Mathematicians prove the triviality of English | Hacker News Reader

Mathematicians prove the triviality of English

theguardian.com

75 pointstokenadult62 comments

Click-bait title much? I don't know about anyone else, but when I read that title I thought it would have something to do with syntax or semantics, not as a monoid word problem[1]. Also, why did they call it "semigroup with identity" instead of monoid?

[1]: https://en.wikipedia.org/wiki/Word_problem_for_groups

Not click-baity but seriously lacking in information. The work relates to the relationship of orthography to pronunciation in English. From the title I was expecting some syntactical or grammatical result. At the very least the title should read, "Mathematicians prove the triviality of English pronunciation" but even then that misses the mark, doesn't it?

Not only that, but I think the conclusions are frankly incorrect. I would not say that because LAM=LAMB (when spoken) that B=1. I would say that LA=LA and M=MB. B is not silent elsewhere, only in combination with M so it is false to discount the B (setting it to 1) without contextualising that B as being alongside an M when this happens, if you see what I mean. This to me seems so obvious that I fear I am missing something huge here because I can't think how otherwise they can assert what they are asserting.

Don't like to be harsh or snarky on HN but in this instance...?

They're not even wrong. [tm]

Some other words: lambda, lambada, lambaste, lambent.

Even if you argue some of those words are foreign loan words, they're still found in English dictionaries. Limiting yourself to very ancient words of Saxon origin won't leave you with much to work on.

The people who build speech synthesizers have already researched this. If pronunciation dictionaries mapping letter patterns to spoken phonemes really were trivial, they'd be very happy.

Unfortunately they aren't.

"Not only that, but I think the conclusions are frankly incorrect."

They can't really be "incorrect", because they made up the rules. It would be more accurate to observe something like this definition of trivial is itself rather trivial and useless.

It's a fun word game, and nothing more. As this appears to be a recreational math column in the Guardian, there's nothing wrong with that. It's no Martin Gardner, but it's not being put forth as some sort of major result or anything either.

A lot of people here seem to be taking it a great deal more seriously than any of the relevant authors.

>A lot of people here seem to be taking it a great deal more seriously than any of the relevant authors.

Lack of context, probably. I fetched the text from my RSS reader - I didnt know where it came from. From the title, I had a false expectation, as other HN readers seemed to have.

People taking this article way to seriously. It is a humorous story about a rigorous analysis of problem that makes false but naively intuitive assumptions to misprove a contradiction. The point is (1) the rigorous analysis is possible and (2) the assumptions are incorrect -- English does not have consistent pronunciation rules.

The entire study is a pun. The joke is that since in most mathematical notations AB is a shorthand for A * B, then all words are actually products of their component letters; based on this assumption it can be proven algebraically that the value of every letter is 1. I don't think there's supposed to be any serious meaning to it at all.

Also, aside from listing the names of the alleged authors, the article doesn't actually cite the study it's based on. This is another indication that it's just trolling.

With regards the 'semigroup with identity', depends what area of research you are in.

Semigroup researchers would naturally talk about many extensions to semigroups, such as 'with identity'.

This is merely a different version of the old linguist joke that "In English, all letters are silent". Leaving this here for the interested: https://en.wikipedia.org/wiki/Silent_English_alphabet

"Queue is just Q followed by four silent vowels"

They're waiting in line...

Thanks for the link. One of the examples is the 'e' in 'like'. In this case, though it is silent, it affects the pronunciation of the word - of the 'i', specifically.

Seems to prove the decided non-triviality of English, really. In fact, it could read as part of a proof that English words can't be read, or written, at all, because essentially it shows that English orthography has basically no rules - all sequences of letters can be pronounced in any way.

Uh that doesn't follow at all. Just means that the rules are a bit more complicated than "this letter is always pronounced that way". Neither does it mean that "all sequences of letters can be pronounced in any way".

Look up Chomsky & HalLe's the sound pattern of English for an attempt at completely describing English phonology with rules.

I'll link to the famous poem ( http://www.i18nguy.com/chaos.html ) and let people decide whether English pronunciation has rules...

Look you can describe anything with rules as long as they are specific enough. In the pathological case each word is its own rule, but English clearly has way more regularity than that, which explains why there is a seminal academic work describing English phonology with a set of rules.

It should be obvious that, since English clearly IS capable of being reliably serialized to and from text, that any proof that it can't be is obviously based on a flawed assumption, and so functions as a proof by contradiction of its assumption's falsity. In this case, it would be the semigroup model for orthography/phonology which is the flawed assuption.

While I agree, it should be noted that there has been serious controversy about the question whether language and language use can and should be characterized by rules, or whether it is based on some kind of emergent phenomenon or dynamical system (connectionism). The controversy is still not really over. Personally I don't think it's an either/or question, but about levels of description.

Oh, there are rules. There's just little relating them. You must learn them all separately.

Proof by contradiction. They are using pronunciation to define algebraic equality. So the minute B=1 and C=1, we can write B=C which is not true under the pronunciation rule. We have a contradiction which indicates the premise being false. In other words, using English pronunciation in that way is wrong. That's good because it seemed pretty stupid on first reading it. Glad my intuition on that was right.

The whole point of the article is to show that the premise is else, in every single case: there is no letter in English that always has unambiguous prononuciation.

I think this is why you don't let linguists do maths.

Why is that? This was actually mathematicians doing "linguistics".

That's a joke ... maybe to show how uncritically journalists will publish anything that comes across as "mathematical".

People are misunderstanding "trivial". Here what they proved is that there is no way to encode any of the English pronunciation rules as a function purely of context-free spelling of phonemes in a way that is consistent across the language. The data shows that all spellings must yield identical pronunciations, unless English JS inconsistent and/or context senstive. Obviously, the latter is true.

Which people? The writers or the readers? 'Cause the context-free interpretation of the headline suggests the writers intended to make the content of an article that fits the definition of trivial sound profound and mathematical in a way that it isn't.

The headline is gimmicky wordplay, like most headlines.

"trivial" is mathematical. The headline makes "trivial" sound non-mathematical. Profundity is nowhere to be be found; the article is about triviality.

>"By phonemic transformation into visual terms, the alphabet became a universal, abstract, static container of meaningless sounds"

-McLuhan

But there are at least 6 ways to write "f" in english (for, off, photon, enough, wife, mazeltov);

I don't think you can include the mazeltov type there - I cannot think of a native english word that uses that pronunciation. Happy to be proven wrong though.

Yeah, and photon is not native english, but greek. But what's native english? All the -gh are old constructs and new english word never follow that pattern, is that modern english? Or a barbarism taken from another, old english language? What about french words, like garage, are that english?

Does it maybe have to do with how long a loan-word has been in use, or how commonly it's used?

Sure.

dvh was commenting the variety of ways the f phoneme can be written down in different english words, and for the purposes of the discussion, elthran's conservative line on where does the english language end is not really super useful. Having 5 or 6 different symbols for a single sound is irrelevant; the point is that the article is considering letters as whole symbols for a sound, when they might be part of a symbol, not the whole symbol, for a sound. If gh is /f/ in a particular word, then using that word as part of the g cancellation is wrong.

Gustav

I don't see the difference between 'for' and 'wife' unless you are considering the 'fe' in the latter to be equivalent to the 'f' in the former. If so, then I wonder if the difference might alternatively be attributed to a silent and possibly 'magic' e (something I just learned about via good_gnu's post: https://en.wikipedia.org/wiki/Silent_e )

My favorite page on the internet about the regularity (or lack thereof) of English orthography; oldie but a goodie: Hou tu pranownse Inglish.

http://zompist.com/spell.html

Well, if we set all letters to equal 1, then what is stopping this from working out?

Haven't they just shown that the product of any combination of 1's is equal to the product of any other combination of ones?

This wouldn't work in eg German, where there's no way to cancel out pronunciation like that.

Yes, because German pronunciation is very close to the written form of words

No dialects in German then?

I am referring to Hochdeutsch

Can you give an example of how this wouldn't work out? I can't even figure out how to make one up, and would like to see how it would look.

Oh, you can only give concrete examples where it would work out. To show that it doesn't work out, you'd have to show that there are no ways to make it work.

I'm sure the poster was intending that it wouldn't be complete, like English. There certainly are some letters that obey this property:

DASS/DAS; S=1

BUND=BUNT; D=T

MANN/MAN; N=1

VIEL=FIEL; V=F

VERBEN=WERBEN; V=W

SIEH/SIE; H=1

GANZ=GANS; Z=S

POPP/POP; P=1

That only gives five letters as the identity (H,N,P,S,Z), but that's how you'd attack it.

The equivalence of D/T, for example, is unsurprising as it's an example of voicing changing; these words are in the process of changing.

If you can do more complex work with bigrams, more are possible, e.g.:

JAHR/JA; HR=1; HR/H; R=1

I'd say that S is idempotent in `dass' vs `das'. Not that s is the identity.

What they proved is that, for certain words, all letters can work as a multiplicative identity element (which if we were talking about numbers it would be 1)

Or, https://en.wikipedia.org/wiki/Ghoti

Not quite, there's also ordering and indirectness. They've proved that all letters in the English alphabet have at least one word where either appending or prepending that letter does not modify the pronunciation, or have at least one word where changing the first or last letter to the one in question does not modify the pronunciation and that these are not disjoint sets.

Someone's silly Friday project, I think.

The whole point is that they didn't immediately set all letters equal to one. The only relations they added (on top of cancellation) was that identically sounding words are equal within the monoid. From this, they were able to prove that the group must be the trivial group, no matter how you try to come up with a multiplication table.

> Well, if we set all letters to equal 1, then what is stopping this from working out?

> Haven't they just shown that the product of any combination of 1's is equal to the product of any other combination of ones?

tl;dr they weren't trying to find a solution, they were trying to characterise the behaviour of all possible solutions; but it turns out that there are no "non-trivial" solutions (i.e. setting all letters to equal 1 is the only way to solve it).

Longer version:

Technically, yes. However, that's not really the goal of this kind of algebra.

In school, we only tend to do algebra where all of the constants and variables are numbers; e.g. 2 * x = 4. Our goal was usually to find a particular value for x which is consistent with the equations, e.g. "solve for x" to get x = 2. Your solution of "set all letters to equal 1" is a perfectly valid way of achieving this kind of goal.

However, for the kind of algebra this article is about, we don't restrict ourselves to working with numbers. Instead, we explicitly avoid talking about "concrete" representations at all. We only focus on the equations we've been given. Some sets of equations are so common that they're given names, like "group (laws)", "semigroup (laws)", "field (laws)", etc.

The quote in the article tells us what the equations are:

> Regard English as a left-cancellative and right-cancellative multiplicative semigroup with identity, i.e. obeying the relations XY=ZY or YZ=YX implies X=Z, and having an element “1” such that 1X=X1=X.

Lots of things satisfy these equations. Some HN-relevant examples:

- Positive integers, where "1" is the number one and multiplication (written as juxtaposition "xy") is integer multiplication. Notice that we can't allow zero, since 1 * 0 = 2 * 0 does not imply that 1 = 2.

- Integers, where "1" is the number zero and multiplication is integer addition.

- Booleans, where "1" is False and multiplication is OR.

- Booleans, where "1" is True and multiplication is AND.

- Sets, where "1" is the empty set and multiplication is set union.

- Lists, where "1" is the empty list and multiplication is concatenation.

- Functions, where "1" is the identity function (i.e. "identity = function(arg) { return arg; }") and multiplication is function composition (i.e. "compose(x, y) = function(arg) { return x(y(arg)); }")

- Commands, where "1" is the no-op command (i.e. it performs no actions) and multiplication is sequencing (i.e. "xy = x; y")

By focusing on the equations and ignoring any particular representation, our results will apply to all representations; making this "universal algebra" a very powerful method.

What these mathematicians have done is impose a whole load of extra equations on top of the semigroup laws, of the form "AISLE = ISLE", etc. They've then shown that this large set of equations is equivalent to the single equation "x = 1".

In other words, they've shown that your solution (AKA the "trivial" solution) is the only solution. In other words, by imposing these extra rules, we've gone from a relatively interesting system which can describe functions, commands, lists, etc. to a relatively boring one, where representations might include:

- The set {one}, where multiplication is integer multiplication.

- The set {zero}, where multiplication is integer addition.

- The set {False}, where multiplication is OR.

- The set {True}, where multiplication is AND.

- The set {{}}, where multiplication is set union.

- The set {[]}, where multiplication is concatenation.

- The set {identity}, where multiplication is function composition.

- The set {no-op}, where multiplication is sequencing.

The title is ironic for me given that I've seen mathematical types push formal specification of software (eg Z, VDM) for over a decade straight with a consistent, English-related justification. They say that English is too imprecise and ambiguous to be sure you can understand what specs mean. It was true in practice enough that a combined formal (math) and informal (English) specification became a requirement for any correctness and security argument for highly assured systems.

And now some are saying English is trivial. Haha.

This wouldn't happen with lojban ;)

No but then again there's barely anyone to talk Lojban with.

Well, at least more fluent speakers of Lojban than fluent speakers of Ithkuil.

I would so love to think/speak ithkuil, but knowing that nobody seems smart enough made me learn lojban instead ^^

.xu do tavla fo la lojban

All it really proves is that English spelling rules is ridiculously inconsistent. It almost seems like an strange cousin of numerology. Where numerology applies significance to to arbitrary numbers, this applies significance to arbitrarily spelled English words.

Wow. And here I thought they were going to discuss vectorization of the English language, akin to how Word2Vec does it.

Instead it's first grader style writing with fake subtraction of letters such. That's a tremendous let down.

That English is trivial is nothing new. That the average US-ian doesn't even know their own language should be more surprising (but isn't). Mark Twain still has the best rant about it, and he knew English.

Seems kind of obvious. When you can set so many products equal to each other, what could each of the terms equal but one?

Misleading title. Someone ought to flag this.

Mathematicians prove there own triviality.

Oops. s/there/their/

there = their => re = ir

LOL

Reply on news.ycombinator.com