Sure,
I will just give a simple example.
Lets say you get the key close to the N half, lets take for example a random point 7fffffffffffffffffffffffffffffff5d1777e13d618b66b81ee5de17cf885a (we pushed for example 120 puzzle to this k) which results in
K = 0269da924844fcc5852ef6808f0e0bc9e851bd7b6b23d6e55d4b152ab0101fedfd
The diff between N/2 and our point is 3ff6921a42c4b727ca4968504b9846
With a big enough hash table which we need to compute as we want (random/incremental), lets say we compute from 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681
A20A0 to 7FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF5D576E7357A4501DDFE92F46681
B20A0 (small example) then our table will contain 0x10000 public keys, but in reality it is 0x20000 (double).
If we discard Y or parity from our K computation when we generate random steps (increments) we have a double change to get it right.
If our random increment difference was bigger than normal hash table size (but not bigger than double) than we will have a match and we can easily compute the right k from the known K.
Dont know if i was clear enough, if not i will try to illustrate it
