5 ms·

Are we sure that the base reveals nothing about the factors if n is composite? I have never seen a proof of that. Usually, zero knowledge proofs also require a

Are we sure that the base reveals nothing about the factors if n is composite? I have never seen a proof of that. Usually, zero knowledge proofs also require a prover who knows the answer (the factors in this case). This is just a primality test that can be performed locally.

ColinWright 10mo 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.
- mathgradthrow 10mo ago
  can you explain this a little better?
  - ColinWright 10mo 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 ...
    - kadoban 10mo ago
      How do I know/prove that you're not just saying any random two colors for whichever edge I choose?
      - ColinWright 10mo 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.
        lordnacho 10mo ago
        This reminds me of the "Where's Waldo (Wally in UK)" example: You can prove that you found Wally with a large piece of paper with a hole in it. You move the hole over Wally, and the person you're sitting with can see you found it, but he's no wiser about where.
        fragmede 10mo ago
        https://youtu.be/5qzNe1hk0oY https://youtu.be/5qzNe1hk0oY for a video if you can't picture that.
        FridgeSeal 10mo ago
        Another way is to get them to put marks/signatures over the back of the blank. Overlay to e blank, and cut Wally out of it where he occurs on the actual page and give them the cutout.
  - fragmede 10mo 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.
- hansvm 10mo 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.
  - ColinWright 10mo 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 https://news.ycombinator.com/item?id=46121137 That gives more detail of the setup, and how it can be implemented digitally.
schoen 10mo ago
We also don't technically have proofs for some of the computational hardness assumptions that popular "real" ZK proof constructions rely on! This might feel different because those assumptions were chosen in part because people had studied them and they certainly seem to be right, whereas perhaps here nobody has really studied this particular random number theory topic one way or the other. But in some sense, there isn't a proof that regular ZK proof methods are actually completely zero-knowledge (against a computationally bounded adversary).
madars 10mo ago
That's a very good question. It all depends on how you pick the witness b: there is a procedure that definitely is not zero-knowledge: say, if prover uses his knowledge of factorization to construct an explicit b that betrays that factorization. For example, if n = p1*p2*...*pk is square-free and not a Carmichael number, then by Korselt's criterion there exists a pi such that pi-1 does not divide n-1 (this also implies that pi>2). Use the Chinese Remainder Theorem to produce b such that b=1 (mod pj) for all j!=i, and b (mod pi) is a generator of (Z/piZ)^*. Then b is a Fermat witness: gcd(b, n) = 1 (because b is non-zero modulo every prime factor) and b^(n-1) != 1 (mod n) because b^(n-1) != 1 (mod pi) (as pi-1 does not divide n-1). However, b "betrays" the prime factorization of n, since gcd(b-1, n)>1 (by construction b-1 is divisible by all pj with j!=i, but not divisible by pi>2), and thus gcd(b-1, n) is a non-trivial factor of n. (I assumed square-free above but if pi^ei (ei>=2) divides n, then b=1+pi^(ei-1) (mod pi^ei), b=1 (mod pj^ej) (j!=i) also would have worked.) On the other hand, it is also known that for non-Carmichael numbers at least half of the bases b with gcd(b, n) = 1 are Fermat witnesses. So if you pick b uniformly at random, the verifier does not gain any new information from seeing b: they could have sampled such a witness themselves by running the same random test. Put another way, the Fermat test itself is an OK ingredient, but a prover who chooses b in a factorization-dependent way can absolutely leak the factors - the final protocol won't be ZK.
deleted 10mo ago
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phkahler 10mo ago
On a related note, I've often wondered what each congruence in the quadratic seive reveals. Once you have enough of them you can factor the number, but what does a partial set of congruences reveal? Its a matrix problem, so each row could be reducing the degrees of freedom of something. But what? And in what space?
- arcastroe 10mo ago
  If: a^2 = b^2 (mod m) Then: a^2 - b^2 = 0 (mod m) (a + b)(a - b) = 0 (mod m) So, (a + b) must be a multiple of one of the factors of m. And (a - b) must be a multiple of the other factor of m. > I've often wondered what each congruence in the quadratic seive reveals. Each congruence reveals that the sum of the bases (a plus b) contains a factor of m. And the difference of the bases (a minus b) contains another factor of m. The only thing you have to watch out for is the trivial case when one of the factors you find through this method is "1" and the other factor is "m". That case isn't very helpful. It's not that each congruence gives you new information. You only have to find one single non-trivial congruence. But the other (trivial) congruences you find along the way only reveal that 1*m=m, which you already knew.
  - dmurray 10mo ago
    > It's not that each congruence gives you new information So it's not that each congruence gives you N bits of information, and you want kN bits in total. It's more like each congruence has a 1/k chance of giving you the full kN bits. But in some information theory sense those are the same! Or concretely, if you were testing a large quantity of numbers in parallel, you would get information from each congruence.
    - arcastroe 10mo ago
      I suppose you're correct. Even if you find a trivial congruence, you do get some information. Mainly: "It's not that one!" :) The same information as trying two bases that don't form a quadratic congruence at all

ColinWright 10mo 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.

mathgradthrow 10mo ago
can you explain this a little better?
- ColinWright 10mo 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 ...
  - kadoban 10mo ago
    How do I know/prove that you're not just saying any random two colors for whichever edge I choose?
    - ColinWright 10mo 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.
      - lordnacho 10mo ago
        This reminds me of the "Where's Waldo (Wally in UK)" example: You can prove that you found Wally with a large piece of paper with a hole in it. You move the hole over Wally, and the person you're sitting with can see you found it, but he's no wiser about where.
        fragmede 10mo ago
        https://youtu.be/5qzNe1hk0oY https://youtu.be/5qzNe1hk0oY for a video if you can't picture that.
        FridgeSeal 10mo ago
        Another way is to get them to put marks/signatures over the back of the blank. Overlay to e blank, and cut Wally out of it where he occurs on the actual page and give them the cutout.
- fragmede 10mo 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.
hansvm 10mo 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.
- ColinWright 10mo 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 https://news.ycombinator.com/item?id=46121137 That gives more detail of the setup, and how it can be implemented digitally.

schoen 10mo ago

We also don't technically have proofs for some of the computational hardness assumptions that popular "real" ZK proof constructions rely on! This might feel different because those assumptions were chosen in part because people had studied them and they certainly seem to be right, whereas perhaps here nobody has really studied this particular random number theory topic one way or the other. But in some sense, there isn't a proof that regular ZK proof methods are actually completely zero-knowledge (against a computationally bounded adversary).

madars 10mo ago

That's a very good question. It all depends on how you pick the witness b: there is a procedure that definitely is not zero-knowledge: say, if prover uses his knowledge of factorization to construct an explicit b that betrays that factorization. For example, if n = p1*p2*...*pk is square-free and not a Carmichael number, then by Korselt's criterion there exists a pi such that pi-1 does not divide n-1 (this also implies that pi>2). Use the Chinese Remainder Theorem to produce b such that b=1 (mod pj) for all j!=i, and b (mod pi) is a generator of (Z/piZ)^*. Then b is a Fermat witness: gcd(b, n) = 1 (because b is non-zero modulo every prime factor) and b^(n-1) != 1 (mod n) because b^(n-1) != 1 (mod pi) (as pi-1 does not divide n-1). However, b "betrays" the prime factorization of n, since gcd(b-1, n)>1 (by construction b-1 is divisible by all pj with j!=i, but not divisible by pi>2), and thus gcd(b-1, n) is a non-trivial factor of n. (I assumed square-free above but if pi^ei (ei>=2) divides n, then b=1+pi^(ei-1) (mod pi^ei), b=1 (mod pj^ej) (j!=i) also would have worked.) On the other hand, it is also known that for non-Carmichael numbers at least half of the bases b with gcd(b, n) = 1 are Fermat witnesses. So if you pick b uniformly at random, the verifier does not gain any new information from seeing b: they could have sampled such a witness themselves by running the same random test. Put another way, the Fermat test itself is an OK ingredient, but a prover who chooses b in a factorization-dependent way can absolutely leak the factors - the final protocol won't be ZK.

deleted 10mo ago

[deleted]

phkahler 10mo ago

On a related note, I've often wondered what each congruence in the quadratic seive reveals. Once you have enough of them you can factor the number, but what does a partial set of congruences reveal? Its a matrix problem, so each row could be reducing the degrees of freedom of something. But what? And in what space?

arcastroe 10mo ago
If: a^2 = b^2 (mod m) Then: a^2 - b^2 = 0 (mod m) (a + b)(a - b) = 0 (mod m) So, (a + b) must be a multiple of one of the factors of m. And (a - b) must be a multiple of the other factor of m. > I've often wondered what each congruence in the quadratic seive reveals. Each congruence reveals that the sum of the bases (a plus b) contains a factor of m. And the difference of the bases (a minus b) contains another factor of m. The only thing you have to watch out for is the trivial case when one of the factors you find through this method is "1" and the other factor is "m". That case isn't very helpful. It's not that each congruence gives you new information. You only have to find one single non-trivial congruence. But the other (trivial) congruences you find along the way only reveal that 1*m=m, which you already knew.
- dmurray 10mo ago
  > It's not that each congruence gives you new information So it's not that each congruence gives you N bits of information, and you want kN bits in total. It's more like each congruence has a 1/k chance of giving you the full kN bits. But in some information theory sense those are the same! Or concretely, if you were testing a large quantity of numbers in parallel, you would get information from each congruence.
  - arcastroe 10mo ago
    I suppose you're correct. Even if you find a trivial congruence, you do get some information. Mainly: "It's not that one!" :) The same information as trying two bases that don't form a quadratic congruence at all