Let's use SET to analyze the highlighted area.

• We can see that Green = Red + {1, 2, ..., 9}

• The arrows on the first row and second row can cancel out. Now, even though the sets of digits are different, but the sum should still be the same.

  • SUM(Green') = SUM(Red' + {1, 2, ..., 9}) = SUM(Red') + 45

• The remaining Red' cells are all in the two digit pills. Because their arrows are three cells long, we know the pills must be 10+x and 20+y.

• Numbers in the pills equals to sum of digits along the arrow, therefore

  • 10+x + 20+y = SUM(r{4,5,6}c{1,9})

  • SUM(Red') + 27 = SUM(r{4,5,6}c{1,9})

• Let's remove r{4,5,6}c{1,9} from Green', and call them Green''

• SUM(Green') = SUM(r{4,5,6}c{1,9}) + SUM(Green'')

• SUM(Red') + 45 = SUM(r{4,5,6}c{1,9}) - 27 + 45 = SUM(r{4,5,6}c{1,9}) + SUM(Green'')

• SUM(Green'') = 18

The minimum of Green'' is indeed 18. So they must all be 1,2,3 triples.

In box 2 and box 9, the two digits long arrow must be 4+5=9

The arrow starting from r7c9, it must be 1+2=3

Also notice that r3c2 and r6c8 must be different, so r6c2 cannot be 1 or 2

r8c2 must be 4 or 5

Because r8c4 cannot be 9, r7c3 can only be 4 or 5, and r8c4 can only be 7 or 8.

For the pills, one of them is 20+y, it can only be 23 or 24.

• In either case, 8,9 must be on its arrow.

Therefore, the red pill is 20+y, the green pill is 10+x.

If we check the candidates on c9, we can see that 4 cannot be in the red pill.

Therefore, the red pill must be 23.

And r1c9 arrow must be 1+6=7

2 and 3 must go to the arrow in box 1. So the arrow must be 5=2+3.

The remaining digits on c1 are 4,6,7,8. And 4 must be on the arrow of green pill. Otherwise 6+7+8=21, but the green pill is 10+x.

Threrefore, r6c2 must be 3.

If we fill in the candidates in box 6, we can place 3, and the other cells are 4,5,7.

This forms a 457 triple on c8. r7c8 is now a naked single, it's 6.

One of r{8,9}c3 is 6. ==> r1c3 is 4 ==> r7c3 is 5 ==> r8c4 is 8

The rest of the puzzle should be straight forward to solve.