My strategy would be to move to positions 1, -2, +4, -8, 16, etc, then we get there in n + 2*(2^0 + 2^1 + ... 2^r) where 2^r < n.
(We always get there having moved n distance from the origin as our last step, otherwise we travel back to the origin spending distances 2, 4, 8, 16, etc. The sum (1, 2, ... 2^r) = 2^(r+1)-1; We double that again to get ~2^(r+2)
An example of the worst case, we lie on a number of the form -(2^r + 1) we might round-trip to +16 before heading down to -9. In that case we'd travel 2,4,8,16,32 + 9 = 71 steps.
Best case, say we lie on a number of the form 2^2r, e.g. 64.
We travel 1, -2, +4, ... , -32, 64.
We've travelled 127 steps.
Overall this takes method travels approximately between 2n and 8n.
I'm not sure how this compares to 1, -1, 2, -2, 4, -4; That has a better worst case.
My strategy would be to move to positions 1, -2, +4, -8, 16, etc, then we get there in n + 2*(2^0 + 2^1 + ... 2^r) where 2^r < n.
(We always get there having moved n distance from the origin as our last step, otherwise we travel back to the origin spending distances 2, 4, 8, 16, etc. The sum (1, 2, ... 2^r) = 2^(r+1)-1; We double that again to get ~2^(r+2)
An example of the worst case, we lie on a number of the form -(2^r + 1) we might round-trip to +16 before heading down to -9. In that case we'd travel 2,4,8,16,32 + 9 = 71 steps.
Best case, say we lie on a number of the form 2^2r, e.g. 64.
We travel 1, -2, +4, ... , -32, 64.
We've travelled 127 steps.
Overall this takes method travels approximately between 2n and 8n.
I'm not sure how this compares to 1, -1, 2, -2, 4, -4; That has a better worst case.