I doubt there is any way to compute the cipher especially since there are so many ambiguous unchecked cells.
As recommended earlier, you should use the fact that the resulting crossing entries are words or phrases or abbreviations (38a is problematic). In many cases there is only one possible letter to choose (you can assume that no letter decodes to itself) or just a couple. 19a is particularly good for this. That will establish a set of decodings that you can use for the other three edges.