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rocdocmac

Cubicle Stack

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20 answers to this question

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There are 12 edges. According to CaptainEd's numbering system, they would be the even cubelets with the exception of 14 (the core). I used cubelet 1 as the corner. 2 and 4 make the same shape. As far as I can tell, 6 and 8 are 2 more. Cubelet 1 is part of 3 faces of the cube so any of the edges on those faces would make one of those 3 shapes. 10, 12, 16, 20, and 22 are out for that reason. 18 gives a 4th shape. 24 and 26 are the same as 18.

So 4 shapes for edge/corner. New total is 22? I'm wishing I had a real cube I could take apart now...

Edited by Thalia

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My question and image seem to have disappeared!

Suppose 27 identical cubes are glued together to form a cubical stack, as illustrated below. If one of the small cubes is omitted, four distinct shapes are possible: one in which the omitted cube is at a corner of the stack, one in which it is in the middle of an edge of the stack, one in which it is in the middle of a side of the stack, and one in which it is at the core of the stack. If two of the small cubes are omitted rather than just one, how many distinct shapes are possible?

image.png.07f16bc4a0809a9cb4138f5dbab5ad24.png

Edited by rocdocmac
Question and image added

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Not sure if my method is the right way to do it...

Spoiler

Now , we can only choose a core , corner , edge or the middle piece from all the pieces.
I made a table where I took 1 piece , and then check the no of ways I can take the second piece.
                                                  corner                edge                 center               core
                       core                       1                          1                        1                        X                 = 3 ways
                       corner                  3                           3                       2                         X                 =8 ways
                       edge                     X                           4                       3                         X                 = 7 ways
                       center                  X                           X                       2                         X            =    2 ways
X is marked for spots that are already counted .
So , total distinct no of shapes =  3+8+7+2 = 20 ways

 

Edited by Slashpuzzler

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23?

I get 3 possibilities if the core is removed. 

Corner (C)
Edge (E)
Middle (M)
2E 2
2M 2
2C 3
EM 4
EC 7
MC 2

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I get 21

Let’s number the cubelets: the plane facing us is

1 2 3

4 5 6

7 8 9

The plane behind it adds 9, and the Back plane adds 18.

Assuming that mirror images are to be counted separately, but rotations are not, I  have categorized them as follows, with example cube pairs.

Two corners 1-3, 1-9, 1-27

One corner, one edge center, 3-2, 3-6, 3-4, 3-7, 25-2, 25-6

One corner, one face center, 3-11, 3-23

Two edge centers 6-24, 6-22, 6-2, 6-20, 6-26

One edge center, one face center 6-15, 6-11, 6-13

Two face Centers 5-15, 5-23

Edited by CaptainEd
Spoiler bracket

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Spoiler

I got 26 possibilities.

My answer is the same answer that Thalia got, but instead of four edge to edge possibilities, there are five. Other than that, everything else is the same.

 

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4 hours ago, Buddyboy3000 said:
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I got 26 possibilities.

My answer is the same answer that Thalia got, but instead of four edge to edge possibilities, there are five. Other than that, everything else is the same.

 

Yep. Counting fail. I counted a reflection as the same shape. 

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Spoiler

No correct answer yet, so revise your own individual counts of possible combinations. Someone ought to get it right soon!

Remember that shapes obtained by rotation do not count, whereas mirror images are fine.

 

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 After recounting...

 

I think CaptainEd had it right but forgot to count the possibilities with the core.

24

3 possibilities if the core is removed
2E 5
2M 2
2C 3
EM 3
EC 6
MC 2

Edited by Thalia

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Spoiler

Close, but no cigar!

Three or more participants have the combinations with respect to some of the possibilities (K+F/C/E, F+F, F+C, F+E, C+C and E+E correct. No-one so far is spot-on with the number of edge versus corner (C+E) combinations, however!

[K = kernel, core or center; F = face or middle, C = corner, E = edge]

 

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I think I got it ....

Spoiler

Using a rubics cube   ;)  , and using rotational symmetry to simplify counting I get,

C- core   ,    M - middle or face ,    E - edge,     R-  as in corner.

CM- 1         MM -2         EE - 5   
CE  -1         ME  -3         ER  -4
CR -1          MR -2          RR -3
       3    +            7      +       12        =    22 ways  

 

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yes I forgot the core. And undercounted CE. And made a typo somewhere. Here’s my new count of CE:

CE  (8) 3-2, 3-6, 3-4, 3-8, 25-2, 25-6 , 25-4, 25-8

So now I agree with 26.

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Spoiler

Well done!

Thalia eventually got the 22 right (albeit with a question mark). Slashpuzzler followed with the correct answer afterwards.

Attaching some images to show the 4 E+C combinations, a second as a summary and a third to show how each fused cube looks like when two cubicles are removed.

 

 

C+E combinations.jpg

22 Shapes.jpg

What they look like.jpg

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Does anyone want to attempt figuring out the number of patterns if three of the cubelets are omitted rather than one or two?

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Here's my argument for 20 distinct shapes.

First, I'm in awe of rocdocmac's images!

The meager sketch below indicates only the visible small cubes.
It does not show the small cube at the center, which I refer to below as B.

The ones that do show are labeled as

  • Corner (C)
  • Edge (E)
  • Face (F)

We agree that C, E, F and B are the four distinct classes of small cubes,
so that removing just
one small cube gives rise to four distinct shapes.
           C
        /    
\   
     E           E
  /               
\
C          F          C
\               /  |
|    E           E    |
|      
\     /       |
E          C          E
|          |          |
|    F     |     F    |   
|          |          |
C          E          C  
  
\       |        /
     E     |     E
       
\  |  /
           C

And now we're removing two small cubes and identifying the equivalence classes.

If we first remove a C,

  • we can remove another C three distinct ways
  • we can remove a E three distinct ways
  • we can remove a F two distinct ways
  • we can remove a B one distinct way

If we first remove a E, having already counted the EC case,

  • we can remove another E four distinct ways
  • we can remove a F three distinct ways
  • we can remove a B one distinct way

If we first remove a F, having already counted the FE and FC cases,

  • we can remove another F two distinct ways
  • we can remove a B one distinct way

There is no BB case, so we're done. And the total is 20.

In summary,

CC 3  
CE 3   EE 4
CF 2   EF 3   FF 2
CB 1   EB 1   FB 1   BB 0

The cases that seem to disagree both involve Edge-cube cases.
Namely,

  • Corner-Edge (CE). Some say 4, I say 3.
  • Edge-Edge (EE). Some say 5, I say 4.

Here are my enumerations:

CE - having removed a Corner, what classes of Edge faces remain? (I claim three.)

  1. Three small cubes E that touch C.
  2. Six small cubes E that do not touch C, but do lie on the same big-cube face.
    (Think of a chess knight move.)
  3. Three small cubes E that touch C'. C' is the small cube diagonally opposite C.

EE - having removed a E, what classes of other Es remain? (I claim four.)

  1. Four small cubes E that touch the first E at one of its corners.
  2. Four small cubes E that touch E' at one of its corners. E' is the small cube diagonally opposite the first E.
  3. Two small cubes E each of which, along with the first E, surround and touch a common F.
  4. The (one) final small cube, E'. Again, E-E' passes through B.

My class descriptions

  • exhaust the 12 (CE) and other-11 (EE) Edge small cubes
  • are stated in a way intended to suggest (at least) that the classes are homogeneous.

I'm eager to hear other class descriptions for these cases.

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OK, I agree with 22. Nice work.

Where my thinking was wrong - I considered left- and right-hand knight moves to be in the same class. (I put all mirror images into the same class.) That's wrong, because mirror image solids (if they lack further symmetry) can in fact be distinct.

Three small cubes? Maybe look at it at some point. rodomac's images provide an advanced start point. That said, it still means finding distinct ways to remove C, E, F and (possibly) B small blocks from 22 different images. :unsure:

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Thank you Bonanova for your input!

To all participants ...

The 3-cubelet removal question has now been submitted as a separate new entry called "Cubicle Stack #2" since the thread sort of "got lost" in the original  Cubicle Stack discussion after having been solved!

Spoiler

Indeed, the images of the 22 solutions posted for the 2-cubelet removal ought to be very handy! But there are a whole lot of more combinations now. Happy playing around, if you have the time of course!

 

Edited by rocdocmac
Rewording

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