This problem combines the familiar Skyscraper puzzle with Sudoku. The board is a basic Sudoku set-up, with each digit from 1 through 9 appearing once in every row, column and each of the nine 3x3 cell blocks. Only one starting number (the '3' in the middle cell of the board), is provided.

The final grid numbers 1 through 9 also represent the unit height (e.g. storeys) of the building sitting on each cell. So in true Skyscraper fashion, the outer numbers provided around the grid indicate the number of buildings that are visible when viewed inwardly across, up or down the respective row or column, or diagonally from the four corners.

Have fun with this one! It's tougher than it might first appear! (Hint: Some digits will be duplicated along the two diagonals, so this isn't a Sudoku 'X' problem!)

4	2	5	2	1	2	3	3	4	4	3
4										4
1										3
3										2
2										4
3					3					2
4										2
9										1
3										3
2										4
3	5	1	2	8	2	2	3	3	2	3

This seems like a good puzzle, but not entirely clear about those major diagonals.  At first glance, the only direct inference would be that row 7, with view counters 9 from left to right and 1 from right to left, must have the row 123456789 (also that any counter of 1 must have a 9 at that end of the row).  This means that the diagonal from upper right to lower left has at least two "3" values.  Two cases: if the consideration of diagonals is not essential (to arrive at a unique solution), Rod should have left that out; and, if it is essential, it should be clearly stated what properties the diagonals have.  Rod does say this is not an "X Sudoku" which is clear.  Since I favor computerized solutions, I note that there are 9! unique rows, each of which has a value applying skyscraper counting from each end.  That would probably exclude certain values from certain squares (for this puzzle), but still way too many combinations for me.  This is my first encounter with the "skyscraper" puzzle, and I read up some by googling that, but I'll wait it out.  For the topology, I think we could simply specify that each building is the full width and breadth of its square, and that all viewings are done from within extensions of the row and column.  

Posted by ed bottemiller on 2009-03-02 11:50:50