Consecutive Decomposition

The green area is composed by a few consecutive pairs / triples.

And, we know that every digits in the red area will appear in the green area.

Can r6c6 be 6, 7 or 8? ==> NO

• If r6c6 is 6 7 or 8, then there will be an 1 on both arrows (r6c{7,8,9} and r{7,8,9}c6).

• There will be three 1s in the red area, so there will be three 1s in the green area.

• Because 1 must be next to 2 in the green area, we can show that it's impossible to put 3 ones in the green area.

Therefore, r6c6 is 9.

• And at least one arrow is made by {2, 3, 4}.

So, there are two 9s in the green area. Both of them need to be next to an 8.

But, there is only one 8 in the red area.

The only way to make it work is placing {9, 8, 9} in the yellow cells.

Because there are white dots on the "9" arrows, both of them contain a 2.

Therefore, there must be a 2 in the purple cells in box 5.

By counting the number of even digitis in box 5, we know that r5c5 must be 2.

Now the purple cells form a {1,2,3} triple, we can't have {2,3,4} on both "9" arrows.

Therefore, one of them is {1,2,6}, the other one is {2,3,4}.

All digits in the red area are:

• {9}

• {1,2,6}

• {2,3,4}

• {1,2,3,4,5,6,7,8,9}

In the green area, we have used,

• {9, 8, 9}

• {1,2,3}

If we consider numbers of odd and even digits we need to place in the green area, we can conclude that the middle cells of blue cells must be odd digits.

If we consider the numbers we need to place in the green area, then the only way to place them is,

• {1,2}

• {6,7}

• {2,3,4}

• {4,5,6}

There are two ways to put them into the green area. And they are mirrored of each other by the diagonal marked in the image.

Now we know that 2s in box 2 and box 4 are either in purple cells or blue cells.

2 in box 1 is locked in 2 cells.

We can't place 2 in r1c3, because the arrow will break.

By knowing where 2 goes, we can place some digits.

1s in row 4 and row 5 force r{7,8,9}c6 be {1, 2, 6}, and we know the order.

r6c{7,8,9} must be {2, 3, 4}.

We can write down more pencil marks.

Now the highlighted arrow is very restricted.

• The circled cell (r7c4) can only be 3, 5 or 7.

• r8c4 can only be 3, 4, or 5.

We can eliminate 3 from r7c4.

r8c3 can only be 1 or 3.

• No matter which, r8c4 must be 4 to make the arrow work.

We can fill in more digits and pencil marks.

We found a {4,8} pair in box 7, which force a {5,9} pair in box 1.

We found a {6, 8} pair in box 1.

Therefore, r2c2 must be 4 or 6 to complete the arrow, and 6 is impossible.

Therefore, the arrow is 2 + 4 = 6.

More digits are placed in.

It's interesting that r2c8 is a naked single.

The 6 in box 8 forces the 6 in box 9 to row 9.

Therefore, the 6 in box 3 must be r3c7.

This also force r4c8 to be 6.

Due to the 78 pair in row 2, r1c8 can't be 8.

On column 8, only r5c8 can be 8.