 1 in box 6 can only be in row 6, and so the rest of the box can be [12358] vs. [4689] coloured.

In box 7, the 3 is in row 7, and can only go in col 3.
If r7c2 was 2, we couldn't place a 1 in the box, so it's a 5.
r7c2 must be the row's 2.
The X-sum in box 7 can't be 2-8 or 3-7, leaving 1-9 and 4-6, but by colouring, 1-9 is impossible.
Box 7's 1 can't be next to the 2, so is in r8c3.
Box 7's 8 can't be next to r9c4, so is in r9c1

In box 1, r2c2 and r2c3 can only be [12] or [23], but [12] is unsatisfiable both ways.
If r3c2 is 3, then r4c1 would be 5, but then we are unable to place a 3 in box 4.
=> r3c2 is 2, and the 3, 3, and 5 can be placed on the line.
=> naked 5 in r4c3.
=> box 1's 8 can only go in r1c3

To avoid the 2 and [4679]-only cells, the 1 in box 4 can only go in r5c1.
=> box 4 can be fully [12358] vs [4679] coloured.
col 2 can be [12358] vs. [4679] coloured, and r1c2 must be the 1.

box 3's 8 is in r3c7
its 2 cannot be next to the 3, so is in r3c7
its 1 goes in r2c9
the X-sum must be [46], leaving a [79] pair.

In box 2, the line must be 5 over 8, and r3c5 must be 3
The 3 in col 6 can only go in r3c5.
The 2 needed in col 6 can't be on the X sum, so is in r1c6.
To avoid 2s, the 1 cannot go in col 6, so is in r3c4.
The X-sum is therefore [46] and the remaining cells a [79] pair.

All [12358] cells should now be placed, and all boxes [12358] vs. [4679] fully coloured.
Make sure you pencil the DubTrip kropke dot values.

From there, it's pure sudoku - follow the [46] and [79] pairs around the grid, some will resolve further between the 4 and the 6, others will resolve the 7 from the 9.