This is a continuation of

viewtopic.php?f=8&t=22729When I saw that topic I suddenly wanted to peel stickers off to get an optimally unstickered cube. But I didn't like how the problem was left unsolved

. The only confirmed solution was 20 removed stickers, and that was done quite randomly IMO. So ... here's my take (complete with steps). I removed 24 stickers. Please confirm if this is indeed a solution.

The problemYou have a 3x3 cube, and a limited number of stickers. You want to save as many stickers as you can. Any two pieces stickered identically are considered indistinguishable. Any state where all faces have one color on them are considered solved states, even if it wasn't the same as the original "solved" state. You have no memory of which sticker were removed, but you do remember the color scheme. For simplicity, its the BOY color scheme. When scrambled, the cube can be in any spatial orientation.

How many stickers can you remove and still keep one unique solution?

Limitations of a legal cube position:Only an even number of edges can have swapped orientation

The orientation twists of corners must add up to a whole number of twists

Only an even number of (edge-swaps + corner-swaps + center axis quarter turns) are allowed

Now here's my solution

CornersI tried to make a huge even-numbered cycle of adjacent corners, since adjacent corners with 1 sticker are interchangeable.

Keeping 1 sticker is better than keeping 2, but keeping none severely limits your options, because any 0-sticker corner can swap with 0- or 1-sticker corners indistinguishably.

I used an even number because that requires an odd number of swaps, which means the edges must have an odd number of swaps as well. I can't get past 6, because there are only 6 colors. An 8-cycle leaves 2 pairs of identical edges.

This is not a unique state, but the other solved state require at least 1 pair of edges to swap. If I fix the edges, then this alternative solution will be visually apparent.

EdgesI started with the 6-cycle for edges, and I made a small modification. Luckily, I came across a state with only 1 visibly solved state. At least I think. Please verify this step - its the weakest link.

There is a 3-cycle of indistinguishable edges, but if you use it, you end up with an odd number of orientation swaps. So it can't be legally solved from there. I think.

CentersOnly 2 centers are needed to uniquely fix the center axis. There will only be 1 visibly correct state.

Corners IIThere are still 2 corners with 3 stickers. You only need 2 stickers to identify any corner. So I removed one from each. As for why I specifically removed a yellow and a blue sticker...you'll see

Final RoundupAll centers are fixed. The corners has 1 alternate state, but that leaves at least 1 swap with the edges. The edges are fixed. So there is only 1 solution.

-12 stickers from the corners

-6 stickers from the edges

-4 stickers from the centers

-2 stickers from the corners (II)

24 stickers removed (30 stickers remaining)

How pretty, every face has an equal number of stickers! (That's why I specifically removed 1 yellow and 1 blue from Corners II)

You can test this cube out using this applet. I can't find any others that lets you blank out stickers and play by dragging.

http://www.randelshofer.ch/rubik/virtua ... tions.htmlSo ... is this the optimal solution? Is this a solution?