A new puzzle - Sudoku X V

Question

If someone is able to find the solution... I found 2,3,8 on line 5, column 3-4-5, but after that I am blocked, it looks too difficult for me.

Answer 1

I believe this does the job (sorry for the crude drawing):

Unfortunately, that's too hard to explain in words (it took about 1.5 hours for me to solve, and it was 1:30 am in my time zone (UTC+3) at the time of posting).

Answer 2

Pull up a chair, this is going to take a while. Especially since I keep using far too many words. There are 19 steps, I kid you not.

The coordinate system used in this answer:

Note that there are four ways to make 10 (1-9, 2-8, 3-7, 4-6) and two ways to make 5 (1-4, 2-3)

Step 1:

First, fill in the obvious X and V deductions. Also, 5 can't be part of an X-pair, so it can only go in R8C6.

Interim Observation 1:

Three of four ways to make 10 will be used for R8's X-pairs, and the fourth's numbers will fall on vertical X-pairs, so those vertical X-pairs will be of the same kind.

Step 2:

We will determine the locations of 3 or 7 in B8, B5, and R8
B8: None of the horizontal X-pairs can be 3-7, to prevent C5 conflicts. Therefore they go in R9C4 and R7C6 in some order.
B5: Neither of the horizontal X-pairs can be 3-7. Column logic disallows 3 or 7 in the vertical X-pair and R4C5. Therefore they go in R4C4 and R4C6 in some order.
R8: The horizontal X-pair in the B8 can't have 3-7. Row logic (with R9) combined with IO1 rules out the vertical X-pairs. Therefore the R8C8-9 must be 3-7 in some order.

Step 3:

Since there are two V-pairs in R5 3 must appear on one of them. It can't be in B5 (pencil marks) or R5C7 (forces a 2 in R5C6, contradiction by column logic). Placing the 3 in R5C3 gives a whole chain of numbers

Step 4:

B8 must have 4 and 6 in the same row (X-pair). Likewise for B9. B9's R9 can't have 4-6 it can't use or not use R9C7. Therefore B8's R9's horizontal X-pair has 4-6 in some order. Then, B5's vertical X-pair can't be 4-6 (row logic) so the remaining X-pair must have 4-6 in some order.

Step 5:

The vertical X-pair in B5 must be 1-9. The 9 can’t be part of the V-pair so the 1 is. After quick propagation there, we also note where the 4 and 6 must go in the bottom-right box and that either 4 or 6 goes in R3C6.

Interim Observation 2:

The two V-pairs in B3 must use 1-4 and 2-3, so none of those four numbers can be in C7.

Step 6:

A 3 can't go in R4C4 since that would make a V with the 2 in R4C5. We can place the C4 and C6 3s and 7s. In C7 3 can't be in B3 (see IO2) or the B9 (pencil marks), and there's only one place for it in B6.

Step 7:

B3 can't have 2 in R7 (see IO2). It also can't go in R7C7, as it would make a V with the 3 in R6C7. Therefore the X-pair in R7 must be 2-8, so the X-pair in R1 must be 2-8 (see IO1). In B8 we can determine which X-pairs are 1-9 and 2-8, as well as place the 2 and 8 with regular column logic.

Step 8:

B3 can't have 2 in R7 (see IO2) and B6 has no room. The 1 in C7 therefore must go in C7R7. Then in B7 the 1 can only go in R9C3 (not R9C2 as that would make a V with the 4 in R8C2). The same box now only has one spot for a 7, R9C2. I also filled in R9's and C5's remaining pencil marks because why not?

Step 9:

A 4 can't go in R7C8 or it would make a V with the 1 in R7C7. This places a 6 and a 4. Neither the 8 or the 6 in B3 can go in C7 (column logic) or on a V-pair, and with our new 6 on the bottom we can place them for sure. I also filled in C7's remaining pencil marks.

Step 10:

The horizontal X-pair in R1 can't be 2-8, 3-7, or 4-6 by row logic, therefore it must be 1-9. I also added all of R1's pencil marks.

Step 11:

The vertical X-pair in C2 can't be 1-9, 3-7, or 4-6 by column logic, therefore it must be 2-8. The 8 can't go in R4C2 as then B4 could not hold a 2. We can place the 2 and 8.

Step 12:

B3's 2 can only go in R3C7 (column logic + can't make an X with the 8). That also places the 3 by V-pair logic. Then the 4 can't go in R2C9 (would make an X with the 6 in R1C9) so it and the 1 can be placed in the V-pair.

Step 13:

R6's 8 can only go in R6C3 (column logic) and its 7 can only go in R6C1 (column logic + can't make an X with the 3 in R6C7). I also write in all the rest of R6's pencil marks.

Step 14:

B7's 3 must go in R7C2 (column logic + can't make an X with the 7 in R6C1). That creates a hidden single to place B1's 3. I also noticed that the 3 in B3 lets B9's 3-7 X-pair be resolved.

Step 15:

Place a few more 2s and 7s, mostly hidden singles but also note that R8C7 can't be a 2 to prevent making a V-pair with the 3 in R8C8.

Step 16:

A 4 in R9C5 forces one in R3C6, and now there is nowhere a 1 can go in C5 without making a V. Therefore R9C5 is 6, and this sets off a chain of easy deductions.

Step 17:

Place the remaining 4s and 6s, mostly hidden singles but at one point a 4 must avoid touching a 6 and creating an X (specifically, between R4C1 and R5C1).

Step 18:

A 9 can't go in R3C4 as it would make an X with the 1 in R3C5. This sets off a chain of deductions in the top 3 boxes.

Step 19/solution:

Hidden singles all the way to the end.

Congratulations on reaching the end!