Solve Overview:
---------------
0) Note the (d) <- b -- (b+d) -- b+2d -- (d) nature of the broken arrows (BAs), including parity
1) Hit the connected shape hard, pencil mark unknowns
2) Fine tune the solo arrows
3) Combine the solo arrows, start colouring when you get to only 2 options in lots of cells.
4) Finish with sudoku

More details:
-------------
0) getting the vibe of the shapes:
The tip of the arrow is [1234], and 4 is only possible as a 4<-1-5-9
Only one of the broken arrows can have the same delta as the base value - between boxes 5 and 6

1) Hit the connected shape
Because the circles are limited, the lines connecting them are too. Pencil everything!
The first thing to notice is that r6c5 cannot be 3, as r5c4 sees both r5c7 and r6c5, so is [24].
Road bump #1 is rejecting 2 from r5c4, but because it constrains the values on the shape so much doesn't require much deep thought.
A 2 there would be a 1<-1-2-3 BA to the right and a 2<-[45]-[67]-[89] to the right. The delta on r3c2 would need to be 1 or 2, and no matter which it is, because of the [67] in r5c3, row 4 gets a 5, 6, and 7 outside box 5, so r5c5 becomes overfilled.
=> r6c5 is 4
=> there are 4<-1-5-9, 2<-2-4-6, and 2<-4-6-8, and 1<-5-6-7 BAs going from box 4 to box 3
=> it's another 4<-1-5-9 in teh BA to the left into box 1
=> don't forget to do sudoku, there are naked digits in that middle band.
This segues gracefulling into the next stage....

2) Fine tune the solo arrows
Whilst doing sudoku, after "where's 1 in box 6?" comes "where's the 5?", and you're looking at the BA in box 9 - more naked digits appear once you've placed the [23] pair on the arrow, but it remains unresolved apart from b+d=5.
Similarly, box 8's BA can be pencilled down to 2 values in each cell.
Likewise, box 2's BA is limited to 2 options, but at least b=2 is fixed
Box 1's BA isn't quite as narrow, it's [13]<-[267]-[578]-[89].
The final important thing is that the glue layer of [37] pairs in the middle band is pencilled, as that's what permits implications from boxes 1 and 2's BAs to affect boxes 8 and 9's BAs.

3) Combinations of arrows
Pressure points on the unresolved arrows include things that could endanger the relation that b and b+2d are of the same parity, and whilst colouring parity doesn't seem to help, trying to find an arrow with such a weakness is the next in, and that's speed bump #2.
Box 2 has no parity question, box 9 has nothing pointing at it, neither really does box 8 - the only possible parity break in is in box 1. The [37] connected to column 9 from below and the [39] in column 9 from the right. Make those 7 and 9 respectively, and the BA breaks totally.
=> r7c9 can't be 7, and is 8
=> the box 9 BA is 3<-2-5-8, and box 8's is 2<-5-7-9
=> sudoku places an 8 in box 3, and a 6 and [79] pair in box 9, which resolves the [37] pairs in the middle band, which leads to the next phase

4) Do more sudoku
column 6 has a [35] pair, so you can place the 9 in r3c6, then make a 1<-2-3-4 BA in box 2.
box 1 needs a 9, forcing r2c9 to be a 3, completing box 3.
column 2 needs a 5, it goes in row 7, so columm 1's 5 is in r3c1, and that completes row 3, and boxes 5 and 2. And 1. And the rest of the grid.

Enjoy the singsong.


Proof of uniqueness:
--------------------
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.DoubleArrow%7ER3C2%7ER4C3%7ER5C3.DoubleArrow%7ER5C3%7ER6C4%7ER6C5.DoubleArrow%7ER6C5%7ER5C6%7ER5C7.DoubleArrow%7ER5C7%7ER4C8%7ER3C8.DoubleArrow%7ER3C8%7ER2C8%7ER1C7.DoubleArrow%7ER6C9%7ER7C9%7ER8C8.DoubleArrow%7ER9C6%7ER9C5%7ER8C4.DoubleArrow%7ER1C1%7ER2C2%7ER2C3.DoubleArrow%7ER1C6%7ER2C6%7ER3C5.Arrow%7ER3C8%7ER4C7%7ER5C7.Arrow%7ER6C5%7ER6C6%7ER5C7.Arrow%7ER5C3%7ER5C4%7ER6C5.Arrow%7ER3C8%7ER2C7%7ER1C7.Arrow%7ER5C3%7ER4C2%7ER3C2.Arrow%7ER6C9%7ER7C8%7ER8C8.Arrow%7ER9C6%7ER8C5%7ER8C4.Arrow%7ER1C1%7ER1C2%7ER2C3.Arrow%7ER1C6%7ER2C5%7ER3C5