# Apocorubik's Cube (10,000 blocks across): How long to solve if a quarter turn takes 2.8 seconds?

Suppose you have a Rubik's cube, $$10,000$$ blocks across (so, $$600,000,000$$ total tiles), scrambled. Assuming an ideal solving algorithm, approximately how long will the cube take to be solved, given one quarter turn can be made every 2.8 seconds?

• Do ou have some reason to think this is known? Or possible to compute with existing technology? – GEdgar Jan 17 '19 at 1:17
• @GEdgar I know that the optimal algorithm for a rubiks cube of arbitrary size might not be known, but I think an approximation is possible. – qazwsx Jan 17 '19 at 1:18
• The so-called "God's Number" —counting the number of moves required to solve the $3\times 3\times 3$ cube— is known to be $20$; the number of quarter turns (as opposed to moves) is reportedly $26$. So, make your best guess about how $26$ scales to a $10000\times 10000\times 10000$ cube, and multiply by $2.8$. – Blue Jan 17 '19 at 1:38
• @Blue But how does the God's Number scale with cube size? Its not just a linear scale. God's number for 2 by 2 by 2s is 14. – qazwsx Jan 17 '19 at 1:52

According to https://arxiv.org/abs/1106.5736 the asymptotic bound for God's Number on higher order cubes is:

$$\Theta(n^2/\log n)$$

But this is only asymptotic, so you can't just plug in $$n$$ and expect to get the exact right answer. But we use it as a very loose approximation. Since God's Number for $$n=3$$ is 26 (using quarter-turn metric), we can do:

$$\frac{10000^2}{\log 10000} * \frac{\log 3}{3^2} * 26 = 34458757$$

So to round off your question take that number, multiply it by 2.8s, and you get about 3.1 years.

• +1. Since approximations are very loose anyway, this is a pointless nit-pick, but ... As mentioned in my comment, God's number counts moves; OP counts quarter turns. So, there's an additional "expected number of quarter turns" factor to consider. – Blue Jan 17 '19 at 2:15
• How far off in either direction could this estimate be? Is it within an order of magnitude? – qazwsx Jan 17 '19 at 2:17
• @qazwsx there is really no way to tell... $\Theta$ notation really only says what will happen in the limit... but before that it can deviate an arbitrary amount. – timidpueo Jan 17 '19 at 2:18
• @Blue thx amended – timidpueo Jan 17 '19 at 2:22
• The end of section 4 of that paper gives an explicit lower bound: an $n \times n \times n$ cube requires at least $$\frac{\left(\left\lfloor \frac{n}{2} \right\rfloor-1\right)^2}{\log_c (6n)} - 1$$ quarter-turns to solve, where $c=\frac{24!}{(4!)^6}$. For $n=10000$, this appears to work out to about $8 \cdot 10^7$ moves or a little over $7$ years. It looks like you could extract an explicit upper bound from the earlier bits of section 4 — it comes from an explicit solving algorithm — but it'd take more work. – Micah Jan 17 '19 at 2:24