How to determine large point for the same x axis in the elliptic curve cryptography?

Question

For a given X value of elliptic curve cryptography there are two Y values. One point is P(x,y) and another point is Q(x1,y1) where P =-Q or Q = -P. Suppose given X value is 103388573995635080359749164254216598308788835304023601477803095234286494993683

Then output Y will be

37057141145242123013015316630864329550140216928701153669873286428255828810018 and 78734948092074072410555668377823578303129767736939410369584297579653005861645

So, P = (103388573995635080359749164254216598308788835304023601477803095234286494993683, 37057141145242123013015316630864329550140216928701153669873286428255828810018) and Q = (103388573995635080359749164254216598308788835304023601477803095234286494993683, 78734948092074072410555668377823578303129767736939410369584297579653005861645)

Also generator point G =(55066263022277343669578718895168534326250603453777594175500187360389116729240, 32670510020758816978083085130507043184471273380659243275938904335757337482424) = 1

-G = (55066263022277343669578718895168534326250603453777594175500187360389116729240, -32670510020758816978083085130507043184471273380659243275938904335757337482424) = -1

or

-G = (55066263022277343669578718895168534326250603453777594175500187360389116729240, 83121579216557378445487899878180864668798711284981320763518679672151497189239) = -1

If G is added to P and Q, what will correct answer between P+G > Q+G or P+G < Q + G? How to identify it from the given x value?

Answer 1

If G is added to P and Q, what will correct answer between P+G > Q+G or P+G < Q + G?

You're essentially asking "given two points $P$ and $-P$ (and thus, $xG$ and $-xG$), how do we determine which is the smaller?

The answer is: we hope that is infeasible.

Here's why: if we have a magic box that, given $P$ and $-P$, determines which is smaller (that is, which is the smaller multiple of $G$), then we can compute discrete logs (and thus break ECC on this curve).

Here's why: when we compute $P$ and $-P$, we are actually comparing $x$ and $n-x$ (where $xG = P$ and $n$ is the order of $G$). If $x < n-x$, then we know that $x < n/2$; if $x > n-x$, then $x > n/2$. That is, it acts as an Oracle to determine whether $P$ is in the 'first half' of the internal of multiples of $G$ or the second half.

So, to compute the discrete log of $P$, we call our Oracle to determine whether $x < n/2$ or $x > n/2$ - let us say that it's in the first half ($x < n/2$). What we would do then is compute $2P$ and pass it to our Oracle, which would tell us if $2x < n$ or $2x > n$ - that is, if it's in the first quarter or the second (and we could do the same if it turned out that $x > n/2$).

Continuing on, we could repeat this, doing binary search, until we find the value of $x$, using $\lceil \log_2 n \rceil$ calls to our Oracle, thus solving the discrete log problem.