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# Cubicle Stack #2

Go to solution Solved by Thalia,

## Question

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?

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@Thalia ...

Spoiler

I've done the exercise of removing three cubelets from the stack many years ago and put it aside after solving the 2-cubelet removal. Even started with a 4-cubelet removal, but I lost interest in that one because of the tediousness involved. Let's not say any of us is/was right or wrong (I most probably am!), but rather agree on the eventual solution. That said, today I am in agreement with you on all 13 of your latest figures! Three more for us to finalize.

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[No content, sorry!]

Edited by rocdocmac
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New number for EEE. But I'm still getting the same result for CEE and CEM.

Spoiler
 MMM (2) 2 CCC (3) 3 CCX (3) 3 CEX (4) 4 CMX (2) 2 EEX (5) 5 EMX (3) 3 MMX (2) 2 CCM (8) 8 CMM (5) 5 EMM (9) 9 EEM (18) 18 CCE (16) 16 EEE  (11) 11 CEE (21) 21 CEM (24) 24

Total is 136.

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I agree with 24 for CEM, but we're still not on par with CEE and EEE. The total for 3-cubelet removal (say T3 = 13?) is surely somewhere between 135 and 140.

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.

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Full guess with numbered cubelets and revision to CEE.

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  (11) 2.4.6 2.4.10 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 (23) 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 1.18.26 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 138.

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Spoiler

EEE <> 11

CEE <> 23

All others correct.

Total <> 138

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@Thalia ...

Spoiler

I actually noticed something in this line yesterday!

T26<>1: If 26 cubelets are removed, one would be left with either X, M (= F), C or E, i.e. T26 = 4 (not "1"!).

We now know that T3 = 139.

Thus T0, T1, T2, T3, ... = 1, 4, 22, 139, ...

The sequence should therefore end with

... T24, T25, T26, T27 = ... 139, 22, 4, 0

Maximum values would be for T13 = T14

Edited by rocdocmac
Expanded response
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37 minutes ago, rocdocmac said:
Spoiler

Tried to update the explanation, but time ran out!

Series starts with T0 = 1, T1 = 4, T2 = 22, T3 = 139, ... and ends with its reversal, i.e. ... T24 = 139, T25 = 22, T26 = 4, T27 = 1.

Only one possibility for T27, which is zero cubelets left.

Edited by rocdocmac
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57 minutes ago, rocdocmac said:

Numbering system used ...

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3 hours ago, rocdocmac said:

@Thalia ...

Hide contents

I actually noticed something in this line yesterday!

T26<>1: If 26 cubelets are removed, one would be left with either X, M (= F), C or E, i.e. T26 = 4 (not "1"!).

We now know that T3 = 139.

Thus T0, T1, T2, T3, ... = 1, 4, 22, 139, ...

The sequence should therefore end with

... T24, T25, T26, T27 = ... 139, 22, 4, 0

Maximum values would be for T13 = T14

When you remove a cubelet, the kind you remove matters because it changes the shape. When you take away 26, it doesn't matter which one is left. There may be 4 different labels but in the end, when it comes to the remaining shape, all you see is a cube.

Another thing. For T27, there is one possibility. But given the phrasing of the original question, does nothing count as a shape?

Edited by Thalia
Clarification
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2 hours ago, Thalia said:

When you remove a cubelet, the kind you remove matters because it changes the shape. When you take away 26, it doesn't matter which one is left. There may be 4 different labels but in the end, when it comes to the remaining shape, all you see is a cube.

Another thing. For T27, there is one possibility. But given the phrasing of the original question, does nothing count as a shape?

Thanks Thalia,

Interesting twist in the tale, for sure!

Yes, surely 26 cubelets removed leaves a single cubelet, but where was the remaining one positionally located/"fixed" before removal of the other 26? Should be one specific remnant shape for each combination and, therefore,  four variants/labels/possibilties.

"Nothing" or "no shape" may be regarded as a "blank" stacked cube.

(1) What's your suggestion for the end of the sequence?

(2) What do our other BrainDenners think?

(3) How many distinct shapes can one expect for a 4-cubelet removal? Just a ball-park figure, perhaps!

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41 minutes ago, rocdocmac said:

Thanks Thalia,

Interesting twist in the tale, for sure!

Yes, surely 26 cubelets removed leaves a single cubelet, but where was the remaining one positionally located/"fixed" before removal of the other 26? Should be one specific remnant shape for each combination and, therefore,  four variants/labels/possibilties.

"Nothing" or "no shape" may be regarded as a "blank" stacked cube.

(1) What's your suggestion for the end of the sequence?

(2) What do our other BrainDenners think?

(3) How many distinct shapes can one expect for a 4-cubelet removal? Just a ball-park figure, perhaps!

2) I wouldn't use "what shapes are left" but instead keeping with "after removing 27 cubelets, how many possibilities are there?" Which is 1.

3) ~950

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I did some quick counting for the easier ones. I've noticed that at least up to this point, the number of combinations of C, E, M, and X is (# cubelets removed+1)^2. There are 25 combinations for 4 cubelets removed. The ones including X would be the same as the counts for 3 cubelets without X. I did not count the combinations that have 2 or more E's or CCEM, CCMM, AND CEMM. I got 191 for the rest. I'm guessing the ones I didn't count are going to be pretty high so maybe around 500?

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Plotting the sequence 1, 4, 22, 139, ... as a scatter diagram, one can fit a 3rd degree polynomial curve with equation y = 14x^3 - 34.5x^2 + 23.5x + 1 (x = 0, 1, 2, 3).  Using this graph as an estimator, one can guess that the number of shapes will be in the region of 439 if four are removed. Therefore the guestimate for 500 is not far out!

T25 appears to be 9 (see attached image) and I agree that T26 is 1. The roots indicate the distance between the two cubelet-middle points in the grid (x = y = z = 1).

Thus, would the sequence end with ..., 9, 1, 0 (the 0 indicating no shape at all)?

Will determine T24 a bit later!

Edited by rocdocmac
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@Thalia ...

Done some counting, but still a lot to do! Nine combinations of 4-cubelet removal still to be done.

Spoiler

I found 190 so far, whereas you got 191, i.e. if we counted the same groups.

Here's my list for checking on your side (*incomplete) ...

Edited by rocdocmac
Moved parts to spoiler
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Sorry to say I didn't save my counts. Slightly burnt out from counting, recounting, and rerecounting the previous ones. lol. Maybe I'll try again when I've had more time to recover.

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@Thalia ...

Virtually completed counting (4-cubelet removal). Still have to verify, but the total seems to be less than 840!

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... and above 800 ...

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Well, there goes your curve estimation. I don't remember the situation but there were certain combinations that I could calculate with some multiplication. I think one of them was a CE combination. But of course the other combinations didn't feel like cooperating. I'll try to work on this over the weekend.

Agree on your pink combinations except for XMEE. Should be the same count as EEM, right?

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That "19" was a typo, sorry! Latest included ...

Spoiler

Total : 839 (max)

Edited by rocdocmac
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@Thalia (re 4-cublet removal) ...

Spoiler

MMEE.xlsx

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@ Thalia (4-cubelet removal) ...

Spoiler

CEEE.xlsx

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@Thalia ... Please ignore my previous post. Use the later spreadsheet in next message.

Edited by rocdocmac
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@Thalia ... revised MMEE

Spoiler

MMEE.xlsx

Edited by rocdocmac

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