Seemingly impossible functional programs (2007)

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

by federicoponzi • 5 years ago

191 45 5 years ago

math.andrej.com

thesz• 5 years ago

The article mentions the inability to decide equality of functions. A paper [1] about proving properties of functional programs shows how to do that modulo halting theorem. E.g., if (\x -> f x == g x) halts we can decide whether f and g are equal.

[1] http://xenon.kiam.ru/~roman/doc/2009-Klyuchnikov_Romanenko--...

Basically, if supercompilation of the expression above results in a huge nested case that has either True as a result (termination) or an application of function (not terminated due to infinite data structures, for example), then functions are equivalent. Otherwise, if there is at least one False, they are not, the path to False provides a counterexample.

bmc7505• 5 years ago

One of the authors of this work apparently now works on supercompilers at Meta and open sourced some nice tools for equivalence checking based on graph rewriting [1] and constructive type theory [2]. If you're interested in that kind of thing I would recommend checking out his work:

[1] https://github.com/ilya-klyuchnikov/mrsc

[2] https://github.com/ilya-klyuchnikov/ttlite

ajkjk• 5 years ago

Are we already calling it 'Meta' unironically?

mycroftiv3• 5 years ago

I found this post and the code and math so fascinating it changed my life when I found it three years ago. I did not previously understand the capability of computers to work with the abstract logic of infinite sets in a meaningful way and it led me into the world of formalization of mathematics in the dependently typed programming language Agda, which is one of the current activities of the mathematician (Martin Escardo) who wrote this guest blog post.

bckr• 5 years ago

Can you mention some downstream effects such that your life was changed?

gijvikvff• 5 years ago

I think they just mean that they wouldn't necessarily have heard about or got into Agda otherwise.

mycroftiv3• 5 years ago

I made the study of mathematical logic and the connections between set theory and type theory and formalization in Agda my full-time focus. Increasing physical disability ended my career as a performing musician a few years ago and I had a life-long interest in philosophical topics connected with infinity. In the past two years I've written over 100,000 lines of Agda code as a product of my learning and research. I'd like to figure out a type-theoretical translation of the set-theoretic large cardinal axioms at the level of measurable and beyond.

theamk• 5 years ago

Maybe I am not a enough of a functional programmer, but I don't see what's impossible here? Any symbolic algebra system (like Wolfram Mathematica) can do such derivations, and much more.

Sure, this is interesting, but in the sense of "look at this emergent behavior -- such a simple system can do unusually complex result", rather than "wow! no one could do such things with computer before"

lmm• 5 years ago

You can do symbolic algebra on symbolic function definitions. What's interesting about this stuff is that it works for "real" functions - plain Haskell functions that we can't inspect the definition of, just call in the normal way (there's no macros or monkeypatching or anything like that going on). It's like being able to use numerical methods but still somehow solve everything exactly.

octachron• 5 years ago

Deciding that symbolic expressions are equal is in fact undecidable for even relatively simple sets of symbolic expressions: this is Richardson's theorem.

So this is an example of carefully constrained conditions where equality between all total functions on an infinite set is decidable.

jiveturkey• 5 years ago

2007

tyilo• 5 years ago

The first thing to note that, given a deterministic program that computes a boolean given an infinite bit string will either run in O(1) time for all inputs or will have an input where it doesn't halt.

Thus if you assume that the function is total and thus doesn't halt, it must run in O(1) time.

To find an infinite input bit string where the function halts, we can just record which k = O(1) bits the function is querying and then try all 2^k possibilities and as calculating the function is O(1) the total time is also O(1). (We don't always have to try all 2^k possibilities).

tempodox• 5 years ago

How is not halting O(1)? Wouldn't that be O(∞)?

scapp• 5 years ago

I think they meant

Thus if you assume that the function is total and thus does halt, it must run in O(1) time.

Since that's what total means.

quocanh• 5 years ago

Can someone explain to me in plain English what this is?

I looked up Cantor Sets so I have a vague understanding... but there are many things I don't understand. Like what does it mean to have a sequence with a bit appended to a Cantor Set? Why is that a Cantor Set? What the heck is a total p - is it a function?

Jtsummers• 5 years ago

p is a predicate, which is a function mapping elements of some type to true/false (booleans). A total function is a function which is defined for all possible inputs. So a total predicate is a function that maps all possible inputs to either true or false.

pvillano2• 5 years ago

For real numbers, f(x)=x² is a total function, but g(x)=√x is not, because g is undefined for x<0

ZoltanAK2• 5 years ago

Cantor space is just the space formed by infinite binary sequences, that is sequences which assumes only the values 0 or 1.

Considered as a topological space, Cantor space happens to have the same structure as the Cantor set, a highly disconnected subset of the real numbers that has some at-first-unintuitive properties.

But you don't have to understand, or even worry about this correspondence to grasp what's going on with seemingly impossible functional programs. Thinking about Cantor space as "the type of infinite binary sequences" is good enough.

Sharlin2• 5 years ago

Considered as a topological space, Cantor space happens to have the same structure as the Cantor set […]

And this is, intuitively, because every element of the (standard) Cantor set can be expressed as a trinary number in the range [0, 1) where every digit is either 0 or 2 (because at every level the middle third is excluded in the construction of the set). That is, a string of exactly two symbols – that is, a bitstring.

zvorygin• 5 years ago

Thank you for asking this, I was having a devil of a time googling what "total" meant in this context.

pvillano2• 5 years ago

Yeah, math is riddled with ungoogleables, for example, graph can mean network of nodes or function plot depending on context

Sniffnoy• 5 years ago

(2007)

recursive• 5 years ago

The Maybe type constructor is predefined by Haskell as [...]

It's hard for me to imagine a reader that understands the rest of this jargon, presented without explanation, who does not know what `Maybe` is.

dan-robertson• 5 years ago

1. People often struggle to target the things they write well towards a specific group of people and that often leads to this sort of thing.

2. I think when I came across this when it was posted to hacker news 5-10 years ago, I didn’t know any Haskell or any ML language and I didn’t know the maths but the program was small enough and direct enough that I think I managed to puzzle through a lot of it just by trying to match the code to the type signature (though this isn’t really sufficient to get a good feel for what is going on)

wyager• 5 years ago

Probably targeted at people who use other ML family languages (although this isn't a large crowd).

maxiepoo• 5 years ago

A mathematician that has only used a type theory like Coq or Agda and never Haskell.

ogogmad• 5 years ago

Same ideas as the above: https://en.wikipedia.org/wiki/Computable_analysis

shikoba• 5 years ago

At first I was doubtful, but when I read the code I understand that the author was just searching exhaustively through a finite set. What a fraud.

anfelor2• 5 years ago

This seems like little more than a parlor trick.

First, a true "exhaustive search" is actually impossible. The cantor space is uncountable (which can be seen by prepending '0.' before every binary sequence which gives the real number interval [0,1).) But an exhaustive search will visit all elements in some order which provides an ordering. Thus its existence implies that the cantor space is countable. Contradiction. The post gets around this by only looking at the first n elements of every binary sequence: f,g and h evaluate little more than the 7th digit. Sure, exhaustively searching 2^7 elements can be done in less than a second, why is this new?

In fact, e.g. the function type Integer -> Integer doesn’t have decidable equality because of the Halting Problem, as is well known. However, common wisdom is not always correct, and, in fact, some other function types do have decidable equality, for example the type Cantor -> y for any type y with decidable equality, without contradicting Turing.

I fail to see how this can be true. We can set y = Integer, since integers have decidable equality. Then we can encode every integer as an element of cantor space by converting it into binary with the least significant digit first (and using the 0th element of the sequence to distinguish positive and negative integers) and padding the sequence with zeros towards infinity. Then two functions f,g : Integer -> Integer are equal iff their transformed representations f . p, g . p : Cantor -> Integer are equal. Thus functions from the cantor space having decidable equality implies functions from the integers having decidable equality which "solves" the halting problem. Again the post gets around this by only looking at the first few elements... but equality of functions f,g : Integer -> Integer is trivially decidable if f and g are zero for integers bigger than some N, so how is this new?

mycroftiv3• 5 years ago

I would suggest you look at more of Escardo's writing and research on related topics. https://www.cs.bham.ac.uk/~mhe/papers/omniscient-journal-rev... is one such paper. The issues of compactness of cantor space in classical and constructive math and the computational interpretation of principles like Brouwer's fan theorem in relation to dependent choice and weak König's lemma is really deep.

saithound• 5 years ago

Your argument is wrong. The function that encodes every Integer as an element of Cantor, the one you call `p`, has signature `p: Integer -> Cantor`. This means that the compositions `(f.p)` and `(g.p)` are not defined. The compositions in the opposite order, `(p.f)` and `(p.g)`, are defined, but have signature `Integer->Cantor`, which is not in the form `Cantor->y`.

Your comment, "the post gets around this by only looking at the first n elements of every binary sequence" is also wrong. This is not at all what the `equal` function does.

anfelor2• 5 years ago

You are right, a decoding function is needed. The decoding for my given encoding function is not computable because you need to look at all elements of the sequence to see if the rest are the padded zeros or part of the number (because there is a '1' somewhere). But here is a computable, surjective decoding function: 1. The first bit tells you whether the integer should be positive or negative. 2. Then we parse blocks of ten bits. The first bit tells you whether this block is part of the number ('1') or if we reached the end ('0'). If it is part of the number, then the next nine bits are part of the binary encoding of the integer.

This will not terminate for some sequences (for example the all '1' sequence), but it terminates for all sequences that denote integers. Wouldn't this encoding be valid? (I don't know too much about the halting problem so I might well be wrong here).

This is not at all what the `equal` function does.

Could you explain what it does do? The way I see it the `find` function constructs a counter-example lazily. The `p` predicate is then applied to this counter-example in `forsome`: `forsome p = p(find(\a -> p a))`. Only if the `p` predicate can determine if the counter-example is valid by looking at a finite amount of digits will the function terminate.

tetromino_• 5 years ago

Let's try fixing grandparent's argument: let p be the decoder function, Cantor -> Integer. We can define it as the length of the initial run of zeroes (treat the initial bit as sign). This almost works.

Where it fails is the infinite sequence of zeroes: p would count forever; f.p and g.p are not decidable.

So grandparent's argument in fact shows that OpenCantor -> y does not have decidable equality (where OpenCantor is Cantor with the zero sequence - or by extension, any one computable element - excluded). One special element makes all the difference!

TheNewAndy• 5 years ago

The cantor space is uncountable (which can be seen by prepending '0.' before every binary sequence which gives the real number interval [0,1).)

You can do the same trick with natural numbers to "prove" that they are uncountable too - just drop the duplicates. 0.0, 0.1, 0.2, 0.3, ... 0.9, 0.11, 0.12, 0.13...

Unfortunately, neither of these sequences contain the real 1/3, so they have gaps.

kevinwang• 5 years ago

I find this very interesting, but I'm unable to understand it, so the result still seems impossible to me.

Can someone explain how find works if the predicate is something like: return true if and only if every other bit is 1

alew1• 5 years ago

The trick is that your predicate can’t be implemented in Haskell, because the predicate itself requires looking at infinitely many elements.

CraneWorm• 5 years ago

if a predicate depends on finitely many bits then you can exhaustively check whether it is total by enumerating all of it's possible inputs

Seemingly impossible functional programs (2007)

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