Category Archives: Algebra and Number Theory

Simultaneous Incongruences

Back to the problems we’ve skipped over, starting with a cool take on the classic “Chinese Remainder Theorem”.

The problem: Simultaneous Incongruences.  This is problem AN2 in the appendix.

The description: Given n pairs of positive integers (a1, b1)…(an, bn), can we find an integer x such that x \not\equiv ai mod bi?

(G&J also add the rule that  ai ≤ bi for all i, but we can easily make that happen by doing division)

Example: I think it’s easier to see this as the actual congruences:

Can we find an x such that:

  • x \not\equiv 1 mod 2
  • x \not\equiv 2 mod 3, and
  • x \not\equiv 3 mod 4?

If we chose x as 4, we’ll see that it works.  For a simple example of a case that fails, we can do:

  • x \not\equiv 1 mod 3
  • x \not\equiv 2 mod 3, and
  • x \not\equiv 3 mod 3

Reduction: I found this reduction in the Algorithmic Number Theory book by Bach and Shallit.  Their Theorem 5.5.7 calls this the “Anti-Chinese Remainder theorem”.  They reduce from 3SAT.

Our formula will have t variables and each clause in our formula is made up of 3 literals, which we’ll represent as Ci = (zai∨zbi∨zci).  For each ai find pai, the aith prime number, and find pbi and pci similarly.  Define ai‘ to be 0 if zai is a positive literal, and 1 if it’s a negative literal, and define bi‘ and ci‘ similarly.   Now for each clause, use the regular Chinese Remainder theorem to find a value xi where:

  • xi \equiv ai‘ mod pai
  • xi \equiv bi‘ mod pbi
  • xi \equiv ci‘ mod pci

Our system of incongruences will be:

  • x \not\equiv 2 (mod 3)
  • x \not\equiv 2,3 (mod 5)
  • x \not\equiv 2,3,4,…pt (mod pi)   (pt is the tth prime number)

The above incongruences are there to force x to be 0 or 1 mod each pi. I think these correspond to true-false values to the variables in the SAT instance (x being 0 mod pi means setting that variable false, making it 1 mod pi means setting that variable true)

  • x \not\equiv xi (mod pai*pbi*pci) for each i

This turns out to be O(n+t3) incongruences by the Prime Number Theorem.

Each clause in the SAT instance is satisfiable unless all 3 literals are false.  By the way we’ve created our ai‘  (and b and c), this means that the variables can’t be set to be equal to all of ai‘, bi‘ and ci‘.  Because of how our xi was chosen, this means that x is not congruent to xi (mod pai*pbi*pci).

Difficulty: 7.  This is a cool short reduction, but the way the x value works is something you don’t usually see.

Quadratic Diophantine Equations

Since this problem’s reduction is basically the same as last week’s, let’s skip ahead to it now.

The problem: Quadratic Diophantine Equations.  This is problem AN8 in the appendix.

The description: Given positive integers a,b, and c, can we find positive integers x and y such that ax2 + by = c?

Example: Let a=1, b = 2, and c=5.  Then we’re asking if positive integers x and y exists to solve x2 + 2y = 5.  This is true when x=1 and y = 2.  If we change b to 3, then there are no positive integers x and y where x2 + 3y = 5.  (y has to be positive, so the only y that has a chance of working is y=1, which would require x to be √2)

Reduction: This is in the same paper by Manders and Adleman that had the reduction for Quadratic Congruences.  In fact, the reduction is almost exactly the same.  We go through the same process of creating the Subset Sum problem, and build the same H and K at the end.  The only difference is the instance: We build the quadratic (K+1)3*2*8m+1*(H2-x2) + K(x2-r2)-2K*8m+1y = 0.  We can multiply out the vales to get a,b, and c.

The rest of the reduction uses similar crazy algebra to the last problem.

Difficulty: 9, for the same reasons last week’s problem was.

Quadratic Congruences

On to the next section!  The “Algebra and Number Theory” section should take us through the end of the year.

The problem: Quadratic Congruences.  This is problem AN1 in the appendix.

The description: Given positive integers a,b, and c, can we find a positive integer x < c such that x2 ≡ a (mod b)?

Example: Let a = 3, b = 7, c = 13. It turns out that for every x < c, x2≡ either 0,1,2 or 4 mod b.   So this isn’t solvable.  But if we change a to 2, then the first c that works is 3.

One thing I learned in doing this problem is that the squares of mod b will always form a cycle with period b.  The proof is a pretty cool algebra problem if you set it up right.

Reduction: The paper by Manders and Adleman uses 3SAT.  Starting from the list of variables, they define a set Σ that holds all 3-element clauses of 3 different literals from that list of variables.  We won’t actually build this set (that’s exponential), but we can list out a number j for each of those clauses.  Then define τΦ to be -1* the sum of 8j for each of the j numbers that correspond to the clauses in our actual formula.

For each variable i, we define fi+ to be the sum of 8j for each j number in ∑ that has that variable positively, and fi to be -1* the sum of 8j for each number in Σ that has the variable negatively.

Let m be|Σ|, and set n=2m +The number of variables, and define a set C wit values c0 through cn:

  • c0 = 1
  • If j is odd(=2k-1) and ≤ 2m, cj = -8k/2
  • If j is even (=2k) and ≤ 2m, cj = -8k
  • If j is > 2m (= 2m+i), then cj = (fi++fi)/2

Set τ = τΦ + The sum of all of the ci‘s + The sum of all of the fi‘s.

This actually turns into a Subset Sum problem: Let C be our set, and count each value positively if we take it, and negatively if we don’t take it.

But we still need to prove this is SOS problem is solvable exactly when the original problem is.

Create a set p0 .. pn of consecutive prime numbers where the first one (p0) is 13.  Then define a set of θj values as the smallest number satisfying:

  • θj≡ ci (mod 8m+1)
  • θj ≡ 0 (mod The product of each (pi)n+1, except for j.)
  • θj is NOT ≡ 0 (mod pj)

Set H = The sum of all of the θj‘s.  Set L=the product of all of the (pi)n+1.  We are finally ready to create our Quadric Congruence instance:

b = 2*8m+1*K.

Let d = The inverse of (2*8m+1+K), mod b.

Then a = d*(K*τ2+2*8m+1 * H2)

c = H.

From here, it’s a bunch of algebra to show that an x exists if and only if the original formula was satisfiable.

Difficulty: 9.  Maybe it should be 10.  I don’t know enough number theory to be convinced that the θ’s exist, and be computed in polynomial time.  The algebra I’m glossing over is pretty complicated as well.

Protected: Comparative Divisibility

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