TL;DR - it's a puzzle: In a previous puzzle, we presented a puzzle inspired by the recent Japanese Sums puzzle series at Grandmaster Puzzles which combined the Japanese sums cluing technique with a Sudoku, with a little bit of Cross the Streams thrown in for flavor. This puzzle makes this hybridization stronger by adding in the path constraints of a Cross the Streams puzzle, as well as more strongly using Cross the Streams cluing tricks like the question mark and asterisk.

The grid below is to be filled out according to the usual Sudoku rules, but with additional constraints:

Japanese Sum constraints - the clues given on the outside of the grid describe the pattern of odd digits in the corresponding row/column. A number indicates the sum of a complete contiguous block of odd digits in the grid. A question mark indicates that a contiguous block of odd digits exists, but does not indicate a sum. An asterisk implies that any number of blocks of odd digits may exist, including zero. The clues are provided in the correct order (left-to-right, or top-to-bottom), and there is at least one even digit between each pair of clued odd-digit blocks.

Cross the Streams constraints - the cells which contain an odd digit form a single orthogonally (with an edge, not just diagonally) connected whole, with no 2x2 block of cells all containing odd digits.

Solver notes: PLEASE, if you are providing a solution, be prepared to present the key logic points, ideally with diagrams of partial progress. This puzzle has what I think are some particularly delicious logic steps...either that or I missed obvious stuff in my test solves :-)

I hope you enjoy!

enter image description here

Penpa link...sorry I forgot it at first!

  • $\begingroup$ Spent a decent amount of time on this yesterday, but got too sleepy before I could finish it. Super fun puzzle, thank you for creating it! $\endgroup$
    – kristinalustig
    Commented Aug 31, 2021 at 16:10
  • $\begingroup$ @kristinalustig Glad you're enjoying, and glad to see you around! $\endgroup$ Commented Aug 31, 2021 at 16:19

1 Answer 1


Final grid:

enter image description here

(Click on any image for a larger, high-res version)

I will clue all completed odd digits as green to make the path a bit clearer.

Starting off near the middle

Row 6 has 4 clues, meaning that the 5 odd numbers will be split into 3 singles, and 1 double. The 6 also must be the double, and must be 1+5. Due to the groups having to be split, there are 3 cells the 1/5 pair can be in, and hence the middle cell MUST be 1 or 5.

However, this cell intersects the column with 7, 17 and 1. This means that there is a 1 towards the bottom of the column, and the cell therefore must be a 5:

enter image description here

Now moving towards the right

Column 8 is interesting as the pairs can only be in 1 way, 3+7, 1 and 9+5. Now the 9/5 can only be on the bottom two cells, and due to the 6 in row 6, the 1 can only be in row 5. This now gives us the pairings for row 5.

In the top right, the 5 can be in one of two places. However, if it in C7, there would have to be a gap to the left which would trap the odd number, so it must be top right with an odd number below. As there is a 6 in row 9, then it must be a 1 below. Below the 1, must be even and this can only be 8. This leaves only one place for the 6 and 9 in this box as well.
Row 7, column 4 can currently be a 3 or a 9. However, if it a 9, there is no room for the 1+3+5 required, meaning it cannot be a 9, and the 9 in column 4 must be in row 5.

enter image description here

Lets take a look at the rows with 16

16 can be created in two ways, 1+3+5+7, or 9+7. The way the rows work with the asterisks, both are possible for all rows, at least to start with.

For row 2, we have a 1 and a 3/7 isolated, meaning this 16 must be 9+7. For row 3, we have a 9 next to a 3/7. This must also then be a 9+7, and hence the top right box can be completed.

For the bottom two rows with 16, at least one of them is 1+3+5+7, as there is a 9 in one of the rows, but due to the 7s being elsewhere in the same box, cannot be next to a 7.

In the middle right box there is now only one place a 5 can go. This still leaves both options for the 16 in that row however.

enter image description here

A few more big deductions

Looking at the top 3 rows, there must be a 1 in the top two left cells. This places a 1 in row 3 column 5, as there is a 1 at the bottom of column 4. This also completes the 6 in row 6, and places another 1 in the middle box.

In row 6, there is now only one place a 4 can be placed.

Now for a big deduction: The 16 in row 4, MUST be a 1+3+5+7. If it was a 9+7, then it would have to be in the left box, however there is already a cell which must be either 9 or 7 in the same box. Hence the 16 must be 1+3+5+7. This also means the 9 must be in the middle left box, fixing a 7, otherwise there are too many evens in the box. In row 6, the missing 3 and 9 can be placed in the right middle box.

enter image description here

Things are beginning to fall into place

In row 5, if the 7 is on the right, that would leave the 179 triangle isolated, so it must be on the left. This gives a 26 pair in the box, solving the 68 pair in column 8.

Now in column 9, the 3 must be under 7 or it will be isolated, and this leaves a 4 at the bottom. Next to the 3 is a 9, or a 5. If it is a 9, then 16 cannot be created, so it must be a 5, solving the 95 pair.

This gives us the two 16s. The bottom 16 must be a 1+3+5+7, as a 7 cant be next to the 9, meaning an 8 must be in that cell instead. To prevent isolation, the 1 must be next to the 5, and a 7 next to that 5 and this solves the 16 in row 8.

enter image description here

Nearly solvable now

In row 7, the first 9 must be a 1+3+5. The 3rd cell must be the 3 as it cannot be 1 and 5 must be isolated in that column. This solves both 35 pairs further up in the grid, and in turn the 135 ordering.

There is now a 1 fixed in the top left cell, completing the 1s, and there must be an 8 to the right of it. In the 1st column, there is only one place for a 4, and that places a 9 in row 4.

The bottom 16 now must also be 1+3+5+7, and the top one a 9+7. The bottom 16 must have a 7 in the 3rd cell, as there is a 35 pair in the middle box.

enter image description here

You can now reach the following using normal sudoku rules, and solve after with the logic given:

enter image description here

To prevent isolation of the 739 in the top middle, the 5 must be in column 5. The 9 at the bottom must be above the 7 or will be isoalted. From there, everything falls into place.

  • 1
    $\begingroup$ Great solve, and thank you for the detailed logic! $\endgroup$ Commented Aug 29, 2021 at 18:52
  • $\begingroup$ What app are you using for the display when solving this? $\endgroup$
    – justhalf
    Commented Aug 30, 2021 at 2:30
  • 1
    $\begingroup$ @justhalf the Penpa link at the bottom of the question, useful sudoku tool $\endgroup$ Commented Aug 30, 2021 at 4:33

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