-4

Imagining a solid cube of $64 \times 64 \times 64$ bits (each of which can have exactly two states), how long would it take to test all possible states of one of these? Let's also assume we're using the world's current most powerful supercomputer, which can perform 33.86 quadrillion math operations per second. (mind you, this means you have to account for how many basic operations a calculation takes up: $2\times3$ is one, $2\times\frac 1 3$ is two, etc.)

I tried asking Wolfram|Alpha, but it timed out: http://wolfr.am/1q6fxd9

I ask this question in the name of encryption security as well as randomization seeding uniqueness, and decided to ask it here instead of StackOverflow because it has more to do with math than computing.

Ky -
  • 1,300

2 Answers2

1

There are $64^3 = 262144$ bits, so that there are only $2^{262144}$ possible cubes, which is around $10^{79000}$.

Thus, it's not very feasible to do any sort of computation involving looping over all possible cubes.

Nathaniel Bubis
  • 33,018
  • Can I have some sort of time estimate? – Ky - Jul 06 '14 at 00:38
  • 1
    @Supuhstar - infinity. Take a computer that runs $10^{16}$ operations a second, Notice how $79000$ is so much bigger than $16$? Now take into account that the entire universe is only $10^{17}$ years old, and you get the idea. – Nathaniel Bubis Jul 06 '14 at 00:48
  • I was hoping for a real answer :/ – Ky - Jul 06 '14 at 04:55
  • 1
    @Supuhstar - Why is this not a real answer? $10^{79000}$ seconds is the approximate time it would take on any machine, no matter how fast, a.k.a infinity. – Nathaniel Bubis Jul 06 '14 at 05:06
  • you're not even showing any work. All I see are disjoint numbers... – Ky - Jul 07 '14 at 22:16
  • 1
    @Supuhstar - Do I really have to spell it out? To get the number of seconds, divide the number of operations by operations per second: $10^{79000} / 10^{16} = 10^{78984}$ Which is, as mentioned, equivalent to never. – Nathaniel Bubis Jul 07 '14 at 22:19
  • you're rounding unnecessarily and giving me vague answers such as "infinity" and "never", when that's not the case – Ky - Jul 08 '14 at 14:48
  • 1
    @Supuhstar - How un-vague do you want? Isn't $10^{78984}$ seconds good enough? – Nathaniel Bubis Jul 08 '14 at 16:52
  • because it's rounded and at an impractical scale – Ky - Jul 08 '14 at 18:02
  • 1
    @Supuhstar - yes, your problem has an impractical scale. It's not the rounding which is the problem. – Nathaniel Bubis Jul 08 '14 at 18:06
0

There are $2^{64\times 64\times 64}=2^{262144} $ possibilities, so at a rate of 33.86 quadrillion per second, it will take $\frac{2^{262144}}{33.86\times 10^{15}}$ seconds.

Esteemator
  • 1,192