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According to case 166 of The Codeless Code:

It is interesting to note that current physics predicts the heat death of the Universe in no less than 10^100 years, with a subsequent Big Bang arising perhaps in another 10^(10^56) years. This means that in order to display our uptime in seconds right before rebooting the Universe, we would need time_t to have (3.32e56 + 30) bits. Since planet Earth only has about 1.33e50 atoms to play with, we’d need about 2.5 million Earths (or roughly one G-type main-sequence star) to build a simple register alone. Now, where can we get a G-type star? Hmmmmm...

Obviously, that's far too big for us to build at our current level of technology. But this is the future, and we've managed to get hold of one spare Earth - and a place in space to store it nearby.

Using only this spare planet, what future possibilities for data storage might we use, without exceeding the limit of the number of atoms in the Earth?

(Yes, as it stands this is idea generation: to nullify that, I will objectively judge the best answer to be the one that uses the fewest atoms.)

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closed as unclear what you're asking by 2012rcampion, bowlturner, James, Jim2B, Frostfyre Jun 16 '15 at 3:20

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ What is it you're asking for here? Max amount of information storable in 1.33e50 atoms? $\endgroup$ – evankh Jun 12 '15 at 8:56
  • $\begingroup$ @knave the methods we would use to minimise the space needs to store 3.32e56 + 30 bits. $\endgroup$ – ArtOfCode Jun 12 '15 at 8:58
  • $\begingroup$ I'm not sure I see the point of the planet then. It says in your quote it requires a G-type star to have enough atoms to build the register. $\endgroup$ – evankh Jun 12 '15 at 9:01
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    $\begingroup$ @ArtOfCode: While it is thinkable that an atom can store more than two states (a bit has two states), current tech still needs multiple atoms to do so (and special ones to start, not every element is usable for our methods of storage). If we assume a star, most atoms are hydrogen in any case. So current tech aside, what can we store in one hydrogen-atom with the ability to store and write states... $\endgroup$ – Mnementh Jun 12 '15 at 9:43
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    $\begingroup$ ...The electron could be in energetic excited state, but usually it will lose that state after some time and emit a photon with the energy. So I see not a way to store even one bit in only one atom reliably, not to speak about multiple. The assumption of at least one atom per bit is an upper bound of thinkable stuff. $\endgroup$ – Mnementh Jun 12 '15 at 9:43
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Once we have reached the heat death of the Universe, the clock will no longer be able to tick. Neither will any form of life (not only human) be able to function, and observe the working of the clock. It is therefore pointless to consider any further dates after this point.

Up to the heat death, we have:

  • years: $10^{100}$
  • seconds per year: $31 \times 10^6$
  • total seconds until HD: $31 \times 10^{106}$ (approximately)
  • number of bits necessary: $\log_2(\text{total seconds})$ = 357 bits
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  • $\begingroup$ Number of bits required for n 0-9 digits is n*log2(10), not log2(n). Note that this minimizes for log2(2), which is binary storage. $\endgroup$ – Dan Smolinske Jun 15 '15 at 15:12
  • $\begingroup$ Thank you - this is a much better answer. I would advise you to take comments into account, like Dan's above me. (I don't know which is right, your version or his, but just worth noting.) $\endgroup$ – ArtOfCode Jun 15 '15 at 15:22
  • $\begingroup$ @DanSmolinske Note log_a(x^b) = b * log_a(x), for all real a,b with a strictly positive. Thus log_2(10^n) = n * log_2(10). $\endgroup$ – ALAN WARD Jun 15 '15 at 17:03
  • $\begingroup$ @ArtOfCode I can see some SE sites can accept humor, others less. OK, point taken. $\endgroup$ – ALAN WARD Jun 15 '15 at 17:06
  • $\begingroup$ @ALANWARD We'll all take humour, but it does have to applied correctly. Thanks for understanding though - I must admit you've taken a deletion in much better spirit than many :) $\endgroup$ – ArtOfCode Jun 15 '15 at 17:10
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There should be no need for spare worlds to accommodate our computational needs.

An $n$ qubit quantum computer can process the equivalent of $2^n$ classical bits of information (while in a fully entangled state of affairs). This means that a $1000$ qubit quantum computer, having $2^{1000}$ possible states, has more than the desired computational power. Compare this to the approximately $10^{80}$ particles estimated to exist in the observable universe.

Of course, according to our current understanding, the catch would be reading its full state without causing it to lose coherence.

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The Bekenstein bound limits the amount of information that can be contained in a system. It states that:

$$ I\leq \frac{2\pi RE}{\hbar c\log 2} $$

Where $I$ is the amount of information (in bits), $R$ is the radius of a sphere enclosing the system, and $E$ is the amount of mass-energy contained in the system.

Obviously in order to increase storage density we must increase energy density. However, general relativity places a limit on how much energy we can store in a given volume before forming a black hole:

$$ E\leq\frac{Rc^4}{2G} $$

Combining these two, we obtain the following:

$$ I\leq\frac{\pi R^2c^3}{\hbar G\log 2}\approx 1.4\cdot10^{69}~\text{bits}\cdot\text{m}^{-2}\times 4\pi R^2 $$

That is, the amount of information contained in a volume is limited by the surface area of that volume.

Fortunately for you, this means that your counter can be theoretically compressed into a sphere just $280~\text{nm}$ in diameter!


Of course, you want the solution with the least number of atoms. I'll assume that your counter can be at most one light-second in diameter (so that it can be updated once per second). We can use the Bekenstein bound again to show that the minimum amount of mass-energy you need with the given volume is:

$$ \frac{3.3\cdot 10^{56}~\text{bits}\times\hbar\log 2}{\pi c\times 1~\text{s}\cdot c}\approx 86~\text{tons} $$


Of course I can trivially saturate your criteria by noting that the information can be stored in non-atoms. With a thin shell of equipment around the volume of the counter you could contain $86~\text{tons}$ of photons (that's $7.7\cdot 10^{21}~\text{J}$), continually processing the outgoing photons and injecting new photons to maintain and update the counter's state.

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