[Guest]

Something that I have spent a little time trying. The challenge is to figure out how many unique ways you can arrange sets of cubes of the same size. There is only one way to arrange a single cube. One way to arrange a pair of cubes (side by side). Two ways to arrange 3 cubes (in a line or at a right angle). Try to figure out how many ways there are for the next few. 3 dimensions can be used. The real challenge though is to see if there is some sort of formula that can calculate the possible number of arrangements.

[EventHorizon]

Something that I have spent a little time trying. The challenge is to figure out how many unique ways you can arrange sets of cubes of the same size. There is only one way to arrange a single cube. One way to arrange a pair of cubes (side by side). Two ways to arrange 3 cubes (in a line or at a right angle). Try to figure out how many ways there are for the next few. 3 dimensions can be used. The real challenge though is to see if there is some sort of formula that can calculate the possible number of arrangements.

the 2-d analog to this problem is interesting as well. I remember working on it way back in 6th grade.

Tetrominos (like tetris pieces), pentominos (5 squares), hexominos, etc. I don't think I came up with a formula for it either. Though I cut them out of paper (up to heptominos (septominos?))

Perhaps working on the 2d version would be better first. Or is it a solved problem?

[Guest]

Something that I have spent a little time trying. The challenge is to figure out how many unique ways you can arrange sets of cubes of the same size. There is only one way to arrange a single cube. One way to arrange a pair of cubes (side by side). Two ways to arrange 3 cubes (in a line or at a right angle). Try to figure out how many ways there are for the next few. 3 dimensions can be used. The real challenge though is to see if there is some sort of formula that can calculate the possible number of arrangements.

I would think that the answer for N cubes would be N * N. You say that there are only 2 possibilities for 3 cubes, but you can use three dimensions, I think your answer must include all 3 dimensions, even when the result looks like another answer. For instance

█ █

█

█ █

   █

are different answers. Each of these examples above can be rotated 90 degrees on the X axis to produce 4 variants in 3 dimensions, giving 8 different arrangements. The ninth is a straight line: █ █ █.

Edited June 23, 2008 by Swervin Kervin

[Guest]

...but by playing with my son's blocks (I'm no mathematician) I found:

4 cubes – 6 unique arrangements

5 cubes – 9 unique arrangements

6 cubes – 13 unique arrangements

**unique meaning that the same pattern counts only once no matter how many different angles you rotate it.

[Guest]

I would think that the answer for N cubes would be N * N. You say that there are only 2 possibilities for 3 cubes, but you can use three dimensions, I think your answer must include all 3 dimensions, even when the result looks like another answer. For instance

█ █

█

█ █

   █

are different answers. Each of these examples above can be rotated 90 degrees on the X axis to produce 4 variants in 3 dimensions, giving 8 different arrangements. The ninth is a straight line: █ █ █.

I was considering rotations of the same arrangement as one solution. Only unique arrangements count. Also, if you go past 3 you will see that the number of solutions grows much faster than n squared

[Guest]

my first post

i got 7 for 4 cubes and 22(maybe more) for 5 this is including 3d shapes without repeating a shape that has been rotated)

[Guest]

...but by playing with my son's blocks (I'm no mathematician) I found:

4 cubes – 6 unique arrangements

5 cubes – 9 unique arrangements

6 cubes – 13 unique arrangements

**unique meaning that the same pattern counts only once no matter how many different angles you rotate it.

I got 7 for 4 blocks.

see attached word doc.

7_shapes.doc

[Guest]

i cant view your pic for some reason so heres my 7 for 4

[Guest]

I found 29 for 5 cubes. And wonder there is the 30th and more combinations that I missed!

I have posted the pictures for 5 cubes here.

Can't imagine how many will it goes for 6 cubes!

cube5.ppt

Edited June 24, 2008 by woon

[Guest]

I found 29 for 5 cubes. And wonder there is the 30th and more combinations that I missed!

I have posted the pictures for 5 cubes here.

Can't imagine how many will it goes for 6 cubes!

cube5.ppt

I disagree that it's 29, because the ones in yellow are the same shape, only flipped and/or rotated. By my count your list is 23, but there's more... (Good analysis, though!)

If there's a mathmatical formula to figure this out, I believe it has to come from the fact that a cube has 6 sides, and can have a maximum of 2, 3, or 4 adjacent cubes (if we are using a maximum of 5 cubes). I'll post this here for someone to add, but the configuration of 23 unique 5-cube arrangements contained in the powerpoint yielded the following:

2 arrangements with a maximum of 4 adjacent cubes to any 1 cube.

7 arrangements with a maximum of 3 adjacent cubes to any 1 cube.

14 arrangements with a maximum of 2 adjacent cubes to any 1 cube.

I took the powerpoint and shaded 1 cube in each, demonstrating the maximum adjacent cubes for that arrangement. I'm sure the key to a mathmatical / geometric formula lies in the sub-grouping of adjacent cubes to any 6-sided cube.

thoughts?

5_blocks.ppt

[EventHorizon]

thoughts?

Working with the 2d analog of the problem, I couldn't find a simple solution. You'd think that the equation for one would be similar to the one for the other (if only because each of the 2d arrangements are included in the 3d ones).

Since this is a problem that I had worked with ages ago and never solved, I may write some code to generate all of them (both 2d and 3d (and 4+ d?)). I think I could write it without too much difficulty, and having the correct numbers would definitely help in determining possible equations.

(I did a similar thing with the towers of hanoi problem with more than 3 pegs.....yes it is easier than with 3, but the question is....how much easier? The answer turned out to be fairly interesting....search for "hanoi" if you're interested)

[Guest]

I disagree that it's 29, because the ones in yellow are the same shape, only flipped and/or rotated. By my count your list is 23, but there's more... (Good analysis, though!)

If there's a mathmatical formula to figure this out, I believe it has to come from the fact that a cube has 6 sides, and can have a maximum of 2, 3, or 4 adjacent cubes (if we are using a maximum of 5 cubes). I'll post this here for someone to add, but the configuration of 23 unique 5-cube arrangements contained in the powerpoint yielded the following:

2 arrangements with a maximum of 4 adjacent cubes to any 1 cube.

7 arrangements with a maximum of 3 adjacent cubes to any 1 cube.

14 arrangements with a maximum of 2 adjacent cubes to any 1 cube.

I took the powerpoint and shaded 1 cube in each, demonstrating the maximum adjacent cubes for that arrangement. I'm sure the key to a mathmatical / geometric formula lies in the sub-grouping of adjacent cubes to any 6-sided cube.

thoughts?

Swervin,

I think you should re consider your disagreement and look back over the cubes (shaded in yellow) they are actually distinct from their 'mirrored partners'. They cannot be rotated in a 3D space to the opposite arrangement. I do like your approach for the n of arrangements vs. n of adjacent cubes. Good luck!

I have yet to come up with anymore than 29 solutions for the 5 cubed 3D space. Has anyone solved the solution for the 2D and 3D formulas?

I will post my data shortly!

[EventHorizon]

I disagree that it's 29, because the ones in yellow are the same shape, only flipped and/or rotated. By my count your list is 23, but there's more... (Good analysis, though!)

...

are you sure there is more than 23? That would mean there is a bug in my code...so let me know if you are certain. I'm almost positive that the 2d numbers are right up to 7 squares.

Here's what I got for totals

number  2d       3d       4d

2       1        1        1   

3       2        2        2

4       5        7        7

5       12       23       26

6       35       112      147

7       108      607      1019

8       369      3811

9       1285

10      4655[/codebox]

Here's the code, executable, and text files it printed out showing all the unique configurations (so you don't have to run it unless you want to).

If you have any questions about what the code is doing, let me know.

The text files give a unique number to each (1 to the first, etc), show the dimensions of the object (length,width,depth,etc), then show 2d slices (third dimension is to the right separated by |, and 4+ d's are down separated by blank spaces).

[Guest]

Swervin,

I think you should re consider your disagreement and look back over the cubes (shaded in yellow) they are actually distinct from their 'mirrored partners'. They cannot be rotated in a 3D space to the opposite arrangement. I do like your approach for the n of arrangements vs. n of adjacent cubes. Good luck!

I have yet to come up with anymore than 29 solutions for the 5 cubed 3D space. Has anyone solved the solution for the 2D and 3D formulas?

I will post my data shortly!

Thank you for kswack helping me to explain this.

Swervin, the most simple item that you can consider is both you are hand. There are mirrored partners but you can't say left hand is same like right hand. No method how you rotate, you can't have the right hand been replace with the left hand.

[EventHorizon]

Thank you for kswack helping me to explain this.

Swervin, the most simple item that you can consider is both you are hand. There are mirrored partners but you can't say left hand is same like right hand. No method how you rotate, you can't have the right hand been replace with the left hand.

My code assumes mirrored (flipped over an axis) objects aren't counted for each mirroring. This is the way I worked with 2d ones back in 6th grade, and that's what I assumed the original poster meant.

If this is not the case, then I'll need to figure out how to find all possible orthogonal rotations of n-dimensional objects.....yay.... :-/

Edited June 26, 2008 by EventHorizon

[Guest]

Thank you for kswack helping me to explain this.

Swervin, the most simple item that you can consider is both you are hand. There are mirrored partners but you can't say left hand is same like right hand. No method how you rotate, you can't have the right hand been replace with the left hand.

Oops, an error due to fast typing. Should be 'both your hand' but not 'both you are hand'. Apologize for this.

[Guest]

My code assumes mirrored (flipped over an axis) objects aren't counted for each mirroring. This is the way I worked with 2d ones back in 6th grade, and that's what I assumed the original poster meant.

If this is not the case, then I'll need to figure out how to find all possible orthogonal rotations of n-dimensional objects.....yay.... :-/

Not all mirrored objects are exactly the same. Some yes and some No. That's make the code assumptions more challenging.

[Yoruichi-san]

Well, I'm not sure if this will be helpful or not, but this reminds me of finding isomers in organic chemistry...so maybe we could look at this with a similar systematic method as finding and naming of isomers:

Consider the longest straight "chain", then the next longest chain, and so on...

For n cubes, there is 1 arrangement with the longest chain of n cubes.

There are (n-1)/2 for odd n and n/2 for even n arrangements with the longest chain of n-1 cubes. (You can put the nth cube on a face of any of the other cubes except the end faces of the end cubes, since that would just be the n cube chain, and putting it on the 1st cube is the same as putting it on the last cube, the 2nd is the same as the 2nd to last, etc. This is equivalent to the rule in OChem that you start counting at the end that will give you the smallest numbers)

For the longest chain being n-2, you have different options. You can either attach a 2 cube "side-chain" or two 1 cube side-chains to the base chain. There are (n-2)/2 for even n or (n-1)/2 for odd n ways of attaching the 2 cube side-chain. For the two 1 cube chains, you can start out putting 1 cube on the first cube in the base chain, then put the second 1 cube side chain either 90 degrees or 180 degrees from it on each of the n-2 cubes in the base chain, for 2(n-2) arrangements. Then move the first cube side-chain and repeat the process until you get to the center cube for another 2(n-3)+ 2(n-4)+...2(n-n/2) for even n or 2(n-(n-1)/2) for odd n arrangements.

Okay, I can see that this is getting complicated really quickly, and as mentioned by others, there is the issue of "chirality" to consider when the side chains are different lengths...but, I dunno, this might be a start or help in some way...I'll keep thinking about it (whether I want to or not...I'm one of those ppl who can't stop thinking about a problem once they get in into their head...X\)

[Guest]

I am very glad to see so many post on this thread. I have some data but I would very much like some one to check it.

My Definition of the problem. n number of cubes yields x number of specific (unique in that dimension) arrangements with n number of cubes. Each arrangement must be unique as it is rotated in the same dimension for which we are writing the formula. (2D, 3D, 4D)

Here are some of my numbers

I was really excited to see the pattern forming in the 2D Symmetrical Arrangements Column

My hunch of a underlying Fibonacci sequence was squandered by 6 cubes (didn't think it could be that simple) but I do think that some very basic sequence will come out of this.

I will continue to work on the 2D formula right now but I am happy to see others putting together the 3D samples and counts. Good Stuff. I am really excited to see what number or ratio we are converging on with the two formulas. We shall see!

I would love for someone to check over the 5 cube and 6 cube arrangements because it is getting late... I think that I am getting closer to a formula but I want to be sure that the 6 cube 2D is accurate before posting!

Best of luck you all!

[EventHorizon]

Not all mirrored objects are exactly the same. Some yes and some No. That's make the code assumptions more challenging.

Depends on whether you count mirrored objects that cannot be achieved by rotation as unique or not. My old code didn't. And it turns out that it is only one extra "if" statement to make it only do rotations. I'll post some of the text files now, and will post code / exe tomorrow.

Here's some data, though I will give a bigger list tomorrow (better computer in the lab...and it's late).

2d -> 1,1,2,7,18,60,196,704,2500

3d -> 1,1,2,8,29,166,1023,6922

Oh, and kswack you show 2 identical ones in your 5 block 2d. The two you are missing are numbers 7 and 15 in my d2_n5.txt file.

xxx

x

x

and

xx

xx

x

Good job with the 6 block 2d ones.

[EventHorizon]

Alright. I've made my code let you choose to find duplicates by only rotation or both rotation and mirroring. I've included a number of text files. The ones named dX_nY_flip.txt are the ones that removed duplicates found by rotations and mirroring.

xominos_both.zip

Here's a table of total arrangements (by dimensions and number of squares/cubes/etc)...

num  2d       3d       4d       5d       2d_flip  3d_flip  4d_flip

1    1        1        1        1        1        1        1

2    1        1        1        1        1        1        1

3    2        2        2        2        2        2        2

4    7        8        7        7        5        7        7

5    18       29       27       26       12       23       26

6    60       166      164      154      35       112      147

7    196      1023     1316     1172     108      607      1019

8    704      6922                       369      3811

9    2500                                1285

10   9189                                4655[/codebox]

One part in the table troubles me. Look at going from 3 to 4 dimensions with 4,5,or 6 cubes. The number actually goes down. Out of curiosity, I just added 5d, and the numbers decrease going from 4d to 5d with 5,6,and 7 hypercubes. I can see how this can happen to specific objects (extra unused dimension allows a rotation that can essentially flip it), but totals would intuitively go up. But my intuition has steered me wrong before...

(It's beginning to make sense....since the number of cubes is so close to the number of dimensions...)

Ideas? Intuition? An all encompassing formula?

Edited June 26, 2008 by EventHorizon

[Guest]

Nice work there Event! Thank you for the corrections. I like the formulas you came up with. I will try to work out the all encompassing formula but I think the couple that you have created seem to work well.

I will let you know what I come up with but I think we should look further into the pattern formed by the number of symmetrical arrangement and that ratio to the number of cubes (all in 2D) I think when we start to increase the number of sides on the object (pentagon, hex, sept, oct ...) I think that each data set for each shape, different number of sides, will show a nice ratio (number of objects: symmetrical arrangements) that converges on some unique number. (Golden, square of 2... something really basic). Just a thought.

Thank you all for the good work on the 'cubes'

Here are my corrected files.

Edited June 26, 2008 by kswack

[EventHorizon]

I just happened to check wikipedia and was able to verify some numbers (2d, 3d, and 3d_flip) on it. (search for polycube or polyomino)

They also say that the problem of finding a function for the number of n-ominos is unsolved. So all that means is that when we solve it we'll be rich and famous!