Calculating the exact number of distinguishable buckyballs is painful but feasible; coming up with an approximate number is easy. The calculation depends on whether two buckyballs that are mirror images (but not proper rotations of each other) are considered different or the same. If we are going to analyse a ball using chemical probes, then we can probably distinguish them. But if we are using spectroscopy, then I don't think we can. (I think this is a question for Gene Salamin: can two mirror compounds be distinguished by spectroscopic methods, perhaps using a circularly polarized probe?) Assuming that mirror images can't be distinguished, the number of distinguishable balls is just slightly north of 2^60 / 120. That is, we lose about 7 bits, and can code a fraction more than 53 bits per ball. If mirror images can be distinguished, that buys us one more bit. The true value exceeds 2^60 / 120 by somewhere in the vicinity of a couple of dozen times 2^32; that is, nowhere near enough to make much of a difference in the storage capacity of a ball. Exact numbers require classifying all the symmetries; I'm comfortable with the proper ones (rotations) but not the improper ones, and would have to draw a bunch of diagrams before I gained confidence. On Wed, Jun 22, 2016 at 10:48 AM, Bill Gosper <billgosper@gmail.com> wrote:
Wow, a whole new meaning for buckybits. So Polya should count for us the number of distinguishable buckyballs with n 0s and 60-n 1s. But can it give us a UID from a 60 bit vector? --rwg
On 2016-06-20 18:37, Eugene Salamin via math-fun wrote:
I have no comment on the practicality of this as a storage scheme. As to the mathematical problem of counting the number of distinct arrangements, this is made to order for the Polya Counting Theorem.
-- Gene
From: Keith F. Lynch <kfl@KeithLynch.net> To: math-fun@mailman.xmission.com Sent: Monday, June 20, 2016 5:44 PM Subject: [math-fun] Data storage in buckminsterfullerene
The recent discussion about data storage in diamonds made me wonder about data storage in another stable form of carbon: C60, buckyballs.
How many bits can be stored in one buckyball, i.e. fullerene molecule? It's a truncated icosahedron with 32 faces (of which 12 are regular pentagons and 20 are regular hexagons), 90 edges, and 60 vertices. The 60 vertices are carbon atoms. Data can be stored by making some of them C12 and some of them C13. But how much data? The base-2 log of the number of distinct arrangements of C12 and C13 atoms. But how many arrangements are there? If the molecules were not free to rotate, there would be 2^60 arrangements, hence 60 bits. But of course the molecules are free to rotate. The rotation group is order 60. So there are at least 2^60/60 arrangements, hence at least 54 bits.
Can anyone think of a practical way to count the arrangements? The obvious way would be to try all 2^60 arrangements and eliminate the ones that are rotations of arrangements that have already been found. But 2^60 is more than 10^18, so that's just barely computationally feasible. Is there a better way? Has someone counted them already? Thanks.
This number might seem to be uselessly small, but it isn't, since data could be stored in a large set of buckyballs. Each molecule could contain a 50-bit sequence number and several bits of data, giving several hundred terabytes of total storage. There would be many duplicates to ensure that at least one molecule with each sequence number is found. Several hundred terabytes may not seem like much by today's standards, but it can store the whole of Wikipedia, plus more books than anyone can read in a lifetime, plus more movies than anyone is likely to view in a lifetime. It's a lot larger than the genome of any organism, so it's a lot more practical than living DNA data storage. And 2^50 buckyballs times 600 duplicates of each of them would mass less than one milligram.
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