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Please find below a variant Sudoku which incorporates elements of Star Battle. The Sudoku is a standard irregular Sudoku, and the clues outside the grid are sandwich clues: the sum of digits strictly between the 1 and the 9 in the appropriate row/column must add up to the sandwich clue. In addition, the 1s and 9s in the grid form a solution to a Star Battle with 2 "stars" using the Sudoku regions; in other words, we have the additional rule that no 1 or 9 can be adjacent to any other 1 or 9, neither orthogonally nor diagonally. I hope you enjoy!

Solve online, but note that the Star Battle adjacency constraint on 1s and 9s is not enforced.

Grid

HINT

After getting the "easy" first 25 digits, there is a fairly elegant argument that setting R6-7C5 to 4/6 leaves no place for a 4 in row 8.

SECOND HINT

After applying the first hint and some follow-on Sudoku logic, the 3s and 4s become constrained in the lower rows. You should be able to show that if the 11-sum in the first column is 4/7, you get a quick contradiction in R3-4C1-2.

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  • $\begingroup$ Small clarification, can a 1 be adjacent to a 9, but not another 1, or can a 1 not be adjacent to a 1 or a 9? $\endgroup$ Jul 26 at 15:14
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    $\begingroup$ @BeastlyGerbil 1s cannot be adjacent to either 1s or 9s. 9s cannot be adjacent to either 1s or 9s. $\endgroup$ Jul 26 at 15:16
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    $\begingroup$ Ahhh, now that makes more sense, I've been stuck for half an hour :P $\endgroup$ Jul 26 at 15:16
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    $\begingroup$ "standard irregular sudoku" LOL! made me laugh $\endgroup$
    – Ankit
    Jul 26 at 16:53
  • $\begingroup$ Can you confirm that you are know that there is a solution? I am pretty sure I found a contradiction. $\endgroup$
    – Ankit
    Jul 26 at 18:24
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(This answer is to show the logic after the initial 'simple' first 25 digits, after which it gets much harder)


So, we have our initial 25 digits, as shown in @Amorydai's answer. The grid looks as follows:

enter image description here

Now the next step of logic, as indicated by the hint, involves the 10 sandwich in the middle column:

The 10 can be made up of two pairs, either 4/6 or 8/2. Lets see what happens if we make it a 4/6:

enter image description here

The 4 leaves only one place a 4 can go in C8. This 4 removes a candidate in the same box in R7, leaving the only 4 in row 7 as the highlighted red cell.

However, now take a look at R8. Every single free cell in R8 is either in the same column, or the same box as a 4, leaving no place for a 4 in R8. Therefore, the 10 pair MUST be the 8/2.

With a bit of normal sudoku logic we get to this point:

enter image description here

The 6 in R7 could only go in one place, as could the 6 in R8. There is now 3/4s in the bottom two rows in 2 boxes, meaning the bottom left box has a 258 triple, and the 8 in C8 must be at the bottom. There are a couple of 3/4 pairs, and some row restrictions towards the bottom.

Now looking at the 11 pair in the left column:

There are two options, it could be 7/4 or 3/8 (or 8/3). Lets see what happens if it's 7/4:

enter image description here

The cells highlighted in yellow are affected and the 3s and 4s in the bottom rows are mostly solved.

However, the interesting bit is C1 and C2. Both have 3 numbers remaining, 2/5/8, highlighted in blue. The problem is the red cell, no matter which number it is, it covers all 3 blue cells in the box below, meaning whatever the red cell is, the box below would not have a place for that number. Hence the 11 pair cannot be 7/4 and is 3/8 or 8/3. And as there is a 2/5/8 triple in the bottom left box, it must be 8/3.

Using these deductions, and with some sudoku logic the digits start to fall into place:

enter image description here

The 3 has a domino effect for the 3s and 4s towards the bottom. The right hand column can be solved, as can the 6s towards the top few rows. The logic gets easier from this point as more numbers are placed.

From here the puzzle can be solved fairly easily:

enter image description here

By going around multiple rows or columns can be solved, and many pairs are established. The bottom right box can be solved now there is a 4 in C6, and solving C6 will pretty much give us everything needed to finish off the puzzle.

Solving with normal sudoku logic from here leads us to the answer:

enter image description here

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Well, I solved it, but now that I go back and look at my notes I am a little confused at some of my deductions. I did keep a lot of the clues in memory, as the program provided doesn't allow for much note-taking, so the process may not be obvious from the pictures. I will provide my process at the 3 places I got stuck. Step1:

First step was to solve the star battle. This step was pretty easy. Row 9 and Column 2 have sum between 1 and 9 as 29, which means there has to be either 5 or 6 other numbers in between. And row 4 has a sum of 31, so there has to be exactly 6 other numbers between. Solution looks like this:
enter image description here

Step 2:

With the help of the sums, I got to the 25 numbers that other commenters mentioned:
enter image description here

Step 3:

I spent a couple of hours here. Ultimately what I had to do was guess at a choice between 2 placements and get a contradiction, this way I would know that the number belonged in the other cell. I had to do this 3 times.
First the 6 could only be in R4C3 or R6C3. I chose R4C3 and got a contradiction. enter image description here Like I said, it's hard to follow this pic because some of the key information is not on the board, and having solved the puzzle over several hours I don't exactly remember what all was goin on at this time. But from this step I determined that the 6 belonged in R6C3.
Next, another 6 is either in R7C8 or R8C8, choosing R8C8 leads to a contradiction, so I determined the 6 belongs in R7C8.
enter image description here
Finally, last guess is 7 between R2C1 and R5C1. Guessing R5C1 leads to a contradiction so R2C1 it is.
enter image description here

Finally, after these guesses I used normal sudoku rules and got the end result:

enter image description here

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    $\begingroup$ At your step 2 image, options for 6 in R7C5 and R8C6 rules 6 out of R7C6. This creates a 567 triple in the lower right box with 8 column clue (column 7). This creates a 234 triple in that box that looks at R7C8. $\endgroup$
    – LeppyR64
    Jul 27 at 11:37
  • $\begingroup$ This is of course the correct answer, but there is a logical path through. I'll add a hint to the puzzle, and encourage you and others to spot the logic. $\endgroup$ Jul 27 at 12:57

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