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If we suppose that W_100 is = (2^37)*G and T_100 is (2^38)*G, than between T_101, T_102, .... , T_2**32 it is no longer possible to find a single distinguished point with the same coordinate of any distinguished point that lies in W_0, W_1, W_2, ....., W_100. For sure.
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I am not sure if I understand you correct. But T_101, T_102, ... etc are not the distinguished points, these are
jump points. At the start only the jump point table is prepared.
It is not possible to know all the distinguished points within the working range.
Distinguished point is a property of point. As soon as the point meets this property, it goes to the hash table:
if tamePosi is distinguished
add (TAME,tamePosi,tamei) to hashTable
if wildPosi is distinguished
add (WILD,wildPosi,wildi) to hashTableThe kangaroo (as wild so tame) will jump permanently by pseudorandom steps from the jump table while reach the Point with patterned X-coordinate (the distinguished point). As reach the distinguished point, this point just saved in the hashtable (kangaroo type, distance, Point) check for coliddion, and kangaroo continues to jump.
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It is impossible to get a collision in the path of the same tame kangaroo too, because the same kangaroo cannot turn back, then this is a pseudorandom sequence very particular ...
The same kangaroo can not go back, but due to symmetry point it so not needed.
P=k*G, you know P and you know that a < k < b. Actually Point P'=k'*G is also suitable for us if k' = b - (k-a) = b + a - k, which is also lies in the range a..b