Rubik's Magic Cube

12. Rubik’s Magic Cube 

Robert Snapp 

snapp@cs.uvm.edu 

Department of Computer Science 

University of Vermont 

Robert R. Snapp © 2012 12. Rubik’s Magic Cube CS 32, Fall 2012 1 / 45

Rubik’s Magic Cube 

Ernö Rubik invented this celebrated puzzle in 1974. 

When completed, each of the six faces displays a 

common color, usually white, yellow, red, orange, blue 

and green. 

Questions: 

1 How many different ways can six 

colors be assigned to the six faces? 

2 How are the colors of each pair of 

opposite faces related at right? 

The cube actually consists of 26 visible cubies, consisting of 

6 single faced, centers, which are stationary. 

12 double faced, edges. 

8 triple faced, corners. 

Rubik’s standard color 

arrangement. 


David Singmaster’s Notation 

David Singmaster 1 published one of the first analyses of the Magic Cube. He 

introduced the following notation: 

U , for the Upper face, 

F , for the Front face, 

D, for the Down face, 

B, for the Back face, 

L, for the Left face, and 

R, for the Right face. 

Note that the Magic Cube can be oriented 24 ways within this coordinate system: 

the upper face can be chosen 6 different ways. 

for each upper face, the front face can be chosen 4 different ways. 

6 4 D 24. 

1. David Singmaster, Notes on Rubik’s Magic Cube, Enslow, Hillside, NJ, 1981. 

L 

F 

Robert R. Snapp © 2012 12. Rubik’s Magic Cube CS 32, Fall 2012 3 / 45 

D 

U 

B 

R

Singmaster’s Operations: U 

Once the cube has been positioned, we define a 

set of rotation operations that maintain the 

orientation of the center cubies. 

For example, U denotes a quarter turn of the 

Upper face in the clockwise direction. 

U 2 denotes a half turn of the Upper face. (N.B., 

U 2 D U U .) 

U 0 denotes a quarter turn of the Upper face in 

the counter-clockwise direction. (N.B., 

U 0 D U 3 .) 


U 

U 2 

U 0

Singmaster’s Operations: F 

F denotes a quarter turn of the Front face in the 

clockwise direction. 

F 2 denotes a half turn of the Front face. (N.B., 

F 2 D FF .) 

F 0 denotes a quarter turn of the Front face in the 

counter-clockwise direction. (N.B., F 0 D F 3 .) 


F 

F 2 

F 0

Singmaster’s Operations: D 

D denotes a quarter turn of the Down face in the 


D 2 denotes a half turn of the Down face. (N.B., 

D 2 D DD.) 

D 0 denotes a quarter turn of the Down face in 

the counter-clockwise direction. (N.B., 

D 0 D D 3 .) 


D 

D 2 

D 0

Singmaster’s Operations: B 

B denotes a quarter turn of the Back face in the 


B 2 denotes a half turn of the Back face. (N.B., 

B 2 D BB.) 

B 0 denotes a quarter turn of the Back face in the 

counter-clockwise direction. (N.B., B 0 D B 3 .) 


B 

B 2 

B 0

Singmaster’s Operations: L 

L denotes a quarter turn of the Left face in the 


L 2 denotes a half turn of the Left face. (N.B., 

L 2 D LL.) 

L 0 denotes a quarter turn of the Left face in the 

counter-clockwise direction. (N.B., L 0 D L 3 .) 


L 

L 2 

L 0

Singmaster’s Operations: R 

R denotes a quarter turn of the Right face in the 


R 2 denotes a half turn of the Right face. (N.B., 

R 2 D RR.) 

R 0 denotes a quarter turn of the Right face in the 

counter-clockwise direction. (N.B., R 0 D R 3 .) 


R 

R 2 

R 0

Restore the Cube: Outline 

Part I: Restore the upper face. 

1. Restore the upper edges. 

2. Restore the upper corners. 

Part II: Restore the middle layer. 

3. Turn the entire cube upside 

down. 

4. Restore the middle edges. 

Part III: Restore the final face. 

5. Invert the upper edges. 

6. Reposition the upper edges. 

7. Reposition the upper corners. 

8. Twist the upper corners. 


Part I: Step 1 — Restore the Upper Cross 

1a Select a color for the upper face (e.g, green), and 

an adjacent color for the front face (e.g., white). 

1b Identify the cubie that belongs in the upper-front 

(uf ) edge, e.g., the green-white edge. It should 

be easy to bring this cubie to the correct 

location. 

1c If this colors of the uf edge need to be flipped, 

then apply the sequence 

F 0 UL 0 U 0 : 

1d Rotate the entire cube one-quarter turn, and 

repeat the above until all four upper edges are in 

place. You should see a green cross. 


Part I: Step 2 — Restore the Upper Corners 

2a For each corner cubie in the Down layer that 

belongs in the Upper layer: 

i Rotate the Down layer (using the D operation) until 

this cubie is directly below its desired postion. 

Rotate the entire cube so that the desired position 

is under your right thumb (upper-right-front 

position). 

ii Apply the operation R 0 D 0 RD one, three, or five 

times, until this corner cubie is in the correct 

position, with the correct orientation. (This will not 

destroy the cross, obtained in Step 1.) 


urf 

drf

Part I: Step 2 — Restore the Upper Corners (cont.) 

2b For each Upper layer corner cubie that is 

incorrectly placed, or incorrectly rotated, 

i Rotate the entire cube until the misplaced cubie is 

under your right thumb. 

ii Place the cubie in the Down layer using R 0 D 0 RD: 

iii Then apply step 2a (above) to move this cubie in 

the correct position. 

2c Apply the above steps until the entire upper layer 

is complete. 


Part I: Step 2 — Restore the Upper Corners (cont.) 


Part II: Step 3 — Turn the Cube Upside Down 

Turn the entire cube upside down, so that the completed 

green layer is the bottom (or down) layer. The 

new upper layer should have a blue center. 


Part II: Step 4 — Restore the Middle Layer 

The key operation is RU 0 R 0 FR 0 F 0 RU 0 which swaps and inverts ul and fr. 

4a Rotate the entire cube until a front-right (fr) edge 

is incorrect, or flipped. (Assume the right edge of 

the white face is incorrect.) 

4b Locate the correct edge (e.g., the red-white 

edge). 

Case A: If the correct edge is in the middle 

layer: 

i Rotate the entire cube so that 

the correct edge is a front-right 

(fr) edge. (Note, the red-white 

edge is in the middle layer.) 

ii Perform the sequence 

RU 0 R 0 FR 0 F 0 RU 0 which will 

place the correct edge in the 

upper layer (at ul ). 

iii Apply Case B. 


ul 

fr

Part II: Step 4 — Restore the Middle Layer (cont.) 

Case B: If the correct edge is on the top layer: 

i Ensure that the misplaced edge 

is still the front-right (fr ) edge. 

ii Rotate the upper layer (using U 

operations) so that the correct 

edge is an upper-left (ul ) edge. 

iii Apply the operation 

RU 0 R 0 FR 0 F 0 RU 0 . 

iv If the correct edge needs to be 

flipped, apply Case C. 

Case C: If a middle edge is flipped in the 

correct location: 

i Apply the operation 

RU 0 R 0 FR 0 F 0 RU 0 twice. 


Part II: Step 4 — Restore the Middle Layer (cont.) 

The top row illustrates two successive occurrences of Case B. The left two diagrams show how 

the red-yellow edge is moved into its correct position with RU 0 R 0 FR 0 F 0 RU 0 . The right two, 

show how the orange-yellow edge is moved into its correct position by the same operation. 

The bottom row illustrates an occurrence of Case B, that leads to a Case C. First the 

orange-white edge is moved into its correct position, but with an incorrect orientation. Applying 

RU 0 R 0 FR 0 F 0 RU 0 moves it back into the top layer, but flipped. A third application, brings 

the orange-white edge into the correct position and orientation. 


Part III: Restoring the Upper Layer 

Now that the bottom and middle layers are complete, every cubie in the upper layer 

has a single blue face. In order to restore the upper face, one needs to 

5. Flip the edge cubies so that the blue face of each 

faces upwards. 

6. Move the edge cubies to their final locations, 

without destroying their orientation. 

7. Move the corner cubies to their final locations. 

8. Rotate the corner cubies (in place) so that the 

blue face of each faces upwards. 


Part III: Step 5 — Flip the New Upper Edges 

5. Orient the cube so that it matches one of the four orientations: 

“Blue Dot” “Blue Corner” “Blue Line” “Blue Cross” 

a. If the ”Blue Cross” is displayed, move on to Step 6. 

b. If the ”Blue Cross” is not displayed, apply the maneuver 

FRUR 0 U 0 F 0 

and repeat Step 5 as many times as required. 


Part III: Step 6 — Restore the New Upper Edges 

At this point of the solution, the bottom two layers should be solved, and a blue cross, 

should appear on the top face. If you are very lucky, the red, white, yellow and orange 

sides of the blue cross match all four of the corresponding center cubies. (Twist the 

upper layer using a succession of U operations, to see if this occurs. If so procede to 

Step 7.) If you are not so lucky, twist the upper layer until exactly one of the sides of 

the blue cross matches its center cubie. Rotate the cube so that the matching side 

cubie is in the front face. In the figures below the matching cubie happens to be red. 

RWYO ROWY RYOW 

Apply the sequence RUR 0 URU 2 R 0 until the sides of the four top edge cubies match. 


Part III: Step 7 — Place the Upper Corners 

We shall now ensure that each upper corner is in the 

correct position. (Don’t worry now about their orientations; 

those will be restored in Step 8.) 

Compare the colors of each upper corner with those of 

the adjacent centers. If all three match, even if the orientation 

is wrong, then this piece is in the correct position. 

In the diagram at right, the upper-left-front (ulf) 

corner (red-white-blue) is in the correct position. The 

upper-right-front (urf ) corner (yellow-orange-blue) is 

not. 

ulb urb 

ulf 

The key sequence of Step 7 is L 0 URU 0 LUR 0 U 0 , which rotates (or cycles) the upper 

three corners (ulf, ulb, urb ), in a clockwise direction, while maintaining the positions 

and orientation of the remaining 23 cubies. 


Step 7 — Place the Upper Corners (cont.) 

7a. If no upper corners are in their correct positions, apply L 0 URU 0 LUR 0 U 0 (once 

or twice) until one is. Then continue. 

7b. If one corner is in its correct position, then rotate the entire cube so that the 

correctly placed corner is near your right thumb, in the upper-right-front (urf ) 

position. Then apply L 0 URU 0 LUR 0 U 0 (once or twice) until all four upper corners 

are correctly placed. 


Part III: Step 8 — Twist the Upper Corners 

At this point every cube is in the correct position. However, two or more corners may 

have an incorrect orientation. 

The key sequence of Step 8 is R 0 D 0 RD, which you already practiced in Step 2. 

8a. Rotate the entire cube until an incorrectly 

oriented (twisted) corner is located near your 

right thumb. (It should be in the urf position.) 

8b. Apply the sequence R 0 D 0 RD (two or four times) 

until this corner cube has the correct orientation. 

Don’t worry about the middle and bottom layers: 

they are temporarily messed up. 


urf 

urf

Part III: Step 8 — Twist the Upper Corners (cont.) 

8c. Now rotate only the upper layer, by applying one 

or more U operations, until the next twisted cube 

is near your right thumb in the urf position. 

8d. Repeat steps 2 and 3 until every corner is 

correctly oriented. 

8e. Finally, restore the cube using one or more U 

operations. 

8f. Fix yourself an ice-cream cone. 


urf 

urf 

urf 

urf

Summary 

Step Operations Goal 

upper 

(green) 

cross 

upper 

(green) 

corners 

flip 

entire 

cube 

middle 

edges 

Use the six basic operations to move the desired edge immediately below 

its home, without moving the other upper edges. Then rotate that 

face one-half turn. 

To flip an inverted edge, apply F 0 UL 0 U 0 . 

Use R 0 D 0 RD to swap (and twist) the urf and drf corners. After each 

misplaced corner has been moved to the down (blue) layer, use the D 

operator to move it immediately below its home. Then apply R 0 D 0 RD 

a sufficient number of times, so that it is correctly placed and correctly 

oriented. 

Easy as pie! Turn the entire cube upside down so that the blue center 

on top and the completed green face is the new down layer. 

Use RU 0 R 0 FR 0 F 0 RU 0 to swap and flip the ul and fr edges, without 

displacing the other cubies on the lower two layers. 


Summary (cont.) 

Step Operations Goal 

orient 

upper 

edges 

restore 

upper 

edges 

place 

upper 

corners 

twist 

upper 

corners 

If the blue facets on the upper face form a corner, rotate the cube so 

that the corner is at ul, u, and ub. If the upper facets of the upper edges 

form a blue line, rotate the cube so that the blue line runs from left to 

right (ul, u, ur). Apply FRUR 0 U 0 F 0 until a blue cross is displayed. 

Apply U until the the uf edge matches the color of the front face. Then 

apply RUR 0 URU 2 R 0 until every upper edge matches the side faces. 

If an upper corner is correctly placed, rotate the entire cube so that 

this becomes the urf corner. Then apply L 0 URU 0 LUR 0 U 0 until each 

corner is correctly placed. 

Apply U until urf is twisted. Then apply R 0 D 0 RD until this urf is correct. 

Repeat until every corner is untwisted. Apply U to restore the 

cube. 


urf

How Many States are in the Cube? 

Claim: A 3 3 3 Rubik’s cube can be placed in exactly 

N D 43; 252; 003; 274; 489; 856; 000 

different configurations, using a sequence of legal moves based on L, R, U , D, B 

and F , more than the number of seconds in 10 billion centuries. 

Counting this number is sort of like counting the number of anagrams that can be 

formed from a given set of letters. We thus count permutations. 

Recall that there are three kinds of cubies: 8 corners, 12 edges, and 6 centers. First 

note that it is impossible to exchange a three-sided corner with a two-sided edge, 

and likewise we can’t exchange a center with either a corner or edge. 


How Many States are in the Cube? 

We will use the multiplication principle to count the number N of configurations that 

can be obtained by a sequence of the operations, L, R, U , D, B and F . 

Let, 

Then, our first estimate of N is 

What is the value of N1? 

N1 D number of configurations of the 6 centers 

N2 D number of configurations of the 12 edges 

N3 D number of configurations of the 8 corners 

N D N1 N2 N3: 


Estimating N1 

Since the locations of the centers are unchanged by each of the six basic operations, 

they are also unchanged by any sequence of these operations. Thus, 

Thus, 

What is the value of N2? 

N1 D 1: 

N D 1 N2 N3: 


Estimating N2 

Since there are 12 locations (cubicles) for each edge, there are 12Š ways to order the 

edges. In addition, each edge can be flipped in two different ways: e.g., the red-blue 

edge can be red-side up, or blue-side up. This suggests that there are at most 

ways to arrange the 12 edges. 

What can we say about N3? 

N2 D 12Š 2 12 D 1; 961; 990; 553; 600 


Estimating N3 

Since there are 8 corner cubicles (locations for the corners), there are 8Š ways to 

order the corners. In addition each corner can be twisted three different ways. This 

suggests that, at most, 

N3 D 8Š 3 8 D 264; 539; 520 

ways to arrange the eight corners. 

Does 

N D 1 .12Š 2 12 / .8Š 3 8 /‹ 


Counting the Configurations of Rubik’s Cube 

This number, 

1 .12Š 2 12 / .8Š 3 8 / D 519; 024; 039; 293; 878; 272; 000 

actually represents (exactly) the number of different ways that Rubik’s cube can be 

reassembled, assuming that the centers are not rearranged. 

Anne Scott (cf., Berlekamp, Conway, Guy, 2004), showed that this value 

overestimates the correct value of N by a factor of 12. 


Invariants 

Consider a “puzzle” that concerns the value of a variable x. Initially, x D 0. Every 

second a coin is tossed. If the coin lands heads then we add 4 to x. If the coin lands 

tails, we subract 2. Here is a sample sequence. 

time (s.) 0 1 2 3 4 5 6 7 8 9 10 

coin toss H T H H T T T T T H 

x 0 4 2 6 10 8 6 4 2 0 4 

Question: Can x ever equal 1? 


Invariants 

Correct! The answer is no. Since x begins as an even number, and every possible 

operation (adding 4 or subtracting 2) preserves evenness, x will always be even. 

In this context, evenness is said to be an invariant property, or an invariant (for short), 

of x. 


Invariants and Loyd’s 14-15 Puzzle 

Sam Loyd (1841–1911) created many popular puzzles, including the celebrated 

14–15 puzzle, shown above. Can you interchange just tiles labeled 14 and 15, by 

sliding tiles horizontally or vertically into the space? (Loyd offered a $1000 prize to 

anyone who could.) 

How many states are realizable? 


Invariants (cont.) 

For the space to wind up in the lower-right corner, there must have been an even 

number of vertical moves, and an even number of horizontal moves. Consequently, 

only permutations that swap and even number of pieces are possible. For Loyd’s 

puzzle, only half of the 16Š states are realizable. 

Anne Scott used invariants to exactly count the number of possible states for Rubik’s 

cube. 


Reexamining the allowed corner twists 

Place a 0, 1, or a 2 on each corner face, 

as shown at right. The initial sums are then 

computed for each face, and recorded under 

column I of the table. Sums are also computed 

following each legal quarter turn. Note 

that ever entry is a multiple of 3. This latter 

property is preserved for every sequence of 

legal operations. 

However, if one were able to twist a single 

corner, one-third of a turn, in either direction, 

the sums of the adjacent faces change 

to numbers that are not multiples of 3. 

Consequently, only one-third of the total 

number of corner twists 3 8 can be realized 

using a sequence of legal operations. 

2 

1 

1 

2 

0 

1 

2 

2 1 

0 

0 0 

0 

1 

2 

0 0 

0 

Face Sums 

Face I L R U D F B 

left 6 6 6 6 6 3 3 

right 6 6 6 6 6 3 3 

upper 0 3 3 0 0 3 3 

down 0 3 3 0 0 3 3 

front 6 3 3 6 6 6 6 

back 6 3 3 6 6 6 6 


2 

2 

1 

1 

1 

2

Reexamining the allowed edge flips 

Place a 0 or 1 on each edge, and construct 

a stationary blue window for each face, as 

shown. The initial sum of the values that appear 

in the blue windows is computed under 

column I in the table. It can be shown that 

the window sum will always be a multiple of 

2, and even number, after every sequence of 

operations. (After F U , for example, it equals 

6.) 

However, flipping any single edge results in 

an odd window sum. Consequently, it is not 

possible to invert a single edge using a sequence 

or rotations. 

Thus only one-half of the 2 12 edge states are 

realizable. 

0 

1 

1 

0 

1 

1 

0 0 

1 

0 

0 

1 

1 

0 0 

1 

1 

1 

0 0 

1 

Blue-Window Sums 

I L R U D F B 

sum 12 8 8 8 8 8 8 


0 

0 

1

How many states are expressible by the cube? 

The final reduction factor is obtained by observing that only one-half of the 12Š 8Š 

permutations of the locations of the 12 edges and 8 corners are realizable. Each 

sequence of operations always moves a multiple of 4 pieces. It is thus impossible to 

interchange just two corners, or just two edges. 

Thus, 

N D 1 

2 

1 

2 

1 

3 

12Š 2 12 

8Š 3 8 

D 43; 252; 003; 274; 489; 856; 000: 


Some Symmetrical States 

Let Fs D FB 0 denote a move called a front slice. Similarly, 

let Rs D RL 0 denote the right slice, and Us D UD 0 denote the upper slice. 

“Dots” “Chessboard” “Cross” 

RmF 0 m R0 m Fm 

The definitions of Rm, R 0 m , Fm, and F 0 m 

F 2 

s R2 s U 2 s 

appear below. 

R0L2F 2 

s U 2R2 2 

s Fs D2R0 Robert R. Snapp © 2012 12. Rubik’s Magic Cube CS 32, Fall 2012 41 / 45

Singmaster’s Operations: Rm 

Start with yellow on top, blue in front, and red at 

right. Rm denotes a quarter turn of the middle 

layer (only) parallel to the direction of R. The 

easiest way to complete this is to rotate both the 

right face, and the middle layer behind the right 

face, one quarter turn clockwise, followed by R 0 . 

R 2 m 

denotes a half turn of the middle layer 

behind the right face. 

R0 m denotes a quarter turn of the middle layer, 

behind the right face, in the counter-clockwise 

direction, i.e., parallel to R0 . (N.B., R0 m D R3 m .) 


Rm 

R 2 m 

R 0 m

Singmaster’s Operations: Fm 

Fm denotes a quarter turn of the middle layer 

(only) parallel to the direction of F . The easiest 

way to complete this is to rotate both the front 

face, and the middle layer behind the front face, 

one quarter turn clockwise, followed by F 0 . 

F 2 m 


behind the front face. 

F 0 m denotes a quarter turn of the middle layer, 

behind the front face, in the counter-clockwise 

direction, i.e., parallel to F 0 . (N.B., F 0 m D F 3 m .) 


Fm 

F 2 m 

F 0 m

Singmaster’s Operations: Um 

Um denotes a quarter turn of the middle layer 

(only) parallel to the direction of U . The easiest 

way to complete this is to rotate both the upper 

face, and the middle layer behind the upper face, 

one quarter turn clockwise, followed by U 0 . 

U 2 m 


behind the upper face. 

U 0 m denotes a quarter turn of the middle layer, 

behind the upper face, in the counter-clockwise 

direction, i.e., parallel to U 0 . (N.B., U 0 m D U 3 m .) 


Um 

U 2 m 

U 0 m

References 

1 Christoph Bandelow, Inside Rubik’s Cube and Beyond, Birkhäuser, Boston, 1982. 

2 Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy, Winning Ways For Your 

Mathematical Plays, Second Edition, Vol. 4, A. K. Peters, Natick, MA, 2004. 

3 John Ewing and Czes Ko´sniowski, Puzzle It Out: Cube Groups and Puzzles, Cambridge 

University Press, Cambridge 1982. 

4 Alexander H. Frey, Jr. and David Singmaster, Handbook of Cubik Math, Enslow, Hillside, 

NJ, 1982. 

5 Martin Gardner, ed., The Mathematical Puzzles of Sam Loyd, Dover, NY, 1959. 

6 David Joyner, Adventures in Group Theory: Rubik’s Cube, Merlin’s Magic & Other 

Mathematical Toys, Johns Hopkins University Press, Baltimore, 2002. 

7 Ernö Rubik, Tamás Varga, Gerzson Kéri, Györgi Marx, and Tamás Vkerdy, Rubik’s Cubic 

Compendium, Oxford University Press, Oxford, 1987. 

8 David Singmaster, Notes on Rubik’s Magic Cube, Enslow, Hillside, NJ, 1981.

Rubik's Magic Cube

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