witzar

You didn't get it. Move "4U" means: move 4th column Up (just one cell). (If you repeat this move 4 times, then you will return to the same position.)

I've played for a while with Anza Power's simulator and I've managed to swap "1" and "2" in original position. This means that it is possible to swap any two horizontally adjacent cells, you just have to: 1. Make a couple of obvious moves to bring these two cells in place where "1" and "2" originally are. 2. Apply the procedure that swaps cells "1" and "2". 3. Undo moves made in step 1 (perform "opposite" moves in reversed order). Exactly the same can be done for every vertically adjacent cells due to the symmetry of the puzzle, you can easily transform procedure swapping "1" and "2" into procedure swapping "1" and "4". Now when you know how to swap any two adjacent (horizontally or vertically) cells, it is easy to solve every possible puzzle: you can just put numbers into their cells one by one. First put 1 in place, then 2 without touching 1, then 3 without touching 1 and 2, etc. Alternatively you can solve first three rows using method described by Anza Power and use above for the last row. All you have to do is to find a sequence of moves that swaps "1" and "2" (possibly by trial and error). And such a sequence exists, I've checked it.

Think for example about a graph in a shape of a cross (or letter X): four arms, each arm made of 100 segments. This graph has 400 edges (segments) and only 4 odd nodes (one at the end of each arm). You can start and finish where You want. Can you traverse it in 402 steps? Can You do better then 600? PS If You don't like "dead ends" (vertices with only one segment)in graph, You can add a triangle (loop made of 3 segments) at the end of each arm.

OK, I made a mistake here: I missed 'X' on hole 1 day 4. It should be: Holes: 1 2 3 4 5 Day1 : - ! - - - Day2 : X - ! - - Day3 : - X - ! - Day4 : X ! X - X Day5 : X X ! X - Day6 : X X X ! X So 6 days is enough. On Day 4 you can check hole 4 instead of hole 2 due to symmetry of X-es and continue by analogy. You can also start with hole 4 instead hole 2 on day 1. This gives total of 4 optimal solutions.