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Someone hands you a $4\times 4\times 4$ cube for solving. It is scrambled as if it were a $3\times 3\times 3$, i.e. without ever separating the $4$ central pieces of each face (which implies that edge pairs remain together too).

If this restriction is lifted now, is it possible that you might be able to solve this cube in fewer moves than you would a regular $3\times 3\times 3$ by utilizing the extra moves that $4\times 4\times 4$ provides?

Equivalently, are there any valid $3\times 3\times 3$ cube positions that are reachable from solved state in fewer moves on a $4\times 4\times 4$?

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  • $\begingroup$ By 'solve in fewer moves,' you mean with perfect play, right? $\endgroup$ – I. Pittenger Jul 15 '19 at 23:00
  • $\begingroup$ I mean generally. Any improvement to any solving stage of any generic 3×3×3 algorithm would imply existence of 3×3×3 positions that can be solved in fewer moves with perfect play on a 4×4×4. In other words, I am interested to know whether 4×4×4 gives any additional "power" to a solver of 3×3×3 positions, compared to a regular 3×3×3. $\endgroup$ – Szczepan Hołyszewski Jul 15 '19 at 23:45
  • $\begingroup$ And the edge-pairs are matched? Then it is just like a 3x3 but could still have a parity error, yes? Could end up with a single edge-pair flipped I think. $\endgroup$ – jdods Jul 16 '19 at 2:13
  • $\begingroup$ In other words I think you're asking whether there are any permutations of the 96 stickers on the 4x4x4 that are possible using only the outer cuts, but are possible in fewer moves when you also allow using the center cuts? (Or the same, modulo permutations of the four center pieces on each side?) $\endgroup$ – hmakholm left over Monica Jul 16 '19 at 2:27
  • $\begingroup$ As far as I know no such positions are known, but it has not been proved they don't exist. There are many examples of 3x3x3 positions that are in the square group (i.e. reachable using only half turns) which are solvable using fewer moves when quarter turns are allowed. $\endgroup$ – Jaap Scherphuis Jul 16 '19 at 8:25
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The answer is YES. There are some "Pseudo $3\times 3\times 3$" $4\times 4\times 4$ positions which, although they can be exclusively solved by turning outer layer faces only (e.g., using a $3\times 3\times 3$ solver), if one applies certain $4\times 4\times 4$ move sequences (which consist of both inner and outer slices), the resulting solution will be fewer moves than optimal solutions retrieved from a $3\times 3\times 3$ solver.

This is a "trivial position" found by Stefan Pochmann in the speedsolving.com thread, Pseudo $3\times 3$ shorter than $3\times 3$?.

As you can see, I was a participant in that thread as well, and I found the following more complicated position. (I was motivated to find what may be considered "non-trivial" positions of this type -- to show that Stefan's wasn't the only $4\times 4\times 4$ position like this.)

Uw2 2R2 Uw2 Rw2 U2 Rw' D' U' R2 D' U' s2 Rw (15 htm, 13 stm)

versus

z2 x' B2 L U D R2 B e' m' B' U' D' L' s' U2 (17 htm, 14 stm)

(You may verify that 17 htm is move optimal from any move optimal 3x3x3 computer solver you wish.)

In fact, based on the eight short PLL parity fixes that I listed in the last post of that thread (I eventually found four more and put all $12$ in this section of the $4\times 4\times 4$ parity algorithms speedsolving wiki page), we can make dozens of such examples by combining two short PLL parity move sequences together.

I never explored the topic further to know if there is another method for finding such positions; and therefore, I cannot claim that my method is the only method for finding them.

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