6
$\begingroup$

enter image description here

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.

$\endgroup$
5
  • 3
    $\begingroup$ What do the X and V symbols mean? Please can you provide a link to the source of this puzzle including the rules for solving it? $\endgroup$ – Rand al'Thor Apr 4 '20 at 20:13
  • $\begingroup$ X means two adjacent numbers add up to 10, and V means they add up to 5. $\endgroup$ – classicalMpk Apr 4 '20 at 20:17
  • $\begingroup$ @Randal'Thor further, note that every possible X and V are placed already, giving the additional information that there is an X or V iff the two numbers adjacent sum to 10 or 5, respectively. $\endgroup$ – El-Guest Apr 4 '20 at 20:39
  • $\begingroup$ And the sum of two adjacent numbers can not be 5 or 10 if there is not the sign X or V $\endgroup$ – perayu Apr 4 '20 at 21:08
  • $\begingroup$ Did you create this puzzle yourself, like the previous ones, or did you find it somewhere? $\endgroup$ – Rand al'Thor Apr 4 '20 at 21:31
7
$\begingroup$

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

enter image description here

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).

$\endgroup$
3
  • $\begingroup$ Congratulations ! It's all good... $\endgroup$ – perayu Apr 4 '20 at 23:16
  • $\begingroup$ Yes, this is the correct solution. The trick to start is to understand that the 5 in row 8 is forced as is the 28 x-wing in rows 8 and 9. Once you have propagated all you can in the 6 bottom rows, you have to start using the rule that you can't have 5 or 10 sums that are not indicated on the grid. Good luck! $\endgroup$ – Stephane Apr 5 '20 at 9:11
  • $\begingroup$ @Stephane Yes, I've basically used the same way, first the obvious numbers written in light blue, then the 5s (pink), then 3s and 7s in columns 4 and 6 (purple), etc. $\endgroup$ – trolley813 Apr 5 '20 at 9:26
5
$\begingroup$

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:

system of rows, columns, and boxes

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.
Step 1

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 2

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 3

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 4

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.
Step 5

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 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

Step 19/solution:

Hidden singles all the way to the end.
Step 19

Congratulations on reaching the end!

$\endgroup$
1
  • 1
    $\begingroup$ Nice! (And I didn't need to pull up a chair since I was already on a chair 😃) $\endgroup$ – justhalf Apr 3 at 5:24

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.