Yes, but it looks promising. 1,50 would be amazing!


I was thinking another thing:

exploiting the symmetry means that we use equivalence classes to increase the probability of a collision (same number of tries, half 'points'); in our case, given a point P(x,y), 'x' decides the width of the jump and 'y' only the direction.
In the secp256k1 there are 2 points that share the same 'x', (x,y) and (x,-y).

But this curve has another symmetry, there are 3 points with the same 'y' .

That means that this curve has about 2^256 points and
 
1) 2^256 / 2 different x-coordinates
2) 2^256 / 3 different y-coordinates

In a interval of consecutive points, all x-coordinates are different, but if we move the interval like we did it becomes symmetric.

If we want to exploit y-coordinates, a jump should be depend on 'y' and not on 'x'.
We can work then with equivalence classes where [P] = {P, Q, R} and yP = yQ = yR

what is the relation between the private key of these points P, Q, R?

If P=k*G, then Q=k*lambda*G and R =k*lambda^2*G (endomorphism)
and P=(x,y) -> Q=(beta*x, y) -> R=(beta^2*x,y)

observation:

if P=k*G, i.e. k is the private key of P respect of the generator G, then
lambda*P = k*lambda*G, i.e. the private key of P'=lambda*P respect of the generator G'=lambda*G is always k.

Besides, if we know that k lies in [a,b], resolving the problem P=k*G is equivalent to resolve the problem P'=k*G'
The interval is the same (from the scalar's point of view) but there actually 3 different intervals of points. All consecutive from a different generator's point of view.

We could work on  3 "spaces" simultaneously :

P=k*G, with all points with generator G   -> interval 1     jumps (1*G, 2*G, ,,,2^(n-1)*G)

P'=k*G' with all points with generator G' -> interval 2     jumps(lambda*G, lambda*2*G, .... lambda*2^(n-1)*G)

P''=k*G'' with all points with generator G'' -> interval 3  jumps(lambda^2*G, lambda^2*2*G, .... lambda^2*2^(n-1)*G)

if the movements of a kangaroo depends only by y coordinates, that means that if a kangaroo reach a point T in the first interval and another kangaroo reachs a point W in the second interval with the same y, we will have a collision!

when they reach a DP (in this case a y-coordinate with some zero bits), we detect this collision.

We work in this way in a space of 3*2^n points (3 intervals), but with only 2^n y-coordinates, then we should have a collision with a higher probability.