• The total sum of the green cells equals to two times the 3 circle cells.

• Therefore, the maximum sum of the green cells is 48,

  • which equals to 2 * (7 + 8 + 9)

• The sum of digits of entire column is 45. It means that the green cell in column to can only be 1, 2, or 3.

• But it cannot be 2, because the sum of the green cells must be an even number.

• The same logic applies to the red cells as well.

• Therefore

  • R5C{2,8} is a 13 pair.

  • For the circles in column 1 and 9, one column is 689, the other column is 789.

  • When the column is 689, in this case, the extra cell in row 5 must be 1.

  • When the column is 789, in this case, the extra cell in row 5 must be 3.

• We cannot place 7 in neither r4c1 nor r4c9.

  • Because, if we put 7 in either of the circle, the extra cell on row 5 must be 3.  But a 3-cells-7-arrow without repeating digits must be 1,2,4. This is a contradiction.

• Next we can look at r5c1. Obviously it cannot be 7, 8, 9. But it cannot be 6 either. If it is 6, r5c2 must be 1. But, in this case, the arrows in column 1 will be 6, 8, 9. And we will have repeated 6s in column 1.

• Therefore, r5c1 can only be 2, 4, or 5. The same logic applies to r5c9.

• Now, if we look at the arrow in the middle of box 5 (r5c{4,5,6}). Not it must have a 2 on it.  Otherwise, the arrow can only be 9=4+5, and both r5c1, r5c9 must be 2.

• Now, r5c1 and r5c9 forms a 45 pair.

• The arrow r5c{4,5,6} must be 8=2+6 or 9=2+7.

• r4c{1,9} cannot be 6 now. And they form an 89 pair.

• The 8 and 9 in box 5 must be in r{5,6}c4.

• We know the sum of the green cells is 17 = 8 + 9.

• The green cells on r5, according to our pencil mark, sum to 13 (1+3+4+5)

• Therefore, the green cells on r6 sum to 4. Therefore they must be a 13 pair.

• The minimum digits we can place on the arrow r6c{5,6} is 4 and 5, which sum to 9. The 8, 9 pair in box 5 is resolved.

• If we look at column 5.  The green cells must sum to some even number. The yellow cell must be an even digit. Therefore, the red cell must be odd, i.e. 5.

• The yellow cell cannot be 2, otherwise the green cells have to sum to 38, and the circles have to sum to 19, which is impossible.

• The yellow cell must be 6, and the green cells have to sum to 34, and the circles have to sum to 17, and they must be a 89 pair.

• There must be a 6 or 7 in r1c1 or r1c9.  Otherwise they will form a 89 pair, and break r1c5.

• Let's assume that one of them is a 6. Then the other one cannot be 7. Otherwise the yellow cells have to sum to 19.

• Can the 6 be paired with 8? The answer is no. Otherwise, the green cells sum to 28 = (6+8) * 2. And the yellow cells have to sum to 17, which can only be a 89 pair.

• Therefore, if there is a 6 in r1c1 or r1c9, then r1c{1,9} will be a 69 pair. And the yellow cells have to sum to 15, and they must be a 78 pair.

• And, if we consider the arrows on column 1 and column 9, r9c{1,9} must be a 78 pair. And the red cells ALSO sum to 15.

• In conlusion, even though we don't know which row is 69 pair, we know that the yellow and the red cells each sum to 15. And r{1,9}c6 forms a 67 pair.

• The yellow cell cannot be 7 anymore. It must be 3.

• r4c5 must be 1, r4c4 must be 7.

• Now r4c5 is 1, r9c5 must be 9 = 2 + 7.

• Therefore, r9c5 is 9, r9c6 is 6. And we can resolve a lot of things.

• Because the arrow on the green cells is 3 = 1 + 2. The red cell cannot be 6.

• And we now resolved the parity of 689, 789 arrows in column 1 and 9.

The rest of the solution path should be straight forward.