Logic Masters Deutschland e.V.

Let’s call the cells that are left out by the loop the outsider cells, and the digits they contain the outsiders. The first rule divides the grid into a checkerboard, we'll call the mosaic where the odds digits must go the odd mosaic, and its counterpart the even mosaic.

GENERAL CONSIDERATIONS

Can you say anything about the digits in the even mosaic ?
Direction: Can all odd digits fit in the odd mosaic ?
Answer: The odd mosaic has 40 cells and there are 45 odd digits in total, that's 5 too many. Hence the even mosaic must contain all the even digits and exactly 5 odds digits.
Direction: Can you find out what these five digits will be exactly?
Answer: The odd mosaic contains space for exactly 8 complete sets of odd digits, which means that all the digits from the 9th set must go in the even mosaic. This is one of each in 12345.

Can you rule out any type of symmetry of the loop ?
Direction: Is the loop compatible with an axial symmetry (vertical, horizontal or diagonal)?
Answer: Consider two places where the loop would cross a symetry axis. If they join on one side, then by symmetry they would join on the other side and any other crossing the axis would belong another loop. But does it rule out all cases ?
Answer: If the axis only contained only two even digits then there would be a possibility. But can you rule it out anyways ?
Answer: That would imply the existence of an even outsider in the diagonal case, at row 2 and column 2 or 8 for example, or seven (odd) outsiders in the same row or column in the vertical/horizontal case.

FINDING THE OUTSIDERS

How many outsiders are there in total ?
Direction: Calculate the length of the loop.
Answer: The loop must visit all 36 even digits and can't double on odds, and as no even digits can be consecutive because of the checkerboard pattern the loop must be 72 cells long with alternating parity. Therefore there are 9 outsiders exactly.

What digits can the outsiders be ?
Direction: how many times can a digit appear in the set of outsider cells ?
Answer: For each digit there is only one row and one column availabe, so a digit cannot be an outsider more than twice. As there are 9 outsiders and five odd digits possible (evens cannot be outsiders), there will be four doubles and one single, all from 12345.

Can the symmetry of the loop help you find some outsider(s) ?
Direction: Can you prove that the center square has to be an outsider square ?
Answer: Only 8 outsider cells can afford to translate onto another one when the rotation of the square grid is applied, so the 9th one has to be the center of rotation.

What can you tell about the odd digits in the even mosaic ?
Direction: Can the loop visit them ?
Answer: All their orthogonal neighbours belong to the odd mosaic due to the checkerboard pattern, so the loop will never be able to visit them without visiting another odd digit just before. They are outsiders.
Direction: Can they see each other ?
Answer: That would imply 6 odd digits in that row or column.

You should now be able to place all the outsiders from the even mosaic if you resolved the kropki dot already, remembering that outsiders must be in their own row or column.

There are obvious things to do with the remaining outsiders.
Direction: Can we identify or place two of them already ?
Answer: The one in column 1 must be an outsider already because it is on its column.
Answer: Use symmetry to place its counterpart on column 9.

Find the possible squares for the last two outsiders and see if you can reduce it down a little.
Direction: Consider the 7 in column 7, can it be in row 8?
Answer: Connected outsiders are forbidden, and we already have one in column 9, so 7 cannot go in row 8.
Direction: Can it be in row 2?
Answer: The configuration of the two outsiders near the corner would force the loop to have a 4 cells long straight segment in order to escape, which is forbidden.

Do Sudoku an see if you an finally place the last outsiders.
Direction: You won't be able to place the 3 and the 7 yet, but something must have popped up. Symmetry will hep.
Answer: You must reach a configuration where there is a naked 37 pair in r4c7 and r6c3, which is the only possible place to put the 3 and the 7, because any other configuration would leave those tho cells without any candidate left.

THE FINISH

Do some more Sudoku to place all the remaining odd digits in the grid. If you're stuck, read the rules again.

It's now possible to draw the loop entirely. There's a tricky part to it, here's a hint:
Direction: Is it possible for the loop to cross between the central 5 and the nearby outsiders 3 and 7?
Answer: It would have to break into separate various loops.

The last one rule that you didn't use yet will let you finish the puzzle.