Message #1164

From: schuma <>
Subject: Duoprism {3}x{3} 3 solved
Date: Tue, 21 Sep 2010 20:33:26 -0000

Yesterday I solved the uniform duoprism {3}x{3} with 3 layers. The solution log file can be found here.

According to the wiki page, no one had solved it before. So I decide to share some experience about it. I solved {3}x{3} before I read Roice’s post about {5}x{5} ( Therefore I didn’t expect parity issues when I started to solve it. The puzzle {3}x{3}-3 is a relatively small puzzle with not many pieces, but it’s full of traps as you start to solve it. So it has been much fun to solve. I really recommend everyone to try it.

Yesterday afternoon, I examined the puzzle and recognized all the pieces. It is like two tori interlocking with each other. It has a nice property that a sticker on one torus always stays on that torus. This constraint separates most of the pieces into two orbits corresponding to the two tori. Then, I constructed some algorithms that did 3-cycles, and some algorithms that simultaneously flipped the orientations of two pieces. I made a plan of solving a certain type of pieces first and another type second, and so on. I thought that I was ready, so I got started.

There are two types of 2C pieces. I call them 2C-side and 2C-top. The first type to solve was the 2C-side pieces. I quickly met the first parity issue. I needed to switch two 2C-side pieces. And there was a 180 deg turn to do it. It was easy to deal with this issue.

After solving some 1C pieces and 3C pieces, I met the second parity issue. I needed to flip a single 3C on one torus, and exchange two 3C pieces on the other torus. Later I realized that, exchanging two 3C pieces on one torus was equivalent to flip a 3C on the same torus, but not to flip a 3C on the OTHER torus. What I needed to do was essentially to flip a single 3C on each torus. This was not anything I prepared. The algorithm I prepared can only flip two 3C pieces in the same torus. After spending some time, I realized that I could do two first layer turns to solve this issue. However, this method affected many solved pieces so that I had to re-solve pretty much everything from the first phase. Although this fact was hard to face, I had to do it.

Since the second parity issue was fixed, I didn’t meet it when I solved the 3C pieces for the second time. I moved on to solve 2C-top pieces. At the end, I met the third parity issue. I needed to flip a single 2C-top pieces on each torus. Again, this was not what I prepared. Again I came up with the solution that affected many solved pieces. I did that and had to start from the first phase for the third time. Believe it or not, I spent nearly 2000 moves and I was still in the very first phase!

This time when I finished solving two types of pieces, I realized I just made a mistake. This was purely my mistake recognizing some colors but not any parity issue. Instead of re-solving again, I decided to give up and take a fresh start.

I took another full scramble. Since I thought that I had seen all parity issues in this puzzle, I made a thorough plan to count the parity of the permutations of all the troubling pieces and the parity of their orientations. I wanted to make sure that all the parities were even in the beginning. I took some time counting them. Then I used only two turns to fix all the parity issues! The remaining part was very smooth. I solved every type of pieces without seeing any parity issue.

Of course, another way to fix the parity issues is to construct algorithms that directly flip only one piece at a time without affecting other solved pieces. But I don’t know such a method. My method of counting parity in the beginning works well enough for {3}x{3}-3. But for {5}x{5}-3, maybe it will be impractical because there are too many pieces to count. That puzzle will be the next for me to try.