Monthly Archives: August 2022

Inequivalence of Loop Programs Without Nesting

This is another one of those “mainly ignore most of the problem” reductions.

The Problem: Inequivalence of Loop Programs Without Nesting.  This is problem PO14 of the appendix.

The description: Given:

  • a set of X variables,
  • a subset Y of X which are “input” variables,
  • a specific output variable x0
  • Two programs P1 and P2 with the commands:
    • x <- y (x gets a copy of y)
    • x <- 0  (x gets set to 0)
    • x<- x + 1 (x is incremented by 1)
    • DO xi {S1 S2..Sn} END  (see below)

.. is there a way to assign the variablse in Y so that P1 and P2 produce different outputs in our output variable x0?

Example: The semantics of the “DO” statement are (they’re better described by Tsichrtizis in his paper) that we execute the loop body a number of times equal to the value of xi (if it is 0, we do not execute the loop at all).  The Si statements can be any program statement except another loop (since we are looking at programs without nesting- even allowing one level of nesting makes this problem undecidable).

So, here are two programs x1 is our input variable and x0 is our output variable:

P1:

x0 <- 0
DO x1
x0 <- x0 + 1
END

P2:

x0 <- x1

These programs are equivalent as long as we do not allow our input values to be negative (which seems to be the convention) because both programs will copy the integer value in x1 into x0 (though P1 does it more slowly).  Suppose we used the alternate P1

P1′:

x0 <- 0
DO x1
x0 <- x0 + 1
x0 <- x0 + 1
END

..this has the effect of making x0‘s value twice the value of x1.  While P1 and P2 will still have the same value in x0 if x1=0, any other value in x1 will result in different outputs, making our programs inequivalent.

Reduction: Constable, Hunt, and Sahni use 3SAT.  The basic idea is to use loops that execute at most once as if statements.  (I don’t think they have a loop that iterates more than once in the whole program).  So for example, they will set up a clause variable Ci that will be set to 1 if any of its literal variables are true with a subprogram like:

Ci <- 0
DO Ci1  (Ci1 is the first literal of clause i)
Ci <- 1 (Technically this is Ci <- 0 followed by Ci <- Ci +1)
END

DO Ci2
Ci <- 1
END

DO Ci3
Ci <- 1
END

..each of the loops will set Ci to 1 if the literal is true, and leave it alone if the literal is false.

So, with that in mind, we take a 3SAT formula and write a program:

  • Our input variables are the variables to the formula.
  • We create the negations of all of the variables in an if statement  (this way we have both positive and negative literals for each variable)
  • We compute all of the Ci variables as above
  • We compute the negations of all of the Ci variables. (we need this because our loop-based if statement can only do things if a variable is 1, not 0)
  • We create a variable P that is initially 1.  If any of the negated Ci variables are 1 (so the regular Ci variable was 0, so the clause was not satisfied), we reset P to 0.  P is our output variable.

This program will output P to 1 if its inputs form a satisfiable arrangement of variables, and will output 0 otherwise.  This is our P1 program.

Our P2 program is a program that always outputs 0.  P1 and P2 will be different if and only if the formula is satisfiable.

Difficulty: 5. The reduction itself is pretty easy- “write a program that sees if the inputs are satisfiable”.  The hard part is realizing that you don’t really use the loop structure as a loop, and that you aren’t searching for an input to make it satisfiable, you’re just using whatever input the SAT instance gives you.

Inequivalence of Finite Memory Programs

Let’s see if I can build my own reduction for this one.

The problem: Inequivalence of Finite Memory Programs.  This is problem PO13 in the appendix.

The description: Given a finite set X of variables, a finite set of constant symbols Σ, two programs P1 and P2, each made with a sequence of instructions I1..Im (possibly of different lengths in both programs)

  • read xi
  • write vi
  • xi <- vj
  • if vi = vj then goto Ik
  • accept
  • halt (“halt” tends to mean “reject”)

Each xi is a variable from X.  Each vj is either a variable from X, a symbol from Σ, or a special “end marker” symbol $.  Ik is an instruction in the program.

Is there an initial assignment of values to all variables in P1 and P2 (the same assignment to variables in both programs) that leads to different output values of variables in both programs?

Example:

Here are two programs:

P1:

I1: X1 <- 2
I2: write X1
I3: accept

P2:

I1: if X1 = X1  goto I4
I2: X1 <- 3
I3: if X1=X1 goto I5
I4: X1 <- 2
I5: write X1
I6: accept

Despite the two programs being different, P2 will always go to I4 in the first if statement, and then both programs set X1 to 2, write the new value, and halt so that both programs will produce the same output on all inputs.

If we change P2’s first instruction to:

I1: if X1=1 goto I4

..then the programs will produce different outputs whenever X1 is not 1.

Reduction (from the paper): The paper by Jones and Muchnick uses Linear Bounded Automaton Acceptance.  They first show that a program in this format can be represented by a “Deterministic Generalized Sequential Machine” (DGSM), which works like a finite automaton that can output any string on a transition.  (I’m getting the description for this from my (old, 1979) copy of Hopcroft and Ullman, section 11.2.).

The DGSM has a state for each pair of:

  • Labels of read statements in the program
  • All possible strings of length m (the number of variables) containing either a symbol from Σ, or $.

In addition to all of those 2-tuples, we have a state qa (the accepting state) and qh the halting state.

The start state is the tuple (instruction of the first read in the program, whatever is in the variables at that point of the program).

The paper goes on to claim that once we have this, the “transition function can be defined in a totally finite manner”, and then claims that once we do this, we can use another result that says that inequivalence for DGSM’s is NSPACE(log(n))-complete.

I’m a little worried that the number of states in our DGSM is exponential in the size of the number of variables.  Maybe that doesn’t matter because this is just proving the existence of the DGSM?

Reduction (my try): So let’s see if I can try to create a reduction myself.

Suppose we have an LBA M, with its starting input x.  We’ll build a program in our language to simulate its behavior.  We’ll store one variable for each tape cell and an extra variable “xt” holding the (integer) position of the tape head.

We can simulate looking at a tape cell with:

if xt = 1 then goto an instruction that looks at x1
if xt = 2 then goto an instruction that looks at x2
.. and so on

This will increase the size of the  program by a linear amount (because the tape cells only use a linear amount of space)

So now we can simulate our machines using these programs and if our programs are inequivalent, then the LBA’s we started with are as well.

I don’t know.  It feels too easy.

Difficulty: 8.  It might be less if you can provide a clear description of what the instructions in the language does.  The paper describes a configuration of a program as an instruction to perform next, and two strings.  One of the strings represents the contents of the variables.  The other string I think is related to what the read and write instructions do, but it is hard for me to see what exactly those instructions are doing.  So I may have missed something.