Quote from: harvysam on January 30, 2018, 07:07:54 AM

Using Elliptic Curve of SECP256K1 of bitcoin, I believe with scalar multiplication of "Point P & Q" we get the "point R",

Correct.

Given a base point of P and a private key scalar value of Q, when you multiply P & Q (using point multiplication on the elliptic curve) the result is public key point R.

Quote from: harvysam on January 30, 2018, 07:07:54 AM

but the value what we get is hash values

Incorrect.

It is not a "hash value". What you get is a pair of very large integers. If it looks like a "hash value" to you then that means that the method you are using to calculate the value is displaying those integers to you in base 16 (also known as hexadecimal).

Quote from: harvysam on January 30, 2018, 07:07:54 AM

but I can convert it to the integer which results in "77 digits in X and Y coordinate".

It was already an integer. You are just converting the integer from base 16 (hexadecimal) to base 10 (decimal) representation.

Quote from: harvysam on January 30, 2018, 07:07:54 AM

I'm trying to learn something that usually in a graph we don't work on large integers,

But in elliptic curve cryptography graphs we DO work on LARGE integers.

Quote from: harvysam on January 30, 2018, 07:07:54 AM

but is there a way to remove the "generator point hash G"

The generator point is not a HASH. It is a POINT. It is a point on the graph with an X coordinate and a Y coordinate. The generator point G IS the point P in your example of "scalar multiplication of Point P & Q we get the point R".

If you remove G then you have nothing to multiply by the scalar private key Q.

Quote from: harvysam on January 30, 2018, 07:07:54 AM

so we can see the actual point values like [0,1] or [10,16] etc.,??

The large integers ARE the ACTUAL POINT VALUES.