I'm new to Dafny and learning it. For the following Dafny code
method Dummy(s: seq<nat>) returns (ret: nat)
requires forall k :: 0 <= k < |s| ==> s[k] > 0
{
ret := 0;
var se := s;
while |se| > 0
{
// Pick a random index of the seq.
var i :| 0 <= i < |se|;
if se[i] == 1 {
// Remove the seq element if it is 1.
se := se[..i] + se[i + 1..];
} else {
// Replace the seq element with 3 occurences of 1.
se := [1, 1, 1] + se[..i] + se[i + 1..];
}
}
return;
}
I'm seeing complaints which are
decreases |se| - 0Dafny VSCode
cannot prove termination; try supplying a decreases clause for the loopDafny VSCode
I know that to prove termination, generally I should provide a decreases clause. I just cannot find what to put in the decreases clause in this case. Simply putting
decreases |se|
there will not work, as in the else part of the if statement the seq size may actaully increase.
If this is a pen and paper proof, I would like to argue that for any element that is 1, it will be removed; and for any element that is greater than 1, it will be replaced by 3 occurences of 1, which will still be removed finally. Hence se will be empty in the end and loop terminates.
How do I turn it to codes that Dafny would agree?
The key is to introduce a function that I call "total potential" that computes how many iterations the loop will take to process the list. Its definition is "for every 1 in the list, add 1 to the total potential, for every non-1 in the list, add 4 to the total potential". Then, we will make the decreases clause the total potential. This way, when a 1 gets removed, the potential goes down by 1, and when a non-1 gets removed and replaced by three ones, the potential also goes down by 1.
There is some additional work to convince Dafny that the total potential actually decreases on each loop iteration. You will need a few lemmas about how the total potential relates to slicing and appending sequences.
Related
Hi for teaching I am setting up a mass of simple dafny questions. Mostly going fine, but... Either I have missed some detail about loop invariants in Dafny or this is a weakness/bug?
method whS(a:int) returns ()
{ var i:int := 0;
while (i<a )
decreases a-i
invariant i<= a
{ assert i<= a;
i:=i+1;
assert i<= a;
}
}
The method fails to verify with error "invariant might not hold on entry" Yet with the invariant commented out the method verifies even with the assertions.
A good way to remember the answer to the question "when must the loop invariant be true?" in Dafny is "whenever the loop condition would be evaluated".
So the loop invariant needs to be true before the first iteration of the loop, even if the body would never execute, because the loop condition will be evaluated to determine whether to enter the loop at all. The invariant also needs to be true after the last iteration, because again, the loop condition will be evaluated to determine whether to execute another iteration.
In your example, if the argument a to the method is negative, then the loop invariant is false before the first time the loop condition is evaluated. An example of an invariant that would work for this loop would be something like a >= 0 ==> i <= a.
many thanks James
I can see that I misunderstood how dafny implemented loop invariants. Wrapping the while in an if statement now verifies and gives me the semantics that I thought while on its own had.
method whS(a:int) returns ()
{ var i:int := 0;
if (i<a) {
while (i<a )
decreases a-i
invariant i<= a
{ i:=i+1;
}
}
}
Thanks again james . I will avoid this mistake when I give my lecture on Tuesday.
I am new to Dafny, and I am trying to write a code to computer 5*m-3*n without using *.
Can anybody tell me what wrong with my code? I think it is the invariant and decrease.
method CalcTerm(m: int, n: nat) returns (res: int)
ensures res == 5*m-3*n;
{
var m1: nat := abs(m);
var n1: nat := n;
res := 0;
while (m1!=0)
invariant m1>=0
decreases m1
{
res := res+5;
m1 := m1-1;
}
if (m<0) { res := -res; }
while (n1!=0)
invariant n1 >= 0
decreases n1
{
res := res-3;
n1 := n1-1;
}
}
But it keep saying that:
A postcondition might not hold on this return path. 29 2
You're right that the issue has to do with loop invariants. I recommend reading the two sections of the Guide on Assertions and Loop Invariants.
Dafny "summarizes" the effect of a loop using only its invariants. So after the second loop in your method, Dafny will only know that n1 >= 0 and that the loop has terminated, so actually n1 == 0. This is not enough information to prove your postcondition: you need a stronger invariant. Here's one that might help you make progress
invariant res == 5 * m - 3 * (n - n1)
This invariant computes the value of res in terms of how many iterations of the loop have executed so far (n - n1). If you add this invariant to the second loop, you'll get a new error (progress!) that says it might not hold on entry. This means that Dafny is able to prove your postcondition, but not able to establish that the new invariant is true after the first loop is done. This is again because the invariant on the first loop is too weak.
Maybe this gives you enough information to try coming up with another invariant for the first loop on your own. Feel free to ask more questions here if you get stuck.
What I want to do in the method is simply to overwrite the previous array and fill it with number that are sorted, however dafny says that the post-condition does not hold and I can't for the life of me figure out why.
I'm guessing I need to add some invariant to the loop but since they are checked before entering the loop I don't know how to put a valid invariant.
method sort2(a : array<int>)
modifies a;
requires a != null && a.Length >= 2;
ensures sorted2(a[..]);
{
var i := a.Length - 1;
while (i >= 0)
decreases i
{
a[i] := i;
i := i - 1;
}
}
predicate sorted2(s: seq<int>)
{
forall i :: 1 <= i < |s| ==> s[i] >= s[i-1]
}
My previous attempt was just to reinitialize a but dafny apparently doesn't allow that inside methods.
You do need a loop invariant in order to prove this program correct.
See Section 6 of the Dafny Tutorial for a general introduction to loop invariants and how to come up with them.
In this case, a good loop invariant will be something like sorted(a[i+1..]), which says that the part of the array after index i is sorted. That way, when the loop terminates and i is zero, you'll know that the whole array is sorted.
You will also need one or two more loop invariants describing bounds on i and the elements of the array itself.
I'd like to write a while() loop in Gforth. Unfortunately, the only tutorial online isn't useful due to a lack of examples, and examples on counted loops (what I'm not looking for) appear fundamentally different.
What are some concrete examples of how to represent something like this?
while (x > 3) { print(x); x--; }
Or really, just some concrete way to represent anything of the form:
while (predicate) { expression(s) }
Your first piece of code translates to:
\ Assuming x is on the top of the stack.
begin dup 3 > while dup . 1- repeat
\ Or if x is in memory.
begin x # 3 > while x ? -1 x +! repeat
And the second:
begin predicate while expressions repeat
in fibonacci series let's assume nth fibonacci term is T. F(n)=T. but i want to write a a program that will take T as input and return n that means which term is it in the series( taken that T always will be a fibonacci number. )i want to find if there lies an efficient way to find it.
The easy way would be to simply start generating Fibonacci numbers until F(i) == T, which has a complexity of O(T) if implemented correctly (read: not recursively). This method also allows you to make sure T is a valid Fibonacci number.
If T is guaranteed to be a valid Fibonacci number, you can use approximation rules:
Formula
It looks complicated, but it's not. The point is: from a certain point on, the ratio of F(i+1)/F(i) becomes a constant value. Since we're not generating Fibonacci Numbers but are merely finding the "index", we can drop most of it and just realize the following:
breakpoint := f(T)
Any f(i) where i > T = f(i-1)*Ratio = f(T) * Ratio^(i-T)
We can get the reverse by simply taking Log(N, R), R being Ratio. By adjusting for the inaccuracy for early numbers, we don't even have to select a breakpoint (if you do: it's ~ correct for i > 17).
The Ratio is, approximately, 1.618034. Taking the log(1.618034) of 6765 (= F(20)), we get a value of 18.3277. The accuracy remains the same for any higher Fibonacci numbers, so simply rounding down and adding 2 gives us the exact Fibonacci "rank" (provided that F(1) = F(2) = 1).
The first step is to implement fib numbers in a non-recursive way such as
fib1=0;fib2=1;
for(i=startIndex;i<stopIndex;i++)
{
if(fib1<fib2)
{
fib1+=fib2;
if(fib1=T) return i;
if(fib1>T) return -1;
}
else
{
fib2+=fib1;
if(fib2=T) return i;
if(fib2>t) return -1;
}
}
Here startIndex would be set to 3 stopIndex would be set to 10000 or so. To cut down in the iteration, you can also select 2 seed number that are sequential fib numbers further down the sequence. startIndex is then set to the next index and do the computation with an appropriate adjustment to the stopIndex. I would suggest breaking the sequence up in several section depending on machine performance and the maximum expected input to minimize the run time.