
Landmark computer science proof cascades through physics and math - digital55
https://www.quantamagazine.org/landmark-computer-science-proof-cascades-through-physics-and-math-20200304/
======
lwneal
The paper[1] claims that MIP* = RE. At first I wasn't sure whether that was
true, because the paper is too long and complicated for me to verify.

But then I found a series of questions and answers written by this quantum guy
Scott Aaronson[2] and by this other quantum guy Kenneth Regan[3], who both
know a bunch of quantum stuff. Each one of their blog posts makes me more
certain about MIP* = RE, to the point that now I'm 99.9% sure it's true, even
though I could never verify it myself.

[1] [https://arxiv.org/abs/2001.04383](https://arxiv.org/abs/2001.04383)

[2]
[https://www.scottaaronson.com/blog/?p=4512](https://www.scottaaronson.com/blog/?p=4512)

[3] [https://rjlipton.wordpress.com/2020/01/15/halting-is-poly-
ti...](https://rjlipton.wordpress.com/2020/01/15/halting-is-poly-time-quantum-
provable/)

~~~
peterburkimsher
I hadn't heard of Aaronson or Regan before, but perhaps they communicate, and
therefore their blog posts might be correlated.

From the new proof though, I'm certain that I can't put an upper or lower
bound on the probability of whether they're right.

~~~
fortydegrees
Start with 100 and 0 and go from there

~~~
peterburkimsher
What about complex numbers in probabilities?

~~~
bergoid
Their modulus squared gives you the probability.

[https://en.wikipedia.org/wiki/Probability_amplitude](https://en.wikipedia.org/wiki/Probability_amplitude)

------
eyegor
For those interested, the arxiv link:
[https://arxiv.org/abs/2001.04383](https://arxiv.org/abs/2001.04383)

Be warned, this is 160 pages of dense math. To anyone complaining that the
quanta article is "unclear" or not descriptive enough, there's no way around
that when summarizing a dissertation.

~~~
darkcha0s
Some nice and easy afternoon reading

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ivalm
From Scott Aaronson:
[https://www.scottaaronson.com/blog/?p=4512](https://www.scottaaronson.com/blog/?p=4512)

~~~
n4r9
Thanks for the link. Does this mean we're closer to experimentally determining
whether infinite-dimensional quantum systems actually exist (rather than being
approximated by very large finite ones) ? I vaguely remember this being a
topic of discussion when I was in research.

~~~
potiuper
"Prior to the new work, mathematicians had wondered whether they could get
away with approximating infinite-dimensional matrices by using large finite-
dimensional ones instead. Now, because the Connes embedding conjecture is
false, they know they can’t." Your question is like asking if limits actually
exist. It might be answered by first proposing what continuity holds for the
quantum system.

~~~
n4r9
I'm not sure what you mean by "proposing what continuity holds for the quantum
system".

There are a number of results about how measurements of an individual
infinite-dimensional quantum system can be approximated to arbitrary degree by
an arbitrarily large finite-dimensional system.

This new result seems to suggest that for joint entangled systems there's a
"buffer" between measurement results obtained by having arbitrarily many
dimensions, and by having infinite dimensions.

I was wondering how practical it could be to actually do an experiment and
verify that the buffer line is crossed.

~~~
AstralStorm
The other version of this is to define or find a problem where the difference
is smallest. (we know it's not vanishingly small)

------
battery_cowboy
From what I gathered so far, this is more of a theoretical win for
mathematicians, computer scientists, and physicists because it basically
proved that if you had two Q-computers with unbounded (read: greater than
infinite, basically) power, they could convince a classical computer in poly
time that they could run a Q-program to completion correctly, in infinite
time.

So basically, we would still need a way to build an unboundedly-powerful pair
of Q-computers that were entangled (whoops, forgot that part!) in order to get
any direct application for this, then we'd have to actually write the software
to run this series of communications the 3 computers exchange from the paper.

So, this is only useful for some unknown, future proof of something else, not
to engineer a real-life system with.

~~~
91aintprime
Hilbert's curve: Is infinite math useful? by 3blue1brown

[https://www.youtube.com/watch?v=3s7h2MHQtxc](https://www.youtube.com/watch?v=3s7h2MHQtxc)

or skip directly to the conclusion;

[https://youtu.be/3s7h2MHQtxc?t=978](https://youtu.be/3s7h2MHQtxc?t=978)

------
dang
A related thread from a few weeks ago:
[https://news.ycombinator.com/item?id=22083935](https://news.ycombinator.com/item?id=22083935)

Also these little ones:

[https://news.ycombinator.com/item?id=22079048](https://news.ycombinator.com/item?id=22079048)

[https://news.ycombinator.com/item?id=22077591](https://news.ycombinator.com/item?id=22077591)

I seem to recall others but can't find them. Anyone?

------
zaroth
TFA starts off a little cheesy, where you might almost think to bounce because
the writing is going to just pummel the subject matter beyond recognition.

But then as you continue, you go deeper and deeper through the rabbit hole.
Expertly guided through a series of analogies and examples through a complex
subject until you arrive at an incredible payoff in perfectly understandable
terms.

I won’t spoil the surprise. But it’ll just say I love how this proof works so,
so much.

In the end it rests on the logic that if A then B but we know !B therefore !A.
It proves that for something to be true, then something else which we _know_
to be false would then have to be true. Therefore that something must not be
true. I love it.

~~~
chasd00
I remember my professors saying proof by contradiction was very powerful. i
guess they were right after all.

------
adhdbrain
Can someone please explain how the math works out to 89%?

 _> > let’s imagine two players, Alice and Bob, and a 3-by-3 grid. A referee
assigns Alice a row and tells her to enter a 0 or a 1 in each box so that the
digits sum to an odd number. Bob gets a column and has to fill it out so that
it sums to an even number. They win if they put the same number in the one
place her row and his column overlap. They’re not allowed to communicate.
Under normal circumstances, the best they can do is win 89% of the time._

~~~
dandanua
Here the explanation and the picture describes the best strategy
[https://en.wikipedia.org/wiki/Quantum_pseudo-
telepathy#The_M...](https://en.wikipedia.org/wiki/Quantum_pseudo-
telepathy#The_Mermin%E2%80%93Peres_magic_square_game)

------
peterburkimsher
Some practical questions:

a. What does this mean about Google's claim of quantum supremacy? [1] Does the
new paper prove that quantum computers with entangled provers are more
versatile than classical computers, because they can work in a larger problem
space? Would that also mean that there is definitely such a thing as quantum
supremacy, regardless of whether we've achieved it yet?

b. How does this affect CERN? That research institute is a melting pot of
physicists, computer scientists, and mathematicians. This proof did not come
from CERN, but I feel like CERN has the best resources to apply the proof.
Need some entangled particles? Sure, let me just go downstairs to the LHC and
pick up a few. Want to play a nonlocal game? Second door on your left. Need
the Grid supercomputing network to verify your proof? You've got it. I hope
that someone will be able to use this to reassure politicians that CERN is
worth funding.

c. Does anyone know whether Tsirelson saw the proof (13 Jan) before he died
(21 Jan), and what his reaction was?

I'm also really impressed at how Vidick & Ito, and Natarajan & Wright were
working independently, and then joined forces to become a genius team. My gut
feeling is that MIP* = RE is the E = MC^2 of quantum computing.

[1]
[https://news.ycombinator.com/item?id=21332768](https://news.ycombinator.com/item?id=21332768)

~~~
baq
what I gathered from it is that infinite quantum computers can't be
approximated by any finite, no matter how unreasonably large, quantum
computers. we're in no danger of possessing an infinite quantum computer, so
probably not that many implications either google or CERN.

------
jansan
Isn't that animation at the top of the page just awesome?

~~~
Insanity
I'm happy someone here mentioned it, I was entertained by it for some time!

------
predictmktegirl
So now when you get stuck on a hard problem, someone can actually prove that
it's because you suck? Thanks science!

------
jhoechtl
In layman terms please?

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gautamcgoel
One of the authors of this paper (Thomas Vidick) is a prof in my department.
Brilliant guy!

------
AstralStorm
If I understand it right, the proof shows the difference between a finite
matrix of recursively enumerable operators and infinite matrix of non-
recursive operators, is that about correct?

------
abductee_hg
wait, does that mean that Rice's theorem is wrong and machines can now solve
nontrivial semantic problems?

~~~
darzu
No, from my reading, this uses the opposite, it uses the fact that the halting
problem is unsolvable to show that MIP* cannot solve harder problems than RE,
which are a class easier than nontrivial semantic questions and the halting
problem.

~~~
bawolff
I think you are mistaken. RE is the class that includes the halting problem.

~~~
bawolff
I know its really bad form to complain about downvotes - but can i ask if
people are disagreeing with me or if there is something else objectionable in
my comment? If its the former, i'm confused as its trivially easy to google
the definition of recursively enumerable, if its the latter I am honestly
curious.

------
umvi
Slight tangent, but it seems to me that humans are not bounded by the halting
problem like computers are. In other words, given a general algorithm to
determine whether a program halts, I can always write a program that fools the
algorithm. But I can't write a program that fools my brain. Ergo, human brains
are not simply biological Turing machines like everyone seems to think?

~~~
codebje
The halting problem is that there is no algorithm possible that can answer,
for all programs and all inputs, whether that program with that input will
halt.

Here is my program:

    
    
        def fool(input):
            while input:
                pass
    

The input given to the program is your answer as to whether it halts given
that input.

That's the gist of the proof that the halting problem is unsolvable. Whatever
the algorithm or method is used to answer the problem, you can take that
method and "embed" it into the program to be verified, and invert the answer.

The halting program doesn't say you can't answer for many or even most
programs and inputs, it just says there'll be at least one counter-example.

~~~
thehappypm
That's not really how it works, though. You could imagine writing an AI
program that can read this program, figure out that it halts on the truthiness
of the input, and handles this case flawlessly.

The real proof is a bit of a head scratcher. Imagine we created such a
function, doesHalt, that can read any program's source code and always figure
out if the program will halt. This program can solve the halting problem -- it
always produces a solution, either a True or a False. Feed it your fool
program, it can always solve it, for any arbitrary input.

So what happens if we run create this little head scratcher function, that
takes no input:

    
    
      def headScratcher():
        if doesHalt(headScratcher):
          while 1: pass
    

This program halts if doesHalt() says it doesn't halt, and does not halt if
doesHalt says it does halt. These are both contradictions -- proving that such
a doesHalt function cannot actually exist.

------
ageitgey
I'm not a physicist or mathematician, but I tried to reduce this article to a
basic level to see if I understood it. Maybe this will be helpful to someone
else:

Assumptions in CS complexity theory:

1\. Turing showed that you can't tell ahead of time if a program will run
forever or eventually stop. This is the halting problem. It proves that some
problems are impossible to solve with computers even with infinite memory and
CPU.

2\. But you can often verify if an answer to a problem is correct more easily
than solving the problem itself.

3\. But some problems are so difficult that even verifying if an answer is
correct is seemingly impossible - like verifying some properties of an
infinite graph of nodes. This is because the graph could be infinitely large,
so you could never have enough memory in your computer to even number the
nodes, let alone inspect them.

4\. Sometimes if you can't verify the solution with your own computer, you can
use your computer to interrogate a more power computer to indirectly verify
it. You ask inter-related questions that are easy to verify to the more power
computer (called the "prover") about it's solution. If you get consistent
answers back more frequently then chance, you can assume the entire solution
is correct.

5\. Going even further, you could have two 'prover' computers and ask inter-
related questions to both about the same solution. Since you can now cross-
check their answers against each other, you are able to indirectly verify even
more complex solutions. This is like a detective checking the stories if
different witnesses against each other.

tl;dr - Some problems are unsolvable, but it may or may not be possible to
verify a single solution. Computer scientists want to quantify the complexity
of verifying these solutions.

Assumptions in Quantum Physics:

1\. Particles can be "entangled". This means that if you observe one particle,
some other particle somewhere else will exhibit the exact same state change,
as though they are connected by a mysterious communication channel.

2\. This is super weird. Einstein thought it was so weird that he thought they
must be missing something in their understanding.

3\. Physicists created two separate mathematical models to describe how
entanglement works: (A) the Tensor Product model, using finite dimension
matrices and (B) the Commuting Operator mode, using infinite dimension
matrices

tl;dr - Quantum Physics is so weird that they came up with two mathematical
models for entanglement

Assumptions in Mathematics:

1\. Because physicists said that the same physical property (entanglement)
could be described either with infinite matrices or finite matrices, maybe you
can approximate an infinite matrix using a finite matrix. This idea is the
'Connes embedding conjecture'.

2\. If finite and infinite matrices are actually interrelated by the Connes
embedding conjecture, maybe it means that solving one means you solved the
other.

tl;dr - The two quantum physics models inspired mathematicians to conjecture
about the relationship between finite and infinite matrices

Back to Physics:

1\. To prove that quantum entanglement exists in real life and isn't just a
math mistake, they came up with the idea of a "nonlocal game". You set up a
seemingly-random test where two 'players' make guesses and the goal is to have
them hit on the same guess more often than chance. If they win higher than
chance, you can assume they are entangled.

2\. Using those tests, they showed that quantum entanglement is real.

Back to Computer Science:

1\. Computational complexity is not just a theory. "The resources that
computers need to solve and verify problems — time and memory — are
fundamentally physical." If you change the rules physics, you can change the
rules of computer science. Our understanding of quantum entanglement as a real
thing thus changes the rules of what a computer can do.

2\. If two entangled particles can behave in a correlated manner, you can (in
theory) use that coordination to verify the solution to harder problems by
using them as two "provers". For example, you could have each particle
randomly pick a node in your infinite graph and inspect a different edge. But
since they are entangled, the two particles always pick the same node. That
means you can use them to indirectly inspect an infinite graph and verify a
previously unverifiable solution.

New CS discovery in the paper:

1\. Entangled provers allow you to verify solutions to the unsolvable halting
problem. That implies that you can also verify the solution to any problem
easier than the halting problem using quantum particles.

2\. This neat and re-writes the rules of CS complexity, but at a glance
appears not very interesting to other fields.

New Implication for Physics:

1\. Wait a sec! Since the problem you are verifying itself is unsolvable, this
implies the two mathematical models of quantum entanglement are NOT equivalent
(more details in article).

New Implication for Math:

1\. This means the Connes embedding conjecture (which said we can estimate an
infinite-dimension matrix from from a finite dimension matrix) is also
definitely FALSE.

So in the process of computer scientists trying to clarify the bounds of
computational complexity theory, they accidentally re-wrote the rules of
quantum physics and invalidated a major mathematical assumption. But the CS
researchers aren't physicists or classical mathematicians, so they don't even
claim to fully understand the cross-discipline implications of what they
discovered.

~~~
einpoklum
> 2\. But you can often verify if an answer to a problem is correct more
> easily than solving the problem itself.

Not necessarily. Or rather, we don't know that's actually the case for the
most important example - machines which work in time polynomial in the length
of their input (P vs. NP).

------
lacker
I am impressed by Quanta's dedication to trying to make cutting-edge
theoretical mathematics comprehensible. I wouldn't say I succeeded in
understanding this, but I feel like I got 10% of the way there.

~~~
gowld
I feel the opposite. They are so dedicated to making math sizzly that their
articles are full of hype paragraphs of "mathiness" to get you excited without
learning much. The clickbait headline, the intricate animation of the
clickbait headline(!), the paragraphs of "human interest" goop. The worst part
is that I know they bury a nugget of actual math in there, but I get
exchausted trying to find it.

~~~
dfdz
> nugget of actual math

If you are seriously looking for a nugget I would suggest trying to understand
the nonlocal game example (which was hidden in the write up)

"But first, to see how the games work, let’s imagine two players, Alice and
Bob, and a 3-by-3 grid. A referee assigns Alice a row and tells her to enter a
0 or a 1 in each box so that the digits sum to an odd number. Bob gets a
column and has to fill it out so that it sums to an even number. They win if
they put the same number in the one place her row and his column overlap.
They’re not allowed to communicate.

Under normal circumstances, the best they can do is win 89% of the time. But
under quantum circumstances, they can do better.

Imagine Alice and Bob split a pair of entangled particles. They perform
measurements on their respective particles and use the results to dictate
whether to write 1 or 0 in each box. Because the particles are entangled, the
results of their measurements are going to be correlated, which means their
answers will correlate as well — meaning they can win the game 100% of the
time."

These notes might help to understand the basic idea:
[https://www.scottaaronson.com/qclec/14.pdf](https://www.scottaaronson.com/qclec/14.pdf)
In particular, after understanding why the best they can is win 89% on normal
circumstances, the easier examples at the beginning of the notes provide
intuition of how a quantum strategy might help for the 3-by-3 grid game. Edit:
also see the write up by ahelwer below.

~~~
taneq
> Imagine Alice and Bob split a pair of entangled particles. They perform
> measurements on their respective particles and use the results to dictate
> whether to write 1 or 0 in each box.

Why not just imagine Alice tells Bob how she's going to choose what to write
in the boxes? That works 100% of the time too.

I don't see a distinction between "Alice and Bob communicate" and "a pair of
entangled particles gives Alice and Bob a piece of shared information that, by
prior arrangement, they both interpret in the same way."

~~~
bscphil
Yep, I quite literally came here to post this. It seems like exactly the kind
of "not giving enough information to _really_ understand a topic" that the OP
was complaining about.

If communication really needs to be forbidden, just let them agree on an RNG
beforehand and share a seed. This _seems_ to eliminate any quantum magic
involved, which suggests I don't really understand the problem, which suggests
that they haven't really explained it.

~~~
dfdz
Suppose that Alice and Bob share a seed to a RNG before the game starts.

Can you explain how this will allow them to always win the game?

~~~
bscphil
No, but as I said in my parallel comment, I don't understand how you can do
any _better_ than having an infinite amount of pre-shared information without
actually communicating. If you can explain that, great (I really would be
interested), but the important thing here is that without explaining it the
article is kind of flawed.

~~~
dooglius
I'll make an attempt, though I am by no means a domain expert. As I said in
another comment, Alice doesn't know which column is Bob's column, and Bob
doesn't know which row is Alice's row. Bell-like inequalities allow something
interesting to happen: Alice and Bob can measure a qubit at different "basis
angles" such that the phase difference in their basis angles is reflected in
the correlation of what outputs they observe from the qubit. You can't do this
classically; what Alice does with a pre-shared bit and her row and what Bob
does with the pre-shared bit and his column, form independent random
variables.

It may help to point out that (someone correct me if I'm wrong here...) this
only works if _both_ Alice and Bob are able to use their hidden information to
choose the bases they're using to measure the qubit; if they were stuck with
measuring at pre-determined angles, this trick doesn't work.

~~~
taneq
OK, so if I'm understanding this right - when you give them a pair of
entangled particles, they can each do some stuff to their particle to
determine what the other party is going to do. How is this not communication?

~~~
JadeNB
I think (but IANAP) the answer is in two parts: (1) you can’t intentionally
send a message—it’s like asking whether ancient civilisations are
communicating with archaeologists, to which the answer is surely “only
metaphorically”; and (2) even if it is communication in some sense, it’s still
surely surprising, no?

~~~
taneq
Point 1) matches my lay understanding of quantum entanglement; information is
transferred somehow between the entangled particles but not in a way that we
can use. So if this new thing IS communication via entanglement (and I can't
see how it's not, given the description) then that's huge, and opens the door
to all sorts of interesting things.

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event
081499

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dkneeland
So is this a proof for P=NP or something else entirely? I found article much
too dense to understand fully

~~~
klyrs
It's seen as a stepping stone towards P vs NP.

~~~
tgflynn
In what way ?

------
z3t4
In high school our math teacher introduced a problem that could not be solved,
which involved a TV-show game where the contender would choose one of 3 doors,
and if there was something behind it, the contender would win it. But the show
host would remove one of the doors without something behind it, and ask the
contender if he/she would like to switch door.

I simply simulated the game on a computer one million times which revealed
that changing the door after the host removed a empty one had a 66%
probability of winning. (rather then 50%) but the teacher wouldn't believe
me...

~~~
computerfriend
This is a famous problem and the solution is not contested.

~~~
brazzy
I'd rather say that the solution is _constantly_ contested by people who can't
shake off their intuition.

~~~
retsibsi
To be fair, the problem is often badly posed. The host always reveals a goat,
and this is crucial information, but people often fail to make it clear. If
the host chose randomly and just happened to reveal a goat, the solution would
be the 'intuitive' one.

~~~
aardshark
No, I don't think that's accurate. In the scenario you pose, sticking and
switching have an equal reward, so both are valid solutions.

~~~
retsibsi
> sticking and switching have an equal reward

Right, and this is the intuitive solution. Nobody argues that switching
reduces your chance of success; the split is over whether it increases it to
2/3 or leaves it at 1/3.

(The problem statement ends with 'Is it to your advantage to switch your
choice?' The right answer under the intended assumption about the host is
'yes', but the intuitive answer is 'no, it makes no difference'.)

~~~
retsibsi
Sorry, I shouldn't have said 'increases it to 2/3 or leaves it at 1/3'. If he
picks randomly and reveals a goat, the probability that your original choice
of door is a winner is now 1/2, as is the probability that the remaining door
is a winner.

------
tus88
> The new proof establishes that quantum computers that calculate with
> entangled quantum bits or qubits, rather than classical 1s and 0s, can
> theoretically be used to verify answers to an incredibly vast set of
> problems.

LOL this was the original promise of Quantum Computers. I would think a proof
would be a quantum computer that actually delivers on that promise.

