12
$\begingroup$

enter image description here

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.

The original and easier version can be found here.
(Adding this as a note to avoid the duplicate warning when posting.)

$\endgroup$
9
$\begingroup$

I believe this is the solution :)

enter image description here

Some information on how I solved it:

The bottom two 10s and 11s are key. Obviously these can't be the left sodoku values since the largest value is 6. So they must be a sum. The top 10 also can't the sum of two 5s otherwise the 11 left of it will have two 5's in the same row. Putting that all together gives you twelve starting values. Then the 9 must be the sum of 6 and 3, left and right respectively, otherwise it will break the rules of the local sodokus. I think I made an early guess that of the three 2s, the middle one was an actual 2 which turned out correctly. I think the next biggest clue was the 5s. The rest started to fall into place from there fairly quickly. Great puzzle by the way.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Can you explain why? For those who don't get it? $\endgroup$ – Mukyuu Jan 17 at 4:50
  • 1
    $\begingroup$ Yes, just added some explanation. Thought I should get my solution in before I get sniped :) $\endgroup$ – Earlien Jan 17 at 4:52
  • $\begingroup$ Yup, this is correct! Well done! $\endgroup$ – athin Jan 17 at 10:35
2
$\begingroup$

There is no need to guess, also proves that solution is unique. L stands for left sudoku, R for the right one. (X,Y) means the X is in L, Y in R, the rest should be self-evident. Formatting of sudokus is crappy but I believe it should be clear enough.

First solve 10s and 11s. Note that 11 is 6+5, this prevents 10s in same rows (= all of them) from being 5+5. Additionally, focus on 10 in the second bottom row - that one has to be 4 in L because 5 and 6 are taken by 11s. This quickly gives solution for all 10 and 11 for both sudokus.

Then

Fill in 6 in the R, using ordinary sudoku rules. This is exactly on the position of left column top 2 of the combined sudoku, this also gives 1,1 for the other 2s. At this point we have the following image:

L/R, horizontal lines are not drawn.

x5x|6xx . . . x6|x4|xx
xxx|xxx . . . xx|xx|xx
x1x|xxx . . . x1|xx|xx
2xx|xxx . . . 6x|xx|xx
xxx|4x6 . . . xx|x6|x5
1x6|x5x . . . 1x|4x|6x

Then

Top left 5 can't be 1,2 or 5 because of L, it cannot be 1 in R, so it only has (3,2) option left. This enables filling in 5 in bottom left rectangle of the L, using ordinary sudoku rules, followed by 4 in same rectangle, 6 in top-left one, then 6 in mid left, 3 in mid left, 2 and 3 in bottom left, 4 in top left.

Solution of L/R at this point is

45x|6xx . . . x6|x4|xx
36x|xxx . . . xx|xx|xx
61x|xxx . . . x1|xx|xx
23x|xxx . . . 6x|xx|xx
523|4x6 . . . xx|x6|x5
146|x5x . . . 1x|4x|6x

Now notice that

2 is in position of 5 of combined sudoku, giving 3 for the R. The other 5 in the 2nd row cannot be 3,4 in L and 2,4 in R, so it has to be (2,3). Fill in the remaining 1 in top-left of L. The 2nd row 4 has to be 4 of the L, because 2 and 3 are blocked in both sudokus, preventing 1+3 or 2+2 combination. Then 9 has to be (6,3) because all other numbers from the L are taken.

At this point we have

451|6xx . . . x6|x4|xx
362|x4x . . . 2x|3x|xx
61x|xxx . . . x1|xx|xx
23x|x6x . . . 6x|xx|3x
523|4x6 . . . x3|x6|x5
146|x5x . . . 1x|4x|6x

Note

Top right 4 of combined sudoku has to be composite, and cannot be (1,3) because of L. This means 3 in top right of R has only one spot left. Fill in the other 3s. Then finish left column of R, then 4, bottom right 1 and 2, 2nd left column and remaining numbers.

This gives

451|6xx . . . 36|54|21
362|x4x . . . 24|31|56
61x|xxx . . . 51|62|43
23x|x6x . . . 62|15|34
523|4x6 . . . 43|26|15
146|x5x . . . 15|43|62

Only one part left now:

Top right 4 is composite and now we know it is 3 in L. This gives all other 3, and 2 in the top row with 1 in the bottom right part, then 2 in the bottom right part, 2 in mid right part. Now we know that the last number 5 of combined sudoku has to refer to the left one, because the remaining option (2,3) is not possible. This now solves the remainder of the sudoku (start with 5, then 4, then 1).

Solution is, as Earlien already found:

451|623 . . . 36|54|21
362|541 . . . 24|31|56
614|235 . . . 51|62|43
235|164 . . . 62|15|34
523|416 . . . 43|26|15
146|352 . . . 15|43|62

| improve this answer | |
$\endgroup$
  • $\begingroup$ This is an impressive work! :o Well done btw! $\endgroup$ – athin Jan 17 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.