Solution #033 - syncro


This post contains a proposed solution to Problem #033 - syncro. Please do not read this solution before making a serious attempt at the problem.

I am excited to tell you that I just released the alpha version of my “Pydont's” book, a book that compiles all my “Pydon't” articles. You can get the book at leanpub:


Congratulations to the ones that solved this problem correctly and, in particular, to the ones who sent me their correct solutions:

  • Filippo M., Italy;
  • Attila K., Hungary;
  • André S., Portugal.

(The list is in no particular order.)


This problem is really hard to solve without a visual representation of what is going on, so let me remind you of what the problem looks like:

Initial configuration.

We want to figure out what is the shortest sequence of circles and squares sends the four white triangles in the four petals into a single flower.

We see that the square rotates the whole thing while the circle swaps the top corners and joins the bottom corners on the bottom left.

Making a couple of smart remarks will make it easier to solve the problem:

  • it makes no sense to start out with a square;
  • two circles in a row are superfluous;
  • the solution uses circle at least three times;
  • the solution ends with circle.

This shows that a hard minimum amount of moves we need is 5, because that is the sequence ○⎕○⎕○, which is the absolute minimum we could need, as we need at least 3 circles and we cannot have consecutive circles. This clearly doesn't work, so the solution is at least 6 steps long.

After that, and playing around a bit, you can figure out that the shortest solution has eight steps and it is


There is, obviously, a more rigorous solution that proves that this is the shortest solution. Doing so in paper involves keeping track of some branching possibilities, so I will do my best to do it here for you, without too much detail:

  1. circle – it makes no sense to start with a square;
  2. square – two circles in a row make nothing, so then we use square. Here is the position we are at:

\[ \begin{bmatrix} 2 & 1 \\ 0 & 1\end{bmatrix}\]

After ○⎕ it is not obvious whether we should use a circle or a square, so let us try both:

  1. circle – we are trying the 3. circle variation;
  2. square – after a circle there is always a square. Here is what we are at:

\[ \begin{bmatrix}1 & 1 \\ 0 & 2\end{bmatrix}\]

  1. square – if we used a circle, then we'd be back at the position we were after the very first circle, meaning we had spent 4 moves doing nothing. Here is our status:

\[ \begin{bmatrix}0 & 1 \\ 2 & 1\end{bmatrix}\]

At this point, we see we cannot complete the puzzle in 3 moves or less, as we would still need to do, at least, ○⎕○ and those moves don't solve the puzzle.

We conclude, thus, that the 3rd step is not circle:

  1. square – circle as step 3. didn't work. We are at this point:

\[ \begin{bmatrix}0 & 2 \\ 1 & 1\end{bmatrix}\]

  1. circle – from the reasoning above, doing square again wastes too much time.
  2. square – after a circle comes a square. We are in this position:

\[ \begin{bmatrix}2 & 2 \\ 0 & 0\end{bmatrix}\]

At this point it becomes clear what must be done to put the two 2s together, and that is ⎕⎕○, so the final sequence is ○⎕⎕○⎕⎕⎕○.

If you have any questions about my solution, found an error (woops!) or want to share your solution, please leave a comment below! Otherwise just leave an “upvote” reaction!

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