Instantly share code, notes, and snippets. jameshfisher/halting_problem_javascript.md Last active Sep 7, 2017

A proof that the Halting problem is undecidable, using JavaScript and examples

Having read a few proofs that the halting problem is undecidable, I found that they were quite inaccessible, or that they glossed over important details. To counter this, I've attempted to re-hash the proof using a familiar language, JavaScript, with numerous examples along the way.

This famous proof tells us that there is no general method to determine whether a program will finish running. To illustrate this, we can consider programs as JavaScript function calls, and ask whether it is possible to write a JavaScript function which will tell us whether a function call will ever return.

Programs, modelled as JavaScript function calls, consist of a JavaScript anonymous function and a list of arguments in parentheses. Anonymous functions look like (function (x, y) { return x + y == 3; }), or (function (x) { return x.length > 0; }). They can take as many parameters as they like. Arguments look like ('foo', 'bar'), or ('blah'). Putting an anonymous function and some function arguments together, we get a function call, like (function (x) { return 0 < x; })(1). This function call is a program, and when run, it returns true, since 0 < 1.

Notice that both JavaScript functions and function arguments can be encoded as strings. The function (function (x, y) { return x + y == 3; }) is simply encoded as the JavaScript string "(function (x, y) { return x + y == 3; })", and the input ('foo', 'bar') is encoded as "('foo', 'bar')".

A function can be run on an input by encoding both as strings, concatenating them, and running eval on that string. For example, to run the function (function (x) { return x == 0; }) on on arguments (3), we encode them as strings "(function (x) { return x == 0; })" and "(3)", concatenate them to get "(function (x) { return x == 0; })(3)", and run eval("(function (x) { return x == 0; })(3)"), which will evaluate to false.

However, eval does not always evaluate to a value. We might pass eval a string that is not syntactically correct (say, "(+("), in which case eval always notices and informs us that the string has a syntax error.

More perniciously, if we define the program badly, the interpreter might not evaluate to anything at all: it will just keep running forever. The simplest example would be a string-encoded program like "(function () { while (true); })()", which will make eval run forever. eval does not notice when we give it a program that will run forever; it simply runs it forever.

Therefore, eval will always do one of the following things:

• complain about a syntax error
• return a value, such as false
• run forever

If eval does not run forever (i.e. it has a syntax error, or it returns a value), we say that it halts. Now, wouldn't it be nice if eval complained when it is not going to halt, just like it complains if we give it a program with a syntax error? Let's think about how we could get it to do that.

We want to write a function h = function (funcString, argString) { ... } which, when given a string-encoded function funcString and string-encoded arguments argString, tells us whether eval(funcString + argString) would halt. For example, h("(function (x,y) { return x < y; })", "(5,3)") must return true because eval would halt, returning false. Also, h("(", "((") must return true, because there is a syntax error. But h("(function () { while (true); })", "()") must return false, because it would cause eval to loop.

How might we implement h? We might start by checking for syntax errors: we can define a parser and parse the strings; if they don't parse, we return true. After that, we need to check whether the parsed program would halt. We could perform simple checks: if the program just has a single return statement, it must halt, so return true. We could perform more intelligent checks: if the program has no loops, it must get to the end of the program and halt; so we return true. We could check if all the loops look like for (var i = 0; i < n; i++) { ... }; if they do, then all those loops will end when i reaches n, and the program will halt when all the looping ends; so we return true. We could even run the program for a little while, and if it halts, then we can return true.

But when can we return false? If we just return false after doing all the checks we can think of, then we might incorrectly identify some halting inputs as not halting. So what we really need is a single, unified way to check whether it halts.

It's fun to think of more powerful ways that we could check for halting. Sadly, though, Alan Turing burst this bubble in 1936, by showing that however you write h, it will always have a bug. Whatever your h looks like, there will either be arguments to it which cause it to incorrectly return true (i.e., your h says it halts, but actually it doesn't), or cause it to incorrectly return false (i.e., your h says it doesn't halt, but actually it does), or cause it to go into a loop (which is a bug).

His proof works by contradiction: assume that you have written a correct version of h, and then show that this leads to an absurdity. Specifically, and strangely, we can show that if your version of h is correct, then it must have a bug! From this reasoning, it follows that your h must always have a bug: if it does have a bug, it obviously has a bug, but if it doesn't have a bug, then, well, it must have a bug!

So let's assume you've written a correct h, which looks like function (funcString, argString) { ... }. Then we can define another function, g, which looks like this:

var h = function (funcString, argString) { ... }; // Your correct solution for h

var g = function (funcString) {
// Notice we pass in funcString as the *arguments* string as well as the function string!
if (h(funcString,funcString)) {
while (true);
}
else {
return false;
}
}

So g is a function which takes a string-encoded function as an argument. Here's what g(funcString) does:

• if funcString(funcString) halts, g loops.
• otherwise, g returns false.

Passing a function to itself is a little mind-bending, so let's work through some examples. What would g("(function (x) { return 2; })") do? Well, it first passes that string to h, which tells us whether this halts:

eval("(function (x) { return 2; })(function (x) { return 2; })")

Try it for yourself: it halts, and returns 2. So h would tells us that this halts, i.e. h returns true. Because of this, g then goes into an infinite loop.

Now let's try g("(function () { while (true); })"). This program would simply loop, ignoring its arguments. Specifically, when passed itself as an argument, it ignores it, and loops forever. So h("(function () { while (true); })", "(function () { while (true); })") would return false. In this case, g returns false.

The function g, like h and everything else, can be written as a string:

var gString = "(function (funcString) { if (h(funcString,funcString)) { while (true); } else { return false; }})";

What is the point of this nonsense function g? Here's where it gets really interesting. A crazy thought: what happens if we run g(gString)? That is, what happens when we call g with the string-encoded version of itself as its own argument? Well, running through it, g passes the string to h as both arguments, like so:

if (h(gString,gString)) {
while (true);
}
else {
return false;
}

Now h either returns true or returns false. It turns out that, in either case, we get a strange contradiction:

• Assume h(gString,gString) returns true. Then, because our definition of h is correct, h is is telling us that eval(gString + gString) halts. What is eval(gString + gString)? It is the same as g(gString). Therefore, h is telling us that g(gString) halts.

But look at the definition of g: when h returns true, g loops! So g(gString) does not halt, i.e., h has given us the wrong answer!

• Assume h(gString,gString) returns false. Then h is is telling us that eval(gString + gString), i.e. g(gString), does not halt.

But by the definition of g, when h returns false, g returns false, i.e. it halts! So g(gString) does halt, and, h has given us the wrong answer!

So whether h returns true or false, it gives us the wrong answer. But this was all based on the assumption that our definition of h was correct: so if h is correct, then we can define a string gString such that h(gString,gString) is incorrect. Therefore, writing a correct version of h is impossible, and this is the halting problem!

oberhamsi commented Dec 24, 2013

 nice one, thanks. the only think i find mildly confusing is that you define h inside of g, which is an unnecessary mind bend :)

jrab89 commented Dec 24, 2013

 very clear explanation, thanks
Owner Author

jameshfisher commented Dec 24, 2013

 @oberhamsi fair enough, you're right that that's incidental; I've moved it above the definition of g :-)

chisophugis commented Dec 25, 2013

 For those interested in some intuition about why the halting problem is undecidable, consider that an algorithm for the halting problem would allow you to "short-circuit" infinite computations, to effectively do an infinite amount of work in a finite amount of time. For example, consider the Collatz Conjecture. It's easy to write a program that just tests all positive integers (all infinitely many of them) looking for a counterexample, and halts if it finds a counterexample. Then you check whether the program halts: If the program halts, then it found a counterexample and the conjecture is false If the program doesn't halt, then there are no counterexample, so the conjecture is true. In general, any problem that can be described by a computer program would be solvable in finite time with an algorithm for the halting problem, whether or not the program would have to do an infinite amount of work to solve the problem. Every problem in mathematics could be solved by an algorithm for the halting problem, since: it is easy to write a computer program that enumerates all possible sequences of bytes a proof can be described by a sequence of bytes (there are various formal languages for writing proofs) a proof can be checked by a computer for correctness Then you can write a program that generates and tests all proofs for a given mathematical statement, and halts if it finds a correct proof. You then just run the algorithm for the halting problem on this program. If it halts, then the statement is true, if not, then the statement is false. The way I like to think of the undecidability of the halting problem is as a statement about computer programs. It is basically telling you that in general you can't know what a program will do without running it. There is in general no way to "summarize" the behavior of a computer program. Also, keep in mind that it's really easy to reduce most questions about the behavior of a computer program to the halting problem (hence those problems are just as hard, i.e. undecidable). For example "does f() return true?" is just as hard as the halting problem (if (f()) { infiniteLoop(); } else { halt(); }). This is the so-called "full employment theorem for compiler writers".
to join this conversation on GitHub. Already have an account? Sign in to comment
You can’t perform that action at this time.