<p>Is your question if, given two points <em>A</em> and <em>B</em>, you can compute <em>x</em> such that <em>A + xG = B</em>, the answer is no.</p>
<p>That would be equivalent to solving ECDLP (elliptic curve discrete logarithm problem), which is assumed to be computationally infeasible: just apply it with <em>A</em> a public key of your own (which you know the private key to), and <em>B</em> the point whose DL you want to compute. The resulting <em>x</em> plus the DL of <em>A</em> is the DL of <em>B</em>.</p>

CC BY-SA 4.0
2021-10-05T00:48:15.480Z
2021-10-05T00:48:15.480Z
Pieter Wuille
<p>Wondering if there is any relative connection between public keys?</p>
<p>I know that they are a function of G, but is there any other connection they share?</p>
<p>Is it possible to calculate how many G are between two public key points?</p>
<p>Thanks in advance.</p>

<public-key><ecdsa>
What is the relativity between ECDSA public keys, if any?
Tanagi
# What is the relativity between ECDSA public keys, if any?

(Asked by: username_0 on 2021-10-05)

Wondering if there is any relative connection between public keys?

I know that they are a function of G, but is there any other connection they share?

Is it possible to calculate how many G are between two public key points?

Thanks in advance.

**Tags:** public-key, ecdsa


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## 1 answers

### Answer by username_1 on 2021-10-05. Score: 1

Is your question if, given two points *A* and *B*, you can compute *x* such that *A + xG = B*, the answer is no.

That would be equivalent to solving ECDLP (elliptic curve discrete logarithm problem), which is assumed to be computationally infeasible: just apply it with *A* a public key of your own (which you know the private key to), and *B* the point whose DL you want to compute. The resulting *x* plus the DL of *A* is the DL of *B*.


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