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May 11, 2022, 6:03:40 AMMay 11

to metamath

Hi all,

I finally put the recording of my talk "Lessons from Metamath" from the Lorentz Center meeting for Machine-Checked Mathematics on YouTube. It is primarily aimed at an audience of people who are familiar with proof assistants but not with Metamath, but I'm sure you all can get plenty out of it as well.https://www.youtube.com/watch?v=OOF4NWRyue4

May 11, 2022, 7:37:00 AMMay 11

to meta...@googlegroups.com

Great talk Mario! Thanks for sharing.

I nearly asked a couple of nights ago when it turned out Glauco and I have both accidentally written Metamath verifiers in TypeScript which probably run in O(n^2), hopefully on our way to writing verifiers that are actually useable! What's a 'good' length of time in big-O notation for a verifier to run in?

In your talk you state that this is O(n) - unless I've wildly misunderstood.

I'm pleasantly surprised by that, and also curious. The best I was hoping for was O(n.log(n)), and my thinking was that as you add more proofs to the system, subsequent proofs sometimes refer back to them, which requires the verifier to do a look up, which is why I was assuming 'n' to make your pass of the file, and 'log(n)' for the lookups. Broadly, how does an O(n) verifier avoid the need for this, please?

Best regards,

Antony

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May 11, 2022, 12:21:33 PMMay 11

to metamath

It's true that O(n) is an approximation of the reality. If you assume that array lookups can be done in O(1) (and there are at most 2^word size theorems total, which is almost certainly the case unless you have a "streaming" model for theorem verification, which is not the case for most proof systems), then you can get O(n) total time on a word RAM computational model. There is another asterisk though which is that theorems have to be O(1) in size for this to work. In general if theorems have size O(m) then verifying one metamath step takes O(m) so total verification time is O(mn), so in the worst case, if you have a single theorem whose statement takes up half of the entire file, you can spend the other half of the file applying that theorem O(n) times for O(n^2) work. This is the actual bound I have established for MM0 verification, but if you make a less pessimistic assumption it looks approximately linear. (There are also complications getting this linear time bound in metamath when *expressions* stop being O(1): you can actually get exponential size terms in O(n) steps by reusing subproofs. MM0 uses dag-like sharing of terms to circumvent this problem, but metamath fundamentally uses token strings for its expressions so you have to get fancy with data structures in order to express such a highly redundant string without losing the ability to freely reassociate token strings.)

Setting aside the issues with large expressions and theorems and returning to the lookup question, the reality is still more complicated. If you don't use a word RAM model and instead use functional data structures, then your array is actually a tree and then you get that O(n log n) bound. If you remember The Myth of RAM (http://www.ilikebigbits.com/2014_04_21_myth_of_ram_1.html) then you will actually want to revise this to O(n sqrt n). But actually, it's a bit better than this because like regular programs, mathematics also has a lot of temporal and spatial locality, and this reflects in the fact that the theorems you access in the proof of a given theorem will often be hit many times and nearby theorems will also likely be accessed. So the usual caching tricks work and it goes down to O(n sqrt s) where s is the working set size, which is difficult to quantify but general considerations of how humans work suggest that this should be roughly constant - each area of mathematics generally has direct connections to O(1) other areas of mathematics. So we're back to O(n) again.To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/CAJ48g%2BCwN2q7qKqi5oh4WezRV75V%2BikgT9DJyT4_tbmO825yyQ%40mail.gmail.com.

May 12, 2022, 5:24:44 AMMay 12

to meta...@googlegroups.com

Wow, great answer, thanks Mario! Mind duly blown.

The claim that contributors to mm files have in effect sorted the proofs in a manner favorable to efficient verification should be scientifically testable by the way (i.e. to a lower standard of 'proof' than what we're used to around here). It's possible to write a program to shuffle an mm file, the prediction being that makes verification a lot less efficient. After that it might be possible to run some sort of regression on the timings.

(I'm glossing over a lot of detail here, but a shuffle algorithm 1. makes a pass of an mm file to construct a DAG, 2. writes out a random axiom, 3. removes the axiom from the DAG which promotes any theorem no longer dependent on any axioms to the status of axiom (though obviously still outputted as a theorem), 4. return to step 2 until the whole file is written out)

Best regards,

Antony

To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/CAFXXJSuz8Q5Hxcu7U3iV6%3DU2KwwbWg6dhiJEVj5ZJSTHpayA_A%40mail.gmail.com.

May 12, 2022, 6:04:20 AMMay 12

to metamath

It's tricky to shuffle an mm file because theorem dependencies put a constraint on allowable reorderings. I would probably do a random shuffle and then do DFS to get a topological ordering in order to get a random allowable ordering.

My prediction would be that it does not make a huge effect on verification speed. Except for the most optimized, Metamath verifiers are typically bottlenecked on things other than loading the theorem into cache; stuff like substituting expressions into theorem statement are probably 20-100 times more expensive except at *really* large scale.To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/CAJ48g%2BDNwi2KhpTfYh%2BgfSTYd%2BV33Qmr_hWNfLSn3O6OnfgBXQ%40mail.gmail.com.

May 12, 2022, 6:35:38 AMMay 12

to meta...@googlegroups.com, Mario Carneiro

Rather than shuffling which sounds complex, a simpler way to
verify that hypothesis might be to count the number of different
chapters/sections each theorem uses, and have a statistical
approach to confirm whether that matches with O(1) as Mario
suggests.

To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/CAFXXJSvYQQn1sDmjt1E_czs_vAuROUTTDVnWZpjC%3DjkaVSTWVA%40mail.gmail.com.

May 12, 2022, 9:36:38 AMMay 12

to Thierry Arnoux, metamath

I got curious about this, so here's some code to do just that. We can calculate the "unified bound" (https://www.youtube.com/watch?v=DZ7jt1F8KKw&t=1178s) explicitly for the access sequence used in set.mm using some stuff on top of metamath-knife (code here: https://gist.github.com/digama0/78c5a2de9589d53b4667b9a75b3ca5f3), and the result is:

Define the score of an access x_j to be the min of (t_ij + |x_i - x_j|) over all previous x_i in the sequence, where t_ij is the number of distinct values in {x_i, ..., x_j-1}. Then we can compute the average score of the access sequence, for different kinds of average weighting:

set.mm commit 3de3ae52

number of theorems: 42879

avg score: 4370145307 / 3075357 = 1421.0

2^(avg log(score)): 2^(10750033.8 / 3075357) = 2^3.4955 = 11.2788

(avg sqrt(score))^2: (12705970.2 / 3075357)^2 = 4.1315^2 = 17.0696

shuffled 1:

avg score: 4452015191 / 3075357 = 1447.6

2^(avg log(score)): 2^(12857388.6 / 3075357) = 2^4.1808 = 18.1359

(avg sqrt(score))^2: (17604771.6 / 3075357)^2 = 5.7245^2 = 32.7695

shuffled 2:

avg score: 4453096882 / 3075357 = 1448.0

2^(avg log(score)): 2^(12873701.1 / 3075357) = 2^4.1861 = 18.2027

(avg sqrt(score))^2: (17659385.1 / 3075357)^2 = 5.7422^2 = 32.9731

completely random 1:

avg score: 4860210535 / 3075357 = 1580.4

2^(avg log(score)): 2^(22395391.0 / 3075357) = 2^7.2822 = 155.6550

(avg sqrt(score))^2: (40220315.5 / 3075357)^2 = 13.0783^2 = 171.0409

completely random 2:

avg score: 4860319828 / 3075357 = 1580.4

2^(avg log(score)): 2^(22396702.0 / 3075357) = 2^7.2826 = 155.7010

(avg sqrt(score))^2: (40224473.6 / 3075357)^2 = 13.0796^2 = 171.0762

The "completely random" data just uses the output of a random number generator with the same range and length as the set.mm access sequence. The value of s is now much larger, 155.7. I believe the operative effect here is the within-proof locality: the same theorems are referenced many times in a given proof, while a random proof will touch a lot more of the library.

All in all I would say this is a pretty small number: s = 11.2 is very far from n = 3075357, so even if it's not constant (obviously difficult to measure) it is still very sub-linear.May 12, 2022, 10:39:03 AMMay 12

to Thierry Arnoux, metamath

Correction: There was a bug in the code that lead to unusually large values for "avg score", due to a single very large data point. The corrected results:

avg score: 75178012 / 3075357 = 24.4

avg lg(score): 10750001.8 / 3075357) = 3.4955 = lg(11.2787)

avg sqrt(score): 12640434.2 / 3075357 = 4.1102 = sqrt(16.8940)

shuffled 1:

avg score: 157481709 / 3075357 = 51.2

avg lg(score): 12857460.5 / 3075357) = 4.1808 = lg(18.1362)

avg sqrt(score): 17549817.2 / 3075357 = 5.7066 = sqrt(32.5652)

shuffled 2:

avg score: 157220284 / 3075357 = 51.1

avg lg(score): 12861417.8 / 3075357) = 4.1821 = lg(18.1524)

avg sqrt(score): 17552661.4 / 3075357 = 5.7075 = sqrt(32.5758)

completely random 1:

avg score: 565223103 / 3075357 = 183.8

avg lg(score): 22394511.3 / 3075357) = 7.2819 = lg(155.6242)

avg sqrt(score): 40151474.7 / 3075357 = 13.0559 = sqrt(170.4559)

completely random 2:

avg score: 565104301 / 3075357 = 183.8

avg lg(score): 22394838.4 / 3075357) = 7.2820 = lg(155.6357)

avg sqrt(score): 40149999.7 / 3075357 = 13.0554 = sqrt(170.4433)

avg score: 75178012 / 3075357 = 24.4

avg lg(score): 10750001.8 / 3075357) = 3.4955 = lg(11.2787)

avg sqrt(score): 12640434.2 / 3075357 = 4.1102 = sqrt(16.8940)

shuffled 1:

avg score: 157481709 / 3075357 = 51.2

avg lg(score): 12857460.5 / 3075357) = 4.1808 = lg(18.1362)

avg sqrt(score): 17549817.2 / 3075357 = 5.7066 = sqrt(32.5652)

shuffled 2:

avg score: 157220284 / 3075357 = 51.1

avg lg(score): 12861417.8 / 3075357) = 4.1821 = lg(18.1524)

avg sqrt(score): 17552661.4 / 3075357 = 5.7075 = sqrt(32.5758)

completely random 1:

avg score: 565223103 / 3075357 = 183.8

avg lg(score): 22394511.3 / 3075357) = 7.2819 = lg(155.6242)

avg sqrt(score): 40151474.7 / 3075357 = 13.0559 = sqrt(170.4559)

completely random 2:

avg score: 565104301 / 3075357 = 183.8

avg lg(score): 22394838.4 / 3075357) = 7.2820 = lg(155.6357)

avg sqrt(score): 40149999.7 / 3075357 = 13.0554 = sqrt(170.4433)

May 18, 2022, 11:24:42 AMMay 18

to Metamath

This was a really nice talk Mario, thanks for taking the time to give it and put it out there. I think the point about the separation of proof authoring tool and proof checking tool is a really important one, I've always liked how the metamath database is very open and separate from anything used to create or modify it.

Is it ok to ask about your progress on MM0/MM1? Where have you got to with it, what have you been working on recently and where are you hoping to get to with it? I think it's awesome what you're doing with it and I'm interested to see what becomes of it. (I guess maybe this should be a new thread but I thought I would ask here).

May 18, 2022, 11:32:20 AMMay 18

to metamath

On Wed, May 18, 2022 at 11:24 AM Jon P <drjonp...@gmail.com> wrote:

Is it ok to ask about your progress on MM0/MM1? Where have you got to with it, what have you been working on recently and where are you hoping to get to with it? I think it's awesome what you're doing with it and I'm interested to see what becomes of it. (I guess maybe this should be a new thread but I thought I would ask here).

I'm currently in the final stages of the PhD, so in a few weeks I will basically have a whole book about MM0. That's what has been occupying me of late. Once the deadlines for that are passed, I will probably go back to working on the MMC compiler lemmas with the goal of making the verifying compiler actually produce full correctness theorems, and then start growing the set of operations that the programming language supports and handle all the trivial bugs that I have noticed while writing about it.

After that comes the implementation of the MM0 verifier in MMC, and then adding all the invariants and proving them, and then it will actually be a bootstrapping system as originally intended.

To view this discussion on the web visit https://groups.google.com/d/msgid/metamath/c1a69dcd-249e-40fc-b85c-ed62d5b4de0bn%40googlegroups.com.

May 21, 2022, 11:49:39 AMMay 21

to Metamath

That sounds awesome. I'd be interested in reading your thesis when you're done, if that's ok, which I hope feels cool as often it's a struggle to even get the committee to read it haha. And yeah looking forward to seeing where you take the system, I think it's got a lot of power and once MM0 is checked in MMC that's a really solid basis I think.

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