Tag Archives: Strong Inequivalence of Ianov Schemes

Strong Inequivalence of Ianov Schemes

I’d never heard of “Ianov schemes” before, but we’ll be using them for the next couple of problems, so let’s take a look!

The problem: Strong Inequivalence of Ianov Schemes.  This is problem PO16 of the appendix.

The description: Given two programs with a single variable x and instructions:

  1. x = f(x) (where f is some function from some given set of functions F)
  2. if p(x) goto I1 else goto I2 (p(x) is a boolean predicate from some given set of predicates P, I1 and I2 are instructions in the program)
  3. Halt

Are the two programs not strongly equivalent: Can we find a function from F for each f, a predicate from P for each p, and an initial value for each variable that gives different final values for the variable in both programs?

Example: Here is one similar to what we will need in a minute:

P1:

1: if p(x) then 4 else 2
2: x =  f(x)
3: Halt
4: Halt

P2:

1: Halt

Suppose the only choice for f is f(x)= x+ 1(so, changing the variable).  Then the programs are different if we have a p(x) that will let us get to line 2 in P1.  If our only choice of p is something like “p(x) = true”, then the two programs are always equivalent.  If our set of P is (p(x) = true, p(x) = false), then we can choose a predicate that makes the programs inequivalent.

Reduction: The paper from Constable, Hunt, and Sahni basically does this in two parts.  G&J say to use 3SAT, but this looks to me like DNF Non-Tautology.

First (this is Proposition 2.11), they show how to take a DNF formula and turn it into a program.  The program works as a giant chain of if statements checking each literal in each clause.  If a literal is checked to be false, we move on to the next clause.  If all of the literals in a clause are true, we make it to the first (“+”) halting state.  If none of the clauses make it to the “+” state, we go to the second (“-“) halting state.  This means that the “-” state is only reachable if we do not start with a tautology.

Next (This is part iii of Corollary 2.14 of Theorem 2.13), they show that they can extend this into an Ianov scheme program by making the “+” state be just a halt command (so the variable never gets changed), and the “-” state be:

1: x= f(x)  (Where f is some function that changes x)
2: halt.

So the only way this program is different from the trivial “just halt” program is if we can reach the “-” instruction, which only happens if the formula is not a tautology.

Difficulty: 5. The program to test the non-tautology is not that hard.  As usual, the hardest part is parsing all of the definitions in the problem