Stitches solution and tips

I've lately played a lot of "Stitches" puzzle game on this site. Slowly going on harder ones, I'm now stuck on a "10x10/3÷ Hard Stitches" puzzle (see screenshots below). I can't see what to do to go further without doing a blind guess.

How can I solve it? What are the tricks involved?

You can find the rules on the left of the site I linked.

Here is the blank puzzle (Puzzle ID: 3,432,163):

Here is the state I'm stuck at: (See "Daniel Mathias"s answer first pic for a slightly more advanced stuck version)

• The red region requires 9 out of 10 cells to be holes (c6 turns out to be the non-hole), so you might start looking there. Commented Sep 13, 2022 at 20:57
• @RobPratt why is c6 the non-hole ? I dont see the clue Commented Sep 14, 2022 at 7:58
• I don’t have a simple explanation for why c6 is a non-hole. That just turned out to be true for the unique solution, which I found by using integer linear programming. Commented Sep 14, 2022 at 11:31
• A deduction that helps a lot: There must be1 bc stitch, which means c56 and c78 cannot both be stitches, thus all other red squares must be holes. Commented Sep 14, 2022 at 18:11
• @Retudin You mean an odd number of stitches between columns b and c because columns a and b contain an odd number ($6+7=13$) of holes? And there are only $2$ candidate bc stitches, so the odd number must be $1$. Commented Sep 14, 2022 at 20:27

Here is the detailed answer: (feel free to make it clearer)

First a bit of naming. As:

• R: the Red block
• TY: the Top Yellow block
• BY: the Bottom Yellow block
• TB: the Top Blue block
• BB: the Bottom Blue block
• TG: the Top Green block
• BG: the Bottom Green block

Part 1:

Without looking at the numbers

• Putting stitches in cells where they can only be (only 3 cells on a border between 2 blocks)
• Crossing the cells that have no borders (or only borders already used)

Part 2:

Now looking at the numbers

• Crossing borders on columns or rows that have only 1 used cell remaining (numbers on the bottom and on the right) => row 1
• Column j has 2 used cells remaining but stitches can't go vertically, so they are 2 stitches going to the column i. That means column i will have 1 used cell remaining, so stitches can go vertically in i, and so on for column h => crossing borders of columns h and i
• Row 8 has 2 used cells remaining and R/BY stitches have only one BY cell to go, so this cell is marked and the 2 other stitches will use the row 8. That means no other stitch can go in row 8 => every other cell of the row is crossed and borders between row 8 and adjacent rows are crossed as well (apart from R/BY borders)

Part 3:

• 3 cells left for the BY/BG border
• Column 8, 3 used cells remaining in 4 cells remaining, forcing cells a2 and a6

Part 4:

Considering the borders on the left side TY/TB + TB/R + R/BY + BG/BB: 4*3=12 stitches => 24 used cells total. Current state makes it impossible for other borders stitches to be in columns a/b/c, which have 6+7+5=18 used cells (numbers on the top and on the left), so there are 6 used cells outside of a/b/c for the above-mentioned borders stitches. For these 6, we can count 8 possible cells: d3/4/5/6/7/8/10 e5. As d4/6 e5 can only stitch with d5, only 1 of those 3 will be used, so we can mark the cells d3/5/7/8/10 and wait for the 3 last.

Part 5:

• 3 cells left for the R/TG border
• Column c, 3 used cells remaining, meaning there can't be both c5/6 and c7/8 stitches, meaning either c6 or c7 will be crossed. However, the Red block have 10 cells and 9 must be used (3 with the TB, 3 with BY and 3 with TG blocks). So all Red cells that are not in column c are used.

Part 6:

• Column d only has 2 used cells remaining and 1 of those will be stitching with d5, so only 1 used cell in the rest of the column => crossing d1/2 border
• Row 5 only have 1 used cell remaining and it must be used for the missing for the missing TB/R stitch, also meaning d5 must be a TB/R stitch => stitching d5 with R block and crossing g/h/i/j5 (forgot to cross c4/5 border in this picture, fixed in the next one)
• Row 10 only has 1 used cell remaining => crossing h/i10 border
• Rows 9/10 only have a total of 4 cells remaining and 2 of them must be used for the BY/BG border, the 2 others can only be i/j9

Part 7:

• Column e has 3 used cells remaining and 4 cells remaining => e2 is marked
• Column f has 2 used cells remaining and 1 of those will be used to stitch with f5 => crossing other borders
• f5 is the only used cell remaining in row 5 possibly stitching up. Let's count total used cells above to know where it stitches : 1+4+5+3=13 => odd number meaning 1 stitch between rows 4/5, which already is e4/5 stitch, so stitching f5/6

Part 8:

• e6 is forced to stitch down
• Row 4 remaining cells force b/c4 stitch, row completed
• Column b has 1 used cell remaining => cross borders
• Column c has 2 used cells remaining and it must be for either c5/6 or c7/8 stitch => crossing c2
• In part 2, we said there must be a stitch between rows h/i and it can only be h/i3

Part 9:

• Row 1 has 1 cell remaining => stitching g1/2
• Column f has 1 cell remaining => stitching f/g7
• Last TG/BY stitch can only be i/j6 => row completed
• It forces a5/6, a/b7 and c7/8 stitches => rows 5/7/8 completed

Part 10:

No need to explain

This puzzle type is new to me. While I cannot (yet) provide a full logical path to the solution, I have some initial observations that may help you get unstuck:

The region D9-H10 must have three stitches. The F9-F10 stitch is forced, as is one hole in each of the other two, as marked.
Column A has 3 x's marked, out of 4 total x's. An x on A2 or A6 would force a fifth x in the column, so both of these must be holes.
Similarly, column B has only two more x's. You may be able to determine where some of the other holes in this column should be.

With a bit of trial and error (which I generally avoid, but this is new to me and needs to be explored a bit) I was able to complete the puzzle. I may return later with a fully logical path.

• Good one, I can't accept it yet tho, not logical until the end Commented Sep 14, 2022 at 8:11