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Monday, July 04, 2016

Is determining if a poly over a finite field is 1-1 hard? Sure seems so.


When I teach cryptography  to High School  students I begin with shift and linear  ciphers which are

x --> x+s mod 26 (s is a shift, x is a letter of the alphabet. Hmm- x really IS a letter of the alphabet!)

x--> ax+b mod 26 (Note that a has to be rel prime to 26.)

I then ask why nobody seems to have ever used

x --> ax2 + bx + c mod 26.

I then tell them that this is because there is no quick test that will, given (a,b,c), tell if
f(x) = ax2 + bx + c mod 26 is 1-1 (and hence onto).

It recently dawned on me that I don't really know that its hard to test.
(ADDED LATER) In fact its not true. Algebra shows that f(x) is NOT 1-1 iff

a(x+y)+b =0 mod 26

has a solution. If a is rel prime to 26 then clearly there is a solution (many in fact).
If a is not rel prime to 26 then I suspect this is not hard.

How hard is the following problem?

Given a poly f(x) of degree d, and n,  is f(x)  mod n 1-1?

We will assume all coefficients are between 0 and n.. We can also assume that c is 0 since f(x) is 1-1 then f(x)-c is 1-1.
 The coefficients are  given in binary so the length of the input is roughly dlog(n).

One can of course compute f(0), f(1),...,f(n-1) and see if there are any repeats, but this takes O(n) steps which is exp in the input of length log n.

I suspect that this is either a well known solved problem (either in P or NPC) or a well known open problem. Any help or references will be more appreciated than usual-- see next paragraph.

I am the new SIGACT News Open Problems Column editor. In the future I will be soliciting people to write columns for me, but the first one I'll do myself and this might be a good topic- if its open! And if it is open, would be good to know references and what is known.


11 comments:

  1. Replies
    1. 2x²+x is 1–1 mod any power of 2.
      Rivest has shown in 1999 that polynomial ∑aᵢxⁱ is a permutation mod 2ⁿ iff a₁ is odd, (a₂+a₄+a₆+⋯) is even, and (a₃+a₅+a₇+⋯) is even.

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  3. These polynomials are known as permutation polynomials in the literature. Testing if f in 1-1 mod m reduces to testing if for every prime factor of p, f in 1-1 mod p and f' has no root mod p (see for instance Theorems 2 and 6 of http://arxiv.org/abs/math/0509523). Both tests can be achieved in deterministic polytime (see http://eccc.hpi-web.de/report/2005/008/ for the first one, and the second one is classical). Thus given the prime factorization of m, the test is deterministic polytime. This does not imply that it is "easy" ;-).

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  4. These polynomials are known as permutation polynomials in the literature. Testing if f in 1-1 mod m reduces to testing if for every prime factor of p, f in 1-1 mod p and f' has no root mod p (see for instance Theorems 2 and 6 of http://arxiv.org/abs/math/0509523). Both tests can be achieved in deterministic polytime (see http://eccc.hpi-web.de/report/2005/008/ for the first one, and the second one is classical). Thus given the prime factorization of m, the test is deterministic polytime. This does not imply that it is "easy" ;-).

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  5. To elaborate a little bit:
    Quadratic polynomials are never 1-1 over a field of characteristic different from 2 (because a non-zero element has 0 or 2 square roots).
    Over GF(2^r): x^2 is 1-1 but unless I am mistaken, a polynomial with a linear term will not be 1-1 (you have 0 or 2 preimages).
    If you start counting mod n for composite n there are other 1-1 polynomials (like in the above example).

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  6. Might this problem somehow be connected to whether the polynomial f(x) has roots mod n? At least in some cases this might lead us to a partial solution:

    Since we assume that the constant c is zero, we know that the polynomial has at least the root 0. If the polynomial has another root y, then it is obviously not 1-1.

    Finding out if the polynomial has roots may be fast in some cases. If n is of the form p^k, where p is a prime, then we may use factorization algorithms for polynomials over finite fields (see for instance https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields). If the degree d of the polynomial is small in relation to n, then we might find an algorithm that is faster than O(n). For instance if n is a prime, then Shoup's algorithm (http://www.shoup.net/papers/detfac.pdf) can factorize the polynomial in time O(n^0.5*(log n)^2*d^(2+eps)). Since the polynomial may have at most d factors, it should be easy to find any linear factors (if d is small in relation to n)

    This is of course only a partial solution, since everything above only works for finite fields (or at least all the analysis is done over finite fields). I'm also sure that there is a polynomial that is not 1-1 and does not have more than one root; an example of a polynomial like this would of course be nice!

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  7. It appears that testing whether a polynomial over a finite field is a permutation polynomial is, in fact, in P.

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  8. It is called permutation polynomials. It is a very important research problem. There are tons of survey on this topics. Start with wikipedia first.

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  9. See the notes and references here. The problem is apparently in ZPP.

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  10. Or indeed in P: http://eccc.hpi-web.de/report/2005/008/download/

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