问题
When using formal aspects to create some code is there a generic method of determining a loop invariant or will it be completely different depending on the problem?
回答1:
It has already been pointed out that one same loop can have several invariants, and that Calculability is against you. It doesn't mean that you cannot try.
You are, in fact, looking for an inductive invariant: the word invariant may also be used for a property that is true at each iteration but for which is it not enough to know that it hold at one iteration to deduce that it holds at the next. If I is an inductive invariant, then any consequence of I is an invariant, but may not be an inductive invariant.
You are probably trying to get an inductive invariant to prove a certain property (post-condition) of the loop in some defined circumstances (pre-conditions).
There are two heuristics that work quite well:
start with what you have (pre-conditions), and weaken until you have an inductive invariant. In order to get an intuition how to weaken, apply one or several forward loop iterations and see what ceases to be true in the formula you have.
start with what you want (post-conditions) and strengthen until you have an inductive invariant. To get the intuition how to strengthen, apply one or several loop iterations backwards and see what needs to be added so that the post-condition can be deduced.
If you want the computer to help you in your practice, I can recommend the Jessie deductive verification plug-in for C programs of Frama-C. There are others, especially for Java and JML annotations, but I am less familiar with them. Trying out the invariants you think of is much faster than working out if they work on paper. I should point out that verifying that a property is an inductive invariant is also undecidable, but modern automatic provers do great on many simple examples. If you decide to go that route, get as many as you can from the list: Alt-ergo, Simplify, Z3.
With the optional (and slightly difficult to install) library Apron, Jessie can also infer some simple invariants automatically.
回答2:
It's actually trivial to generate loop invariants. true is a good one for instance. It fulfills all three properties you want:
- It holds before loop entry
- It holds after each iteration
- It holds after loop termination
But what you're after is probably the strongest loop invariant. Finding the strongest loop invariant however, is sometimes even an undecidable task. See article Inadequacy of Computable Loop Invariants.
回答3:
I don't think it's easy to automate this. From wiki:
Because of the fundamental similarity of loops and recursive programs, proving partial correctness of loops with invariants is very similar to proving correctness of recursive programs via induction. In fact, the loop invariant is often the inductive property one has to prove of a recursive program that is equivalent to a given loop.
回答4:
I've written about writing loop invariants in my blog, see Verifying Loops Part 2. The invariants needed to prove a loop correct typically comprise 2 parts:
- A generalisation of the state that is intended when the loop terminates.
- Extra bits needed to ensure that the loop body is well-formed (e.g. array indices in bounds).
(2) is straightforward. To derive (1), start with a predicate expressing the desired state after termination. Chances are it contains a 'forall' or 'exists' over some range of data. Now change the bounds of the 'forall' or 'exists' so that (a) they depend on variables modified by the loop (e.g. loop counters), and (b) so that the invariant is trivially true when the loop is first entered (usually by making the range of the 'forall' or 'exists' empty).
回答5:
There are a number of heuristics for finding loop invariants. One good book about this is "Programming in the 1990s" by Ed Cohen. It's about how to find a good invariant by manipulating the postcondition by hand. Examples are: replace a constant by a variable, strengthen invariant, ...
来源：https://stackoverflow.com/questions/2935295/what-is-the-best-way-of-determining-a-loop-invariant