Rubiks Cube/Numbers - Revision history

Admin: /* Number of permutations of 3x3 */

2019-04-22T21:44:05Z

Number of permutations of 3x3

← Older revision		Revision as of 21:44, 22 April 2019
Line 6:		Line 6:

	* 3 possible rotations of 8 corners, 7 corners determine the 8th = <math>3^7</math>		* 3 possible rotations of 8 corners, 7 corners determine the 8th = <math>3^7</math>
	* 2 possible rotations of 12 edges~~, 11 edges determine the 12th~~ = <math>2^{12}</math>		* 2 possible rotations of 12 edges = <math>2^{12}</math>
	* 8 different corner pieces to be distributed to 8 locations = <math>8!</math>		* 8 different corner pieces to be distributed to 8 locations = <math>8!</math>
	* 12 different edge pieces to be distributed to 12 locations = <math>12!</math>		* 12 different edge pieces to be distributed to 12 locations = <math>12!</math>
			* Half of the cube permutations (those requiring center faces to be swapped) cannot be reached without disassembling the cube = <math>\frac{1}{2}</math>
	~~Furthermore, some~~ of ~~these possibilities are not possible. Only half of the parity cases can occur on~~ the ~~3x3~~ cube~~, due~~ to the ~~fact that the centers are fixed, cutting the number of ways of distributing the corners and edges in half.~~

	Total number of permutations of a 3x3 cube:		Total number of permutations of a 3x3 cube:

	<math>		<math>
	N = 3^{7} \times 2^{11} \times ~~\dfrac{ 8~~! \times ~~12!~~ }{ 2 } = ~~43,252,003,274,489,856,000~~ \sim 10^{19}		N = 3^{7} \times 8! \times 2^{12} \times 12! \times \frac{1}{2} = 86504006548979712000 \sim 8.65040 \times 10^{19}
	</math>		</math>

	or about 43 quintillion permutations.		or about 86 quintillion permutations.

	==Number of permutations of 4x4==		==Number of permutations of 4x4==

Admin: /* Number of permutations of 5x5 */

2019-04-22T19:29:36Z

Number of permutations of 5x5

@@ Line 71: / Line 71: @@
 <math>
-T = 3^7 \times 8! \times 48! \times \left( \dfrac{24!}{12!} \right)^2 \times 12! \times 2^{12} = ...
+T = 3^7 \times 8! \times 48! \times \left( \dfrac{24!}{12!} \right)^2 \times 12! \times 2^{12}
 </math>

Admin: /* Number of permutations of 4x4 */

2019-04-22T19:27:31Z

Number of permutations of 4x4

← Older revision		Revision as of 19:27, 22 April 2019
Line 42:		Line 42:

	<math>		<math>
	T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 \times 24! = ...		T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 \times 24! = 91793597096746925044380089227138852984017051539472384000000000 \sim 9.17935 \times 10^{61}
	</math>		</math>

Admin: /* Number of permutations of 4x4 */

2019-04-22T19:25:22Z

Number of permutations of 4x4

← Older revision		Revision as of 19:25, 22 April 2019
Line 22:		Line 22:
	==Number of permutations of 4x4==		==Number of permutations of 4x4==

	On a 4x4 cube, we use the same technique as the 3x3, counting permutations due to both placement and orientation of each type of piece. Now each face has four corner pieces, four double edges for eight edge pieces, and four ~~corners~~. Counting:		Note: dedge = double edge

			On a 4x4 cube, we use the same technique as the 3x3, counting permutations due to both placement and orientation of each type of piece. Now each face has four corner pieces, four double edges for a total of eight edge pieces, and four corner pieces. Counting:

	Placing cubies:		Placing cubies:

	* Placement: 8 corners being placed, for total of <math>8!</math>		* Placement: 8 corners being placed, for total of <math>8!</math>
	* Placement: 12 left~~-double-edge~~ pieces that can go in 24 possible double-edge locations, for a total of <math>\dfrac{24!}{12!}</math>		* Placement: 12 dedge left wing pieces that can go in 24 possible double-edge locations, for a total of <math>\dfrac{24!}{12!}</math>
	* Placement: 12 right-double-edge pieces that can go in 24 possible double-edge locations, for a total of <math>\dfrac{24!}{12!}</math>		* Placement: 12 dedge right-double-edge pieces that can go in 24 possible double-edge locations, for a total of <math>\dfrac{24!}{12!}</math>
	* Placement: 24 center squares, which can be arranged in any configuration, for a total of <math>24!</math>		* Placement: 24 center squares, which can be arranged in any configuration, for a total of <math>24!</math>

Line 34:		Line 36:

	* Orientation: 3 orientations are possible on the 8 corners, placement of 7 determines the last, for total of <math>3^7</math>		* Orientation: 3 orientations are possible on the 8 corners, placement of 7 determines the last, for total of <math>3^7</math>
	* Orientation: interestingly, the position of a ~~double-edge~~ piece uniquely determines its orientation. If a left~~-double-edge~~ piece is placed in a left~~-double-edge~~ position, it will be oriented "~~right-side up~~". If a left~~-double-edge~~ piece is placed in a right~~-double-edge~~ position, it will be oriented "~~upside-down~~". And likewise for right~~-double-edge positions. That means that~~ if a ~~double edge cubie is made part of~~ a ~~double edge, it has only two possible positions on that double edge~~. ~~Therefore~~, there are 0 degrees of freedom due to orientation of double edge pieces.		* Orientation: interestingly, the position of a dedge piece uniquely determines its orientation. If a dedge left wing piece is placed in a left dedge position, it will be oriented "correctly". If a dedge left wing piece is placed in a right dedge position, it will be oriented "backwards". And likewise for dedge right wing pieces being "correct" if placed in a right dedge spot and "backwards" if placed in a left dedge spot. Thus, there are 0 degrees of freedom due to orientation of double edge pieces. It's orientation is determined by the type of wing piece and the position being occupied.
	* Orientation: Like the double-edge pieces, the center squares get 0 degrees of freedom due to orientation. Distributing 24 center squares among 24 center square positions uniquely determines the state of those center squares.		* Orientation: Like the double-edge pieces, the center squares get 0 degrees of freedom due to orientation. Distributing 24 center squares among 24 center square positions uniquely determines the state of those center squares.

Line 40:		Line 42:

	<math>		<math>
	T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 = ~~147947189226886589978930309391974400000 \sim 10^{38}~~		T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 \times 24! = ...
	</math>		</math>

	~~(TODO: fix above number, missing 24 factorial due to 4 x 6 = 24 center squares)~~

	==Number of permutations of 5x5==		==Number of permutations of 5x5==

Admin: /* Counting Permutations */

2019-04-22T19:21:12Z

Counting Permutations

@@ Line 4: / Line 4: @@
 To count number of permutations of 3x3, count permutations due to both the placement and orientation of each type of piece (corners and edges; centers are fixed):
-Furthermore, some of these possibilities are not possible. Only even parity cases can occur on the 3x3 cube, cutting the number of ways of distributing the corners and edges in half.
+* 3 possible rotations of 8 corners, 7 corners determine the 8th = <math>3^7</math>
 Total number of permutations of a 3x3 cube:
@@ Line 22: / Line 23: @@
 On a 4x4 cube, we use the same technique as the 3x3, counting permutations due to both placement and orientation of each type of piece. Now each face has four corner pieces, four double edges for eight edge pieces, and four corners. Counting:
 * Placement: 8 corners being placed, for total of <math>8!</math>
 * Placement: 12 left-double-edge pieces that can go in 24 possible double-edge locations, for a total of <math>\dfrac{24!}{12!}</math>
 * Placement: 12 right-double-edge pieces that can go in 24 possible double-edge locations, for a total of <math>\dfrac{24!}{12!}</math>
-* Placement: 24 center squares, which can be arranged in any configuration
+* Placement: 24 center squares, which can be arranged in any configuration, for a total of <math>24!</math>
 * Orientation: 3 orientations are possible on the 8 corners, placement of 7 determines the last, for total of <math>3^7</math>
 * Orientation: interestingly, the position of a double-edge piece uniquely determines its orientation. If a left-double-edge piece is placed in a left-double-edge position, it will be oriented "right-side up". If a left-double-edge piece is placed in a right-double-edge position, it will be oriented "upside-down". And likewise for right-double-edge positions. That means that if a double edge cubie is made part of a double edge, it has only two possible positions on that double edge. Therefore, there are 0 degrees of freedom due to orientation of double edge pieces.
@@ Line 35: / Line 42: @@
 T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 = 147947189226886589978930309391974400000 \sim 10^{38}
 </math>
 ==Number of permutations of 5x5==
-On a 5x5 cube, we apply the above techniques.
+On a 5x5 cube, we apply the above techniques. We have fixed centers, so we don't consider those a type. The other cubie types are face cubies, corner cubies, and tredges, composed of tredge centers (the middle piece of the three pieces that make up a tredge) and tredge wings (the two end pieces of the three pieces that make up a tredge). There are 8 corner cubies, 9 face cubies on each face (and since 1 is fixed, actually 8 per face) for a total of 48 face cubies that can move, and 12 triple edges, for a total of 12 tredge centers, 12 tredge left wing pieces, and 12 tredge right wing pieces.
 =Counting Cycles=

Admin: /* Number of permutations of 4x4 */

2019-04-22T18:39:39Z

Number of permutations of 4x4

← Older revision		Revision as of 18:39, 22 April 2019
Line 35:		Line 35:
	T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 = 147947189226886589978930309391974400000 \sim 10^{38}		T = 3^7 \times 8! \times \left( \dfrac{24!}{12!} \right)^2 = 147947189226886589978930309391974400000 \sim 10^{38}
	</math>		</math>

			==Number of permutations of 5x5==

			On a 5x5 cube, we apply the above techniques.

	=Counting Cycles=		=Counting Cycles=

Admin at 02:45, 11 January 2018

2018-01-11T02:45:56Z

← Older revision		Revision as of 02:45, 11 January 2018
Line 47:		Line 47:


			=Flags=

			{{RubiksFlag}}




	~~[[Category:Rubiks]]~~

Admin: /* Number of permutations of 3x3 */

2018-01-02T00:52:45Z

Number of permutations of 3x3

← Older revision		Revision as of 00:52, 2 January 2018
Line 14:		Line 14:

	<math>		<math>
	N = 3^{7} \times 2^{11} \times \dfrac{ 8! \times 12! }{ 2 } = 43,252,003,274,489,856,000 \sim 10^19		N = 3^{7} \times 2^{11} \times \dfrac{ 8! \times 12! }{ 2 } = 43,252,003,274,489,856,000 \sim 10^{19}
	</math>		</math>

Admin at 00:52, 2 January 2018

2018-01-02T00:52:31Z

@@ Line 1: / Line 1: @@
 ==Number of permutations of 3x3==
 * 3 possible rotations of 8 corners, 7 corners determine the 8th = 3^7
 * 2 possible rotations of 12 edges, 11 edges determine the 12th = 2^11
@@ Line 16: / Line 19: @@
 or about 43 quintillion permutations.
-==Length of Cycles==
+==Number of permutations of 4x4==
 One of the interesting features of a Rubik's Cube is that, starting from a solved cube, if any sequence of moves is repeated long enough on the cube, will eventually result in the cube returning to the solved state.

Admin: /* Number of permutations */

2018-01-02T00:05:52Z

Number of permutations

← Older revision		Revision as of 00:05, 2 January 2018
Line 1:		Line 1:
	==Number of permutations==		==Number of permutations of 3x3==

	* 3 possible rotations of 8 corners, 7 corners determine the 8th = 3^7		* 3 possible rotations of 8 corners, 7 corners determine the 8th = 3^7
Line 11:		Line 11:

	<math>		<math>
	N = ~~\dfrac{~~ 3^{7} \times 2^{11} \times 8! \times 12! }{ 2 } = 43,252,003,274,489,856,000		N = 3^{7} \times 2^{11} \times \dfrac{ 8! \times 12! }{ 2 } = 43,252,003,274,489,856,000 \sim 10^19
	</math>		</math>