Comment by ColinWright

Comment by ColinWright 4 days ago

My understanding is that there is a difference between the concept of a Zero-Knowledge Proof (ZKP), and then the applications that such a thing is possible.

In the example given, I can prove that N is composite without revealing anything (well, almost anything) about the factors. But in practice we want to use a ZKP to show that I have specific knowledge without revealing the knowledge itself.

For example:

You can give me a graph, and I can claim that I can three-colour it. You may doubt this, but there is a process by which I can ... to any desired level of confidence ... demonstrate that I have a colouring, without revealing what the colouring is. I colour the vertices RGB, map those colours randomly to ABC, and cover all the vertices. You choose any edge, and I reveal the "colours" (from ABC) of the endpoints. If I really can colour the graph then I will always be able to reveal two different colours. If I can't colour the graph then as we do this more and more, eventually I will fail.

So you are right, but the message of the post is, I think, still useful and relevant.

hansvm 4 days ago

Suppose the graph admits only 4+ colorings, but when attempting a 3-coloring it's possible for only one edge to be misaligned. Then (A) you need O(n_edges) calls to the oracle to gain any confidence about the 3-colorability of a 3-colorable graph (else you might be easily duped by the one misaligned edge), and (B) in so doing, you learn almost all of the structure of the graph (since you have way more random calls than there are edges).

Restating, not only is the ZK algorithm slow, but by the time you have confidence in the ZK proof you also have additional knowledge about the structure whose properties you're proving.

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ColinWright 2 hours ago

In the version I'm discussing, both parties already know the graph, so your point (B) seems irrelevant. The interrogator needs to know the graph in order to specify an edge (and to know that they've done so), so I'm not sure of the precise version you are thinking of.
Put point (A) is relevant ... certainly each call only provides a small amount of additional confidence, so a lot of calls might be required. Even so, the system seems sound to me, and I'd appreciate any details of ways in which it is not.
See also my comment here: https://news.ycombinator.com/item?id=46121137
That gives more detail of the setup, and how it can be implemented digitally.

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mathgradthrow 4 days ago

can you explain this a little better?

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fragmede 4 days ago

The key insight is that Colin can show you a red-green-blue coloring of the graph, and flip the whole graph secretly, so it's blue-red-green instead when you look at an individual section, but really the graph is yellow-pink-orange colored. Even after showing you all the intersections of the graph individually in the red green blue coloring to satisfy that he can 3-color it, you still have no idea what is yellow pink or orange on his copy of the graph.

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ColinWright 4 days ago

I can certainly explain it more, a question of "better" is debatable!
Here's the process:
(A) You give me a graph to 3-colour;
(B) I claim I can 3-colour it;
(C) You demand that I prove it;
(D) I colour it with colours ABC and cover the vertices;
(E) You point at an edge;
(F) I reveal the colours of the vertices at the ends of the edge;
(G) If I have coloured the graph then the colours revealed will always be different;
(H) We repeat this process with a permutation of the colours between each trial;
(I) If I'm lying then eventually you'll pick an edge where either the vertices are not coloured, or the have the same colour.
(J) This process reveals nothing about the colouring, but proves (to some level of confidence) that I'm telling the truth.
So ... what's unclear?
Instructions on how to email me are in my profile if you prefer ...

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- mathgradthrow 2 days ago
  
  Ah, I see. This is not an example of a ZKP, because you are relying on a third party who has full knowledge of the coloring, which is wherever you have drawn your coloring.
  
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  ColinWright a day ago
  
  No, that is not the case. The process does not rely on a third party.
  Person A provides person B with the graph.
  Person B claims to have coloured it.
  Person A demands that they prove it.
  Person B hides the colouring
  Repeatedly:
  * Person A points at an edge
  * Person B reveals that the endpoints are differently coloured
  * Person B re-hides the colouring and permutes the colours
  If person B does not have a colouring, with probability 1 this process will fail and person A will know that person B does not have a colouring.
  But if person B does have a colouring then each step will succeed, and by repeating the process person A can achieve any desired degree of confidence that person B must, indeed, have a colouring.
  This process can be made digital rather than physical, and no third party need be involved. As a sketch of one step:
  * Person B colours the graph
  * For each vertex, person B generates a long random string, pre-pends the colour, applies a cryptographically strong hash function to that, and sends the result to A. This "Fixes and hides" the colouring
  * Person A asked for two "colours" to be revealed
  * Person B provide the associated "colour and random string"s, the pre-images of the requested hashes
  * Person A checks the hashes and now knows the colours of those two vertices.
  Should I write this up "properly"? It's already discussed elsewhere on the 'net.
  
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- kadoban 4 days ago
  
  How do I know/prove that you're not just saying any random two colors for whichever edge I choose?
  
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  ColinWright 4 days ago
  
  The version I'm describing has it physically sitting in front of you at the time, so you can see that the colours haven't been changed "on the fly" after you pick an edge. In this version:
  (A) I colour it;
  (B) I cover the vertices so you can't see any of them, but I can no longer change them;
  (C) You choose the edge, and I reveal the endpoints.
  Converting this to a digital version requires further work ... my intent here was to explain the underlying idea that I can prove (to some degree of confidence) that I have a colouring without revealing anything about it.
  So just off the top of my head, for example, I can, for each vertex, create a completely random string that starts with "R", "G", or "B" depending on the colour of the vertex. Then I hash each of those, and send you all of them. You choose an edge and send me back the two hashes for the endpoints, and I provide the associated random strings so you can check that the hashes match.
  
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