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354 lines
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Markdown
354 lines
14 KiB
Markdown
---
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date: "2021-01-12T00:03:10-08:00"
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description: Using thermal physics, cosmology, and computer science to calculate
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password vulnerability to the biggest possible brute-force attack.
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outputs:
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- html
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- gemtext
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tags:
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- security
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- fun
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title: Becoming physically immune to brute-force attacks
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---
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This is a tale of the intersection between thermal physics, cosmology, and a tiny
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amount of computer science to answer a seemingly innocuous question: "How strong does
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a password need to be for it to be physically impossible to brute-force, ever?"
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[TLDR]({{<ref "#conclusiontldr" >}}) at the bottom.
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*Note: this post contains equations. Since none of the equations were long or
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complex, I decided to just write them out in code blocks instead of using images or
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MathML the way Wikipedia does.*
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Introduction
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------------
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I realize that advice on password strength can get outdated. As supercomputers grow
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more powerful, password strength recommendations need to be updated to resist
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stronger brute-force attacks. Passwords that are strong today might be weak in the
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future. **How long should a password be in order for it to be physically impossible
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to brute-force, ever?**
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This question might not be especially practical, but it's fun to analyze and offers
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interesting perspective regarding sane upper-limits on password strength.
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Asking the right question
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-------------------------
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Instead of predicting what tomorrow's computers may be able to do, let's examine the
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*biggest possible brute-force attack* that the laws of physics can allow.
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A supercomputer is probably faster than your phone; however, given enough time, both
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are capable of doing the same calculations. If time isn't the bottleneck, energy
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usage is. More efficient computers can flip more bits with a finite amount of energy.
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In other words, energy efficiency and energy availability are the two fundamental
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bottlenecks of computing. What happens when a computer with the highest theoretical
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energy efficiency is limited only by the mass-energy of *the entire [observable
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universe](https://en.wikipedia.org/wiki/Observable_universe)?*
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Let's call this absolute unit of an energy-efficient computer the MOAC (Mother of All
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Computers). For all classical computers that are made of matter, do work to compute,
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and are bound by the conservation of energy, the MOAC represents a finite yet
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unreachable limit of computational power. And yes, it can play Solitaire with
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*amazing* framerates.
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How strong should your password be for it to be safe from a brute-force attack by the
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MOAC?
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### Quantifying password strength.
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A good measure of password strength is **entropy bits.** The entropy bits in a
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password is a base-2 logarithm of the number of guesses required to brute-force
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it.[^1]
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A brute-force attack that executes 2<sup>n</sup> guesses is certain to crack a
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password with *n* entropy bits, and has a one-in-two chance of cracking a password
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with *n*+1 entropy bits.
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For scale, [AES-256](https://en.wikipedia.org/wiki/Advanced_Encryption_Standard)
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encryption is currently the industry standard for strong symmetric encryption. As the
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name suggests, its keys have 256 bits of entropy; if your password has more than 256
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entropy bits, then the AES-256 encryption algorithm is the bottleneck.
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To calculate the entropy of a password, I recommend using a tool such as
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[zxcvbn](https://www.usenix.org/conference/usenixsecurity16/technical-sessions/presentation/wheeler)
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or [KeePassXC](https://keepassxc.org/).
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The Problem
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-----------
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Define a function `P`. `P` determines the probability that MOAC will correctly guess
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a password with `n` bits of entropy after using `e` energy:
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P(n, e)
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If `P(n, e) ≥ 1`, the MOAC will certainly guess your password before running out of
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energy. The lower `P(n, e)` is, the less likely it is for the MOAC to guess your
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password.
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Caveats and estimates
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---------------------
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I don't have a strong physics background.
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When estimating, we'll prefer higher estimates that increase the odds of it guessing
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a password; after all, the point of this exercise is to establish an *upper* limit on
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password strength. We'll also simplify: for instance, the MOAC will not waste any
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heat, and the only way it can guess a password is through brute-forcing. Focusing on
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too many details would defeat the point of this thought experiment.
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I won't address any particular encryption algorithms; this is just a pure and simple
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brute-force attack given precomputed password entropy. Furthermore, quantum computers
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can use [Grover's algorithm](https://en.wikipedia.org/wiki/Grover%27s_algorithm) for
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an exponential speed-up; to account for quantum computers using Grover's algorithm,
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calculate `P(n/2, e)` instead.
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Obviously, I'm not taking into account future mathematical advances; my crystal ball
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broke after I asked it if humanity would ever develop the technology to make anime
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real.
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Finally, there's always a non-zero probability of a brute-force attack guessing a
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password with a given entropy. Literal "immunity" is impossible. Lowering this
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probability to statistical insignificance renders our password practically immune to
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brute-force attacks.
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Computation
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-----------
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How much energy does MOAC use per guess during a brute-force attack? In the context
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of this thought experiment, this number should be unrealistically low. I settled on
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[`kT`](https://en.wikipedia.org/wiki/KT_(energy)). `k` represents the [Boltzmann
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Constant](https://en.wikipedia.org/wiki/Boltzmann_constant) (about
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1.381×10<sup>-23</sup> J/K) and `T` represents the temperature of the system. Their
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product corresponds to the amount of heat required to create a 1 nat increase in a
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system's entropy.
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A more involved approach to picking a good value might utilize the [Plank-Einstein
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relation](https://en.wikipedia.org/wiki/Planck%E2%80%93Einstein_relation).
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It's also probably a better idea to make this value an estimate for flipping a single
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bit, and to estimate the average number of bit-flips it takes to make a single
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password guess. If that bothers you, pick a number `b` you believe to be a good
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estimate for a bit-flip-count and calculate `P(n-b, e)` instead of `P(n, e)`.
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What's the temperature of the system? Three pieces of information help us find out:
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1. The MOAC is located somewhere in the observable universe
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2. The MOAC will be consuming the entire observable universe
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3. The universe is mostly empty
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A good value for `T` would be the average temperature of the entire observable
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universe. The universe is mostly empty; `T` is around the temperature of cosmic
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background radiation in space. The lowest reasonable estimate for this temperature is
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2.7 degrees Kelvin.[^2] A lower temperature means less energy usage, less energy
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usage allows more computations, and more computations raises the upper limit on
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password strength.
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Every guess, the MOAC expends `kT` energy. Let `E` = the total amount of energy the
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MOAC can use; let `B` = the maximum number of guesses the MOAC can execute before
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running out of energy.
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B = E/(kT)
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Now, given the maximum number of passwords the MOAC can guess `B` and the bits of
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entropy in our password `n`, we have an equation for the probability that the MOAC
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will guess our password:
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P(n,B) = B/2ⁿ
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Plug in our expression for `B`:
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P(n,E) = E/(2ⁿkT)
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### Calculating the mass-energy of the observable universe
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The MOAC can use the entire mass-energy of the observable universe.[^3] Simply stuff
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the observable universe into the attached 100% efficient furnace, turn on the burner,
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and generate power for the computer. You might need to ask a friend for help.
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Just how much energy is that? The mass-energy equivalence formula is quite simple:
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E = mc²
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We're trying to find `E` and we know `c`, the speed of light, is 299,792,458 m/s.
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That leaves `m`. What's the mass of the observable universe?
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### Calculating the critical density of the observable universe
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Critical density is smallest average density of matter required to *almost* slow the
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expansion of the universe to a stop. Any more dense, and expansion will stop; any
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less, and expansion will never stop.
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Let `D` = critical density of the observable universe and `V` = volume of the
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observable universe. Mass is the product of density and volume:
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m = DV
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We can derive the value of D by solving for it in the [Friedman
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equations](https://en.wikipedia.org/wiki/Friedmann_equations):
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D = 3Hₒ²/(8πG)
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Where `G` is the [Gravitational
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Constant](https://en.wikipedia.org/wiki/Gravitational_constant) and `Hₒ` is the
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[Hubble Constant](https://en.wikipedia.org/wiki/Hubble%27s_law). `Hₒd` is the rate of
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expansion at proper distance `d`.
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Let's assume the observable universe is a sphere, expanding at the speed of light
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ever since the Big Bang.[^4] The volume `V` of our spherical universe when given its
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radius `r` is:
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V = (4/3)πr³
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To find the radius of the observable universe `r`, we can use the age of the universe
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`t`:
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r = ct
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Hubble's Law estimates the age of the universe to be around `1/Hₒ`
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### Solving for E
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Let's plug in all the derived values into our original equation for the mass of the
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observable universe `m`:
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m = DV
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Remember when I opened the article by saying that none of the equations would be long
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or complex?
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I lied.
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m = (3Hₒ²/(8πG))(4/3)π(ct)³
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m = c³/(2GHₒ)
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E = mc²
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E = c⁵/(2GHₒ)
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Yay, we found an expression for the total energy the MOAC can consume!
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Final Solution
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--------------
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P(n,E) = E/(2ⁿkT)
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P(n, c⁵/(2GHₒ)) = c⁵/(2GHₒ*2ⁿkT)
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Let's copy and paste the values for those constants from Wikipedia and Wolfram Alpha:
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- c = 299,792,458 m/s
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- G ≈ 6.67408×10<sup>-11</sup> m³/kg/s²
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- Hₒ ≈ 2.2×10<sup>-18</sup> Hz (uncertain; look up the Hubble tension)
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- T ≈ 2.7 K
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- k ≈ 1.3806503×10<sup>-23</sup> J/K
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Plugging those in and simplifying:
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**P(n) ≈ 2.21×10<sup>92</sup> / 2<sup>n</sup>**
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Here are some sample outputs:
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- P(256) ≈ 1.9×10<sup>15</sup> (password certainly cracked after burning 1.9
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quadrillionth of the mass-energy of the observable universe).
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- P(306.76) ≈ 1 (password certainly cracked after burning the mass-energy of the
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observable universe)
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- P(310) ≈ 0.11 (about one in ten)
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- P(326.6) ≈ 1.1×10<sup>-6</sup> (about one in a million)
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If your threat model is a bit smaller, simulate putting a smaller object into the
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MOAC's furnace. For example, the Earth has a mass of 5.972×10²⁴ kg; this gives the
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MOAC a one-in-ten-trillion chance of cracking a password with 256 entropy bits and a
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100% chance of cracking a 213-bit password.
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Sample unbreakable passwords
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----------------------------
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According to KeePassXC's password generator, each of the following passwords has an
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entropy between 330 and 340 bits.
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Using the extended-ASCII character set:
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¦=¦FVõ)Çb^ÄwΡ=,°m°B9®;>3[°r:t®Ú"$3CG¨/Bq-y\;
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Using the characters on a standard US QWERTY layout:
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%nUzL2XR&Tz5hJfp2tiYBoBBX^vWo3`g6H#JSC#N6gWm#hVdD~ziD$YHW
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Using only alphanumeric characters:
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tp8D69CGWE5t5a9si5XNsw32CKyCafh8qGrKWLwE6KJHpGyUtcJDWpgRz5mFNx
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An excerpt from a religious text with a trailing space:
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I'd just like to interject for a moment. What you’re referring to as Linux, is in fact, GNU/Linux,
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Don't use actual excerpts from pre-existing works as your password.
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Conclusion/TLDR
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---------------
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Question: How much entropy should a password have to ensure it will *never* be
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vulnerable to a brute-force attack? Can an impossibly efficient computer--the
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MOAC--crack your password?
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Answer: limited only by energy, if a computer with the highest level of efficiency
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physically possible is made of matter, does work to compute, and obeys the
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conservation of energy:
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- A password with 256 bits of entropy is practically immune to brute-force attacks
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large enough to quite literally burn the world, but is quite trivial to crack with
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a universe-scale fuel source.
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- A password with 327 bits of entropy is nearly impossible to crack even if you burn
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the whole observable universe trying to do so.
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At that point, a formidable threat would rather use [other
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means](https://xkcd.com/538/) to unlock your secrets.
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Further reading: an alternative approach
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----------------------------------------
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Check out Scott Aaronson's article, [Cosmology and
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Complexity](https://www.scottaaronson.com/democritus/lec20.html). He uses an
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alternative approach to finding the maximum bits we can work with: he simply inverts
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the [cosmological constant](https://en.wikipedia.org/wiki/Cosmological_constant).
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This model takes into account more than just the mass of the observable universe.
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While we previously found that the MOAC can brute-force a password with 306.76
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entropy bits, this model allows the same for up to 405.3 bits.
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Another well-known approach to calculating physical limits of computation is
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[Bremermann's limit](https://en.wikipedia.org/wiki/Bremermann%27s_limit), which
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calculates the speed of computation given a finite amount of mass. This article's
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approach disregards time, focusing only on mass-energy equivalence.
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Acknowledgements
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----------------
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Thanks to [Barna Zsombor](http://bzsombor.web.elte.hu/) and [Ryan
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Coyler](https://rcolyer.net/) for helping me over IRC with my shaky physics and
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pointing out the caveats of my approach.
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My notes from Thermal Physics weren't enough to write this; various Wikipedia
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articles were also quite helpful, most of which were linked in the body of the
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article.
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[^1]: James Massey (1994). "Guessing and entropy" (PDF). Proceedings of 1994 IEEE
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International Symposium on Information Theory. IEEE. p. 204.
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[^2]: Assis, A. K. T.; Neves, M. C. D. (3 July 1995). "History of the 2.7 K
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Temperature Prior to Penzias and Wilson"
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[^3]: The MOAC 2 was supposed to be able to consume other sources of energy such as
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dark matter and dark energy. Unfortunately, Intergalactic Business Machines ran out
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of funds since all their previous funds, being made of matter, were consumed by the
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original MOAC.
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[^4]: This is a massive oversimplification; there isn't a single answer to the
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question "What is the volume of the observable universe?" Using this speed-of-light
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approach is one of multiple valid perspectives. The absolute size of the observable
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universe is much greater due to the way expansion works, but stuffing that into the
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MOAC's furnace would require moving mass faster than the speed of light.
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