The minimum of the sum of the green cells is 45 + 1 + 2 = 48

The maximum digits can be placed in the circles of the green arrows are 7, 8, 9, which sum to 24, which is exactly half of 48.

Therefore, those 3 green arrows must have 7, 8, 9 in the circle. The green cells on column 4 are 1 and 2 in some order.

The same logic applies to the red cells.

The total of green cells is at least 17 = 10 + 7 (4 cells in column 5, 2 cells in row 5)

The total of green cells is at most 17 = 8 + 9 (the circle of the arrows are both in column 5)

Therefore, the arrows must be 8, 9 pair in the circles, and all green cells must be as small as they can be (12 pair or 34 pair)

One of the red cell is 3

• In this case, the green cells in the same column must be 45 pair, and the corresponding yellow circle must be 9.

• Therefore, one of the yellow circle is 9.

The blue cells see both yellow cells.

• Therefore, there is no 9 in either of the blue cells.

The sum of green cells equal to ((2 * yellow cell) + blue cell)

The minimum of the sum is:

• 1 (row 6) + (21 - 4) (row 7) + 6 (row 8) = 24

Therefore, the yellow cell cannot be 8, thus must be 9.

• If it is 8, the total is 2 * 8 + 7 = 23, which is not enough.

Therefore, the yellow cell must be 9. The green cell in r6 must be 1, 2 or 3.

• The total is at most 2 * 9 + 8 = 26

The same logic applies to the red cells.

r1c3, r9c7 must be different.

The green cells in box 2 sum to 9.

• Because they cannot contain 1 nor 2, it must contain 3 or 4.

The green cells in box 2 need to appear in the box 8.

It cannot be in r{8,9}c4, otherwise r7c5 will collide with r3c5.

Therefore, the green cells in box 8 must be on column 5.

The same logic applies to the red cells.

• They sum to 9, in box 2, they are on column 5.

The marked yellow cells can only be 7, 8 or 9.

• The yellow cell in box 5 must be 7.

We know that r1c3, r9c7 must be 7 and 8 in some order.

If we hide the arrows on green and red cells for a moment, then we can see that the board has rotational symmetry.

Therefore, if we ignore the green and red cells, then we can "guess" r1c3 is one of 7 or 8. And when about to put something that interact with the arrows, in such way that only one direction will work, then we will know the direction of the arrows.

So before we start, let's remind ourselves:

• The green and red cells are two arrows, either both of them pointing up, or both of them pointing down.

• We also know that the middle of the green cells and the middle of the red cells must not be 7 or 8.

Let's begin...

Let's put 7 in r1c3

Now the green arrow is 8, red arrow is 7. And r9c7 is 8.

I still pencil marked the 8 and 7 in green and red area. Note that they cannot be in the middle cells.

Remember that we marked some yellow cells that can only be 7 8 or 9 in STEP 5? We can make some progress now.

We know that there is a 34 pair on the red arrow, and a 34 pair on the blue green arrow. And we know the green arrow is a 9 arrow, the red arrow is a 8 arrow. So we can place the 1 and 2 in box 5.

The total of digits on yellow arrows are 7 + 9 * 2 = 25.

• According to the computation in STEP 4, the yellow cell in box 6 must be 1.

Consider the 7 arrow in box 6 (currently we represent it as a killer case, because we don't know the direction). 

• It cannot be a 16 pair because r4c9 is 1

• it cannot be 34 pair because there is a 34 pair in box 5 pointing to it.

• Therefore it must be a 25 pair and r5c7 must be 2 or 5.

Let's check the remaining cells in column 7. They are 1,3,4,6.

Therefore, r7c6 cannot be 2, otherwise the 8 arrow is broken.

We know resolve the 12 pair in box 2 and box 8.

The green 7 arrow now must have a (1,5) or (2,4) in column 3.

The green 9 arrow cannot be (2,2,5), otherwise it breaks the green 7 arrow.

Also note that we cannot place 1 in the 8 arrow in box 4 either. Otherwise it will need a 7 on the arrow.

The green cells forms an X-wing of 1s.

So the red cell cannot be 1 any more.

Therefore, the cell below the red cell cannot be 7 (r3c8 cannot be 7).

We can fill-in more 7, 8, 9 in the grid now.

On row 7, 5 can only be in the yellow cell.

Also consider the other yellow cell, we know that the red cell must be 1 or 2.

Remember the X-wing of 1s in previous step? The red cell must be 2, and the other yellow cell must be 2 as well.

Because the 1, 2 in the box 4, the green cells (remember that it is an 8 arrow?) must be 3,5,8.

This resolves the 7 arrow above it.

And then, we can very quickly fill in more cells.

We should end up like this.

Note that the direction of 3,5,8 is forced, and it's not the direction we want...

We can place the 7 arrow in box 6 in the same direction, and all cells will be fill in. 

We can place the arrows back, but in another direction.

Now, if we rotate everything 180 degree, we will get a solution for the original puzzle.