Not even a single mention of the halting problem?
In case other HNers haven't read Scooping the Loop Snooper, I (re)submitted it:
Due to the fact that a program is a composition of programs which at the lowest level are loops and instructions this also means that it is impossible to determine whether or not a loop halts.
It is easily provable that this is theoretically possible, but the required time is too long for it to be applicable. Suppose your memory has N states. Run your program through N+1 computation steps and if it's not already finished, then it will never halt.
Thing is, N is very large for real memory, so this straightforward approach is not applicable. However, there's much room to improve this algorithm and I assume that's what the MIT folks did.
True halting problem is about computational devices with infinite memory.
Let's consider both computers (yours and 3rd party) as a single computational device, with a state represented by a pair of symbols, where the first represents your computer and the second the 3rd party's. Suppose your computer has N states and the other M. It's enough to run N * M + 1 computation steps -- if your program hasn't ended so far, it will never, for there's a certain state (a, b) that must have occured twice, by the pigeonhole principle.
Please keep in mind that this is all purely theoretical, and not at all practical. It's just a counterargument to other purely theoretical argument from unsolvability of halting problem.
def is_collatz(n):
s = set()
while n not in (1, 2, 4):
if n % 2:
n = 3 * n + 1
else:
n /= 2
if n in s: # Non-Collatz loop
return False
s.add(n)
return True
Secondly, you're making the logic error A => B implies not A => not B. If state spaces as large as modern memory limits allow are too large, that doesn't imply that state spaces far below modern memory limits are small enough to handle.
Thirdly, your program trivially terminates doing nothing.
Pretend that an OOM condition is the same as a False answer. :3
Here's a program, building on the previous one, which either terminates or doesn't.
from itertools import count
for i in count(1):
if not is_collatz(i):
breakReal-life computers have bounded memory, so they don't have any more computational power than, say, regular expressions matchers -- every computational task performed by real computer can be represented by appropriately complicated regular expression. That's why one needs to be especially careful when arguing from computability when talking about real-life problems.
The more serious issue is a treatment of I/O.
I can tell you now that your code terminates. The set s is strictly increasing in size, inevitably going towards an OOM condition if the loop doesn't terminate by other means. Note that we are talking about determining halting or not halting, not with which value the program halts.
This is not a troll. You tried to respond with something witty. I just pointed out that in this case things like the collatz conjecture don't apply, because it is a conjecture over all natural numbers. When you are working with finite memory, there is no way to encode this conjecture into a program (e.g. by making it halt if the conjecture is true and making it loop when the conjecture is false).
What few people realize, is that most completely arbitrary programs are very uninteresting. More interesting programs have characteristics that would very often allow to statically prove that all of their loops that should terminate indeed will terminate. And even interesting loops for which it would not be in the first time possible to determine that often gain in clarity from being slightly changed so that the tool can do its work.
IIRC MS has a tool that does sort of that for windows drivers.
Given a program and an input, determine whether that program, given that input, will ever halt, or if it will loop infinitely.
One of the fundamentals of academic computer science is learning that it is mathematically impossible to solve the halting problem - which is why the grandparent comment simply notes "the halting problem" with no further explanation; it's a famous problem. (Don't feel bad, everyone has to learn something the first time!)
The poem linked elsewhere in this comment tree is a simple and pithy description of the mathematical proof.
For (many) more details, Wikipedia has plenty of information.
A couple accessible examples:
1. Constant Propagation - we know that we'll never detect all compile-time constants because of Rice.
2. Exception frequency - I can write a function that has "raise new FooException()" in the code, but the function never raises, and you won't be able to prove it never raises.
However, the actual halting problem is moot in most of the references I see to it, as the halting problem is usually misused by the half-educated to make claims that 'nothing useful can ever be inferred about the behavior of any program, ever'.
For example, we built a tool to either estimate a reasonable ceiling on the stack usage (whole-program) or give up; it works very well and can detect when the structure of the program prevents this tool from operating correctly (nasty stack mischief or stack adjustments appearing in 'unusual' places). Quite useful for us, especially because we control the source code and know that we don't do naughty things to our own stack.
Entertainingly, on the way to build this tool, we encountered at a couple screeds about how this problem just plain can't be solved because it's equivalent to 'solving the halting problem'. Well, yes - in theory, but no - in practice.
We have a 'little bit' of recursion where we know that we won't go around the recursion more than once and we've hand-hacked that it. We have similar hacks to deal with indirect calls.
I'm not saying we've solved the problem for arbitrary code; I'm saying that solutions that fall very far short of solving problems for arbitrary codes are still enormously useful. Many compiler optimizations don't work for 'arbitrary code' either, for example, and know just enough to bail out when they see an irreducible flowgraph or a bizarre indirect jump, etc.
I had to write software which proved a method optionally yielded, which isn't just undecidable, but unrecognizable. For non-CS folks, that means it's actually harder than the halting problem. Yet my software works on nearly all Ruby code I throw at it - it only really has difficulty on complex delegation patterns. Unsurprisingly, the code people write to create Block-Optional methods is highly structured.