Aside from the disassembly/assembly problem there was another problem that
I have not yet seen answered satisfactory. The question is: what is the
maximum number of stickers that can bee peeled of such that there is still
an unique solution for the cube (i.e. the remaining stickers must match in
color on a face). The only solution I have seen was along the lines (this
is from memory, but I do not think there are any mistakes):
1. The total rotation of the corner cubes is 0, so there is one corner
cube that can have all its stickers removed; the remaining corner
cubes need at least one sticker. Suppose this is the FUR cube.
2. You can remove two stickers (F, R and/or U) from each of FRD, BRU, FLU;
they still remain distinguishable. (6 stickers.)
3. Of the remaining corner cubes (DBL, DRB, ULB, DLF) you cannot *now*
(emphasis mine) remove two stickers because the cube will become
indistinguishable from one of the cubes handled in step 2. You can
remove sticker R, U and F from DRB, ULB and DLF respectively. No
other stickers can be removed. (3 stickers.)
4. Because of flip parity you can remove two stickers from (say) FU.
5. You can remove the F sticker from all of FR, FL and FD. (3 stickers.)
6. Now, because of the product parity of corner cubes permutation and
edge cube permutation you can make either two corner cubes identical or
two edge cubes. You must nevertheless still be able to observe both
the corner twist parity and the edge flip parity. This means you may
a. Remove a single sticker from any edge cube that still has two stickers.
b. Remove a single sticker from the DLB cube.
(You can not remove two stickers from the DLB cube. Say you remove the L
and B sticker. Let us denote removed sticker by lower case letters.
In that case Dlb is indistinguishable from Dfr, which is not a problem.
But the DLf cube can now be put in the Dlb position, leading to a 3-cycle.)
7. You can remove the sticker from the front center cube. (1 sticker.)
This leads to a total of 19 removable stickers.
This is not maximal. There are, for instance, other ways to do corner cubes:
1. Remove all F stickers. (4 stickers.)
2. From all Fxy cubes, remove the y sticker. (4 stickers.)
3. Also from all Bxy cubes, remove the y sticker. (4 stickers.)
4. From one Bxy cube remove also an x sticker. (1 sticker.)
5. From one Fxy cube remove also an x sticker. (1 sticker.)
(Fxy and Bxy named clockwise.)
This leads to 14 stickers for the corners and a total of 21.
Are there other ways leading to more? Are there better ways that we can
remove more center stickers?
dik t. winter, cwi, amsterdam, nederland