You may have heard of a type of Sudoku called an XV-Sudoku. In such a Sudoku, cells connected with an "X" must sum to 10 and cells connected with a "V" must sum to 5. In this W-Sudoku (W is right between V and X), cells connected with a "W" must sum either to 5 or to 10. Unconnected cells cannot sum to 5 or 10. Other than that, normal Sudoku rules apply. Enjoy!

enter image description here

Inspiration: I first saw this type of Sudoku in a video from Cracking the Cryptic

  • 5
    $\begingroup$ This looks like it was inspired by a recent video on Cracking The Cryptic. I believe you should acknowledge the source. $\endgroup$
    – happystar
    Oct 25, 2020 at 5:42
  • 1
    $\begingroup$ @happystar Done! $\endgroup$
    – Jens
    Oct 25, 2020 at 15:52
  • $\begingroup$ If you were inspired by CTC, you might want to consider sending it to them. It seems up their ally, and they seem to appreciate puzzles made for/inspired by them. $\endgroup$
    – Chipster
    Oct 26, 2020 at 4:46

3 Answers 3


Step 1.

Let's start with:

Pencil marks.


step 1

Step 2.

However, the basic rule of any sudoku says:

You can't repeat numbers in the same row, column and 3x3 box.


step 2

Step 3.

However, this is not a normal sudoku. So:

We discard all the pencil marks that share a W-edge with a given number and does not sums 5 or 10. And also all the non-W-edges with a given number that does add to 5 or 10.


step 3

Step 4.

Now, the things get harder. But let's see that:

No W-connected edge can have the number 5. A 5 can't W-connect to zero (because there is no zero) nor to another 5 because this would make two 5's in the same row or column.


step 4

Step 5.

If we have three W-connected cells in the same 3x3 box, as A-B-C, we have that $A+B=5$ and $B+C=10$ (we can perhaps swap A and C). The result is that B can't be 5 or higher because this would necessarily throw A to zero or negative. We can't have $A+B=B+C$ because this would lead to $A=C$ which is impossible in the same 3x3 box.


If we have four W-connected cells in the same 3x3 box, as A-B-C-D, we can have that $A+B=5, B+C=10, C+D=5$ or $A+B=10, B+C=5, C+D=10$. However, the former is impossible, because $A \in [1,4]$ would give $C \in [6,9]$ which makes $D$ being impossible to fill and $A \in [6,9]$ makes $B$ being impossible to fill.


If we have four W-connected cells in the same 3x3 box, as A-B-C-D, then $A+B=10, B+C=5, C+D=10$, thus $B, C \in [1,4]$ and $A, D \in [6,9]$.


step 5

Step 6.

We have a:

Hidden single 5 in the 3rd row 7th column.


step 6

Step 7.

More usual chaining of single candidates.


step 7

Step 8.

The very first cell can't have a 5, otherwise, both the second row and the 2nd 3x3 box would lack a 5. This also trigger some cascading.


step 8

Step 9.

Let's pay more attention to the four-cell-W-edge chains. In the 4th row, we can't have a 3 in either the 7th or the 8th row due to the lack of a 2 in them. The remaining 1 and 4 could only go with 6 and 9. This also creates two naked pairs: 1-4 and 6-9.


step 9

Step 10.

Looking in the box with a given 2, the 8 can't be anywhere else than below the given 2, since we lack the very same 2 to pair up with it in the other two cells in the same box. This also gives a naked pair 3-7.


step 10

Step 11.

We can now look through W-edges eliminating impossible combinations.


step 11

Step 12.

Let's apply the single candidates and naked pairs down the stream:

step 12

Step 13.

We can't have 1 as part of the four-cell-W-edge chain in the third 3x3 box because we also lack the 4 between them. Removing it created naked pairs:

step 13

Step 14.

The last column in the third row can't have 1 because the cell above it can't feature a 4 or a 9:

step 14

Step 15.

Let's chain up the W-edges once again, starting by the four-cell-W-edge in the last box. Also, note that the 9 in the 6th column of the 2nd row can be eliminated as part of this.

step 15

Step 16.

9 is a hidden single in the 2nd box. This also forces 1's in the neighbouring w-edges and also creates another 9 as a hidden single in the first box.

step 16

Step 17.

Remember that two cells sharing a non-W edge can't sum to 5 or 10? We can apply this to the 8 and 9 in the middle box, to the 4 in the last box and to the 1 in the 7th box. This solves the last box. And we can use this again from the revealed 1 in the last box.

step 17

Step 18.

There are the hidden singles 4 and 9 in the 7th row and a hidden single 9 in the 9th row and we can discard the 1 below that 4, which makes the remaining 1 in the same box as a hidden single.

At this point, all the 1's, 4's, 6's and 9's would be already revealed.

step 18

Step 19.

Now, it gets something hard. I'll need an advanced technique:



Let's look at the 3rd column in the 5th row. It is either 3 (green) or 2 (blue). It forces a 2 or 3 in the cell wight above it which in turn forces a 8 or 2 the cell in the left and also a 2 or 8 above it again. Whatever is the choice, this consumes both the 2 and the 8 for the 2nd column, which leaves the 3 as a single candidate in it's 8th row and forces a 7 above it. This also eliminates a few pencil marks.

step 19

Step 20.

Let's continue with single candidates and naked pairs from the previous step. This also solves all the remaining 5's.

step 20

Step 21.

Once again, two neighbours sharing a non-W edge can't sum to 5 nor 10. So, we can use this in the 2 in the second row to eliminate the 3 in its left. Also, in the 3rd row, a 7 can't share a w-edge with either a 2 or an 8.

step 21

Step 22.

7 is a hidden single in the 3rd row. This cascades everything easily to the full solution.

step 22

  • 5
    $\begingroup$ Well done! I really like your step by step explanations which are easy to follow. For this reason, I'm giving you the checkmark. Congratulations! $\endgroup$
    – Jens
    Oct 25, 2020 at 16:06
  • $\begingroup$ @Jens. Thanks. It was a nice challenge! $\endgroup$ Oct 25, 2020 at 23:35
  • $\begingroup$ saw a alternative logic around the "hidden single 5" step: some of th 3x3 sections got only 1 cell without any W endge which are forced fives due to some earlier reasoning $\endgroup$
    – masterX244
    Oct 26, 2020 at 7:32
  • $\begingroup$ Do you normally post step-by-step sudoku solutions in text format? I think CTC's video format is much better for this. $\endgroup$
    – user253751
    Oct 26, 2020 at 13:13
  • 1
    $\begingroup$ @user253751 Normally I do step-by-step puzzle solutions because I think that they are much more understandable, specially for people who have difficulty in understanding what is going on. But for sudoku-like puzzles, this my first and so far only one in this site. I think that making a video is a lot of effort and I have doubts if it is really worth. $\endgroup$ Oct 26, 2020 at 20:47

Here's how I did it:

Step 1:

The first thing we can do is assign 5 to any cell that is not connected by a 'w' as there would be no corresponding digit to sum to either 5 or 10. Looking at the middle top 3x3 square, all the cells are connected by 'w' except for R1C6 (which is 5). Hence, 6 must go in the cell in R3C5, since its only corresponding partner is a '4'. Similar logic can be applied to the right middle 3x3 square, where the '8' must be connected to the '2'. Again, looking at the right middle 3x3 square, '7' cannot appear in cells R4C7 and R4C8 since those cells are connected by 'w' twice and 7 does not have 2 partners. '7' also cannot be in cells R5C6 and R4C9 since this would force a '3' in either R4C7 or R4C8 and '3' would require '2' as its other partner. However, '2' is already used so the cells in row 6, columns 7,8 must be 3,7 because 3 and 7 cannot appear anywhere else within the 3x3 square without causing a contradiction.


Step 2:

Trying to place a '6' in R4C9 would mean the following: R4C8 is '4', R4C7 is '1', R5C7 is a '9' and R5C6 is a '1'. However, as there is already a '1' in row 5, this leads to a contradiction. So, the only number that can be placed there is a '9'. This allows for some further deductions, resulting in the following grid:


Step 3:

In row 6, '6' and '4' must belong to columns 2 and 3 as they cannot appear anywhere else in the same row. Furthermore, the cells connected by 'w's in row 4-5, columns 3-4 must be occupied by 3s and 7s. In particular, cells in R5C3, R5C4, and R4C4 must be a '37' pair. Placing either '3' or '7' R5C3 would lead to '2' being in either R4C3 or R4C2. This means that 2 must appear in the top row of middle left 3x3 square and therefore 2 must appear in row 6, column 4. Looking at the right top 3x3 square, '9' cannot appear in R1C7 or R2C7 since those are connected by 'w' twice. It also cannot be in R1C8 or R2C8, since that would force a '1' into R1C7 or R2C7 and it needs a '4' as its other partner. However, '4' is already used for C7, so this placement leads to a contradiction. Therefore, the other deduction we can make is that 9 must be in R3C8. Additionally, this forces a '19' pair must be in R7C7 and R8C7 of the bottom right 3x3 square:


Step 4:

'8' cannot appear in R8C9 and placing a '6' would use '4' for R9C9 and '1' for R9C8. Since we know that '1' must appear in either R7C7 and R8C7, this placement cannot be true. So, there must be a '7' there instead. Using the knock-on deductions, this allows us to get here:


Step 5:

Now, look at column 6. There must be '28' and '37' connected in either the top 2 squares (R2C6, R3C6) or the bottom 2 squares (R7C6, R8C6). Trying to place '37' in the bottom 2 squares leads to the following: the top squares must be '28', the cells in R2C7, R2C8 must be 3 and 7 respectively, the cells in R1C7 and R1C8 must be '2' and '8' respectively. However, because of the '37' pair in R4C4 and R5C4, none of the cells in column 4 of the middle top 3x3 square can be a '3' or a '7'. So, we have no way to place '3' or '7' in the middle top 3x3 square and so, the bottom two squares in R7C6 and R8C6 must be a '28' pair.

Placing '28' in the bottom squares leads to the following: the cells in R2C6 and R3C6 must be a '37', the cells in R2C7 and R2C8 must be a '2' and '8' respectively and the cells in R1C7 and R1C8 must be a '3' and a '7' respectively. 2 cannot appear in column 4 of the middle top square, so this results in only 1 way to resolve the middle top 3x3 square. From there, using the 'w' connected cells, we can make a few other deductions, which results in:


Step 6:

Now, focus on the bottom left square. '6' and '4' must appear in column 1. If '4' is not part of the connected squares, then it leads to a contradiction. If '6' is not part of the connected squares, then '4' has to be connected to a '1' instead and '1' already appears in R1C5. Therefore, '6' appears in R8C1 and '4' in R9C1. This information leads to some knock-on deductions: '4' must be in R7C4 and '6' must be R9C4 by sudoku rules, and this leaves only R8C4 available for '5' and R9C5 available for '7'.


Step 7:

Next, looking at column 2, the only remaining number are '2378'. Since '28' occurs in R4C2 (and also in column 4), that means '37' must occur in rows 7 and 8. '7' is already used for row 8, so therefore, 7 appears in R7C2 and 3 appears in R8C2. This leads to placing '3' in R7C5 as it is the only place it can be and '1' must be in R8C5 as it is the only digit left. '9' must also be in R7C3 since it cannot be in R8C3 due to the '9' in row 8 and cannot be in R7C1 due to the '9' in column 1. This leads us to:


All that is left is the mass of cells near the centre-left. Placing a '7' in R5C3 will eventually lead to a contradiction (it will lead to placing '3' or '7' in R3C1, which cannot be true as those would already be occupied by the cells in R4C3 and R5C3), so we try a '3' there, and the whole sudoku is resolved from there.


  • $\begingroup$ Nice! I must admit I find it very hard to follow your logic, though. Eg, I don't see how in your very first step you can deduce that the 4 connects to a 6 (and not a 1) or that the 2 connects to an 8 (and not a 3). But good job getting the right answer! +1 $\endgroup$
    – Jens
    Oct 25, 2020 at 16:02
  • $\begingroup$ @Jens That was because after placing the '5' the rest of the squares in the 3x3 were connected by a w. So 6 must be connected to a 4, and there is only one place for it to go. Same for the '8'. I edit it later when I have the time. $\endgroup$
    – Alaiko
    Oct 25, 2020 at 22:29
  • $\begingroup$ Ahhh, thank you! I get it now. $\endgroup$
    – Jens
    Oct 26, 2020 at 0:46

Nice puzzle, I certainly bumped my head along the way quite a few times. I hope you like the steps I took to solve it, as I enjoyed figuring it out.

Step 1

5 cannot add with any other number to be 5, and any two connected 'w' tiles are adjacent, so lets start by filling in the empty 5 on r5c8, and the other 5s that follow in columns 7 and 9 enter image description here

Step 2

An interesting thing to note is that a chain of 2 'w' connections in a single block/row/column will require that the middle of those 2 connections be less than 5. Proof by contradiction, if a square connected by two 'w's in the same block/row/column has a number > 5 (e.g. 8), then it only has one corresponding number to pair up with (e.g. 2). This would force two of those corresponding numbers to end up in the same block/row/column, which is illegal.

Now follow this logic with chains of 3 connections, we can then assert similarly that the two squares in the middle must both be less than 5. This helps us get some easy possible pairs in the 3rd block of the center row. The number 2 is ruled out, and likewise it's counterparts 3 and 8 are also ruled out from this 4-square chain.

This forces the chain to be built upon 14 in the center, with the outer set being a 69 enter image description here

Step 3

Continuation of the w chains into the middle block, from r5c7, means that r5c6 must be a 14 - however, we already have a 1 in row 5, therefore, it must be a 4, which forces the rest of the sets of the chain to resolve. enter image description here

Step 4

The remainder of squares in the center left column give us an 8 and a 37 pair as we only have the options of 378 left, and we must make a w pair with one of those sets (only 37) enter image description here

Step 5

There is another 4 chain in the top right block. The 2 squares in c7 must be a 14 pair or a 23 pair, but there is a 4 in the column already, so it must be a 23 pair. Likewise, their paired tiles in c8 must be an 87 pair. This retroactively forces the 37 pair in r6 to resolve to 7 3, as 3 is no longer permitted in r6c7 enter image description here

Step 6

Row 6 has only 4 remaining numbers, a pair of which must sum 5 or 10, and the options are 2456. Of those, the obvious answer is that the 46 must pair. Given that the 4 is tied up here at the bottom, the number paired with 1 on r5 is a 9.

Additionally a late follow-up to the logic in step 1. We know that the 5 in c1 must be in r4 or r6, as no other squares in that block permit a 5. Therefore, the bottom left square must have a 5 in c3, therefore the top left square must have a 5 in c2. The only unentangled spot in c2 is r2. Additionally, there must be a 5 in r1c6, just by nature of it being the only available square. enter image description here

Step 7

The only remaining numbers on row 5 also happen to tie into a web of entangled w-connections. These numbers are 378, of which 8 is the outlier. Following that chain of connections gets us the following options. enter image description here

Step 8

r4c3 is tied twice within the same row and block, therefore must be a low number. It also forces r4c4 and r5c3 to be the same value. This doesnt do much, but it does guarantee us that a 3 is in c3 at r4 or r5. It also forces a 2 into either r4c2 or r4c3. At this point in r4 we definitely have a 2 and a 3 in the entangled squares. If we focus in on r4c5, we can see that its not a 2, 3, or 7, and must be a 5. This forces the 25 pair in row 6 to resolve. This means that r4c1 must be a 7 or 8. enter image description here

Step 9

Lets resolve r4c1 - rules state that unentangled squares cannot sum to 5 or 10. If r4c1 was an 8, r4c2 would be forced to be 2, an illegal move as it is adjacent to an 8 unentangled. This forces r5c3 to be a 3, as it is the only remaining option, and forces the whole entangled set to resolve. Note, we still have a 38 unresolved in r3c1. enter image description here

Step 10

Lets detour back to the bottom-right corner real fast. With c7, 2, 3, and 4 are already consumed. This leaves only one option, a 19 pair. It gets more interesting when we then consider that there is a chain of 4 entangled squares in there as well. This chain must use a pair of low numbers to sum to 5 as the center two squares. The only remaining low numbers being 2, 3, and 4, it must consume 23. This resolves quickly as c7 only has 1 unaccounted for number - 8. This resolution of numbers forces the empty entanglement on c8 to be 46, conveniently forcing the 49 pair on r3c8 to be just 9. Before we stop here, we also must note that there are only 3 remaining numbers in c9: 1, 4, and 6. There's also one entanglement left, and it can be resolved by 14 or 46. 4 cannot go into one of the entangled tiles, so that leaves a 4 in r3c9, and a 16 pair across the rest. enter image description here

Step 11

Back at the top left, there's an unresolved chain of 3 entangled squares; however, we conveniently can rule out 2 and 3. So the chain must be either a 146 chain or a 419 chain. We know that the 4 must be at the top, for sure. Both r1c3 and r2c3 cannot be a 46, as that would break the fixed pair on r6 (at c3). Additionally, r1c2 cannot be a 9.

So, starting with r1c3 being 14, with the given restictions, it chains out as follows: enter image description here

Step 12

Doesn't take a keen eye to see the conflict on r3c4, resolving the 19 to be a 1, as r3 already has a 9. This forces the resolution to chain backwards. This also resolves the 46 pair in r6. enter image description here

Step 13

Notably, we can also now force the 9 to be r1c1, 7 to be in r3c3, and a 38 pair in r2c1, by sudoku. The 1 and 6 also resolve the 16 entanglement in the top right. Lastly, the 6 in the top center is forced to be paired with the 4, both by sudoku and by rules of entanglement eliminating the option of 1. enter image description here

Step 14

r1c4 is now critical, and will resolve all remaining entanglements in the top blocks. Note that the only remaining numbers for top-center are 2378, akin to the top-right's entanglements. Well, c4 already has 2, 3, and 7. r1c4 can only be 8, forcing it's pairing to be a 2, and resolving the chain on the right, which will resolve the 38 pair on the left afterwards. The final empty entanglement in the top center also resolves as 37. enter image description here

Step 15

Fantastic, we're almost done. Lets turn our attention to c1 and c2's remaining numbers and entanglements. C1 needs numbers 2, 4, and 6, while c2 need 1, 3, and 7. C1 can only support the entanglement of 4 and 6, 2 forced to r7c1. C2 can only entangle 3 and 7, 1 forced to r9c2. Additionally, r9c1 and r9c2 must be entangled as well - of 4 and 6, only 4 can entangle with 1. Plus, we have a 7 on r8 which forces the 37 pair to be oriented. enter image description here

Step 16

The previous position of 46 also fixes the 46 in the bottom right. The last entanglement left is in the bottom center, which should help us get everything we need. C6 requires numbers 2, 8, and 9 - of which 28 must be entangled. 9 must go in r9c6. enter image description here

Step 17

R9 still needs 5, 6, and 7. 7 is ruled out of c3 and c4, therefore it fills r9c5. 6 is ruled out from c3, so it fills r9c4, leaving 5 in r9c3. Sudoku then forces 5 into r8c4. r8 is still missing 1289, despite having some related entanglements.. However, r8c5 cannot be 9, 8, or 2, as c5 contains these. It must be 1, fixing the 19 pair on c7. enter image description here

Step 18

By noting that r8 requires a 28 only, and c3 already has a 2, this puzzle's remaineder is solved by natural sudoku rules. enter image description here

  • 1
    $\begingroup$ Very nice! +1 However, your pictures in steps 5 and 6 don't seem aligned with your text and the last picture is missing the grid. $\endgroup$
    – Jens
    Oct 26, 2020 at 1:12
  • 1
    $\begingroup$ @Jens Thanks for letting me know - I had some trouble with doing some of the screenshots... the "missing grid" is from the layers in the paint program not doing a deep copy (my fault, different hotkey). I had to delete and reupload a ton of pictures when I realized I messed that up - guess I missed another. I'll re-evaluate steps 5 and 6, and see if I can correct the imagery! $\endgroup$
    – Serge
    Oct 29, 2020 at 6:07

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