The New Sudoku Players' Forum

[Guest]

Is there a way these puzzles can be extended to 3D? Same idea but now the grid is 3x3x3.
Has anybody looked into this?!?

[scrose]

Yes.

[geoff]

The easiest extension is a 9*9*9 cube with each of the 27 planes a suduko grid. If you want to have lines squares and 3 dimensional boxes each containing each digit the smallest size is 64*64*64 with 8*8 squares and 4*4*4 boxes. That's 262144 cells!

[coloin]

I get the impression the first row of 3d boxes are easy

If every 3x3x9 is a rolling version of a suduku grid - and this holds all the way round the 3-d cube then there isnt a problem here.

- but then if it breaks down - and I get the feeling that it will - maybe there wont be many solutions at all.

But there are an awful lot of ways to put 9 valid sudukus on top of one another. [Upper limit [Bertram]^9]

When you try to work out the individual constraints there are all pottentially increased by ^2.

So we have a very big number with a very large number of constraints.

I think it is one for the programmers as I cant even get on to the third layer on non see thru paper!

I think it is enough to show that it is possible - rather than working out the numbers !

[scrose]

According to The Daily Telegraph, it has a unique solution.

[coloin]

Well Ill be blowed

What a shame

It goes all the way round - not even difficult.

That make 6*3*3*3 and possibly up to 9! more ways to vary this grid and it still holds.

I can hear computors whirring everywhere in an effort to produce a non-symetrical solution [- which perhaps wont have the 6*3*3*3 variations] [But it would have the 9!]

[Ukodus]

Just do a rubiks cube its easier

[Francisco]

Is it just me, or is this example (the one in scrose link) pretty basic?

It's just using a very simple rotation of the same numbers in each box...
I finished 1 slice, and I'm almost finishing another... but it's pretty logic that the slices I haven't touched yet will suffer from the same "disease"...

This is no fun at all

[scrose]

I haven't attempted the Dion Cube myself, but everything I have read is similar to your remark. The puzzle-makers just shuffled three numbers about in each block on each slice. Working towards the solution offers little (if nothing) in the way of surprise. I was only offering this as an example of a 3x3x3 grid, albeit a poor puzzle.

I wonder if it is possible to construct a Dion Cube that doesn't contain such an obvious pattern?

[JOTHI]

Is there a way these puzzles can be extended to 3D? Same idea but now the grid is 3x3x3.
Has anybody looked into this?!?

Hi Sir,

From an INDIAN. I've solved the 3-d puzzle (of course after having created one..!) .can any one give me an idea of 4th dimension?

reply me at mailtojothi@gmail.com

[cjlt]

I already published a 4 dimensional variant. See the thread "Proposed variant puzzle" much further down this forum.
Essentially, I treat the existing diagram not as a 9x9 square but as a 3x3x3x3 hypercube. This, I claim, makes a more elegant and more difficult puzzle, as every 3x3 plane in the hypercube must contain each digit once.
In that thread, I include a sample 4D puzzle. Someone has posted the solution, so, if you want to try it, don't scroll down too far.
Hope you enjoy it.

Chris Lusby Taylor

[johnw]

I already published a 4 dimensional variant. See the thread "Proposed variant puzzle" much further down this forum.

The proposed variant puzzle is what another thread called "restricted sudoku" and is also mentioned in the thread
http://forum.enjoysudoku.com/viewtopic.php?t=564

I agree it is quite an enjoyable variation.

[cjlt]

Thanks for pointing that out. Looks like the variant I thought I'd invented was already well known. Not too surprising, but I wonder why none of the British newspapers uses it? It seems to me much superior to the standard Su Doku.
Chris

[Condor]

I wonder if it is possible to construct a Dion Cube that doesn't contain such an obvious pattern?

It may be that the only way to find out if it is possible would be to investigate it with a computer, if someone is willing to spend the time it will take.

But there are an awful lot of ways to put 9 valid sudukus on top of one another. [Upper limit [Bertram]^9]

Obviously that is much larger by many orders of magnitudes than the real figure.

There is one way I know to reduce the number of possiblities.

A Dion Cube consists of 27 smaller cubes which are made up of boxes in there respective planes. When 1 plane in a box is filled then there is only 40 ways to fill up the remaining 2 planes within the rules of Sudoku.

So the total number of possiblities for the first 3 planes is equal to [Bertram]*40^9, and that is only accounting for the boxes - not row and columns as well.

[Edit 30 March]
Looking back at what I wrote, I realise that it probably was not clear. So I will try and explain it with following diagram.

Code: Select all

.  .  .    .  .  .    .  .  .
.  .  .    .  .  .    .  .  .
.  .  .    .  .  .    .  .  .

.  .  .    .  .  .    .  .  .
.  .  .    .  .  .    .  .  .
.  .  .    .  .  .    .  .  .

X  X  X    .  .  .    .  .  .
X  X  X    .  .  .    .  .  .
X  X  X    .  .  .    .  .  .

Shown is the top slice of a Dion Cube with one box marked with X's. What I was trying to say was to fill the corresponding boxes in the next 2 slices can only be done in 40 ways. As there are 9 boxes in each slice that makes 40^9 ways of filling the next two slices

[Lardarse]

So likely the limit is Bertram^3 ?

Edit: Mis-read your formula. Maybe it's Bertram for a 3x3x3 cube...