Jaap's Puzzle Page

Dual Circle

Dual Circle

Dual Circle consists of two tightly linked rings. Each ring has six sections. If a ring is rotated about its axis, the section on the other ring that is located in the hole of the first ring will be rotated too. The rings are coloured so that the orientation of the 12 sections is visible.

This puzzle is produced by Hanayama, and was invented by Bram Cohen and Oskar van Deventer. Bram came up with the idea of how the pieces move and affect each other, but had them arranged in two parallel rows. He called that puzzle Rotabram. Oskar came up with the design using two linked rings.

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

JavaScript Dual Circle

The number of positions:

There are 12 rotating ring sections, each with 6 possible orientations. This gives a maximum of 612 positions. There is however a twist 'parity' restriction, so only 611 = 362,797,056 positions are possible.

Solution:

Phase 1: Solve the red ring sections

  1. Rotate the red ring to bring any unsolved section into the middle of the blue ring.
  2. Rotate the blue ring until that red ring section is solved.
  3. Repeat steps a-b until the red ring sections are all solved.

Phase 2: Solve the blue ring sections

  1. Rotate the red ring so as to solve the blue section currently inside the red ring.
  2. Choose a blue ring section that you want to solve.
  3. Rotate blue ring to bring the chosen section into the centre of the red ring.
  4. Rotate the red ring one step.
  5. Rotate the blue ring back to its original position.
  6. Rotate the red ring back one step to its original position.
  7. Repeat steps c-f until the chosen blue section is solved. Note that this will have disturbed one red ring section.
  8. Repeat steps b-f for each unsolved blue section (except for the blue section inside the red ring). Note that in steps d and f you must twist in the same direction as the previous times, so that it is always the same red section that gets disturbed. Once all the blue sections are solved, that disturbed red section should be automatically correct.
  9. Rotate the blue ring so as to solve the red section currently inside the blue ring.

Useful sequences

Phase 2 in the above solution can take quite a number of moves. It may therefore be quite useful to have some further move sequences in your arsenal to speed things up. Below is a complete set of move sequences that only affect one or two pieces in the blue ring. Once you have solved the red ring, you will need to apply no more than four of these sequences to solve the blue ring.

Mentally label the 6 sections of the blue ring with the letters A to F, where A is the one that lies inside the red ring. A letter followed by a + or a - indicates that you have to turn that section one way or the other by one click. Two or three plusses or minuses obviously means it has to move two or three clicks.

EffectMove sequenceEffectMove sequenceEffectMove sequenceEffectMove sequence
B++, C++ B-- C++ B-- B--, C-- B++ C-- B++
B++, C- B+ C- B+ B--, C+ B- C+ B- E-, F++ F+ E- F+ E+, F-- F- E+ F-
B+++, D+ B- C- D+ C+ B-- B+++, D- B+ C+ D- C- B++ D+, F+++F- E- D+ E+ F-- D-, F+++F+ E+ D- E- F++
B+++, D+++B+++ C+++ D+++ C+++ D+++, F+++F+++ E+++ D+++ E+++
B++, E+ B++ A- F- E+ F+ A+ B--, E- B-- A+ F+ E- F- A- C+, F++ F++ A- B- C+ B+ A+ C-, F-- F-- A+ B+ C- B- A-
B++, E-- B+ E- B+ E- B--, E++ B- E+ B- E+ C--, F++ F+ C- F+ C- C++, F-- F- C+ F- C+
B+, F+ F+ A- B+ A+ B-, F- F- A+ B- A-
B++, F++ F++ A-- B++ A++ B--, F-- F-- A++ B-- A--
B+++, F+++F+++ A--- B+++ A+++
C+++ B+++ C+++ B+++ E+++ F+++ E+++ F+++
C+++, D++ C- D+ C- D+ C- C+++, D-- C+ D- C+ D- C+ D++, E+++E- D+ E- D+ E- D--, E+++E+ D- E+ D- E+
C+, E+ E+ A- C+ A+ C-, E- E- A+ C- A-
C-, E++ E+ C- E+ C+, E-- E- C+ E- C++, E- C+ E- C+ C--, E+ C- E+ C-
C++, E++ E-- C++ E-- C--, E-- C++ E-- C++
C+++, E+++C+++ D+++ E+++ D+++
D++ D+ A- D+ A+ D-- D- A+ D- A-

The twist constraint

This puzzle has 611 positions rather than the 612 you might expect. I will prove this here by finding an invariant. On most other puzzles invariants are relatively simple. The Rubik's Cube for example has three invariants - the permutation parity, the total edge flip, and the total corner twist (modulo 3). Any legal move on the puzzle leaves these invariants unchanged. For any mixed position these invariants can be calculated, and if they are different than what the solved position would have, then the puzzle is unsolvable. The Dual Circle puzzle also has an invariant - some expression involving the amount of twist of its twelve pieces that remains unchanged (modulo 6) by any move. It is however much more complicated than the total twist.

Lets use the letters a to l to denote the orientations of the twelve pieces. Here a to f are the orientations of the six pieces of the red ring, labelled clockwise looking on the red side, starting from the piece inside the blue ring. Similarly, let g to l be the same for the six blue pieces. These variables have values ranging from 0 to 5, representing the number of steps each piece has been rotated.

Now consider the following five expressions:
Sr = a + b + c + d + e + f
Sb = g + h + i + j + k + l
Vr = 0·a + 1·b + 2·c + 3·d + 4·e + 5·f
Vb = 0·g + 1·h + 2·i + 3·j + 4·k + 5·l
T = Vr + Vb - Sr·Sb
All these are calculated modulo 6, since rotating a piece by 6 steps is equivalent to doing nothing.

If you shift the blue ring, the variables g to l get cyclically permuted, and a gets incremented. The effect of this on the five expressions is this:
Sr gets incremented by one (modulo 6).
Sb remains unchanged.
Vr remains unchanged.
Vb was equal to (0·g + 1·h + 2·i + 3·j + 4·k + 5·l) and becomes (0·l + 1·g + 2·h + 3·i + 4·j + 5·k) so is increased by g + h + i + j + k - 5·l = Sb modulo 6.
Therefore T remains unchanged.

If you shift the red ring instead, T also remains unchanged. This shows that at most 1/6th of the positions are reachable. Since there is a solution method that solves pieces one by one except for the last (which is automatically correct due to that invariant), there are 611 reachable positions.

I don't know if this kind of invariant has ever been described in the mathematical literature on Wreath products of groups. If it has, then please let me know.