I'm going over the proof for The Halting Problem in Intro to the Theory of Computation by Sipser and my main concern is about the proof below:
If TM M doesn't know when it's looping (it can't accept or reject which is why a TM is Turing Recognizable for all strings), then how would could the decider H decide if M could possibly be in a loop? The same problem will carry through when TM D does its processing.
After reading this and trying to visualize the proof I came up with this code which is a simplified version of the code in this answer to a related question:
function halts(func) {
// Insert code here that returns "true" if "func" halts and "false" otherwise.
}
function deceiver() {
if(halts(deceiver))
while(true) { }
}
If halts(deceiver) returns true, deceiver will run forever, and if it returns false, deceiver will halt, which contradicts the definition of halts. Hence, the function halts is impossible.
This is a "proof by contradiction", a reductio ad absurdum. (Latin phrases are always good in theory classes... as long as they make sense, of course.)
This program H is just a program with two inputs: a string representing a program for some machine, and an input. For purposes of the proof, you simply assume the program H is correct: it simply will halt and accept if M accepts with w. You don't need to think about how it would do that; in fact, we're about to prove it can't, that no such program H can exist, ...
BECAUSE
if such a program existed, we could immediately construct another program H' that H couldn't decide. But, by the assumption, there is no such program: H can decide everything. So, we're forced to conclude that no program defined as we defined H is possible.
By the way, the reductio method of proof is more controversial than you might expect, considering how often its used, especially in Computer Science. You shouldn't be embarrassed to find it a little odd. The magic term is "non-constructive" and if you feel really ambitious, ask one of your professors about Errett Bishop's critique of non-constructive mathematics.
Related
I recently came across the halting problem contradiction proof.
In the proof, we have to feed the Turing machine a copy of the program and a copy of the input to decide whether that program halts on the input. In the contradiction, why does it have to be the program as the program and the input? Sorry if it sounds confusing. I can simply feed the machine with a program and a random input and come to the same conclusion.
Can anyone tell me why? Is there a specific reason I didn't think of?
First let me come back on the proof itself.
HALT_TM is undecidable
Assume that any machine has a description which takes the form of a string. Let HALT_TM = {<M, w>| M is a TM and M halts on input w}, and A_TM = {<M,w>| M is a TM and accepts w}. Here I assume that we know that A_TM is undecidable (the proof can be done by diagonalization and realizing that as there are more languages than Turing Machines, and as a given TM only decide one language, then some language are not decided).
Assume by contradiction that HALT_TM is decidable, meaning that we dispose of a decider D for this language. Then we would be able to build a machine M which decides A_TM. On input <M', w>, M does the following:
Run D on input <M',w>
If D reject, reject, otherwise run M' on w (until it halts, which we know because D did not reject!)
If M' accepts, accept, if it rejects, reject.
There we see the contradiction with our assumption
Universal Turing Machines
Now the core of your question: you actually feed M any valid machine description M', not necessarily <M> itself. Remember that a TM and "a program" are actually equivalent: See this nice answer for more details. Quoting this same answer: "A Turing Machine is the formal analogue of an algorithm".
One power of Turing Machines is that they can be encoded as a string, allowing another Turing Machine (called an "Universal Turing Machine") to execute them. Because the given machine is an algorithm, you see that you are actually feeding your "top-level" TM a program, and a input of your choice.
Sometimes the value of a variable accessed within the control-flow of a program cannot possibly have any effect on a its output. For example:
global var_1
global var_2
start program hello(var_3, var_4)
if (var_2 < 0) then
save-log-to-disk (var_1, var_3, var_4)
end-if
return ("Hello " + var_3 + ", my name is " + var_1)
end program
Here only var_1 and var_3 have any influence on the output, while var_2 and var_4 are only used for side effects.
Do variables such as var_1 and var_3 have a name in dataflow-theory/compiler-theory?
Which static dataflow analysis techniques can be used to discover them?
References to academic literature on the subject would be particularly appreciated.
The problem that you stated is undecidable in general,
even for the following very narrow special case:
Given a single routine P(x), where x is a parameter of type integer. Is the output of P(x) independent of the value of x, i.e., does
P(0) = P(1) = P(2) = ...?
We can reduce the following still undecidable version of the halting problem to the question above: Given a Turing machine M(), does the program
never stop on the empty input?
I assume that we use a (Turing-complete) language in which we can build a "Turing machine simulator":
Given the program M(), construct this routine:
P(x):
if x == 0:
return 0
Run M() for x steps
if M() has terminated then:
return 1
else:
return 0
Now:
P(0) = P(1) = P(2) = ...
=>
M() does not terminate.
M() does terminate
=> P(x) = 1 for a sufficiently large x
=> P(x) != P(0) = 0
So, it is very difficult for a compiler to decide whether a variable actually does not influence the return value of a routine; in your example, the "side effect routine" might manipulate one of its values (or even loop infinitely, which would most definitely change the return value of the routine ;-)
Of course overapproximations are still possible. For example, one might conclude that a variable does not influence the return value if it does not appear in the routine body at all. You can also see some classical compiler analyses (like Expression Simplification, Constant propagation) having the side effect of eliminating appearances of such redundant variables.
Pachelbel has discussed the fact that you cannot do this perfectly. OK, I'm an engineer, I'm willing to accept some dirt in my answer.
The classic way to answer you question is to do dataflow tracing from program outputs back to program inputs. A dataflow is the connection of a program assignment (or sideeffect) to a variable value, to a place in the application that consumes that value.
If there is (transitive) dataflow from a program output that you care about (in your example, the printed text stream) to an input you supplied (var2), then that input "affects" the output. A variable that does not flow from the input to your desired output is useless from your point of view.
If you focus your attention only the computations involved in the dataflows, and display them, you get what is generally called a "program slice" . There are (very few) commercial tools that can show this to you.
Grammatech has a good reputation here for C and C++.
There are standard compiler algorithms for constructing such dataflow graphs; see any competent compiler book.
They all suffer from some limitation due to Turing's impossibility proofs as pointed out by Pachelbel. When you implement such a dataflow algorithm, there will be places that it cannot know the right answer; simply pick one.
If your algorithm chooses to answer "there is no dataflow" in certain places where it is not sure, then it may miss a valid dataflow and it might report that a variable does not affect the answer incorrectly. (This is called a "false negative"). This occasional error may be satisfactory if
the algorithm has some other nice properties, e.g, it runs really fast on a millions of code. (The trivial algorithm simply says "no dataflow" in all places, and it is really fast :)
If your algorithm chooses to answer "yes there is a dataflow", then it may claim that some variable affects the answer when it does not. (This is called a "false positive").
You get to decide which is more important; many people prefer false positives when looking for a problem, because then you have to at least look at possibilities detected by the tool. A false negative means it didn't report something you might care about. YMMV.
Here's a starting reference: http://en.wikipedia.org/wiki/Data-flow_analysis
Any of the books on that page will be pretty good. I have Muchnick's book and like it lot. See also this page: (http://en.wikipedia.org/wiki/Program_slicing)
You will discover that implementing this is pretty big effort, for any real langauge. You are probably better off finding a tool framework that does most or all this for you already.
I use the following algorithm: a variable is used if it is a parameter or it occurs anywhere in an expression, excluding as the LHS of an assignment. First, count the number of uses of all variables. Delete unused variables and assignments to unused variables. Repeat until no variables are deleted.
This algorithm only implements a subset of the OP's requirement, it is horribly inefficient because it requires multiple passes. A garbage collection may be faster but is harder to write: my algorithm only requires a list of variables with usage counts. Each pass is linear in the size of the program. The algorithm effectively does a limited kind of dataflow analysis by elimination of the tail of a flow ending in an assignment.
For my language the elimination of side effects in the RHS of an assignment to an unused variable is mandated by the language specification, it may not be suitable for other languages. Effectiveness is improved by running before inlining to reduce the cost of inlining unused function applications, then running it again afterwards which eliminates parameters of inlined functions.
Just as an example of the utility of the language specification, the library constructs a thread pool and assigns a pointer to it to a global variable. If the thread pool is not used, the assignment is deleted, and hence the construction of the thread pool elided.
IMHO compiler optimisations are almost invariably heuristics whose performance matters more than effectiveness achieving a theoretical goal (like removing unused variables). Simple reductions are useful not only because they're fast and easy to write, but because a programmer using a language who understand basics of the compiler operation can leverage this knowledge to help the compiler. The most well known example of this is probably the refactoring of recursive functions to place the recursion in tail position: a pointless exercise unless the programmer knows the compiler can do tail-recursion optimisation.
If I have code for some function f (that takes in one input for simplicity), I need to decide if the input x affects the output f(x), i.e, if f is a constant function defined below.
Define f to be constant function if output of f is invariant w.r.t x. This should hold for ALL inputs. So for example, if we have f(x) = 0 power x, it may output 0 for all inputs except for x = 0, where it may output error. So f is not a constant function.
I can only do static analysis of the code and assume the code is Java source for simplicity.
Is this possible?
This is obviously at least as hard as solving the Halting Problem (proof left as an exercise), so the answer is "no", this is not possible.
It is almost certainly possible. In most cases. Where there aren't weird thing going on.
For normal functions, the ordinary, useful kind that actually return values rather than doing their own little thing, yes.
For a simple function, not recursive, no nastiness of that sort, doing it manually, I would probably make the static-analysis equivalent of a sign chart, where I examine the code and determine every value of x that might possibly be a boundary condition or such (e.g. the code has if (x < 0) somewhere in it, so I check the function for values of x near 0). If this sort of attempt is doomed to fail please tell me before I try to use it on something.
Using brute force to grind away at it could work, unless you are working with quadruple precision x values or something similarly-sized, because then brute force could take years. Although at that point its not really static-analysis anymore.
Static-analysis generally really means having a computer tell you by looking at the code, not you looking at it yourself (at least not very much). Algorithms exist for doing this in many languages, wikipedia has such a list, including some free or even open source.
The ultimate proof that something can be done is for it to have been done already.
Since you'd call a non-terminating function non-constant, here's the reduction from your problem to the halting problem:
void does_it_halt(...);
int f(int x) {
if(x == 1) {
does_it_halt();
}
return 0;
}
Asking if f is constant is equivalent to asking if does_it_halt halts. Therefore, what you're asking for is impossible, since the halting problem is undecidable.
I have a homework assignment, and i am finished other then one question (see title)
For the life of my, i cannot figure this out... so i started to think it was a trick question.
the current answer that i will submit is:
L1 = {a^n b^n: n>=1} is deterministic. And the reverse,
L2 = {b^n a^n: n>=1} is also deterministic.
However, since all deterministic languages are a subset of Non-deterministic languages, L2 can be considered non-deterministic.
On a side note, the only other example i was trying to make work is:
L3= {{a,b}a}
This seems possible because forward there is non-determinism, since the input could be either a, or b as long as its followed by an a.
and in reverse there is determinism since it will accept only an 'a'. But, it introduces new non-determinism since the second input could be either a or b.
any help / guidance would be great.
I know the deadline has passed, but somebody might find this useful in the future.
(a+b+c)*WcW^R, where W is in (a+b)+; this is non-deterministic because you don't know where the "WcW" bit starts.
W^RcW(a+b+c)*, where W is in (a+b)+; this is deterministic because you can write a deterministic PDA to accept simple palindromes of the form "W^RcW" and modify the accepting state to loop to itself on any of a, b and c.
The trick here is that PDAs have to read input from left to right.
I am a Mechanical engineer with a computer scientist question. This is an example of what the equations I'm working with are like:
x = √((y-z)×2/r)
z = f×(L/D)×(x/2g)
f = something crazy with x in it
etc…(there are more equations with x in it)
The situation is this:
I need r to find x, but I need x to find z. I also need x to find f which is a part of finding z. So I guess a value for x, and then I use that value to find r and f. Then I go back and use the value I found for r and f to find x. I keep doing this until the guess and the calculated are the same.
My question is:
How do I get the computer to do this? I've been using mathcad, but an example in another language like C++ is fine.
The very first thing you should do faced with iterative algorithms is write down on paper the sequence that will result from your idea:
Eg.:
x_0 = ..., f_0 = ..., r_0 = ...
x_1 = ..., f_1 = ..., r_1 = ...
...
x_n = ..., f_n = ..., r_n = ...
Now, you have an idea of what you should implement (even if you don't know how). If you don't manage to find a closed form expression for one of the x_i, r_i or whatever_i, you will need to solve one dimensional equations numerically. This will imply more work.
Now, for the implementation part, if you never wrote a program, you should seriously ask someone live who can help you (or hire an intern and have him write the code). We cannot help you beginning from scratch with, eg. C programming, but we are willing to help you with specific problems which should arise when you write the program.
Please note that your algorithm is not guaranteed to converge, even if you strongly think there is a unique solution. Solving non linear equations is a difficult subject.
It appears that mathcad has many abstractions for iterative algorithms without the need to actually implement them directly using a "lower level" language. Perhaps this question is better suited for the mathcad forums at:
http://communities.ptc.com/index.jspa
If you are using Mathcad, it has the functionality built in. It is called solve block.
Start with the keyword "given"
Given
define the guess values for all unknowns
x:=2
f:=3
r:=2
...
define your constraints
x = √((y-z)×2/r)
z = f×(L/D)×(x/2g)
f = something crazy with x in it
etc…(there are more equations with x in it)
calculate the solution
find(x, y, z, r, ...)=
Check Mathcad help or Quicksheets for examples of the exact syntax.
The simple answer to your question is this pseudo-code:
X = startingX;
lastF = Infinity;
F = 0;
tolerance = 1e-10;
while ((lastF - F)^2 > tolerance)
{
lastF = F;
X = ?;
R = ?;
F = FunctionOf(X,R);
}
This may not do what you expect at all. It may give a valid but nonsense answer or it may loop endlessly between alternate wrong answers.
This is standard substitution to convergence. There are more advanced techniques like DIIS but I'm not sure you want to go there. I found this article while figuring out if I want to go there.
In general, it really pays to think about how you can transform your problem into an easier problem.
In my experience it is better to pose your problem as a univariate bounded root-finding problem and use Brent's Method if you can
Next worst option is multivariate minimization with something like BFGS.
Iterative solutions are horrible, but are more easily solved once you think of them as X2 = f(X1) where X is the input vector and you're trying to reduce the difference between X1 and X2.
As the commenters have noted, the mathematical aspects of your question are beyond the scope of the help you can expect here, and are even beyond the help you could be offered based on the detail you posted.
However, I think that even if you understood the mathematics thoroughly there are computer science aspects to your question that should be addressed.
When you write your code, try to make organize it into functions that depend only upon the parameters you are passing in to a subroutine. So write a subroutine that takes in values for y, z, and r and returns you x. Make another that takes in f,L,D,G and returns z. Now you have testable routines that you can check to make sure they are computing correctly. Check the input values to your routines in the routines - for instance in computing x you will get a divide by 0 error if you pass in a 0 for r. Think about how you want to handle this.
If you are going to solve this problem interatively you will need a method that will decide, based on the results of one iteration, what the values for the next iteration will be. This also should be encapsulated within a subroutine. Now if you are using a language that allows only one value to be returned from a subroutine (which is most common computation languages C, C++, Java, C#) you need to package up all your variables into some kind of data structure to return them. You could use an array of reals or doubles, but it would be nicer to choose to make an object and then you can reference the variables by their name and not their position (less chance of error).
Another aspect of iteration is knowing when to stop. Certainly you'll do so when you get a solution that converges. Make this decision into another subroutine. Now when you need to change the convergence criteria there is only one place in the code to go to. But you need to consider other reasons for stopping - what do you do if your solution starts diverging instead of converging? How many iterations will you allow the run to go before giving up?
Another aspect of iteration of a computer is round-off error. Mathematically 10^40/10^38 is 100. Mathematically 10^20 + 1 > 10^20. These statements are not true in most computations. Your calculations may need to take this into account or you will end up with numbers that are garbage. This is an example of a cross-cutting concern that does not lend itself to encapsulation in a subroutine.
I would suggest that you go look at the Python language, and the pythonxy.com extensions. There are people in the associated forums that would be a good resource for helping you learn how to do iterative solving of a system of equations.