Cubicle Stack #2

[rocdocmac]

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. If two of the small cubes are omitted rather than just one, twenty-two distinct shapes are possible (see previously submitted Cubicle Stack at BrainDen.com).

Now, if three of the cubelets are omitted, how many distinct shapes are possible?

 

[Thalia]

A couple more revisions because I can't tell the difference between counting to 4 and counting to 6 at night. 

Spoiler

MMM (2)	5.11.23	5.11.15
CCC (3)	1.3.7	1.3.27	1.9.21
CCX (3)	1.3.14	1.9.14	1.14.27
CEX (4)	1.2.14	1.6.14	1.8.14	1.14.18
CMX (2)	1.5.14	5.14.21
EEX (5)	2.4.14	2.8.14	2.14.16	2.14.18	2.14.26
EMX (3)	2.5.14	5.12.14	5.14.20
MMX (2)	5.11.14	5.14.23
CCM (8)	1.3.5	1.3.13	1.3.17	1.5.9	1.9.11	1.9.13	1.9.23	1.5.27
CMM (5)	1.5.11	5.7.11	5.9.11	5.11.25	1.5.23
EMM (9)	2.5.11	4.5.11	5.6.11	5.8.11	5.11.16	5.11.18	5.11.26	2.5.23	5.10.23
EEM (18)	2.4.5	2.5.8	2.5.10	2.5.12	2.5.16	2.5.18	2.5.20	2.5.22	2.5.24	2.5.26
	5.10.12	5.10.16	5.10.20	5.10.22	5.10.24	5.10.26	5.20.22	5.20.26
CCE (16)	1.3.2	1.3.4	1.3.6	1.3.8	1.3.16	1.3.18	1.3.22	1.9.2	1.9.4	1.9.10
	1.9.12	1.9.20	1.9.22	1.2 .27	1.27..8	1.18.27	1.24.27
EEE  (13)	2.4.6	2.4.10	2.4.12	2.4.16	2.4.18	2.4.20	2.4.22	2.4.24	2.4.26	2.8.20
	2.8.22	2.16.24	2.18.22
CEE (22)	1.2.4	1.2.6	1.2.8	1.2.12	1.2.16	1.2.18	1.2.20	1.2.22	1.2.24	1.2.26
	1.6.8	1.6.12	1.6.16	1.6.18	1.6.22	1.6.24	1.6.26	1.8.12	1.8.18	1.8.24
	1.8.26	1.18.24
CEM (24)	1.2.5	1.4.5	1.5.6	1.5.8	1.5.10	1.5.12	1.5.16	1.5.18	1.5.20	1.5.22
	1.5.24	1.5.26	2.5.19	4.5.19	5.6.19	5.8.19	5.10.19	5.12.19	5.16.19	5.18.19
	5.19.20	5.19.22	5.19.24	5.19.26

 

New total is 139.

Thanks for checking all these. That's a lot of numbers!

On 1/23/2018 at 12:55 AM, rocdocmac said:

  Hide contents

I wonder whether an equation exists that one could use to predict the next number (4-cubelet removal or T4) and higher T-numbers in the sequence. Such a sequence should start with:

T0, T1, T2, T3,  ..., (i.e. 1, 4, 22, 13?, ..., with T0 = 1, indicating a solid cube resulting from 0-cubelet removal),

and end with:

..., T23, T24, T25, T26, T27 (i.e. 13?, 22, 4, 1, 0 with T27 = 0, which is equivalent to the removal of all 27 cubelets leaving nothing.

The maximum value would be for T13.

T26 would leave 1 cube. But T25 leaving 2 cubelets leaves more than 4 shapes assuming the two cubelets don't have to be connected. . .

 

[Molly Mae]

I'm going to stop.  My answer is 231.

First, I tried mapping out all of the possible relationships between two cubelets (22, as established previously).  I lettered them from A to V.  I was going to permutate them against each other in such a way as to return a three letter unique output that forms a triangle.

Example Relationships:
Core to Edge: A
Core to Corner: B
Edge to Nearby Edge: C
Edge to Distal Edge: D
Edge to Opposite Edge: E
And so on.

As long as the output generates a legitimate unique triangle, the pattern is accepted.  So AAC is a triangle that connects the core to two edges, both of which are near each other.  AAA and ABC, for example, are invalid outputs, since they don't form a triangle.  ACA and CAA would also be invalid, since this is an implementation of AAC.

Before beginning the problem, I had estimated that the answer would be around 90 and easily countable.  I should have known better since the likes of Thalia, Mr Ed, and Bonanova were struggling with counting the previous iteration with 22 unique configurations.  I figured if I could name unique connections, it wouldn't be so difficult.  After a short period of that, I feel now that the answer is probably less than 250, though I have no solid proof.

[rocdocmac]

Nice approach!

I'll wait for more solutions to come in before starting to comment.

Spoiler

Watch out for shapes that can be obtained by rotation, thereby duplicating some of them!

 

[Thalia]

Pretty sure these counts are wildly inaccurate but here goes...:

Spoiler

C	C	C	3
C	C	E	21
C	C	M	5
C	E	E	24
C	E	M	24
C	M	M	6
E	E	E	12
E	E	M	20
E	M	M	10
M	M	M	2
C	C	X	3
C	E	X	4
C	M	X	2
E	E	X	5
E	M	X	3
M	M	X	2
			146

 

[rocdocmac]

Spoiler

6/16 correct!

 

[Molly Mae]

6 hours ago, Thalia said:

Pretty sure these counts are wildly inaccurate but here goes...:

  Hide contents

C	C	C	3
C	C	E	21
C	C	M	5
C	E	E	24
C	E	M	24
C	M	M	6
E	E	E	12
E	E	M	20
E	M	M	10
M	M	M	2
C	C	X	3
C	E	X	4
C	M	X	2
E	E	X	5
E	M	X	3
M	M	X	2
			146

 

A few adjustments:

CCC 2
MMM 1

[Thalia]

4 hours ago, Molly Mae said:

 

  Hide contents

A few adjustments:

CCC 2
MMM 1

 

For MMM, you can have the middle out of opposite sides and a middle between them or you can have a middle out of the top of the cube and two out of sides that are next to each other. I will provide the numbers using CaptainEd's system later.

[Thalia]

Revision to CCC and EEX with numbers.

Spoiler

MMM (2)	5.11.23	5.11.15
CCC (2)	1.3.7	1.3.27
CCX (3)	1.3.14	1.9.14	1.14.27
CEX (4)	1.2.14	1.6.14	1.8.14	1.14.18
CMX (2)	1.5.14	5.14.21
EEX (6)	2.4.14	2.6.14	2.8.14	2.14.16	2.14.18	2.14.26
EMX (3)	2.5.14	5.12.14	5.14.20
MMX (2)	5.11.14	5.14.23

 

[Thalia]

Another stab at it.

Spoiler

MMM (2)	5.11.23	5.11.15
CCC (2)	1.3.7	1.3.27
CCX (3)	1.3.14	1.9.14	1.14.27
CEX (4)	1.2.14	1.6.14	1.8.14	1.14.18
CMX (2)	1.5.14	5.14.21
EEX (6)	2.4.14	2.6.14	2.8.14	2.14.16	2.14.18	2.14.26
EMX (3)	2.5.14	5.12.14	5.14.20
MMX (2)	5.11.14	5.14.23
CCM (8)	1.3.5	1.5.9	1.5.19	1.5.21	1.5.25	1.5.27	5.19.21	5.19.27
CMM (5)	1.5.11	5.7.11	5.9.11	5.11.25	1.5.23
EMM (8)	2.5.11	4.5.11	5.6.11	5.8.11	5.11.18	5.11.24	2.5.23	5.10.23
EEE  (11)	2.4.6	2.4.10	2.4.12	2.4.16	2.4.18	2.4.20	2.4.22	2.4.24	2.4.26	2.8.20	2.8.22
EEM (18)	2.4.5	2.5.8	2.5.10	2.5.12	2.5.16	2.5.18	2.5.20	2.5.22	2.5.24	2.5.26
	5.10.12	5.10.16	5.10.20	5.10.22	5.10.24	5.10.26	5.20.22	5.20.26
CCE (15)	1.2.3	1.2.7	1.2.9	1.2.19	1.2.21	1.2.25	2.7.9	2.7.19	2.7.21	2.7.25	2.7.27
	2.9.19	2.9.21	2.9.25	2.9.27
CEE (21)	1.2.4	1.2.6	1.2.8	1.2.12	1.2.16	1.2.18	1.2.20	1.2.22	1.2.24	1.2.26
	1.6.8	1.6.12	1.6.16	1.6.18	1.6.22	1.6.24	1.6.26	1.8.12	1.8.18	1.8.24	1.8.26
CEM (24)	1.2.5	1.4.5	1.5.6	1.5.8	1.5.10	1.5.12	1.5.16	1.5.18	1.5.20	1.5.22	1.5.24	1.5.26
	2.5.19	4.5.19	5.6.19	5.8.19	5.10.19	5.12.19	5.16.19	5.18.19	5.19.20	5.19.22	5.19.24	5.19.26

 

New total is 134. At least one of those is wrong though according to having 6/16 correct the first time...

 

[Molly Mae]

5 hours ago, Thalia said:

 

  Hide contents

For MMM, you can have the middle out of opposite sides and a middle between them or you can have a middle out of the top of the cube and two out of sides that are next to each other. I will provide the numbers using CaptainEd's system later.

 

Oh, jeez, you're right.  I dunced that pretty good.

[rocdocmac]

 

Thalia, I'll check your current numbering later, but ...

Spoiler

... a big apology!

Your first score was actually 7/16! Now you've changed 2 of the correct ones to incorrect and 2 other incorrect ones to correct.

Thus, as your latest stab count stands, you again have 7/16.

Sorry for my initial miss checking!

 

 

[Thalia]

Think I found one of the 2 I changed from correct to incorrect. But I think that would leave one of these as still wrong though I can't figure out which. . .

Spoiler

MMM (2)	5.11.23	5.11.15
CCC (3)	1.3.7	1.3.27	1.9.21
CCX (3)	1.3.14	1.9.14	1.14.27
CEX (4)	1.2.14	1.6.14	1.8.14	1.14.18
CMX (2)	1.5.14	5.14.21
EMX (3)	2.5.14	5.12.14	5.14.20
MMX (2)	5.11.14	5.14.23

 

[rocdocmac]

34 minutes ago, Thalia said:

 

34 minutes ago, Thalia said:

Think I found one of the 2 I changed from correct to incorrect. But I think that would leave one of these as still wrong though I can't figure out which. . .

  Hide contents

MMM (2)	5.11.23	5.11.15
CCC (3)	1.3.7	1.3.27	1.9.21
CCX (3)	1.3.14	1.9.14	1.14.27
CEX (4)	1.2.14	1.6.14	1.8.14	1.14.18
CMX (2)	1.5.14	5.14.21
EMX (3)	2.5.14	5.12.14	5.14.20
MMX (2)	5.11.14	5.14.23

Spoiler

Yes, you have indeed found one of them!

Have another look at XCE and XEE. If you add the other two newly combinations found by you, you will have at least 10/16 correct.

Hope it helps!

We can later deal with the more difficult ones!

 

 

 

Edited January 16, 2018 by rocdocmac
Moved to spoiler

[Thalia]

So 6/7 in the last post and check CEX?

[Thalia]

I see what you mean about CEX although I'm not sure how to show that with the numbering system. That throws a wrench in my counting. . .

Spoiler

MMM (2)	5.11.23	5.11.15
CCC (3)	1.3.7	1.3.27	1.9.21
CCX (3)	1.3.14	1.9.14	1.14.27
CEX (3)	1.2.14	1.6.14	1.14.18
CMX (2)	1.5.14	5.14.21
EEX (5)	2.4.14	2.8.14	2.14.16	2.14.18	2.14.26
EMX (3)	2.5.14	5.12.14	5.14.20
MMX (2)	5.11.14	5.14.23

 

[Thalia]

Nevermind the CEX numbers. I really need to find a cube I can hold...

[rocdocmac]

By the way, Thalia ...

Spoiler

... one of your "high" numbers for multiple combinations is correct. Found that after checking, rechecking, counting, and recounting. Thus, add one more to your score!

[Thalia]

Lol. I'm not sure what my score is anymore. Can you score this?

MMM (2)     

CCC (3)   
CCX (3)    
CEX (4)    
CMX (2)      
EEX (5)     
EMX (3)       
MMX (2)      
CCM (8)
CMM (5)   
EMM (8)
EEE (11)
EEM (18)
CCE (15)
CEE (21)
CEM (24)

[Thalia]

Recount. EEM and EEC intentionally left out.

Spoiler

MMM (2)

CCC (3)

CCX (3)

CEX (4)

CMX (2)

EEX (5)

EMX (3)

MMX (2)

CCM (8)

CMM (5)

EMM (9)

EEE  (12)

CCE (16)

CEM (24)

About CEX

Spoiler

When I had 4, you said to check it. Unless I'm missing something, the shapes involving the core should have the same counts as in the two cubelets question. CE made 4 shapes. CEX should make 4 as well...

 

[rocdocmac]

@Thalia

Spoiler

10/16

 

[rocdocmac]

@Thalia

Spoiler

Count before your latest was 10/16. After recount with EEM and EEC left out ... 11/14

 

[rocdocmac]

@Thalia

Spoiler

About CEX : I'll get back to you on this one. Maybe the "four" in the two-cubicle removal is incorrect. Please have another look!

 

[Thalia]

CEX

I picked an edge and went through the 8 corners from there. The core doesn't really matter at this point so if we look at it as removing CE...

The corners are 1, 3, 7, 9, 19, 21, 25, and 27. So starting with cubelet number 2 as the edge (numbered starting with the side facing us), you can remove corner 1. But if you number from the top face, that shape can be called 1,3. Each time you remove an corner, there is a second corner that will give you the same shape. 2,7 from the side is 2,21 from the top. 9 pairs with 19. 25 pairs with 27. So I'm counting 4. What am I missing?

[rocdocmac]

@Thalia

Spoiler

Sorry, I was wrong and you're not missing anything! The 2-cubelet removal definitely gives 4 CE combinations, so you're right with the center additionally removed (CEX) too! Your score now 12/14.

 

Edited January 19, 2018 by rocdocmac
typo

[Thalia]

Only 13 this time.

Spoiler

MMM (2)

CCC (3)

CCX (3)

CEX (4)

CMX (2)

EEX (5)

EMX (3)

MMX (2)

CCM (8)

CMM (5)

EMM (9)

EEM (18)

CCE (16)