It's perfectly solvable.
There are 10^50 odd positions. Each position, by my reckoning, requires a minimum of 64 round bytes to store (each square has: 2 affiliation bits, 3 piece bits). Once they are collated, the positions that are checkmates can be identified and positions can be compared to form a relationship, showing which positions lead to other positions in a large outcome tree.
Then, the program needs only to find the lowest only one side checkmate roots, if such a thing exists. In any case, Chess was fairly simply solved at the end of the first paragraph.
