Logic Masters Deutschland e.V.

And here comes the main serve... For the first part, see this link.


This is rather a risky move of myself here as I've only had about one or two testing attempts for this one so far but oh well, let's see how this goes:

Rules:

- Shade some cells in the entire grid so that all of the unshaded cells together form an orthogonally connected area, while each group of the shaded cells has to connect to the grid's perimeter.

- In addition, divide the entire grid into smaller regions and enter into each cell a number that's equal to the size of its region (i.e the number of cells in its region). No two regions of the same size can touch each other orthogonally (i.e sharing an edge).

- In the finished solution, every region must be either entirely shaded or entirely unshaded, and the sizes of all of the shaded regions must share the same odd/even parity while the sizes of all of the unshaded regions must share the other odd/even parity. Which parity belongs to which type of shading needs to be deduced.

- A number in a cell with arrow[s] is equal to the combined amount of cells in the indicated direction[s] that share the opposite type of shading (or opposite parity) of that arrow cell.



Puzzle:  Penpa plus  -  Sudokupad



Good luck and have fun solving!