The maximum sum of length 3 arrow is 27 (9 + 9 + 9). Therefore, the yellow cells must be 1 or 2.

The minimum sum of length 3 arrow is 6 (1 + 2 + 3) if all cells on the arrow see each other. Therefore the green cells must be 6, 7, 8, or 9.

If the r9c5 arrow sums to twenty something, then it must be 23 or 24. In this case, r9c{1, 6, 7, 8, 9} can only be 6789, which is impossible.

Therefore, r8c5 must be 1.

The same argument applies to r5c8.

Now we can get a bunch of 1s and 2s on the grid.

The red cells satisfy: x = a + b + c

The blue cells satisfy: 10 + y = d + e + f

Therefore, we must have:

x + a + b + c + y + d + e + f is even.

Therefore, r9c9 must be odd. Because all digits on a row sums to 45.

The same argument applies to r1c9.

Therefore, r1c9, r9c9 is a 79 pair.

r1c9 can't be 7, otherwise we need to put a 2 on the arrow, which collides with the 2 in r1c5.

Therefore, r1c9 is 9, r9c9 is 7.

r5c2 must be 3 or 4. Therefore, the arrow must be 689 or 789, there is always 8 and 9 on the arrow.

r9c1 must be 6 or 7, and it must be 6.

The red and blue arrows satisfy the following equation:

10 + x = a + b + c

Therefore, 10 + x = (a + b + c + x + 10) / 2.

By substituting x, a, b, c into the formula, we can find x for red and blue arrows.

Notice that there is a 1358 quadruple in box 3.

The remaining digits in c1 is 1345. And we have 135 triple in r1, therefore, r1c1 is 4.

The 1358 quadruple gives us an 1 in box 3, and we can remove 8 from r4c9.

The 1 in box 5 can only be r6c6.

There is a 35 pair on row 4.

r4c{3,4,5,6} is on the same arrow clue, which satisfies 10 + x = a + b + c.

To make the equation work, there must be 0 or 2 or 4 even numbers in {x, a, b, c}.

• Obviously it can't be 0 even numbers.

• Having 4 even numbers doesn't work either, if we apply formula

  • x = (10 + x + a + b + c) / 2

• Therefore, there are 2 even numbers in r4c{3,4,5,6}, and r4c{7,8} must both be even.

We found a 248 triple in box 6. And 8 must be r4c7 or r4c8. By using the formula x = (10 + x + a + b + c) / 2, we can find that r4c{7,8} must be a 48 pair, and r4c3 must be 7.

Now we have r6c9=2, r8c9=4, r7c3=4, r6c4=4.

And the remaining cells in box 6 are odd numbers.

One of r5c4, r5c6 is 8.

r5c5, r6c5 must be odd numbers. r6c7 is an odd number as well.

Therefore, r5c6 must be an odd number, which is not 8.

Now we have r5c4=8. {r5c5, r5c6, r6c5}=357 triple ==> r6c7=5, r4c9=3, r4c1=5

One of r{7,8}c7 is 3. But it can't be r7c7. So it must be r8c7.

Now 3 must be on column 2 in box 7, which disambiguates 36 pair in box 4.

r1c{2,3,4} is a 678 triple. And 8 must be on the arrow, otherwise the maximum sum is 6 + 7 + 9 = 22, which doesn't work.

r1c3=8, r1c2=7, r1c4=6 ==> r2c2=9 ==> r2c5=3

r7c7 must be 2 or 6 ==> r7c8 or r8c8 is 8. The rest of the puzzle is almost a normal Sudoku, should be quite straightforward.