# Message #658

From: matthewsheerin <damienturtle@hotmail.co.uk>

Subject: Greetings

Date: Tue, 31 Mar 2009 18:11:05 -0000

Hello everyone, my name is Matthew and I have recently returned to the problem of the 4D cube. I found them sometime last year after the advent of my cube obsession, but was too scared by what I saw to try to solve any!

It took a few days of messing around to figure out the mechanics and get some ideas kicking around about how to solve it, and i had a practice at the 2^4 first. My first (and currently only) 3^4 solution took a few days to work through, due to the time I had available, and also because I was figuring out a method as I went along, so it wasn’t incredibly efficient. I then moved onto the 4^4, and it was simply a matter of adding a few steps onto my 3^4 solution in order to solve it.

I tried the all-purpose reduction tactic. First I solved the 1C pieces in blocks of 2, then I paired up the 2C pieces, first into 2 pairs from the same face, then into a block of 4. I then paired up 3C pieces like i would on a 4^3. Melinda commented on my log file, saying I luckily avoided parity, but from my (perhaps naive) understanding of the cycles involved here, I don’t think the 3D parities are seen on 4D.

I then used an improvement of my 3^4 solution to solve the rest. This involves first solving one of the cubes in columns of 3 pieces, apart from the last column consisting of a 3C and two 4C pieces which is left unsolved. I then solve what I call the "middle cube" (all the pieces not on the central or outside cube) by using the freedom from the unsolved column to first solve the 8 2C pieces above and below, then making more comumns of 3 in the central cube and inserting them. This is equivalent of a keyhole F2L on a 3D cube I believe. I then solved the block of 6 pieces to just leave the central cube, and a few moves before placing these 6 pieces left all 2C pieces facing the correct way. I then treat what is left almost like a 3x3x3, with a few setup moves required to help flip pieces in 4D using 3D algorithms to create simple cycles. After the pieces all have the sticker matching the 1C piece in the remaining cube, I solve the rest as a 3^3. This is done with the realisation that as long as you use the same adjacent cube to perform the face turns, the rest of the puzzle will be unscrambled by this as long as these are overall a multiple of 4 turns, which is easy to account for on the LL, all that is required is a turn on the central cube between moves to change which face is being turned by the adjecent cube. I believe that the nature of the 4D cube eliminates the need to double odd permutations like the T-perm which would require an odd number of moves on the adjacent cube and thus prevent the outer pieces from being solved as the adjacent cube must be moved in mutilples of four.

I have now started looking at the 5D cube to get the hang of the mechanics, and though I wasn’t going to, I am solving the 5^4 slowly in the background, between 5D and schoolwork.

As a new member, any comments would be appreciated, though please no spoilers on the 5D puzzles and their solutions. I can have a mental breakdown of my own accord.

Matthew S