[seqfan] Re: Brick sequences
victor miller
victorsmiller at gmail.com
Sun Oct 25 14:15:08 CET 2009
Peter Winkler gave a very nice talk on the problem of stacking bricks
at the joint mathematics meeting in January (here's a link to the
manuscript: http://math.dartmouth.edu/~pw/papers/maxover.pdf ). In
the paper he gives the criterion for a structure of stacked bricks to
be stable. It turns out to be a feasibility criterion in linear
programming. They used it in computer experiments to find
configurations with large overhang.
Victor
On Sat, Oct 24, 2009 at 7:44 AM, <franktaw at netscape.net> wrote:
> I came up with an idea for a sequence. Actually, it's several
> sequences, some of which I'm not sure how to define mathematically.
>
> Consider a connected structure of bricks, idealized as congruent
> rectangles. The bricks on the bottom are at integer offsets; those on
> the next row up are at half-integer offsets, the next row is again at
> integer offsets, etc. The structure has to stand up.
>
> For the simplest case, we require that each brick be supported by two
> bricks under it. If we regard mirror images as the same, this sequence
> starts, from n=1 (hand-calculated):
>
> 1,1,2,2,4,6,10
>
> (I suppose a(0) = 1, but I'm not including it here.)
>
> For example, the two structures for n=4 are:
>
> |=|=|=|=|
>
> and
>
> .|=|
> |=|=|=|
>
> (Imagine the lines on the = are farther apart than most fonts will make
> them.)
>
> If mirror images are different, the sequence starts:
>
> 1,1,2,3,5,9,15
>
> I'm pretty sure that neither of these is in the OEIS: there are
> sequences matching either, but none with even a vaguely related
> description.
>
> Note that, for these two sequences, we could just as well use hexagons
> as bricks. This won't work for the next one.
>
> -----
> But now things get more interesting. The structure:
>
> .|=|
> |=|
>
> is not stable; the center of gravity of the top brick is right over the
> edge of the brick below it. A small perturbation will send it
> tumbling. However,
>
> .|=|
> |=|=|
> .|=|
>
> is stable; the weight of the top brick holds the center of gravity for
> the weight supported by the bricks under it to be over the base brick.
>
> On the other hand, consider:
>
> .|=|=|
> |=|.|=|
>
> Now the two top bricks are braced against each other. I think that, by
> the definition I want to use, this is actually unstable; perturb the
> top two bricks a bit to the right, and the left one will start to tilt,
> pushing the right one further right until the left brick falls.
>
> At this point I'm not sure how to proceed. How do you determine
> whether a given structure is stable? If you measure the center gravity
> of the weight supported by a brick, it must be over another brick; this
> is certainly a necessary and sufficient condition. But how do you
> distribute the weight for a brick that has two bricks under it? My
> second thought (the first thought was clearly wrong) is that the weight
> should be distributed according to the position of the center of
> gravity of the weight supported by the brick; if this is 1/4 of the way
> across the brick, distribute the left-most 3/4 of the weight to the
> left and the remaining 1/4 to right. I'm pretty dubious that this is
> correct in general, however.
>
> Once this question is settled, there will again be two sequences: one
> for mirror images considered the same, and the other considering them
> different.
>
> Franklin T. Adams-Watters
>
>
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