Difference between revisions of "K4"

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'''Brief Summary of the Method'''
 
'''Brief Summary of the Method'''
  
After two opposite centers are solved, this method takes advantage of the remaining four middle centers to pair up three cross-edges of the same color without the inconvenience of destroying and fixing back previously solved centers as is edge pairing in reduction.  Two adjacent corners are solved between these three cross-edges to form a 1x3x4 block about one of the initially solved centers.  Then the four middle centers are carefully solved without destroying the 1x3x4 block (this can be done as efficiently as solving the middle centers in reduction, according to Thom Barlow in "Step 3" of his guide).  Next, the last two corners of the first layer are inserted as well as inserting the last cross-edge (which may either be inserted one winged-edge at a time or by first pairing up the dedge and then inserting it).  (Here is a link to java applets demonstrating possible scenarios of completing the last dedge [http://rachmaninovian.webs.com/step3b.htm]).  Now the first layer is complete along with all six centers.  The remaining dedges in the first three layers are directly solved using various commutators (they all can really be solved with one commutator, but knowing more commutators enables the cuber to solve them with speed).  Solving the last layer which is broken into two-three steps:  1) Solve the corners with COLL or other preferred 3x3x3 corner solving methods, 2) use commutators to place the unsolved winged edges in their solved positions and 3) perform an odd permutation algorithm to the cube if (odd) parity exists ("permutation/even parity" usually does not occur due to the fact that the last-layer dedges are not "paired up" but rather directly solved in pieces).
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After two opposite centers are solved, this method takes advantage of the remaining four middle centers to pair up three cross-edges of the same color without the inconvenience of destroying and fixing back previously solved centers as is edge pairing in reduction.  Two adjacent corners are solved between these three cross-edges to form a 1x3x4 block about one of the initially solved centers.  Then the four middle centers are carefully solved without destroying the 1x3x4 block (this can be done as efficiently as solving the middle centers in reduction, according to Thom Barlow in "Step 3" of his guide).  Next, the last two corners of the first layer are inserted as well as inserting the last cross-edge (which may either be inserted one winged-edge at a time or by first pairing up the dedge and then inserting it).  (Here is a link to java applets demonstrating possible scenarios of completing the last dedge [http://rachmaninovian.webs.com/step3b.htm]).  Now the first layer is complete along with all six centers.  The remaining dedges in the first three layers are directly solved using various commutators (they all can really be solved with one commutator, but knowing more commutators enables the cuber to solve them with speed).  Solving the last layer is broken into two-three steps:  1) Solve the corners with COLL or other preferred 3x3x3 corner solving methods, 2) use commutators to place the unsolved winged edges in their solved positions and 3) perform an odd permutation algorithm to the cube if (odd) parity exists ("permutation/even parity" usually does not occur due to the fact that the last-layer dedges are not "paired up" but rather directly solved in pieces).
  
 
'''Difficulties of the Method'''
 
'''Difficulties of the Method'''

Revision as of 07:27, 19 February 2010

The K4 Method was created by Thom Barlow. His guide can be found here [[1]].


Brief Summary of the Method

After two opposite centers are solved, this method takes advantage of the remaining four middle centers to pair up three cross-edges of the same color without the inconvenience of destroying and fixing back previously solved centers as is edge pairing in reduction. Two adjacent corners are solved between these three cross-edges to form a 1x3x4 block about one of the initially solved centers. Then the four middle centers are carefully solved without destroying the 1x3x4 block (this can be done as efficiently as solving the middle centers in reduction, according to Thom Barlow in "Step 3" of his guide). Next, the last two corners of the first layer are inserted as well as inserting the last cross-edge (which may either be inserted one winged-edge at a time or by first pairing up the dedge and then inserting it). (Here is a link to java applets demonstrating possible scenarios of completing the last dedge [2]). Now the first layer is complete along with all six centers. The remaining dedges in the first three layers are directly solved using various commutators (they all can really be solved with one commutator, but knowing more commutators enables the cuber to solve them with speed). Solving the last layer is broken into two-three steps: 1) Solve the corners with COLL or other preferred 3x3x3 corner solving methods, 2) use commutators to place the unsolved winged edges in their solved positions and 3) perform an odd permutation algorithm to the cube if (odd) parity exists ("permutation/even parity" usually does not occur due to the fact that the last-layer dedges are not "paired up" but rather directly solved in pieces).

Difficulties of the Method

One of the biggest challenges of this method is to master solving the last layer. One can easily solve the last-layer corners using COLL or regular 3x3x3 corner solving methods, but solving the last-layer dedges requires commutators. Thom Barlow covers this portion of the method in "Step 7" of his guide. He has provided an algorithm for each of the 28 possible cases (which have the same probability) for moving three winged edges at a time (3-cycles) as well as parity algorithms and 2 2-cycle algorithms ("permutation parity").

With 28 different possible cases just for 3-cycles, solving the last layer can be overwhelming. However, memorizing all 28 3-cycle cases is not a prerequisite for being able to solve the last-layer dedges with confidence: only a handful of commutators is required because one can use variations of those to attack any of the 28 cases by adding necessary preliminary moves to those commutators (and undoing them after the commutator is complete).

For convenience, member cmowla has made a complete table of all 28 3-cycle cases for the last layer, each accompanied by Thom Barlow's algorithm for that specific case.


File:K4LL3cyc1.PNG File:K4LL3cyc2.PNG


Reducing the Number of 3-Cycle Algorithms to Learn

One can reduce the 28 different cases into 6 groups. In each group, there is an algorithm and its reflection. As mentioned before, all 28 cases can be derived from one algorithm. However, in order to minimize the number of moves to tackle each case, it is better to have a handful of algorithms to derive the 28 as opposed to only one (not all of the algorithms will be efficient). The way to derive "weird cases" from "common cases" is by adding preliminary moves to the "common case" so that the desired pieces to cycle are in the positions of the "common case" one is already familiar with. The 6 groups are listed below.


File:Case A.PNG

Algorithm [A]

L B y' z r U' R' U r' U' R F' z' y (by cmowla)

Algorithm [A Reflection]

R' B' y z' r U L U' r' U L' F z y' (by cmowla)


File:Case B.PNG

Algorithm [B]

L B y' z l' U' R' U l U' R F' z' y (by cmowla)

Algorithm [B Reflection]

R' B' y z' l' U L U' l U L' F z y' (by cmowla)

File:Case C.PNG

Algorithm [C]

x' l' U L' U' l U L U' x

Algorithm [C Reflection]

x' r U' R U r' U' R' U x

File:Case D.PNG

Algorithm [D]

x' r U L' U' r' U L U' x

Algorithm [D Reflection]

x' l' U' R U l U' R' U x

File:Case E.PNG

Algorithm [E]

l' y' L2 U r U' L2 U r' U' y l (by cmowla)

Algorithm [E Reflection]

r y' L2 U l' U' L2 U l U' y r' (by cmowla)

File:Case F.PNG

Algorithm [F]

l' D r U' L'2 U r' U' L'2 U D' l

Algorithm [F Reflection]

r D' l' U R2 U' l U R2 U' D r'


Additional Information

For theory on why there are exactly 28 3-cycle cases, visit this page [3]. (In that thread, there is also discussion on the number of cases for higher order cycles as well, such as 4-cycles, 6-cycles, 8-cycles, etc., and multiples of lower cycles, such as 2 2-cycles).


External Links

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