Auzzle Solution

In this post I continue the analysis of Auzzle, the magnetic puzzle invented by Ilya Osipov and attempt to come up with a simple to remember solution involving only one algorithm. See my first post for the background and initial analysis.

After practically living for three days at my friend’s house and learning more than I ever wanted to know about 3d printing, I’ve finally got myself an Auzzle to play with. The picture below shows the assembly process:

The goal

There are four pairs, or 8 pieces total in the puzzle. The goal is to move them around until we get to one of the combinations (there are multiple) where each pair is aligned vertically. In the picture below, piece #1 should be matched with #3 and #0 with #7, so swapping two pieces either on the top layer or the bottom one would solve both pairs. Obviously on a real Auzzle the pieces are not marked with colors or stickers, so our solution should not assume that, instead we need to deduce it just by looking at the magnets formations.


Lets try to adopt a common notation usually used in Rubik’s cube. The difference is our R and L in Auzzle always rotate by full 180 degrees, so we won’t need R’ and L’, as they are identical to R and L.

Magic algorithm

As I’ve shown in my previous post, it is always possible to get to one of the solved state within 10 moves (in fact most cases can be solved within 5-7 moves), however our goal here is different, what we would like is to come up with one basic, intuitive and easy to remember algorithm, which, when applied properly, would help us solve the puzzle. There are multiple ways to go about it, but here one of them:

What does it do? It just swaps top left and top right pieces on the front. There is a side effect, top and bottom pieces on the right back are being swapped too. Left back is being left completely untouched. Which means, if we have some pieces already paired, keeping them in the back and performing the algorithm will not mess up their pairing, and we pretty much can ignore the side effect.

Phase 1: Getting to one pair

We don’t really need any algorithm for this. We will just rotate the top (T) and will likely get at least one pair matched, if we don’t, after making a full circle, we rotate right side (R) and then rotate (T) again, we WILL eventually get some pairs matched.

If we are really lucky and we got all 4 matched, the puzzle is solved. If we got two pairs matched, we skip to phase 3. Otherwise we need to get us a second pair first.

Phase 2: Getting the second pair

Remember how we said our magic algorithm doesn’t really mess up anything in the back? So we can put our matched pair in the back left or right quadrant and try performing the algorithm to swap pieces at the front, until we get our second pair matched. The problem is, it won’t always work :) In order to explain why, I will use 4 different colors, as before, to represent 4 different matching pairs. Auzzle with one matched pair generally falls within one of the two cases above:

Note: If you look closely, A1 and A2 is actually the same case, just rotated 90 degrees. Similarly B2 is just a rotated B1.

Case A: This is the easy one, for each unsolved pair, one piece is located on the top and one on the bottom. It doesn’t matter where we put our first matched (red) pair, as long as it is on the back it will remain untouched. But the moment we perform our magic algorithm, second pair (either green or blue) is going to get matched.

Case B: Here matching pieces are colocated on the same top or bottom planes. Which means performing the magic algorithm will not result in any new pairs being created.

How can we know whether we are in case A or B? That is actually easy, we can rotate the top layer one click right (T’) or left (T). If another pair (our green or blue ones) is getting matched, we are in case A, otherwise we are in case B.

Phase 3: Last two pairs

Ok, we got 2 pairs matched, just bring them both to the back, and lets work with the unmatched pieces at the front. There are two ways how 2 pairs can be potentially unmatched (see diagram below). On a real Auzzle we obviously won’t see green and red segment colors like in the diagram below, but we can see the magnets interaction and infer the case we are dealing with:

Case A: The easiest one is when we notice an asymmetrical configuration of the magents. The pieces are crossed! We can just perform our magic algorithm to swap to top left and right pieces and the puzzle is solved!

Case B: This one is a bit trickier. The left and the right side are clearly symmetrical. That tells us the pieces are not crossed. Swapping two top ones will not help us much, we will end up exactly where we started. We need to break the “parity” of the puzzle and get to the Case A we love so much.

There are also two magnet configuration which don’t tell us whether we are in A or B. How can we know? Lets try to rotate top layer (T’). If the this is case A (crossed), we will immediately see the pair matched, we are in Case A, rotate the top layer back (T) and perform the magic algorithm. Otherwise we are in Case B.

In fact, if you would rather not think at all about Case A or Case B and dont' care about few extra twists, you can always try to perform the algorithm for A, and if the puzzle is not solved continue to B (i.e

I am sure there are more effective ways to solve the puzzle with less (5-7) moves, even just memorizing and using all the mirror variations of our magic algorithm above would make a huge difference and save tons of unnecessary moves. Phase 2 should be improved too. It’s a start, stay tuned for updates.

Auzzle Puzzle

Below are my notes and a quick brute force scanning of the rotating puzzle called Auzzle by Ilya Osipov.

Disclaimer: I do not own one. All the work below is my attempt to analyze it after having the pleasure to see and play with a prototype for a few minutes.


At first glance the puzzle looks quite overwhelming. It consists of 24 magnets, arranged in pairs, and depending on the polarity of the magnets in the pair, they are either pulled towards the middle, or pushed towards the outer edges. The wooden or plastic pieces holding these magnets can be rotated in multiple directions, constantly changing the matching between the magnets. The puzzle is solved only when all 12 pairs of magnets are pulled towards the middle.

After further inspection, however, it becomes clear that the things are not as bad as they seem. There are actually only 8 unique solid wooden or plastic pieces, each one containing 3 magnets. In other words, only 4 pairs of 2 pieces each. The placement of the magnets within each piece is unique, we can actually represent it as a binary encoding, 1 if the magnets ‘+’ is pointing towards the center, and 0 if ‘+’ points towards the outer edge. Unsurprisingly, 3 bits is the exact number required to encode 8 unique pieces: #0=000, #1=001, #2=010, #3=011, #4=100, #5=101, #6=110, #7=111.

Therefore, each piece has only one and only one matching piece resulting in all 3 pairs of magnets pulling towards the middle.

If we keep thinking in binary, in order for a pair of pieces to be each others matching counterparts, the following condition has to be true:

A xor reversed(B) == 7

It is easy to overlook the "reversed" part, but it is important! We encode every piece in the same way, 1 always means '+' pole is oriented towards the center. Pieces will change their orientation, sometimes they will face up and sometimes down, but when faced each other, we can't operate directly on their encodings, we must remember the top one is upside down (i.e piece 001 (1) will atract magnets in the 110 (6) formation, however the actual encoding of that opposing top piece is 011 (3))

In other words, in order to solve the puzzle, piece #0 must be placed against piece #7, #4 against #6, #1 against #3 and #2 against #5.

Total combinations

Total amount of permutations for 8 unique pieces is 8! = 40,320

We also need to multiply it by 2 to account for two possible states of the outer shell in relation to the internal mechanism.

However if we take any state of the puzzle, whether we rotate it by 180 degrees along X or rotate it in any of the 4 orentations along axis Y, it is still the same state, we don’t want to count each state 8 times, so we divide by 8 respectively.

Total number of combinations: 8! x 2 / 8 = 10,080

What about possible solutions?

The picture above shows only one possible solution of the puzzle. However it immediately becomes clear that the puzzle has many possible solutions: a) within each pair, the two pieces can swap places; b) the position of the pairs relative to one another can change and the puzzle would still remain solved.

Permitations to place 2 matching pieces within a pair: 2

We have 4 of these pairs, so total number of permutations is 2^4

Number of ways to place pairs in relation to one another: 4!

Similarly to how we did it before, we multiple by 2 to account for 2 possible states of the outer edge.

And, again, divide by 8 to account for rotation of the whole construction and avoid counting each state as 8 different ones.

Total number of solutions: 4! x 2^4 x 2 / 8 = 96

Observation #1: It is possible to screw up the assembly of the puzzle and insert magnets in a way that will result in pieces with identical encodings existing within the same puzzle. The number of possible solved combinations will increase, the total number of combinations will decrease. Ouch!

Observation #2: In a proper puzzle that actually has 8 uniquely encoded pieces, it is IMPOSSIBLE to get to a state where all the 12 pairs are pushed towards the outer edges. Piece #0 (000) will only push all 3 magnets when paired with another #0, and since there should be only one #0 piece in the puzzle that should never happen. This is true for any encoding that is a ‘palyndrome’ in a 3bit range (i.e #0, #2, #5, #7) (see 4bit variation for a way to possibly do it with 4 bits)


In order to validate my theory and also calculate the God’s Number (maximum number of moves God has to perform to get to a solved state), I’ve written a simple simulator that tries to scan all possible states by doing it in reverse, starting with a solved state and traversing the tree of all the possible states. The code can be found in It starts at the at the initial solved state shown in the picture above, then performs a BFS’ish scan of the whole space by performing all possible operations. For every state, to avoid double counting rotations (remember the division by 8), it rotates the whole puzzle to a canonical state where piece #0 is inside the lower left quadrant.

Total combinations: 10080, Solutions: 96, Max path to solution: 10
Number of paths of length 0: 96
Number of paths of length 1: 192
Number of paths of length 2: 288
Number of paths of length 3: 480
Number of paths of length 4: 768
Number of paths of length 5: 1824
Number of paths of length 6: 2688
Number of paths of length 7: 3264
Number of paths of length 8: 192
Number of paths of length 9: 192
Number of paths of length 10: 96

I am sure it can be done much more efficiently, but that was a quick and dirty exercise and it proves my theoretical calulations (the numbers match!). It also reveals the God’s Number for Auzzle, that number is 10! 10 moves is all it takes to get from the worst position to one of the 96 solved states.

Actually solving it

Given the total number of combinations and the total number of solutions, it is actually pretty easy to get to one of the solved states just by fidgeting with the puzzle enough time.

If we mark the matching pieces with 4 different colors like in the picture above, the puzzle is trivial to solve systematically. In fact, we can think of it as a subset of a 2x2x2 cube where one axis rotates only by 180 and another axis doesn’t rotate at all.

However, individual pieces are not marked and the only thing that can be observed externally by looking at the magnets is the result of the interaction between two adjustent pieces (representing A xor reversed(B)). This makes it much more challenging.

Update: I’ve 3d printed an Auzzle and played with it a bit, please see my next post on actually solving it.



In this variant, we have more possible encodings (16) than needed to represent 8 unique pieces, however that allows us to skip encodings that require duplicated pieces:

a) Avoid encodings #0, #6, #9 and #15 (#0 will only fully push away another #0, etc)

b) Avoid encodings #5, #3, #10, #12 (#5 will only fully pull another #5, etc)

That leaves with 8 encodings that do not have any of the issues above, each one of them has a proper counterpart for pulling all 4 of its magnets and and another one of pushing all 4. These encodings are: #1, #2, #4, #7, #8, #11, #13 and #14. And the actual matching pairs for both solutions are as follows:

All magnets pulled inside: (#1, #7), (#8, #14), (#2, #11), (#4, #13)

All magnets pushed outside: (#1, #8), (#2, #4), (#7, #14), (#11, #13)

It would be logical to assume the total amount of combination will not change (the amount of pieces did not change, it is still 8). The amount of solution states should probably double to 192 (96 inside and 96 outside), and the God’s number should probably decrease by 1 as well (worst case should probably be only 9 moves away from one of the solved states). To test it, I’ve duplicated the simulation and modified it to work with the 4bit encodings above, the code can be found


Total combinations: 10080, Solutions: 192, Max path to solution: 9
Number of paths of length 0: 192
Number of paths of length 1: 368
Number of paths of length 2: 512
Number of paths of length 3: 800
Number of paths of length 4: 1184
Number of paths of length 5: 2496
Number of paths of length 6: 2816
Number of paths of length 7: 1616
Number of paths of length 8: 48
Number of paths of length 9: 48

Yeap, we guessed right!