EDIT: I might have understood your point. YES you could divide the key by 2 since you have 50 % 50 % that its even or odd and if you're lucky enough if the original private key ends with 2, 4, 8 or 10 (even numbers) yes you could divide and cut a little bit of its length and size but then what? What's next? The new key is still big as hell to be found and searched. And if you try to keep dividing the divided keys by 2 you will definitely f*ck up with a key and divide a key that ends with a odd number ( 1 , 3 , 5 , 7 , 9 ) and then good morning u will end with a random key on the curve that is totally unknown.
It doesn't help.
When one goes down that road (is it even, is it odd?) then the decision tree has the same size as the problem to be solved.
It's basically the same thing as writing:
k = 2*(2*(2*(2*(?/2 + ?)/2 + ?)/2 + ?)...)
which is the same thing as the representation of the problem.
Same thing as hoping that 2
k mod N ends in a 0, and create a strategy off that. But that only always happens when 2
k is inside an infinite field, like [1, 2, ... infinity], not [0, 1, ... N-1]