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.

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