All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
Rubik's Swap (Posted on 2021-06-01) Difficulty: 3 of 5
You own a traditional 3x3x3 Rubik's cube, with nine stickers on each of its six faces.

When you're not looking, a mischievous sprite randomly selects two stickers, peels them off and swaps their places on the cube. What is the probability that the cube can still be solved?

No Solution Yet Submitted by tomarken    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
moves and swaps | Comment 4 of 9 |
I wanted to pin this down a little further. 

In the following I distinguish between "switching" of stickers and "swapping" positions of cubelets. 
I break the answer down like this: 

The sprite's sticker-switch is one of: 

0) inconsequential (switching two stickers of the same color, or switching a sticker with itself, if that is a thing, i.e. Charlie's 54th option)

1) detrimental type 1 (swap labels of different colors between two different components: center, edge and corner). This creates new, illegal cubelets that will never fit anywhere. 

2) detrimental type 2 (the most natural type sabotage - switching stickers of the same components: center, edge and corner, but of different colors.

We now show that detrimental type 1 and type 2 are irreversible. 

When you exclude the possibility of switching a sticker with itself, #1 or #2 will happen randomly (53-8)/53 = 45/53 of the time. One of the 54 stickers is pulled off and another of the 53 is pulled off with 45/53 chance of it not being the same color as the first. They are switched. 

To show that #1 and #2 are irreversible, consider moves and swaps. If you picture the cube as made of 3x3x1 planer slices, then there are 9 different possible slices, depending on which dimension you cut through. The 6 face slices are made of 3x3 cubelets including the center cubelet, while the 3 interior slices are 3x3 but with an unseen center. Each slice has 4 corner cubelets (call them C1, C2, C3, C4) and 4 edge cubelets and with the face slices having a visible 9th center cubelet. C1, ... C4 refer to the cubelets themselves, not their positions, except that they will always be corner cubelets.   We represent the positions of cubelets by their place in the (N, E, S, W) 
4-tuple. They start in the positions C1, ..., C4 start off in the configuration (C1, C2, C3, C4).  

There is really only one single elemental move possiblethe turn of a slice 90 degrees, let's say CCW.. The effect of this is to change the position of all the cubelets of the slice (except the center) to a new position within the cube. The four corners change from (C1, C2, C3, C4) to (C2, C3, C4, C1). This can be seen as resulting from swapping their positions pairwise. There have been 3 swaps made to change the corners locations: C1<->C2, C2<->C4, C3<->C2, where the arrows indicate changed positions of the cubelets, performed in that order. Likewise, an identical 3-step swap pattern applies to the 4 edge cubes. So, each move causes 3+3=6 swaps. Since each move makes 6 swaps (an even number of swaps), and the sprite's sabotage is a swap of 1 pair (an odd number of swaps), it is irreversible: applying any moves to remedy the situation will always leave an odd number of cubelets swapped. 

I see I left out "rotations" where a cubelet only rotates in place after a number of moves. There is a series of moves that will rotate a cubelet in place, so that the three stickers facing outward will have each rotated to face outward on a new face (left, forward, or up). A series of such moves to execute this likewise completes an even number of swaps. 

Therefore, if the sprit exchanges only a pair of stickers on a single cubelet, this is irreversible, since through subsequent user moves, all three stickers must be rotated together.

A single swap set the cube into a different parity of the state of a cube. One pair of switched stickers (of different colors taken from the same component: center, corner, edge) render a three by three cube unsolvable.) 

So, assuming the sprite worked randomly (in the dark?) and actually switched two stickers, the odds are 8/53 the stickers were the same color and thus the cube is still solvable (or still already solved), 45/53 against (hopeless).  


Edited on August 23, 2022, 1:22 am
  Posted by Steven Lord on 2021-06-02 17:55:35

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information