# Message #241

From: Don Hatch <hatch@plunk.org>

Subject: 2^d and 3^d solve function

Date: Mon, 15 May 2006 04:21:10 -0700

Hi folks,

Sorry I’ve been out of the loop, I have not tackled my spam problem yet

and have been simply avoiding my mailbox :-(

But Melinda showed me Roice and Charlie’s 5d program the other day

when I was visiting… awesome, you guys :-)

Attached is a general function to solve the n^d puzzle

(length n = 2 or 3, any number of dimensions d >= 3).

It should be easy to plug this into

Melinda’s and/or Roice&Charlie’s program, if you are so inclined.

(I wrote it in java this time to make it easy on you :-) )

A couple of surprising facts I discovered while doing this–

not sure whether they are already well-known?

- The notion of parity (i.e. whether the solution will have

an even or odd number of twists)

does not exist for 2^d puzzles with d >= 4;

that is, for such puzzles, there is a sequence of an odd number

of 90 degree twists that will bring the solved puzzle

back to solved state (unlike the 2^3, and n^d with n odd,

all of which have a parity restriction).

Not sure whether this is true for n^d for other even n

(I didn’t look at trying to solve anything with n>3).

- There is no "twirl modulus" restriction on corner cubies

for n^d puzzles with d >= 5:

that is, you can cycle 3 stickers

on a single corner cubie (without having to anti-cycle

3 stickers on some other corner cubie like you have to do

in 3 or 4 dimensions).

This is fairly easy to see if you think about it–

it’s exactly the same reasoning that lets you

twirl a single 3-sticker cubie on the 3^4 puzzle.

The crux of the matter is that

in >=5 dimensions you can’t tell the difference

between a cycle and an anti-cycle (of 3 stickers on a corner cubie);

that is, one can be rotated to the other.

So, to cycle 3 stickers a,b,c on a single corner cubie:

1. cycle a,b,c and anti-cycle 3 stickers on some other cubie

2. twist a face containing a,b,c and not the other cubie so that:

a goes to a

b goes to c

c goes to b

d goes to e

e goes to d

(where d,e are two more stickers on the same corner cubie

as a,b,c – this is where d>=5 is needed)

(you can’t just swap b and c without swapping some other d and e,

since that would turn the cubie inside out)

3. undo 1.

4. undo 2.

The result is that a,b,c got forward-cycled twice and nothing else

on the puzzle was touched– i.e. they got backward-cycled once.

So do all of the above in reverse to cycle a,b,c forward, as desired :-)

Don