Guest Posted February 12, 2011 Report Share Posted February 12, 2011 On a related note to bushindo's prove that this cube is unsolvable. Sorry for the bad image. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 (edited) I wish I could find the correct terminology / math to explain it fully, but I know from experience (and a few pranks on cubers [including myself]) that this cube cannot be solved based on the alorithms. If just (1) more piece was out of place, it would be solvable. for the most part, the number of incorrect pieces (as far as edge pieces are concerned) must be a multiple of 3 if all corner pieces are correct.. if I am remembering correctly. I'll edit this post when I remember the correct math.... Ok, I slightly cheated, but the way this site explained it was better than I could have (not so good with putting my thoughts into words... I'm a very visual person): As it turns out, every cube state reachable by legal moves can always be represented by an even number of swaps, and at the same time cannot be represented by an odd number of swaps (the two are mutually exclusive). Since the above cube has an odd number of swaps ("one" swap), this state cannot be reached. To understand why this is so, we need to realise that each legal move always performs the equivalent of an even number of swaps. No matter how many moves you perform, the number of accumulated swaps will therefore always remain even. For example, consider the turning of one face by 90 degrees: C1 E1 C2 E4 E2 C4 E3 C3 → C4 E4 C1 E3 E1 C3 E2 C2 The new corner state can be obtained via 3 swaps (swap C1/C4, swap C1/C3, swap C1/C2). Similarly, the new edge state can be obtained via 3 swaps. All together, this is 6 swaps which is even. Therefore, no matter how many moves you perform, always an even number of swaps will have been performed. Since exactly half of the conceivable permutations are even and the other half are odd, only half of the cube's permutations (ignoring orientation) are reachable by legal moves. Edited February 13, 2011 by Parabola Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 And I did not remember correctly- As stated on that quote, must be even swaps. So (3) out of place would be 2 swaps, hence why I was thinking multiples or 3... but if 4 were out of place, it would be 2 or 4 swaps, depending on the situation / algorithm.... My mistake. It was corner pieces that need to be multiples of 3, not edges. Quote Link to comment Share on other sites More sharing options...
0 Smith Posted February 13, 2011 Report Share Posted February 13, 2011 (edited) Hope I get this spoiler thing right - it's my first post... The cube is unsolvable because the relationship between the out-of-place red face to the out-of-place yellow face is the opposite of the required solution locations. That is, in calling the almost completed red face "Conter-Clock-Wise" of the yellow, the single yellow spot is also CCW of the single red spot, and given that the blue side is complete, the two out-of-place faces are connected to the same planar axis so they can never be rotated into a juxtaposed position. Edited February 13, 2011 by Smith Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 not the traditional way but... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 (edited) Hope I get this spoiler thing right - it's my first post... The cube is unsolvable because the relationship between the out-of-place red face to the out-of-place yellow face is the opposite of the required solution locations. That is, in calling the almost completed red face "Conter-Clock-Wise" of the yellow, the single yellow spot is also CCW of the single red spot, and given that the blue side is complete, the two out-of-place faces are connected to the same planar axis so they can never be rotated into a juxtaposed position. Welcome to BrainDen, Smith! Unless I'm mistaken, I believe that your answer does not prove that the cube is unsolvable. For example, if it were two corners that were switched, with the correct rotation, the cube could be solvable. But I may have misunderstood you. Edited February 13, 2011 by sayalzah Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 I wish I could find the correct terminology / math to explain it fully, but I know from experience (and a few pranks on cubers [including myself]) that this cube cannot be solved based on the alorithms. If just (1) more piece was out of place, it would be solvable. for the most part, the number of incorrect pieces (as far as edge pieces are concerned) must be a multiple of 3 if all corner pieces are correct.. if I am remembering correctly. I'll edit this post when I remember the correct math.... Ok, I slightly cheated, but the way this site explained it was better than I could have (not so good with putting my thoughts into words... I'm a very visual person): As it turns out, every cube state reachable by legal moves can always be represented by an even number of swaps, and at the same time cannot be represented by an odd number of swaps (the two are mutually exclusive). Since the above cube has an odd number of swaps ("one" swap), this state cannot be reached. To understand why this is so, we need to realise that each legal move always performs the equivalent of an even number of swaps. No matter how many moves you perform, the number of accumulated swaps will therefore always remain even. For example, consider the turning of one face by 90 degrees: C1 E1 C2 E4 E2 C4 E3 C3 → C4 E4 C1 E3 E1 C3 E2 C2 The new corner state can be obtained via 3 swaps (swap C1/C4, swap C1/C3, swap C1/C2). Similarly, the new edge state can be obtained via 3 swaps. All together, this is 6 swaps which is even. Therefore, no matter how many moves you perform, always an even number of swaps will have been performed. Since exactly half of the conceivable permutations are even and the other half are odd, only half of the cube's permutations (ignoring orientation) are reachable by legal moves. And I did not remember correctly- As stated on that quote, must be even swaps. So (3) out of place would be 2 swaps, hence why I was thinking multiples or 3... but if 4 were out of place, it would be 2 or 4 swaps, depending on the situation / algorithm.... My mistake. It was corner pieces that need to be multiples of 3, not edges. You got it right, but the corner pieces do not need to be multiples of three, but when they are put in the correct location, their needed CW rotations has to total to a multiple of three. Hope that made sense. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 13, 2011 Report Share Posted February 13, 2011 Solved Quote Link to comment Share on other sites More sharing options...
0 Guest Posted February 15, 2011 Report Share Posted February 15, 2011 You got it right, but the corner pieces do not need to be multiples of three, but when they are put in the correct location, their needed CW rotations has to total to a multiple of three. Hope that made sense. Lol, made sense to me. it's been a long (and I do mean long... like 7 years) time since I've had to think about the cube that way. However.. I still cube daily! My most recent acquisition is a MegaMinx.... look into it. I managed to get 96% done before I got stuck... It's really fun though.... 20 sided rubics... lol Quote Link to comment Share on other sites More sharing options...
0 phaze Posted February 15, 2011 Report Share Posted February 15, 2011 Could we please see the other sides of the cube? Quote Link to comment Share on other sites More sharing options...
0 Molly Mae Posted February 15, 2011 Report Share Posted February 15, 2011 Could we please see the other sides of the cube? I agree. Are we to assume that the orange and white sides are solved? Quote Link to comment Share on other sites More sharing options...
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On a related note to bushindo's prove that this cube is unsolvable.
Sorry for the bad image.
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