People seemed to enjoy my last Sudoku variant puzzle - Samurai Pseudoku, so I spent this week making another! This one is going to require using logic you've never used before...

The last one was a little plain, don't you think? After all, it was just a flat puzzle, only a couple of variants and don't even get me started on the colour scheme...

So this time, this is Sudoku in the third dimension! But wait, I just hit you with a double meaning; not only is this puzzle 3D, it has 3 different dimensions to it, in the form of 3 different Sudoku variants: 3D, Vudoku and 147.

The rules will be explained under the grid, if any clarifications are needed please just ask in the comments!

Oh.... and I forgot to mention. The last puzzle had way too many digits, so this time I'm only giving you one digit:

$$\huge\text{3D VUDOKU } \huge{\times} \huge\text{ 147}$$

                        enter image description here



  • Each box obeys normal Sudoku rules.
  • On top of this, every 9 cell path from an edge must contain each of the digits $1$-$9$ once (normal sudoku). As a result, each cell is part of two paths, and one box.
    • A path starts from the edge and ends at a thick black line.
  • Paths look like the following:
    enter image description here


  • The purple arrowheads connect three cells together. The middle cell is either the sum or the difference of the two outer cells.
    • For example, looking at the top-left arrowhead on the puzzle, if there was a 2 in the bottom left of the box, and a 5 in the middle, the middle left cell could be either a 3 or a 7.
  • Towards the centre you may notice there is a shape that looks like an elongated '$\color{purple}{\text{]}}$'. This should be treated as two separate arrowheads, one pointing North-East, and one pointing South-East.


  • A cell containing a $\color{#8F9}{\text{1, 2 or 3 is green}}$, a cell containing a $\color{#89F}{\text{4, 5 or 6 is blue}}$ and a cell containing a $\color{#F88}{\text{7, 8 or 9 is red}}$.
  • Blank cells can be any as long as they abide by previous rules.

There is a single solution that can be deduced logically with no guesses.

A solution will be accepted with:

  • A complete grid
  • At least a little bit of explanation (preferably with some images) e.g. where you started, route you took etc.
  • If possible, the time it took you to solve! (This is once again just for my personal interest)

The previous puzzle was probably too large and as a result became tedious towards the end, so this is a bit smaller. This is still a very hard puzzle however, and it's going to require some different ways of thinking.

Good luck and enjoy!

  • $\begingroup$ The really tough part here is how to fill in the numbers in this 3D so it looks nice ;-) $\endgroup$
    – Jens
    Commented Aug 17, 2020 at 19:05
  • $\begingroup$ @Jens yeah no excel file for this one sadly :P I couldnt find any 3D sudoku software for this either so I'm afraid I can only provide the image! $\endgroup$ Commented Aug 17, 2020 at 19:06
  • $\begingroup$ There's an ambiguity at the second level at the lower right of the first box. A purple arrowhead connects four boxes. $\endgroup$
    – Jens
    Commented Aug 17, 2020 at 22:06
  • $\begingroup$ @Jens ah I forgot to mention that in the rules. It's intentional, and should be treated as two individual arrowheads $\endgroup$ Commented Aug 17, 2020 at 22:07
  • $\begingroup$ Are all possible arrow heads drawn? Does the lack of an arrowhead mean there is no sum/difference? $\endgroup$ Commented Aug 18, 2020 at 4:44

2 Answers 2


Completed grid (cube?):



There were a LOT of deductions to make here. I'm not going to give step-by-step, but I will show several milestones along the way. I'll explain some of my favorite (read: head-bangy complicated) deductions and some common deduction types I used a lot. Unfortunately, my pictures and my favorite deductions do not always match up.

Initial deductions

My first deduction was in the box with a 4 entered. An arrow between two greens can give a 1, 2, 3, 4, or 5. But since this box had 3 green cells marked, the arrow couldn't give a green. It also couldn't give a 4 because there was already a 4. So it gave a 5. The box "below" that one has two greens at the center of arrowheads connecting same-color boxes. Greens such as that can only be 1 or 2, so the remaining green must be a 3. Now the 2 and 3 making the 5 can be placed. The blue below that 5 must be a 6 (can't be 4 or 5). The blue below that 6 must be a 4 (can't be 5 or 6). The only blue that makes an arrowhead with a 2 and 4 is 6. And so on and so forth.

Picture after I got the starting area:


Moving out

Picture after expanding solved area:

next step

A favorite deduction

One of my favorite deductions has already passed, sadly. (I was re-doing this off my notes after running into a contradiction the first time through. So I didn't get stuck until later, and forgot to take the proper pictures). I'll paint the scene for it anyways. See the box with 1-9-3 in the middle row, 2 to the right of the starting box? The 1 was already there. And see the 5^3^8 a box below? The 5^3^? was there. I had to figure out whether the ? was a 2 or an 8. In the "column" with the aforementioned 1 and ?, there are 2 greens that were currently blank. The 2 for this column must be placed in one of these greens. So the ? cannot be 2, it must be 8!

Another favorite deduction (this one occurred right before the picture was taken)

This deductions is about placing the 1 in the middle box of the bottom left. Said box had two columns finished (as you can see). The 1 cannot go in the bottom row, since there is already a 1 in that row. Also, it can't go in the top row, since that box is red and 1s are green. Therefore it must go in the middle row, as shown.

Where I got stuck

Picture of the state at which I made this deduction:

next step

The (obscenely complicated) deduction:

We're going to work with the box that has a 1 and a 5, and the 1 has 3 ^s sticking out of it. Got it? "Yes bobble, we got it, get the the deduction now!" you groan in the background. Fine, fine.

There are 3 sets of 2 numbers connected to the 1 by a ^. Each must be 1 apart. Since there are 2s in the bottom two rows, the 2 must go on the top row. It will be ^ed to a 3 in some order. The 4 must go below the 1, since its two possible ^ partners (3 and 5) are otherwise occupied. Now we have two ^ pairs for either side of the 1 (6 & 7, 8 & 9). The column to the left of the 1 has three blues higher up. Therefore no blues can occur lower in the right column, so 6 & 7 must go on the left on the 1 and 8 & 9 to the right. The orientation of these pairs can be determined by numbers on the left of the row. 5 numbers placed! Woo hoo!

Raise the roof! /go upwards

Right before this deduction:

next step


We're going to look at two boxes next to each other, the ones with 5^8^3. The remaining two greens in the left box must be 1 and 2. So the arrow between the two blues must be 9 (= 4 + 5), since it can't be a subtraction. Therefore the center box must be 4 or 5. It can't be 5, since there is a 5 in that row. So it's a 4.

The end is in sight!

A picture of near the end

near the end

The next deduction:

I'll be honest, I had trouble figuring out an interesting deduction to showcase near the end. I was pretty used to the logic by now so I didn't really get stuck. This one is neat but quick.

See the ^ connecting 3 green boxes? The left two boxes are 2 & 3 in some order. The middle of this ^, however, needs to ^ with the 1. If it was 2, then the other part of the ^ would have to be 3 (3 - 2 = 1). But there's already a 3 in that row. So the 3 must go above the 2.

Common deductions

  • Regular sudoku logic
  • Elimination of numbers by an incompatible color (1 can't go on a red or blue)
  • Forcing a colored cell to have a certain number, since the other two numbers of that color are eliminated by sudoku logic (a green cell that can 'see' a 1 and a 2)
  • With two known numbers of a ^, having only 1 possible number (4^6^? can only be 2, since 10 isn't a valid number)
  • With two known numbers of a ^, eliminating one of the possibilities by color/sudoku logic (2^5^? where the ? can 'see' a 3)
  • Combinations of the above

All in all, this took me about 4 and a half hours of actual solving time if I remove breaks and don't do my math wrong.

  • 1
    $\begingroup$ Nicely done! I got stuck at the same place as you, i.e. it took me quite a while to deduce how to proceed. Impressive that you could do it in 4 and half hours (took me about 6 hours) :-) $\endgroup$
    – Jens
    Commented Aug 18, 2020 at 16:52

It seemed people appreciated my 'The making of' post for the previous sudoku variant, and as I thought this one was much trickier and complex to make I thought I'd post another insight into the creation process.

This answer may spoil some of the reasoning that bobble will update their answer with. Click the link on any image to see a better resolution version.

Wrap-up: The Making Of 'Sudoku in the third dimension'

This is not a solution to the puzzle, but provides notes from its poster. This type of answer has been approved by the community.

Caution: This post contains spoilers.


The inspiration is almost identical to the inspiration of the previous puzzles, except for the puzzle variants.

It was actually bobble who suggested the idea of a 3D Sudoku, and I very much liked the idea of that.

I also decided to try and play on 'three dimensions' by adding 3 Sudoku variants to the puzzle. However, I didn't want it to be overly complicated, so with 3D already being one of the Sudoku variants, I needed two that would work easily together while not being too complex.

Using the same app as last time, I went through all the variants this time and found 'Vudoku' and '147' simple but fun variants that also wouldn't get in the way of each other.

Once I had found the 3 I wanted I got started.

Creation (and resources)

Starting off

I couldn't find a 3D sudoku editor online or even the 3D Sudoku type I wanted, so the base image in this puzzle is actually a modified screenshot of a 3D sudoku that was close enough to what I was looking for.

The issue I had from the start was the lack of software to make this puzzle. I couldn't use Excel as it had no 3D option, so my only option was to use a photo editor (I use an app called 'Sketchbook'). This meant I knew this would take much longer to make.

Creating a solution

Creating the solution to this proved much harder than creating the solution to the previous puzzle.

The amazing thing about 3D sudoku, is that no matter how you create the 'collapsed' cube, every single cell will be part of one box and two paths. The problem when creating a solution however, is that filling in certain cells in two different places may make it impossible further down the line, so there was a lot of backtracking in the creation.

I started off simply and filled in the left and at this point, I just had to make sure I obeyed 3D rules, as there was little chance of a contradiction. I managed to fill in the left hand side with just one contradiction which luckily was easily fixable at the bottom middle.

enter image description here

The second half became the issue. I encountered a lot of contradictions, but managed to figure out a way of entering which would let me pick up the contradiction before it happened. I did this simply by looking at what each path needed, and making sure there was no cell in an upcoming box that had all 3 of the needed numbers in the other path that crossed over that cell. This allowed me to have to backtrack less and I managed to fill in the solution:

enter image description here

Creating the clues

To start entering clues I set up a blank grid next to the solution which would slowly get filled in with clues over time. I would also 'solve as I go' to see what numbers still needed to be clued:

enter image description here

I realised at this point that it would be a huge hassle entering and then skewing numbers for clues, so this is where the idea that I would give very little numbers came from. The original idea was to actually give 0 numbers.

The only way (I'm sure there's more however) I could think of being able to clue a number was to find a box where there was a $1$ and a $2$ in an arrowhead between two numbers of the same colour and give all that information.

With this information, it could be deduced that the remaining green must be a $3$, as the max difference between two of the same colour is $2$ (and it cannot be $4$-$9$-$5$ as the centre is a green, so the other greens must be $1$ and $2$.

This scenario only occurred in one box, the one below the box with the four. I filled in this information and it allowed me to clue another $3$ by putting the places of greens in the above box where the $3$ ruled out 2 of them. I realised at this point that it probably wasn't possible to clue the whole puzzle without at least one number.

I managed to work out that a $4$ in the box it is in would force a $5$ between the two other greens and make the greens resolvable, as it could not be $3$-$4$-$1$, $3$-$2$-$1$ or $3$-$1$-$2$, leaving $3$-$5$-$2$. This was a MAJOR breakthrough in clueing this puzzle.

From there I could add some more clues to be able to resolve some more numbers and reached this stage:

enter image description here

After cluing some more so that several boxes could be filled in, I came across a stumbling block. I couldn't actually clue the top left with the information given. I didn't want to give any more numbers so instead I decided the 'solve path' would be to go round the bottom, and up round the right and hope that it was clue-able such that you could go round the grid and the top left box would be one of the last solved. Luckily this was the case!

I got round the bottom clueing without any major issues, but then I got to the top right. At the top right of the grid, you have to 'turn' around a thick black line, meaning you only have one path that is providing any information. This was incredibly difficult to clue and I had to look further down the path before I noticed anything that could resolve a number.

It took a while and I had to provide a lot of clues, but I knew that was OK, as I could tell it was still difficult based on how difficult it was to clue.

I managed to sketch the clues in that I needed:

enter image description here

and a quick double check verified my solution matched.

Cleaning up

I didn't want the actual puzzle to look scruffy and hand-drawn, so I saved my clues and opened a blank grid and started filling in the clues neatly. I also decided to just fill in the cells with the appropriate colour instead of doing a little dot in the corner.

Cleaning up took forever, and I initially missed a lot of clues I had hand-drawn so had to double-check four or five times to make sure it was right. Once I was happy I had all the clues in, I just needed to double check.

Double checking

Aside from double checking my solution and my clues, I needed to make sure the puzzle was solvable. I printed some blank versions, and managed to solve without any problems. I also got my parents to solve just to make sure it was solvable.

This also allowed me to gauge the difficulty, and I was happy to find it was quite hard and required some very tricky logic!


Some things I have taken away from creating this puzzle: (a lot of the previous points from the last takeaway also apply)

  • Double check everything - I mentioned last time how important double checking was, but this time it was even more so. I missed a lot of handdrawn clues when cleaning up, which could have ruined the whole puzzle.

  • Don't rush posting - I was eager to post this after spending so long creating it, but in doing so I didn't take my time typing up the question body and making sure everything was covered and hence missed some points that then needed querying in the comments.

  • If possible, always try and get test solvers - test solvers are incredibly useful for both checking the puzzle for errors, and for judging the difficulty. If there is no-one who can test solve for you, then at the very least solve it yourself and see what you find from doing so!

  • Keep persevering! - there were many points when creating this in which I thought it wasn't possible to clue it, but spending enough time thinking about what I could do allowed the puzzle to be created, almost exactly how I wanted it to be. If you get stuck going down one route, keep trying other routes!

I hope this was helpful and the puzzle enjoyable. I personally think this was much harder than the last one (it was certainly harder to make!!) but hopefully it was just as fun.

I may make one more of these to complete the trilogy, but it may be a while as I want to come up with an idea that's a bit more concise but with even better logic :)

Any feedback on this would be much appreciated! (negative or positive, how this compared to the last, any improvements etc), and thank you all for solving!

  • $\begingroup$ There are other ways to get a number from no clues: Gur frira va gur tevq orybj naq gb gur evtug bs gur gbc tevq pna or qrqhprq vzzrqvngryl, gunaxf gb gur gjb neebjf naq terra fdhner haqrearngu vg sbepvat na nevguzrgvp cebterffvba jvgu gur gjb obggbz pbearef. Fantastic puzzle! $\endgroup$ Commented Aug 19, 2020 at 2:33
  • $\begingroup$ @AxiomaticSystem true, clever thinking! Alternatively rot13: N oyhr-erq-oyhr frdhrapr jbhyq unir gb unir n 9 va gur zvqqyr gbb! I could probably have done this with no numbers but I quite like how there’s just one :) $\endgroup$ Commented Aug 19, 2020 at 7:57

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