OK i will try this.
But if you don't have a translation of -(k2-k1)/2 on the wilds (or (k2-k1)/2 on the tames), you get a worst case when the private key of P is at the end of the range.
I think there's a paper about this already:
https://www.iacr.org/archive/pkc2010/60560372/60560372.pdfYou probably need to detect frutiless cycles with this method (stuck kangaroos in a loop without distinguished points).