Refreshing my memory of the halting problem for the thread in the math forum led me to pages about Rice's Theorem, and it took me a while to wrap my head around that proof so I want to check here if I basically understand what's going on.

First of all, we know from the halting problem proof that there can be no program H(n) which always determines whether program N halts with input n. (I'm going to use capital letters for programs and lower case letters for the strings representing those programs. Also all programs take one input, which I don't always specify.)

So then suppose we have a program P(n) which determines whether (the partial function computed by) program N has some non-trivial property (i.e. a property that at least one program has and at least one program doesn't have). P outputs 1 for inputs with the property and 0 for inputs without the property.

We can assume wlog that P(n)=0 when N doesn't halt on any input. We can make this assumption because P(n) is equivalent to Q(n)=1-P(n) in the sense that they partition the set of programs into the same two subsets, so we can use whichever one outputs 0 for never-halting programs.

Because the property determined by P is non-trivial, let a be a string such that P(a)=1.

For any n, we can construct a program TN(m) which

1) Runs N(n)

2) Computes A(m)

(TN is a different program for each n, which takes m as its only input.)

But then if we define H(n) so it

1) Constructs tn as (the string corresponding to the program) described above

2) Outputs P(tn)

then H(n) is a halting oracle.

---

Is that basically correct?

I was getting confused by the first step of H, I think because I was picturing a Turing machine rather than a program written in a more human readable language. I think I get it now that I imagine H just concatenating the relevant strings of programming language.

## Rice's Theorem comprehension check

**Moderators:** phlip, Moderators General, Prelates

- Xanthir
- My HERO!!!
**Posts:**5215**Joined:**Tue Feb 20, 2007 12:49 am UTC**Location:**The Googleplex-
**Contact:**

### Re: Rice's Theorem comprehension check

I think so, yes. The proof sketch in Wikipedia makes it particularly clear to me - if you claim to have a program that can tell whether another program squares its numeric input, then I can feed it a program that does some arbitrary computation, throws the result away, then returns the square of its input. The only thing that can stop my program from returning the square of its input is if the arbitrary computation never halts; thus, your algorithm must be able to tell whether the arbitrary computation loops or halts in order to give a proper answer. It's thus trivially obvious that it must be a Halting Oracle, and is thus can't exist, per the Halting Theorem.

(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

- gmalivuk
- GNU Terry Pratchett
**Posts:**25789**Joined:**Wed Feb 28, 2007 6:02 pm UTC**Location:**Here and There-
**Contact:**

### Re: Rice's Theorem comprehension check

Yeah I saw the wiki page, but was still confused until I typed it up in my own words here.

### Who is online

Users browsing this forum: No registered users and 3 guests