Jaap's Puzzle Page

The Orbik

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.

If your browser supports JavaScript, then you can play the Orbik by clicking the link below:

JavaScript Orbik

The number of positions:

The Orbik has 12 wheels, each with 4 possible states. It is possible to change a wheel without affecting any others, so the number of positions is 412 = 16,777,216.

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
01
116
2181
31,349
47,423
536,352
6130,102
7388,672
8885,732
91,579,229
102,364,511
112,759,590
122,839,626
132,339,075
141,674,665
15987,625
16492,348
17201,604
1867,198
1917,821
203,564
21484
2244
234
Total16,777,216
Moves# Positions
01
112
278
3364
41,353
54,224
611,440
727,456
859,268
9116,336
10209,352
11347,568
12534,964
13766,272
141,024,464
151,281,280
161,501,566
171,650,792
181,703,636
191,650,792
201,501,566
211,281,280
221,024,464
23766,272
24534,964
25347,568
26209,352
27116,336
2859,268
2927,456
3011,440
314,224
321,353
33364
3478
3512
361
Total16,777,216
Moves# Positions
01
11
22
34
47
513
622
739
864
9113
10182
11316
12504
13860
141,363
152,277
163,577
175,816
188,995
1914,153
2021,366
2132,399
2247,334
2369,015
2496,893
25135,667
26182,094
27244,653
Moves# Positions
28312,868
29403,080
30490,098
31604,954
32698,521
33825,120
34904,197
351,020,376
361,060,726
371,140,688
381,124,136
391,147,527
401,070,342
411,030,704
42906,384
43814,906
44669,552
45552,026
46415,188
47303,251
48199,541
49119,642
5062,019
5124,809
527,693
531,078
5460
Total16,777,216

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.

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. Repeat steps b-d as often as possible. Eventually, when only about 4 non-red wheels remain, step b is no longer possible.
  6. 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.)
  7. 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.
  8. 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
  9. Repeat f-h for each incorrect wheel.