.... This example has DD (degree of difficulty) = 1, which means that the computer claims that all cell values are implied, so no trial and error should be required. Hopefully Bill can confirm this!
So far I have gotten nowhere with this DD1 puzzle, but that probably proves only that I am a neophyte at this skyscraper stuff.
I have, however, been able to make a rather peculiar observation.
If there were no 9's anywhere in the puzzle, then as soon as you have found "the" solution, you can create another valid solution just by adding 1 to the values in each cell.
Likewise, if there are no 1's, you can create an additional valid solution by subtracting 1 from each value.
Therefore, by uniqueness, there must be at least one 1 and at least one 9 somewhere in the puzzle.
Proofs by uniqueness are generally considered, by the purist in all of us, to be a scandalous dirty trick. One feels that one of the duties of a (human) solver is to establish uniqueness, not to assume it in order to reach a solution.
The possibility of being able to use a uniqueness argument comes up all the time in Kakuro and Sudoku. Suppose, for example, that you have already established that r2c5,r3c5,r6c5 all must be 1, 2, or 4 (and that all of these are in the same word), and that r2c6,r3c6 must both be either 1 or 2. Then you can conclude that r6c5 cannot be a 4, because then all four cells in the rectangle r2c5,r3c5,r2c6,r3c6 would have to be either 1 or 2, which means that, given one solution, you could create another by replacing the two 1's with 2's and vice versa. (Please note that everything in this entire paragraph applies to both Kakuro and Sudoku.)
Anyway, the existence of the uniqueness assumption temps me to try to solve this latest SSX puzzle by first looking for someplace a 1 can be, and someplace a 9 can be. If there is only one of each, then I'm on my way. But that would be a dirty trick!