[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.
Graphene might be a lot easier to deal with. Graphene Turing Machine with *hexagonal* "tape squares" ! Alternatively, you could build a "wire recorder" with carbon nanotubes -- perhaps doubling as a space elevator ! At 05:44 PM 6/20/2016, Keith F. Lynch wrote:
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.
For all of these schemes (diamond, bucky-balls, graphene) I have yet to hear a plausible technology for reading out the stored data (or for producing the isotope crafted molecule). As I have been quoted saying, biology is the nanotechnology which works. DNA is superior in most ways to these schemes, since it can be replicated, is highly stable, and easily both read and written. The ultimete storage would likely be to encode data in DNA, and modify the genome of some organism to contain it. A very useful, but quite difficult project would be to add even simple forms of error correction to the sequence. Evolution hasn’t done that (and likely won’t) because evolution needs the carefully controlled mutation rate available from un-checked duplication. Leave the bucky balls and isotopically crafted diamond to Merkel and Drexler.
On Jun 20, 2016, at 8:53 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Graphene might be a lot easier to deal with.
Graphene Turing Machine with *hexagonal* "tape squares" !
Alternatively, you could build a "wire recorder" with carbon nanotubes -- perhaps doubling as a space elevator !
At 05:44 PM 6/20/2016, Keith F. Lynch wrote:
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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On 21/06/2016 01:44, Keith F. Lynch wrote:
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?
You're looking for the Polya enumeration theorem. I always find this painfully fiddly to apply and am likely to get the following wrong, but it goes like this. The number of configurations is the average, over all elements of the symmetry group, of m^c where m is the number of "colours" (2, in this case) and c is the number of cycles in the permutation produced by the group action. In this case our group is A_5 and it goes like this: - the identity has 60 (trivial) cycles - there are 15 elements with 30 cycles of length 2 - there are 20 elements with 20 cycles of length 3 - there are 24 elements with 12 cycles of length 5 giving 1/60 (2^60 + 15.2^30 + 20.2^20 + 24.2^12) different configurations. If I haven't made a mistake, which I probably have. -- g
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. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Henry Baker -
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Tom Knight