Seemingly impossible functional programs (2007)

http://math.andrej.com/2007/09/28/seemingly-impossible-functional-programs/

by ogogmad • 6 years ago

87 23 6 years ago

math.andrej.com

amicin• 6 years ago

I took Martin Escardo’s course on functional programming at university. Nothing but positive things to say about the man. A fantastic teacher with a real care for his students.

Makes me happy to see him on HN.

m_j_g• 6 years ago

You may be interested in his brand new course on HoTT in Agda

https://www.cs.bham.ac.uk/~mhe/HoTT-UF-in-Agda-Lecture-Notes...

cvoss2• 6 years ago

I think I have a half-baked understanding of what's going on here. Let me talk through it.

Take a function f : unit_interval -> bool. I believe that if there is no bound on the execution time of f, then f must not be a total function. (So conversely, if f is total, then it has bounded execution time, and is therefore only considering prefixes of the digit streams of up to some bounded length. From there, searching for an example input x s.t. f(x) = true is easy.)

Naively, it would seem plausible that there exists a total function which, nevertheless, takes longer and longer to run on a particular sequence of inputs. So let such a sequence be given, where the runtime of f(x_i) tends to infinity, even though it is finite for each x_i.

But from topology, we know that a sequence of points in a compact space, like the unit interval, has a convergent subsequence. The interesting thing happens when you look at f(y) where y is the limit point of such a subsequence. It is the case that as i increases, x_i and y increasingly look the same for longer and longer prefixes of their decimal (binary) expansions. Therefore, f has to do more and more work to tell apart x_i and y, which gives us an argument that there is no bound on the runtime of f(y); i.e., f doesn't terminate for that particular input. (Except, you have to get nitpicky about redundant representations of reals like 0.0099999... vs 0.010000..., but we can always choose the representation of y that suits us: the one that looks more like the x_i's.)

contravariant2• 6 years ago

Yeah I think there are some small problems with your argument, unless it was only meant to illustrate the idea.

For the sake of argument let's say we simply input both decimal representations of each number. A program that simply tries to find the first decimal that isn't 9 for a number in [0,1] would take an arbitrary amount of time, but is still total because 1 can be identified in a single step.

The problem here is that while the function is total it isn't continuous and so the compactness of the topology is irrelevant. As far as I can tell the article uses some trickery to ensure that computable functions _are_ continuous, in which case it does work.

cvoss2• 6 years ago

You're right, it's not a precise line of reasoning. I was considering only computable functions (else "runtime" isn't a meaningful concept) and treating the input domain as actually being the Cantor space, where 0.9999... and 1.0000... are actually distinct values. I just found it helpful to think about the Cantor space in terms of the unit interval.

JadeNB3• 6 years ago

The example you suggest has a subtle issue, since the number 0.999… is the number 1; so a program fed numbers (rather than, say, decimal strings, in which case the domain is no longer [0, 1]) cannot meaningfully distinguish the former from the latter, and hence cannot terminate on 1.

dang• 6 years ago

If curious see also

2015 https://news.ycombinator.com/item?id=10575626

Kednicma2• 6 years ago

One way to imagine this space is as an indexed bank of light switches. Each switch starts out unset, and our goal is to find a configuration on a finite number of switches which will satisfy our search. The search can only check one switch at a time and must commit after each check. This allows a basic nemesis to compute a configuration by running the search repeatedly and rescuing it after each time it fails by checking the last switch checked.

A gorgeous immediate consequence is that each search is looking for a solution to a SAT problem! Each SAT solution describes the edge of a region of Cantor space whose elements all have the searched-for property. And thus, while the search can proceed in finite time, it must sadly be NP-complete.

ridiculous_fish3• 6 years ago

Here's my attempt to translate. I think it's probably wrong and welcome corrections.

You have a function `p` which accepts an "element of the Cantor space" (infinite sequence of bits) and returns true or false. You want to find some value where it returns true, or prove that it returns false for every value.

Crucially `p` is not a math function, it is a Haskell function. So you can't say "p is true iff its argument is pi" because you can't write that in Haskell. In fact, all `p` can do is look at finitely many bits - if it looks at infinite bits, there is an input for which it will never return. And because it looks at finitely many bits, there is necessarily a largest bit-index which it looks at.

Now a generic-imperative programmer might attempt:

    for (bignum x = 0; x < infinity; x++)
        if (p(x)) return x;
    return NotFound;

but the problem is knowing when to stop, so as to return NotFound.

However, if we could detect when `p` demands more bits than we have produced, then the search will terminate:

    int bits = 0;
    bignum x = 0;
    while (x < (1 << bits)) {
      try {
         if (p(x)) return x;
         x++;
      } catch (BitAccessOutOfBounds) {
         bits++;
      }
    }
    return NotFound;

Haskell's laziness enables it to "drive the loop" but the idea is not so different.

Kednicma2• 6 years ago

Yes, you have it exactly right, including the usage of non-local control flow to interrupt and restart the search. [0] is another implementation in an imperative language.

[0] https://github.com/monte-language/typhon/blob/master/mast/fu...

shachaf3• 6 years ago

Laziness isn't the essential thing here -- the article's construction specifically doesn't rely on laziness, and would work in a strict language (almost verbatim -- you might need to write "lambda i: f(i)" instead of "f" in a few places).

I'd say that the thing that makes the article's version work is that you can pass a closure to the predicate, which is a channel for communicating extra information that makes it not quite a black box. In an imperative language, you could imagine passing in a function that mutates some state to get information out (or throws an exception, the way you set it up). The article shows that even that isn't necessary: You can pass in a sequence (represented as a function from index to value) that closes over the predicate itself, and recursively uses it to build a bit sequence that satisfies it, if one exists.

By the way, there's a clearer (I think) version of the core idea here that uses conatural numbers (the naturals + infinity) that people don't usually present, but might be worth writing up.

ridiculous_fish3• 6 years ago

In the article `p` really does take a list not a function! Haskell lists may be "backed by" functions but these functions must be pure - no back-channels. I'm not sure if this fact illuminates or obscures the point.

Anyways you can't pass a closure to the predicate, and the Haskell version has no "back channels" other than termination. The idea is to arrange the recursive search so that the predicate (not the search) demands the bits. This is what terminates the recursion, and it's hard to arrange without laziness (except by introducing more functions as you say).

panic• 6 years ago

Really nice explanation! I'd also add that it's not enough that p is continuous (only looks at a finite number of bits for each input) -- it must also be uniformly continuous (the number of bits it looks at can't become arbitrarily high for arbitrary inputs). For example, if you exclude the sequence of all ones 111111... from the Cantor set, you can write this function p which always returns false:

    function p(bits) {
      n = 0
      while bits(n) == 1 {
        n++
      }
      return false
    }

This function would loop forever on the all-ones sequence. But since that sequence was excluded, p is a total function which looks at more and more bits depending on how many leading ones there are. While each particular input only demands a finite number of bits, there's no finite number of bits which lets you search the entire space. The compactness of the Cantor set ensures that all continuous functions are also uniformly continuous, but if you remove points from the set, it can stop being compact, and the magic spell can be broken!

ogogmad2• 6 years ago

On a similar vein, you can write functional programs that can do exact Riemann integration: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.54....

billforsternz• 6 years ago

Sadly this goes over my head. But, I've got to say, this is a very nice looking website. It's nicely organised on github too.

ogogmad2• 6 years ago

Here is an application of the technique to proving Conway's cosmological theorem: https://www.cs.cmu.edu/~kw/pubs/conway.pdf

Seemingly impossible functional programs (2007)

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