Cube stacking puzzle
Evan Whitney
posted. I made up this version myself, but take no credit for the physics :-).
You wish to build a symmetrical, marginally stable arch by stacking two columns
of perfect 1-cubic foot cubes. The columns are separated 5 feet at the base
of the arch, and the cubes are stacked one on top of the other and offset
such that the two columns gradually meet at the top. Except for the bottom
pairs of cubes, each cube touches exactly two other cubes. The two top cubes
touch only each other and the cubes immediately below them; the two bottom
cubes touch only the ground and the cubes immediately above. All other cubes
touch only those immediately below and above them.
__ __
_|__|__|_
_|__| |__|_
Rough idea -> |__| |__|
|__|<- 5ft ->|__|
Each column must stand under its own weight without toppling. The cubes at
the top barely touch; i.e., the two columns are not leaning against each other
for support. Nothing may be used to cement the cubes together.
Each cube has a length, width, and height of 1 foot, smooth surface, positive
mass, uniform density, and is extremely sturdy (won't deform).
What is the minimum number of cubes necessary to build this arch?
Steve Thomas
>You wish to build a symmetrical, marginally stable arch by stacking two columns
>of perfect 1-cubic foot cubes. The columns are separated 5 feet at the base
>of the arch, and the cubes are stacked one on top of the other and offset
>such that the two columns gradually meet at the top.
> __ __
> _|__|__|_
> _|__| |__|_
> Rough idea -> |__| |__|
> |__|<- 5ft ->|__|
>What is the minimum number of cubes necessary to build this arch?
This is a simple variant of the `coin-stacking' problem. The amount
of overhang given a single column is H_n - 1, where H is the harmonic
series (1 + 1/2 + 1/3 ... to n terms). To solve the problem as
stated, we need to find the smallest n such that H_n >= 3.5; H_19 is
about 3.54+, and H_18 is 3.49+.
We need 38 cubes to build the required structure.
The thing I find most fascinatingly counter-intuitive about this whole
exercise is that it's possible to bridge an arbitrarily large gap (though
you do need an exponentially large number of cubes.
Steve
Jeff Kenton
>
> You wish to build a symmetrical, marginally stable arch by stacking
> __ __
> _|__|__|_
> _|__| |__|_
> Rough idea -> |__| |__|
> |__|<- 5ft ->|__|
>
>
> What is the minimum number of cubes necessary to build this arch?
I assume from your conditions that the physics of this is equivalent to
building a single unsupported column which extends out 2.5 feet past the
base cube.
If so, what we are trying to do is find out how far out we can extend each
succeeding cube without it falling over.
Base cube -- supported by table
2nd cube -- extend out 1/2 foot
3rd cube -- extend another 1/3 foot
4th cube -- extend another 1/4 foot
. . .
Continue this process until the sum >= 2.5 feet, and double the result for
two columns. In theory, you could extend this to any distance if you build
high enough, since the sum does not converge. (It grows as ln n).
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
jeff kenton --- temporarily at jke...@pinocchio.encore.com
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Robert D. Silverman
:posted. I made up this version myself, but take no credit for the physics :-).
:
:You wish to build a symmetrical, marginally stable arch by stacking two columns
:of perfect 1-cubic foot cubes. The columns are separated 5 feet at the base
:of the arch, and the cubes are stacked one on top of the other and offset
:such that the two columns gradually meet at the top. Except for the bottom
:What is the minimum number of cubes necessary to build this arch?
I can give a quick, approximate answer: e^2.5.
Proof:
To maintain stability the minimal overlap of the n'th cube from the
bottom must protrude 1/n of the way from the cube it rests on. i.e.
the 2nd cube protrudes 1/2 way out from the first, the 3rd another 1/3
of the way out etc.
Now: 1 + 1/2 + 1/3 + ... + 1/n = log(n) + gamma + o(1)
where gamma is Euler's constant. We must therefore have:
1 + 1/2 + 1/3 + ... + 1/n > 2.5 [for each arch], so
log(n) ~ 2.5 and n ~ e^2.5.
The exact answer is the first n such that 1 + 1/2 + .. + 1/n > 2.5
--
Bob Silverman
#include <std.disclaimer>
Internet: b...@linus.mitre.org; UUCP: {decvax,philabs}!linus!bs
Mitre Corporation, Bedford, MA 01730
Jeff Kenton
>
> The exact answer is the first n such that 1 + 1/2 + .. + 1/n > 2.5
>
I believe that the problem required 5 feet *between* the bases of the
columns, so you need to remove the first (and largest) term in the series.
In your series, I believe n = 7 works. Without the first term 19 cubes
are required (times 2 for both columns).
Dave Dodson
>What is the minimum number of cubes necessary to build this arch?
38 if you require all of the cubes to have the same orientation.
Now, how many are required if the distance between the base cubes
is increased to 10 feet? 20 feet?
----------------------------------------------------------------------
Dave Dodson dod...@convex.COM
Convex Computer Corporation Richardson, Texas (214) 497-4234
Dean Hickerson
says:
>
>You wish to build a symmetrical, marginally stable arch by stacking two
>columns of perfect 1-cubic foot cubes. The columns are separated 5 feet at
> __ __
> _|__|__|_
> _|__| |__|_
> Rough idea -> |__| |__|
> |__|<- 5ft ->|__|
>
>Each column must stand under its own weight without toppling. The cubes at
>the top barely touch; i.e., the two columns are not leaning against each other
>for support. Nothing may be used to cement the cubes together.
>What is the minimum number of cubes necessary to build this arch?
168. As several people have mentioned, this is equivalent to an old
classic. (I think it was discussed in one of Martin Gardner's columns many
years ago.) But the postings I've seen all missed a factor of 2: the
overhangs of the individual cubes are 1/2, 1/4, 1/6, ..., not 1/2, 1/3, 1/4,
... Also, the maximal overhang occurs at the top of the stack, not the
bottom.
Suppose there are N cubes in each half. Then the maximum extension past
the edge of the bottom cube is obtained as follows. The top cube extends
1/2 foot past the edge of the 2nd, the 2nd extends 1/4 foot past the edge
of the 3rd, ..., the (N-1)'th extends 1/2(N-1) foot past the edge of the
bottom cube. So for the problem at hand we need
1 1 1 1
- + - + - + ... + ------ > 2.5
2 4 6 2(N-1)
In other words, the (N-1)'th harmonic number H(N-1) must be greater than 5.
According to Macsyma, H(82) < 5 < H(83). So we need N>=84. The total
number of cubes needed is twice that, namely 168.
-------
Dean Hickerson, Dept. of Math., Penn State Univ., University Park, PA 16802
h...@psuvm.psu.edu
Evan Whitney
to stack objects at 1/2 + 1/3 + 1/4 + 1/5 ... + 1/n, they topple
immediately at the second term (1/3), and even a slight adjustment will
not balance.
According to my calculations of centers of mass and experimental verification,
the progression is 1/2 + 1/4 + 1/6 + 1/8 + ... + 1/2n. For n=83 this totals
2.501, but you have to add 1 for the base, so a total of 168 cubes is required
to build the arch. I should also note that the progression proceeds from
the top of the arch down, not the bottom up as someone suggested. Four bricks,
books, or Kleenex boxes (what I used) is all that is necessary to verify
the progression.
But, there is a twist (literally) to this puzzle that will allow you to
build the arch using fewer cubes. Stack the cubes along their surface
diagonals. This multiplies the effective span by SQRT(2). By doing so
n=19 sums to 2.509, so there are 20 cubes per column and a total of
40 cubes for the arch.
I think it is wonderfully coincidental that n=19 is the right answer but
was mutually achieved incorrectly. I would be happy to email my analysis
of centers of mass to support my conjecture to anyone interested.
Evan
Karl Heuer
>From article <563...@hpscdc.scd.hp.com>, by ev...@hpscdc.scd.hp.com (Evan Whitney):
>> You wish to build a symmetrical, marginally stable arch by stacking
>> two columns of perfect 1-cubic foot cubes. . . .
>
>building a single unsupported column which extends out 2.5 feet
That's what everyone's been assuming so far, but it seems to me that the top
cubes can lean against each other.
Karl Heuer ka...@haddock.ima.isc.com rutgers!harvard!ima!haddock!karl
m...@sunb8.cs.uiuc.edu
ev...@hpscdc.scd.hp.com said:
>But, there is a twist (literally) to this puzzle that will allow you to
>build the arch using fewer cubes. Stack the cubes along their surface
>diagonals. This multiplies the effective span by SQRT(2). By doing so
>n=19 sums to 2.509, so there are 20 cubes per column and a total of
>40 cubes for the arch.
neato!! this was a cool variation to the "standard" answer. it was
particularly cool because it didn't violate anything the original poster
specified. in fact, if you treat the sketch as not just a sketch but a
specification, then it's probably closer to the truth! (except there would
be lines visible in the middle of each block for the cube-edges that are
pointing toward us in the picture.)
> __ __
> _|__|__|_
> _|__| |__|_
> Rough idea -> |__| |__|
> |__|<- 5ft ->|__|
>
karl heuer said:
>That's what everyone's been assuming so far, but it seems to me that the top
>cubes can lean against each other.
from the original posting:
>Each column must stand under its own weight without toppling. The
>cubes at the top barely touch; i.e., the two columns are not leaning
>against each other for support.
mag