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 Post subject: "Twist parity" of helicopter cube corners
PostPosted: Sun Mar 25, 2012 7:58 pm 
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Joined: Wed Jan 14, 2009 6:37 pm
Location: The Great White North
So some time ago, I picked up the helicopter cube from my twisty collection and started playing with it again, and eventually thought I finally mastered the basics of solving from a jumbled state.

And then the very next day, I jumbled it again and resolved it, only to find that the entire cube is solved except for a single corner that has the wrong twist. The problem is, during that resolve there was one point when I felt the cube "expanding" as though it were about to pop, and I managed to keep it in one piece. So now I'm not sure if that single mis-twisted corner is actually a real, reachable state if jumbling moves are allowed, or if it somehow acquired the wrong twist during that almost-popping state where perhaps I may have accidentally pushed it into an impossible state.

So... is it possible to twist a single corner on the helicube? If so, I'd like to know how to fix it, 'cos I've tried resolving, all the jumbling moves that I know (which aren't very many), and still the wrong twist persists. Am I just going crazy here??!


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 Post subject: Re: "Twist parity" of helicopter cube corners
PostPosted: Sun Mar 25, 2012 8:14 pm 
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Joined: Wed Nov 24, 2010 11:12 am
Location: Hong Kong/Beijing
As I know this situation should not happen. Actually through jumbling not the corners are being rotated but the edge pieces. So normally I'll solve it with putting the corners back to right position first. With this , I never face your situation before. It did happen once that there're only two corners need to swap on the same edge piece with the same direction. But still, two corners, not single. Hope this help....


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 Post subject: Re: "Twist parity" of helicopter cube corners
PostPosted: Sun Mar 25, 2012 10:23 pm 
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Joined: Wed Jan 14, 2009 6:37 pm
Location: The Great White North
Hmph. I'm 99% sure I got into the present state through valid moves (though jumbling is included, but I don't know if that factors into it). There is a chance that during that almost-popped state I unknowingly twisted a corner piece into an impossible state, but I'm unsure about that. It seems too big a change to go unnoticed.

What I need to know is either a proof or disproof that such a position is reachable using valid moves.


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 Post subject: Re: "Twist parity" of helicopter cube corners
PostPosted: Sun Mar 25, 2012 10:39 pm 
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Joined: Mon Aug 18, 2008 10:16 pm
Location: Somewhere Else
The mass produced helicopter cube's mechanism makes it rather easy to twist one corner in place without moving it.

One twisted corner cannot occur on the HC, although I have no proof of this.


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 Post subject: Re: "Twist parity" of helicopter cube corners
PostPosted: Wed Mar 28, 2012 7:18 am 
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Joined: Fri May 06, 2005 10:13 am
Location: Norway
Jumbling move or not - a basic edge turn involves 3 swaps of corner faces, resulting in an odd "sticker state". A single twisted corner involves a sticker 3-cycle, or 2 swaps. This makes up a partial proof at least :wink:

Per

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 Post subject: Re: "Twist parity" of helicopter cube corners
PostPosted: Sat Jul 14, 2012 11:57 pm 
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Joined: Tue Jul 13, 2010 12:50 pm
A single corner rotation is impossible. Helicopter corners behave like standard 3x3 corners with even permutations of the (rotational) centers. This is most obvious with the curvy copter, because it has its centers visible. Jumbling has no effect on the orientational states of the corner pieces, nor its permutational states. The reason 2-swaps and 4-cycles are possible in the heli is because center pieces do not have orientation. The cube is in an even perm in its solved state(0 rotations), and the first turn moves that to an odd state(1 rotation), move an adjacent side(2), and you have a 3 cycle of corners. If it was possible to solve in an odd state, then you could do 2 swaps from an even state to an even state, and be able to rotate all four corner the same direction at once. Twisting a single edge would then be trivial, just do a 4-rotate from a sune.


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 Post subject: Re: "Twist parity" of helicopter cube corners
PostPosted: Sun Jul 15, 2012 12:16 am 
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Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
A single twist of a Helicopter Cube corner is impossible. This is trivial to prove for all non-jumbling scrambles:

Assign each corner a twist number from {0, 1, 2} for each orientation in each position. That is, each corner will have eight 0s, eight 1s, and 8 2s, one for each position it could be in. Choose any random state to start, add up the twist value for each corner in whatever position they happen to be in, mod 3. The resulting value is constant. For each of the 12 possible moves, apply it, recompute the total twist, and see that the value does not change. Since no move can change the total twist, it is easy to see that no sequence of moves and therefore no random scramble can ever change the total twist.

I'm not sure how to begin proving it for jumbling scrambles.

It is also true on the Unbandaged Helicopter Cube. If you have a twisted corner than you either have another twisted corner or you have a twisted triangle (which causes it to stick out of the surface of the puzzle).

The Helicopter Skewb and Unbandaged Helicopter Skewb can have a single twisted corner because Skewb moves do not maintain overall twist.


Edit: my proof doesn't work because it gives you too much freedom to choose the value of the twist for each piece in each orientation in each position. Instead you have to pick a "zero sticker" for each piece. It's easiest if you just pick the top face color as the zero sticker for the top 4 corners and the bottom face color as the zero sticker for the bottom corners. Then pick (in any way you would like) the 1 and 2 sticker positions. When the 0-sticker is in the 0 spot sticker spot for whatever position + orientation it happens to be, it has a twist value of 0. When the 0-sticker is in the 1 spot, the piece has a twist of 1, etc. The proof will work if you number in this way.

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