The Orbik

The Orbik is a fairly rare puzzle. It is a black disk, about 7 cm diameter, which has 12 small windows arranged around the front face of the disk. Behind each window is a small drum wheel that shows one of four coloured shapes - a red square, a green oval, a blue triangle, or a yellow wave. When you turn the front face of the disk clockwise, three of the drum wheels will advance once, showing the next coloured shape. Three of the windows are marked with dots, to indicate that they are the ones that will change. If you turn the front face anti-clockwise no changes occur, except of course that the marked windows will have moved to different positions. The aim is of course to get all the windows to show the same coloured shape.

There is a sticker in the centre of the front face, bearing a triangular logo. If you hold it so that the triangle points upwards, and think of the front as a clock face, then the three marked windows lie at 12 o'clock, 3 o'clock, and 7 o'clock.

The Orbik was invented by Shih-Hung Juang, and has patent US 4,752,074 filed 25 September 1986, granted 21 June 1988 (also European patent EP 262,251).

The Orbik I have is one (of about 200?) that the late Edward Hordern brought to the 1992 International Puzzle Party as exchange gifts. Later the puzzle was distributed for a while by James Dalgety.

I have used a computer to calculate God's Algorithm. The results are in the three tables below. Each table uses a different definition of what constitutes a single move.

Any number of clicks in either direction is one move

Every clockwise click counts as a move

Every click both directions counts as move

Moves	# Positions
0	1
1	16
2	181
3	1,349
4	7,423
5	36,352
6	130,102
7	388,672
8	885,732
9	1,579,229
10	2,364,511
11	2,759,590
12	2,839,626
13	2,339,075
14	1,674,665
15	987,625
16	492,348
17	201,604
18	67,198
19	17,821
20	3,564
21	484
22	44
23	4
Total	16,777,216
Avg Depth	11.605

Moves	# Positions
0	1
1	12
2	78
3	364
4	1,353
5	4,224
6	11,440
7	27,456
8	59,268
9	116,336
10	209,352
11	347,568
12	534,964
13	766,272
14	1,024,464
15	1,281,280
16	1,501,566
17	1,650,792
18	1,703,636
19	1,650,792
20	1,501,566
21	1,281,280
22	1,024,464
23	766,272
24	534,964
25	347,568
26	209,352
27	116,336
28	59,268
29	27,456
30	11,440
31	4,224
32	1,353
33	364
34	78
35	12
36	1
Total	16,777,216
Avg Depth	18

Moves	# Positions
0	1
1	1
2	2
3	4
4	7
5	13
6	22
7	39
8	64
9	113
10	182
11	316
12	504
13	860
14	1,363
15	2,277
16	3,577
17	5,816
18	8,995
19	14,153
20	21,366
21	32,399
22	47,334
23	69,015
24	96,893
25	135,667
26	182,094
27	244,653

Moves	# Positions
28	312,868
29	403,080
30	490,098
31	604,954
32	698,521
33	825,120
34	904,197
35	1,020,376
36	1,060,726
37	1,140,688
38	1,124,136
39	1,147,527
40	1,070,342
41	1,030,704
42	906,384
43	814,906
44	669,552
45	552,026
46	415,188
47	303,251
48	199,541
49	119,642
50	62,019
51	24,809
52	7,693
53	1,078
54	60
Total	16,777,216
Avg Depth	37.275

Solution:

The mathematical theory that lies behind this puzzle is linear algebra. If you think of a move as a clockwise turn of one step, then there are twelve possible moves you can do (as there are 12 ways the marked windows can be arranged). The order in which these moves are done does not matter, and so linear algebra can be applied to solve it just like the Lights Out or Rubik's Clock. See the Mathematics of Lights Out for details.
The solution below is loosely based on Hordern's solution, published in CFF 29.

Notation:

In some of the steps below I shall use the letters A and C to denote anti-clockwise and clockwise twists respectively, always followed by a number to indicate how many steps to move in that direction.

Choose one of the four colours. You will try to make all the wheels show this colour. In the rest of the solution I will assume you chose red.
First we will use a simple method to make all but a few wheels red. Rotate the windows anti-clockwise, until the marked windows show one of the following
1. Three wheels, of which two have the same non-red colour (the third may be red).
2. Three wheels, all different non-red colours.
Rotate the windows clockwise one step, and then anti-clockwise one step (i.e. do the moves C1 A1). This should have changed all three marked windows.
In case 1, repeat step c until the pair of equal wheels both show red. This will then have increased the number of windows showing red. In case 2, a single application of step c will have turned a wheel red.
Repeat steps b-d as often as possible. Eventually, when only about 4 non-red wheels remain, step b is no longer possible.
Choose one of the remaining incorrect wheels. Figure out if it needs 1, 2 or 3 turns for it to be the correct colour. (The colour cycle is red-green-blue-yellow-red, so assuming we want everything red, then yellow needs one twist, blue two, green three.)
Hold the puzzle with incorrect wheel at the 4 o'clock position, and turn anti-clockwise until the front logo is upright, so that the marked windows are at 12, 3, and 7 o'clock.
To move the incorrect wheel once, twice, or three times, do one of the following:
1. A1 C4 A4 C4 A3 C5 A1 C4 A1 C1 A1 C1
2. C8 A8 C8 A4 C1 A1 C1
3. A1 C1 A1 C5 A1 C4 A4 C4 A2 C3
Repeat f-h for each incorrect wheel.

Jaap's Puzzle Page

The Orbik

JavaScript Orbik

The number of positions:

Solution:

Notation: