# First Check-in — Placement of Numbers on a Grid

Beginning with 2 by 2 grids was surprisingly simple, with only 24 possible positions. Each grid contained four digits, and the goal was to return each digit to the original position using the same number of moves as the distance from the original position.

However, the 3 by 3 grids have proved a different challenge. My first clue to this was the realization that, with now 9 numbers on the grid, there are over 40,000 possible positions. As I solved more examples, I realized how important it became to view each “move” as a set of disjoint cycles between the numbers. (In other words, I could switch any numbers or move them in any circle, as long as different circles weren’t touching.) Each cycle had to be carefully chosen to move the numbers in the right direction, not to move numbers in the opposite direction, and not to interfere with each other.

My biggest problem is justifying the number of moves by the distance from the original position. If any number is four away from its original position, then I must have a minimum of four moves. That’s simple. Yet I have a lot of examples that I solved in four moves, though the numbers are all a maximum distance of 2 from their original position. Am I not solving them in the most efficient way? Can they all be solved in 2 moves? How can I know which may require more? How can I justify that? Perhaps the distance from original position should be counted using only the perimeter of the grid?

So I continue with my trial-and-error examples in hopes of answering any of these questions. So far, I have a few theories. It seems that if 7 or more of the numbers are a distance of 2 or greater from their original position, the solution will require one more move than the greatest distance. Though I may have found some contrary examples, I am hoping I can solve them to fit in my theory. Additionally, it seems like some distances should be measured using the perimeter. For example, if all of the outer middle digits are switched, they may only be two away from their original position, but they cannot all use the middle path. Thus, they must use the outer path along the perimeter, which uses four moves. But again, these are theories, which could likely be proved wrong with my very next example.

Hopefully, my next update will come with more answers and stronger theories. Back soon!