4 ms·

Interesting. I wonder, if you can solve 5x5x5 cube, then you can solve them all. How good an approximation is using the optimal methods for solving 5x5x5 to sol

by js8 9y ago

Interesting. I wonder, if you can solve 5x5x5 cube, then you can solve them all. How good an approximation is using the optimal methods for solving 5x5x5 to solve the NxNxN case?

dtech 9y ago
If the NxNxN case is NP-complete, then that currently known optimal method will still require an N^k amount of steps.
Extigy 9y ago
I'm not sure that's true. I much prefer solving my 5x5x5 rather than my 4x4x4 because I always forget the algorithms for the extra parity that even cubes provide. I don't think I could solve a 6x6x6 with the 5x5x5 algorithms I know.
- Zanni 9y ago
  It's been a while so maybe I'm misremembering, but can't you solve a 4x4x4 by reducing it to a 3x3x3 and solving that? Pretty sure that's how I used to do it. Treat each 4 as a 3 with a fat middle: 1-2-1. Then solve the centers (2x2) and the edges (2x1) with standard moves and you've got, essentially, a 3x3x3.
  - timjver 9y ago
    A 4x4x4 definitely has parity issues that don't arise on a 3x3x3.
  - captn3m0 9y ago
    4x4x4 has parity problems, same as 5x5x5, but as the GP posted, the number of parity cases for 4x4x4 is higher, which makes it slightly more difficult. (I prefer solving the 5x5x5 for exactly the same reason).
- deleted 9y ago
  [deleted]
tripa 9y ago
What optimal methods are you referring to? The paper doesn't really define one, and real world cubers sure don't use any. Assuming you mean real life cubes, the most common [1] method to solve anything bigger than 3×3×3 is to first reduce to 3×3×3, then solve as a normal 3×3×3 using outer slice turns. On even-sided cubes, that reduction process is sensitive to a number of so-called "reduction parities", meaning in a nutshell that you might be recreating edges [2] in a configuration that isn't possible/solvable on a 3×3×3. This can't happen with odd cubes, because the center piece on faces and edges disambiguates from the start. So really, most of what you need to know about solving big cubes is there on a 4×4×4, and not on a 5×5×5. To make it fully pedantry-proof [3], you need to know standard methods for both 4×4×4 and 5×5×5, both of which already require solving a 3×3×3. [1] There are alternate methods that are actually used for 4 or 5 and don't have the "parity weakness", but they don't scale well enough to anything bigger. [2] Or, obviously, faces, but in real life reducing faces in a valid configuration is considered a prerequisite. [3] It kind of depends on who you ask whether or not the reduction methods are "the same" moving from 4×4×4 to 5×5×5. They're a direct extension, but they do involve pieces with no counterpart.

by js8 9y ago

Interesting. I wonder, if you can solve 5x5x5 cube, then you can solve them all. How good an approximation is using the optimal methods for solving 5x5x5 to solve the NxNxN case?

dtech 9y ago

If the NxNxN case is NP-complete, then that currently known optimal method will still require an N^k amount of steps.

Extigy 9y ago

I'm not sure that's true. I much prefer solving my 5x5x5 rather than my 4x4x4 because I always forget the algorithms for the extra parity that even cubes provide. I don't think I could solve a 6x6x6 with the 5x5x5 algorithms I know.

Zanni 9y ago
It's been a while so maybe I'm misremembering, but can't you solve a 4x4x4 by reducing it to a 3x3x3 and solving that? Pretty sure that's how I used to do it. Treat each 4 as a 3 with a fat middle: 1-2-1. Then solve the centers (2x2) and the edges (2x1) with standard moves and you've got, essentially, a 3x3x3.
- timjver 9y ago
  A 4x4x4 definitely has parity issues that don't arise on a 3x3x3.
- captn3m0 9y ago
  4x4x4 has parity problems, same as 5x5x5, but as the GP posted, the number of parity cases for 4x4x4 is higher, which makes it slightly more difficult. (I prefer solving the 5x5x5 for exactly the same reason).
deleted 9y ago
[deleted]

tripa 9y ago

What optimal methods are you referring to? The paper doesn't really define one, and real world cubers sure don't use any. Assuming you mean real life cubes, the most common [1] method to solve anything bigger than 3×3×3 is to first reduce to 3×3×3, then solve as a normal 3×3×3 using outer slice turns. On even-sided cubes, that reduction process is sensitive to a number of so-called "reduction parities", meaning in a nutshell that you might be recreating edges [2] in a configuration that isn't possible/solvable on a 3×3×3. This can't happen with odd cubes, because the center piece on faces and edges disambiguates from the start. So really, most of what you need to know about solving big cubes is there on a 4×4×4, and not on a 5×5×5. To make it fully pedantry-proof [3], you need to know standard methods for both 4×4×4 and 5×5×5, both of which already require solving a 3×3×3. [1] There are alternate methods that are actually used for 4 or 5 and don't have the "parity weakness", but they don't scale well enough to anything bigger. [2] Or, obviously, faces, but in real life reducing faces in a valid configuration is considered a prerequisite. [3] It kind of depends on who you ask whether or not the reduction methods are "the same" moving from 4×4×4 to 5×5×5. They're a direct extension, but they do involve pieces with no counterpart.