Minor correction - the method the paper describes to 'dump' the table is a bit different.
It sounds like you're proposing that they iterate through all possible keys, check if the key is in the table, and if it is, dump that key and delete it. This would take O(2^b) time, where b is the number of bits in the key. Obviously, for reasonably-sized b, this can get a bit out of hand.
The paper takes a different approach. It looks for a cell with Cs = 1. Cs = 1 means Ks and Vs represent exactly a key-value pair. Dump that (Ks,Vs) pair, then delete (Ks,Vs) from the table. In the process of deleting, there will probably be some other cell that goes from Cs = 2 to Cs = 1. continue from there. Eventually, you should get all Cs = 0. If you ever get stuck, you're unable to dump the entire list.
That method takes O(n), where n is the number of elements in the table, and the fastest you could expect to dump any kind of list.
It sounds like you're proposing that they iterate through all possible keys, check if the key is in the table, and if it is, dump that key and delete it. This would take O(2^b) time, where b is the number of bits in the key. Obviously, for reasonably-sized b, this can get a bit out of hand.
The paper takes a different approach. It looks for a cell with Cs = 1. Cs = 1 means Ks and Vs represent exactly a key-value pair. Dump that (Ks,Vs) pair, then delete (Ks,Vs) from the table. In the process of deleting, there will probably be some other cell that goes from Cs = 2 to Cs = 1. continue from there. Eventually, you should get all Cs = 0. If you ever get stuck, you're unable to dump the entire list.
That method takes O(n), where n is the number of elements in the table, and the fastest you could expect to dump any kind of list.