[Prof. Templeton]

I've been giving more thought to my original stacking blocks problem where we figured out how to make a 4 layer structurally stable square pyramid with fewer blocks by leaving some areas hollow. Instead of 30 total blocks ([4x4=16] + [3x3=9] + [2x2=4] + [1x1=1] = 30) we could make what appeared on the outside to be a solid square pyramid with only 26 blocks ([(4x4)-4+1=13] + [(3x4)-4=8] + [(2x4)-4=4] + [1x1=1] = 26. But what if we expand this idea to a larger pyramid? What if we had, say, 1300 blocks? If the stucture were solid we could make a square pryamid with 15 levels (15² + 14² + 13² + 12² ... + 1²) which would take a total of 1240 blocks. We only need enough blocks to keep the structure from caving in and support the upper levels so some areas inside the pyramid can be left hollow. How many levels can we create with the 1300 blocks and can a general rule be formed for all square pryamids?

[Prof. Templeton]

Correct link to the original stacking block topic here.

Some clarification: [(4x4)-4+1=13] this represents the four sides of the 4x4 layer minus the corner blocks plus the one block in the center that helps support the 3x3 layer. [(3x4)-4=8] represents the four sides of the 3x3 layer minus the corner blocks.

Edited January 10, 2009 by Prof. Templeton

[Guest]

Correct link to the original stacking block topic here.

Some clarification: [(4x4)-4+1=13] this represents the four sides of the 4x4 layer minus the corner blocks plus the one block in the center that helps support the 3x3 layer. [(3x4)-4=8] represents the four sides of the 3x3 layer minus the corner blocks.

Just for some additional clarification, your formula (roughly, but not quite this... [(layer x 4)-4]) solves for the number of blocks at the perimeter of a layer. When I first read it, I was trying to visualize "the four sides of the (4x4 layer minus the corner blocks)" instead of "(the four sides of the 4x4 layer) minus the corner blocks".

Edited January 10, 2009 by Phatfingers

[Prof. Templeton]

Just for some additional clarification, your formula (roughly, but not quite this... [(layer x 4)-4]) solves for the number of blocks at the perimeter of a layer. When I first read it, I was trying to visualize "the four sides of the (4x4 layer minus the corner blocks)" instead of "(the four sides of the 4x4 layer) minus the corner blocks".

Sorry for the confusion. Yes those are the blocks needed for the perimeter. Notice that at layer 4 we start to need additional block(s) on the inside of the perimeter to help support the layer(s) above. Basicly I'm asking what is the minimum amount of blocks we need inside the perimeter to hold the whole thing up? Is there a formula that can be made for N levels?

[Guest]

Sorry for the confusion. Yes those are the blocks needed for the perimeter. Notice that at layer 4 we start to need additional block(s) on the inside of the perimeter to help support the layer(s) above. Basicly I'm asking what is the minimum amount of blocks we need inside the perimeter to hold the whole thing up? Is there a formula that can be made for N levels?

Part of the equation might involve this. If you have an equation for a solid pyramid of blocks f(height), then the outer shell would be f(height)-f(height-2).

[HoustonHokie]

At the end of your last puzzle, the solution involved a single block in the center of the 4th layer, rotated 45 degrees to the rest of the pyramid. That single rotated block could become the apex of an internal pyramid below it, and the outer one could remain hollow. I think the general solution involves a series of rotating internal pyramids, with a new one starting every 4th level. So:

Levels 1 - 3 have one outer rim of blocks

Levels 4 - 6 have two rings - the outer one square and hollow and the inner one equivalent to rotating levels 1-3 by 45 degrees

Levels 7-9 contain 3 rings - the outer one square and hollow, the other two equivalent to those found in Levels 4-6, just rotated by 45 degrees.

And so on...

There's definitely a way to create a recursive solution to the problem:

f(1) = 1

f(x) = (x-1)*4 + f(x-3) - of course, f(x-3) is only defined for x >= 4

I think this gives a complete outer shell of the pyramid with minimal internal reinforcement. Does that make sense?

[Guest]

At the end of your last puzzle, the solution involved a single block in the center of the 4th layer, rotated 45 degrees to the rest of the pyramid. That single rotated block could become the apex of an internal pyramid below it, and the outer one could remain hollow. I think the general solution involves a series of rotating internal pyramids, with a new one starting every 4th level. So:

Levels 1 - 3 have one outer rim of blocks

Levels 4 - 6 have two rings - the outer one square and hollow and the inner one equivalent to rotating levels 1-3 by 45 degrees

Levels 7-9 contain 3 rings - the outer one square and hollow, the other two equivalent to those found in Levels 4-6, just rotated by 45 degrees.

And so on...

There's definitely a way to create a recursive solution to the problem:

f(1) = 1

f(x) = (x-1)*4 + f(x-3) - of course, f(x-3) is only defined for x >= 4

I think this gives a complete outer shell of the pyramid with minimal internal reinforcement. Does that make sense?

I can visualize that the four corners of the (2x2) layer of the first inner pyramid would brace the middle two blocks of each side of the (4x4) layer above it. Nice.

One layer below that, a (3x3) layer of the first inner pyramid has four corners each bracing one inner block of the (5x5) layer above it. If I'm not mistaken, that leaves two blocks on each side of the (5x5) outer layer unsupported. I'm still personally at a loss as to what the inner scaffolding looks like if it's not as you described.

If you telescoped a solid (n x n) base, it would match the perimeters of all the even or odd layers. So, if you telescope the bottom two layers of a solid pyramid, and interleaved them, it would make a complete outer shell. Therefore, a simple nonrecursive function for the outer shell can be derived from adding the number of blocks of the bottom two solid layers:

n^2 + (n-1)^2, where n is the height of the pyramid (or the number of blocks at the side of the base).

[Guest]

I tried modelling the problem with my kids' duplos. My strategy was to place scaffolding blocks first to each support two blocks from the perimeter of the layer outer pyramid immediately above it. Then, I would add any additional blocks necessary to support the scaffolding from that same layer. What unfolded was a strange fractal-like progression. Here is a photo of what I came up with. Note that I removed four green blocks from the center so you can see underneath.

I thought I had a good progression of 1, 4, 9, ... but then the next layer was 12, so I'm still scratching my head on this one.

Edited January 11, 2009 by Phatfingers

[Guest]

I thought I had a good progression of 1, 4, 9, ... but then the next layer was 12, so I'm still scratching my head on this one.

A different scaffolding pattern yielded a viable structure of 1, 4, 9, 9, 14.

An inner pyramid rotated 45 degrees can only fit in an outer pyramid shell for a few layers deep. The outside corner of the inner pyramid grows exponentially [ sqrt(2 * side^2)/2 ], while the inside flat of the outer pyramid has linear growth [ (side-2)/2 ]. It outgrows the outer pyramid's inner boundary by the third layer.

[HoustonHokie]

A different scaffolding pattern yielded a viable structure of 1, 4, 9, 9, 14.

An inner pyramid rotated 45 degrees can only fit in an outer pyramid shell for a few layers deep. The outside corner of the inner pyramid grows exponentially [ sqrt(2 * side^2)/2 ], while the inside flat of the outer pyramid has linear growth [ (side-2)/2 ]. It outgrows the outer pyramid's inner boundary by the third layer.

You're right. I was in the middle of drawing it out when I saw your post. I had come to the same conclusion and was trying to see if there was any way to salvage my idea. No such luck so far

[HoustonHokie]

A different scaffolding pattern yielded a viable structure of 1, 4, 9, 9, 14.

Not sure about that scaffolding pattern, but I found a 1, 4, 9, 16 pattern that works. I also found the 1, 4, 9, 12 pattern that you saw. Here's the formulae:

If L is the layer, the number of scaffolding blocks on that layer is

1. (L-3)²

2. L even: (L-3)²

L odd: [(L-2)² - 1] / 2

I've taken both out to 9 layers deep and they work. I just can't get my pictures to upload because I've blown my attachment space . Here's how I would write a description for each:

1. Solid scaffolding except the diagonals are removed. If L is odd, you will get a perfect X shaped diagonal. If L is even, the center of the diagonal will be 2x2 instead of 1x1. In this case, add a central block rotated 45 degrees. When viewed from below, it looks like you're creating 4 small triangular pyramids inside the larger square pyramid.

2. Exactly the same as #1 on the even layers. For the odd layers, use a checkerboard pattern.

I think the professor had pattern #1 in mind when he wrote the OP because 1300 blocks makes exactly 16 layers that way. But pattern #2 is more efficient and requires only 1160 for 16 layers and can accomplish 17 layers with 1336 blocks.

We may be able to improve on this if we can figure out a good checkerboard pattern for the even layers.

I'll see if I can improve the efficiency of my graphics and may post my pictures later .

It may be good to define exactly what "stable" is for this exercise. I've taken it to mean that one of the following conditions is true:

1. a block is completed supported from below

2. a block has one side completed supported* and at least one point on the opposite side supported

3. a block has two adjacent sides completely supported*

* when I refer to a side as completely supported, I mean the actual edge plus some offset (generally half the width of the block, but I suppose it could be less)

[HoustonHokie]

Couldn't upload the images here, so I've uploaded the images there:

Blocks 1

Blocks 2

[Prof. Templeton]

Holy Cow! You put a lot of work into this. I knew the first pattern would work, but I confess I didn't know if it could be improved upon. I appreciate all the effort you've put into your proof. 1000 Kudos points!*

*Kudus points are non-transferable and only redeemable at approved locations. Certain restrictions apply.

[HoustonHokie]

Holy Cow! You put a lot of work into this. I knew the first pattern would work, but I confess I didn't know if it could be improved upon. I appreciate all the effort you've put into your proof. 1000 Kudos points!*

*Kudus points are non-transferable and only redeemable at approved locations. Certain restrictions apply.

YEESSSSS!!!! I'm going to Disneyworld! Or something like that... Got to read the fine print

[Guest]

YEESSSSS!!!! I'm going to Disneyworld! Or something like that... Got to read the fine print

Ha ha! I've been obsessing over this one, too. I really enjoyed how you detailed it out! I arrived at a similar solution that's as optimal as I can get it while still being easy to build. I experimented a little with positioning 3 blocks in the same space as four. It was slightly precarious, but could, as HoustonHokie pointed out, meet someone's definition of stable. The solution proposed here does not do that, but is a natural extension of the original puzzle.

Each layer "n" (after the 5th layer) has three components to it:

1. The "shell", which is the visible outer layer.

2. The "support", which is a single layer of blocks adjacent to the shell, staggered such that each block always supports two blocks of the shell above it. This was necessary because the rotated pattern would otherwise always leave some blocks from the shell unsupported.

3. The "fill" which is a checkerboard staggering of blocks all facing the same direction per layer, but rotated 45 degrees from the layer above. Each even layer contains the diagonal set of blocks, which are positioned above the voids left by the layer above or below. The odd layers hold a checkerboard pattern which is always open on all four corners.

The Professor gave us a formula for the "shell": 4n-4

The formula for the "support" is: 2(n-3)

The formula for the "fill" is: ((n-4)^2 - (n mod 2))/2

A block count for any layer, starting at layer 5, is [ shell + support + fill ].

Layer	Shell	Support	Fill		Layer	Total

1					1	1

2					4	5

3					8	13

4					13	26

5	16	4	0		20	46

6	20	6	2		28	74

7	24	8	4		36	110

8	28	10	8		46	156

9	32	12	12		56	212

10	36	14	18		68	280

11	40	16	24		80	360

12	44	18	32		94	454

13	48	20	40		108	562

14	52	22	50		124	686

15	56	24	60		140	826

16	60	26	72		158	984

17	64	28	84		176	1160

18	68	30	98		196	1356

[/codebox]

This has been a really fun puzzle to noodle over! Thanks for putting it out there.