algorithm,logic,tail-recursion,eiffel , Prove two algorithms are identical

Prove two algorithms are identical


Tag: algorithm,logic,tail-recursion,eiffel

I am supposed to show that two algorithms execute identical statements in identical order. One is a tail recursive version of the other. They are written in Eiffel.

tail_rec(x:instance_type):result_type is
        if b(x) then
            Result :=c(x)

Then the non-tail recursive version.

non_rec(x:instance_type):result_type is
        from until b(x) loop
            x:= d(x)
        Result:= c(x)

where b(x), c(x), and d(x) are any functions of type BOOLEAN, result_type, and instance_type respectively.

How are these two algorithms similar and how do they execute identical statements in identical order?


By substituting all flow-of-control constructs with gotos, (essentially turning the code from Eiffel into pseudocode,) and allowing if statements to execute only gotos, it can be shown that both functions end up consisting of the exact same set of instructions.

Let me begin by copying the original tail_rec here for convenience:

tail_rec(x:instance_type):result_type is
        if b(x) then
            Result := c(x)
            y := d(x);
            Result := tail_rec(y)

First, get rid of Eiffel's silly Result := construct and replace it with return for convenience. (Otherwise, we are going to have to add more gotos, and frankly, the fewer the better.)

tail_rec(x:instance_type):result_type is
        if b(x) then
            return c(x)
        y := d(x);
        return tail_rec(y)

Substitute the if-then-end with if-then goto:

tail_rec(x:instance_type):result_type is
        if not b(x) then goto label1
        return c(x)
        y := d(x);
        return tail_rec(y)

Replace tail recursion with another goto:

tail_rec(x:instance_type):result_type is
        if not b(x) then goto label1
        return c(x)
        x := d(x);
        goto label0

Replace if not b(x) with if b(x):

tail_rec(x:instance_type):result_type is
        if b(x) then goto label1
        x := d(x);
        goto label0
        return c(x)

Similar transformations to tail_rec should turn it into the exact same thing.


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