Compoundoku wants REALLY BIG!

Question

Rules of Compoundoku:

• Solve both left and right Sudokus.
• In addition, the board below them is the Compound Board of both Sudokus.
• Each number on the Compound Board should tell either: (1) the number on the left Sudoku, or (2) the sum of both numbers on the left and right Sudokus; in the respective position.

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^{Credit to @athin for the original idea of this sudoku variant.
The original and 'BIG' version can be found here and here.}

Answer 1

SOLUTION

REASONING

The first step is to address the smaller and larger numbers in the compound grid first. The compound 1 has to correspond with a 1 in the left grid. The compound 2's in the grid must either be 2's in the left grid, or 1/1 (a 1 in both the left and right grids). Since there is already a 1 in the fourth row/third column of the left grid, this forces the compound 2's in the middle-left and center squares to be 2's in the left grid.

Similarly, the two compound 18's must be 9/9. Now look at the compound 16's in the bottom middle square. Since there are three of them, they must be 7/9, 8/8, and 9/7 in the left/right grids. But there is already a 9 in both right columns, so this one must be 8/8. Using similar logic, the compound 16's in the lower left and lower right squares must also be 8/8. Moving to bottom-center, the compound 12 in this square must be 6/6, since 789 are already obligated in this square both grids. The same logic then forces the compound 10 in this square to be 5/5. At this point, our grid looks like:

Some additional small deductions:

Note that the compound 2 in upper right square cannot be 1/1. If it were, then neither compound 3 in the lower right square could be either 1/2 or 2/1, forcing both to be 3's in the left grid. So we must have a 2 in the left grid in this location. In the middle-left square, the compound 3 must be a 3 in the left grid, since it can be neither 1/2 nor 2/1. This then forces the compound 3 in the upper left square, upper left cell to be 1/2.

Things just start taking off:

Compound 12 in upper-middle must be 6/6, since all other sums require 5, 8, or 9 which are all barred. Compound 3 in middle-right square must be 3 left since it can be 1/2. Compound 10 in lower right must be 7/3: all other pairs of summands are blocked by Sudoku rules, and we cannot have 3/7 since one of the compound 3's in this square must be a 3 in the left grid. Center square, center cell has to be 6, since bottom-middle must be 8 or 9. Compound 7 in bottom left square, bottom right cell must be 3/4. This forces the compound 3's in the lower-right square, and then normal Sudoku logic lets us finish the bottom row in the left grid. We then use the compound grid and normal Sudoku rules to finish the bottom row of the right grid. The grid at this point is:

The rest of the solution is similar to the previous analysis, just bouncing standard Sudoku rules against the possibilities for the compound grid...I don't remember there being any major logic leaps at this point.