Is the Quantum Randi Challenge (QRC) valid?

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Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Tue Feb 11, 2014 12:53 am

Unlike Gill's invalid "CHSH Challenge" where no concrete example is given that can violate it, the QRC does have a simple example that shows CHSH can be violated. See Figure S6 in this paper. So in theory, Joy Christian's model should be able to beat the QRC. Perhaps it is time for a deeper analysis to see if the scheme in the QRC is really valid? Refer to Figures S1 to S6 in the paper.

After some discussion I will put up a poll.

A note about some comments in the paper: In no way does Joy's model invalidate quantum theory as the author insinuates. Joy's model actually explains and supports quantum correlations in a local realistic way. And shows that "entanglement" is indeed just an illusion based on a lack of knowledge.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Tue Feb 11, 2014 10:46 am

Hmm... this post seems to have stopped the other discussions. Or perhaps everyone is worn out for now? :-) Let me re-iterate; There is no doubt from my viewpoint that Joy's model does reproduce the predictions of quantum theory regarding the EPR-Bohm scenario so that if it can't beat the QRC then there is a flaw with the QRC. Let's find out exactly what that is if so.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Wed Feb 12, 2014 12:52 am

The whole point of my old CHSH challenge (reference below) was that it cannot be won (with classical computer programs), though the physics lab parallel of it can, in principle, be won by QM. Its purpose is pedagogical. No one has ever yet won my challenge. No implementation of Joy's model exists which wins my challenge.

Similarly, the whole point of Vongehr's QRC is that it cannot be won....

The presently definitive description of the QRC is at

http://arxiv.org/abs/1308.6752

Annals of Physics 339 (December 2013) pp 81-88
DOI: 10.1016/j.aop.2013.08.011

Exploring Inequality Violations by Classical Hidden Variables Numerically, by
Sascha Vongehr

My challenge, which is a one-on-one bet, is described in

http://arxiv.org/abs/quant-ph/0110137

Accardi contra Bell (cum mundi): The Impossible Coupling

pp. 133-154 in: Mathematical Statistics and Applications: Festschrift for Constance van Eeden. Eds: M. Moore, S. Froda and C. Léger. IMS Lecture Notes -- Monograph Series, Volume 42 (2003). Institute of Mathematical Statistics. Beachwood, Ohio

Anyone who is interested in taking it up should contact me. We have to agree on protocol (there a number of non-negotiable items, from my side) and on wager (I suggest 4000 Euro) and we have to recruit a mutually acceptable team of say three adjudicators.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Wed Feb 12, 2014 1:33 am

Richard, please give at least one concrete example with math or spreadsheet or whaever, that your CHSH challenge can be violated by anything including QM. It can't and it is pointless for anyone to even try. At least the QRC has an example that at least something can violate the real CHSH.

Here is the Mathematica notebook file for QRC with the S6 violation example.

Here is a PDF copy of that for those that don't have Mathematica.

Now, the whole point of this exercise is that if the QRC is valid, Joy's model will show violation of the CHSH part once the programming is figured out. But first, we should be able to do some simple analysis to see if there might be a flaw in the QRC that might prevent that.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Wed Feb 12, 2014 2:24 am

Fred, Please could you tell me exactly what it is you want to see. I gave you a spreadsheet containing outcomes of Alice and Bob's measurements - both the actually performed measurements and the other "not performed" measurement - together with the actual settings they used, such that the CHSH inequality was violated when one selects from that big spreadsheet, exactly the part which is observed in experiment: one setting each for Alice and Bob, one outcome each for Alice and Bob, 100 times.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Thu Feb 13, 2014 12:32 am

OK, here is a PDF file of the QRC with the S3 HV; photon polarization vector. It is easy to see that the Tables for A[j] and B[j] are not setup right because this should give 100 percent anti-correlation. So these two lines are wrong for that particular example.

Table[A[j] = If[Random[] < (Cos[Alpha[j] - H[j]])^2, 1, 0], {j,n}];

Table[B[j] = If[Random[] < (Cos[Beta[j] - (H[j] + Pi/2)])^2, 1, 0], {j,n}];
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Thu Feb 13, 2014 12:55 am

The file you are looking at is a template. You are supposed to modify certain sections so that, with your modifications, it does win QRC. Vongehr has supplied, by way of example, some ways of filling in the template, but he has not supplied a succesful solution.

If he had already won his own challenge he would hardly present it to the world!

You are challenged to change the code in the sections where you are allowed to do so, such that you do win the challenge.

I suggest you take the time to carefully read the paper "Exploring Inequality Violations by Classical Hidden Variables Numerically", Sascha Vongehr, Annals of Physics 339 (December 2013) pp 81-88; http://arxiv.org/abs/1308.6752. You need to understand the rationale behind this challenge - a challenge which cannot be won.

There is an interactive web version of the challenge at http://fmoldove.blogspot.nl/2013_08_01_archive.html
Last edited by gill1109 on Thu Feb 13, 2014 1:06 am, edited 1 time in total.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Thu Feb 13, 2014 1:05 am

That is not the point here. I am just trying to analze what he did. If the example for photon polarization is wrong that he is presenting then are there other things wrong with the QRC? Anyways, the S3 photon polarization vector example would be the starting place to build on for Joy's model I would think. But the A[j] and B[j] tables need to be set right to start with so that you get 100 percent anti-correlation. With those set basically correct so that you get 100 percent anti-correlation, doesn't mean that you will violate Bell at that juncture. But 100 percent anti-correlation is one of the conditions of Bell when b = a. So you start by always being true to that condition and build from there. Got it?

There is no need to look at any other kinds of program versions of the challenge. The Mathematica version is very simple and does the job so far. It is basically a Bell model tester.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Thu Feb 13, 2014 1:07 am

Sure, that would be a sensible strategy.

I am still perplexed by what you mean with "valid".

The challenge stands. It seems to be quite unambiguous. You can take it up: try to win it. Or you could analyse the challenge and find out if there is any point on spending time trying to win the game. Or at a higher level still you could discuss the rationale of quantum Randi challenges.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Thu Feb 13, 2014 11:57 am

"Is it valid?" just means is it really a true Bell model tester? Nothing really perplexing about that. So far, it seems pretty limited since you can only have 3 different angles for a, a', b, and b'. Actually that is not quite right; you can have two angles for a and a' and two angles for b and b'. Is that sufficient for a true test?
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Thu Feb 13, 2014 12:49 pm

In the QRC, Alice has two angles, Bob has two angles. One of each of their two angles is the same.

The QRC is based on the original Bell inequality - the inequality from Bell's landmark 1964 paper.

Bell showed that local realism plus perfect anti-correlation led to a "three variable inequality" E(A, C) - E(B, A) - E(B, C) <= 1
A, B, C stand for the binary outcomes observed (+/-1) when measuring according to settings a, b, c.
You notice that Alice uses settings a and b, Bob uses settings a and c.

One must also verify that E(A, A) = -1

It is easy to find settings a, b, c such that the singlet correlations violated the inequality while obviously at the same time manifesting perfect anti-correlation when equal settings are used.

In laboratory experiments, one would not expect to observe *perfect* anti-correlation. This fact led Clause, Horne, Shimony, Holt to derive the famous CHSH inequality.

Bell in his later writing espoused this powerful extension of his earlier work.

The challenge of QRC is to violate the three variable Bell inequality *and* exhibit perfect anti-correlation when using the same settings in each wing of the experiment.

My CHSH challenge is *easier* to win. Also in the lab we are close to seeing a loohole free CHSH experiment, but we won't ever see a loophole free Bell-original-inequality experiment, because experiments will never be entirely free of noise.

Put another way: anyone who wins the QRC has also won my challenge since the QRC experiment is actually the same as the CHSH experiment: each party choosing between two settings, four correlations observed. The QRC puts a stronger requirement on the four correlations. If the three correlation Bell-inequality is violated, and the fourth correlation is perfect, than the CHSH inequality is violated.

Neither challenge is particularly novel. Both are of course "trick" challenges: they are designed so that they can't be won. Their purpose is pedagogical and sociological.

Sasha Vongehr argues that his challenge has a number of major advantages over mine. Their nature is different, too. Mine is a one-on-one, once-off, bet, with referees and an audience. There is a chance I could lose. Vongehr's is an open challenge, anyone who writes the programs who "wins the challenge" automatically becomes world-famous (and wins the Nobel prize). Vongehr or anyone else is not personally involved as a kind of opponent to the challenger. The succesful challenger simply posts his programs on internet and achieves undying fame, quantum physics is rewritten, Joy Christian is rehabilitated, I eat my hat, ...
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Thu Feb 13, 2014 2:13 pm

Your "challenge" is impossible to win so pointless for anyone to try. That has already been proven. If the same situation is proven for the QRC, then they are both junk and don't mean anything.

But since the only angles that can be chosen in the QRC for a is 0 or 3pi/8 and for b is 0 or pi/4, I am wondering if that is really a good enough test. We will eventually find out.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Thu Feb 13, 2014 2:17 pm

The QRC is (I believe!) impossible to win. That is the whole point of it!

But QM can in principle. Similarly, for my challenge. They are both simple versions of Bell's theorem - the original Bell theorem (QRC) and present day Bell theorem (CHSH).

QM can do what local realism can't (I believe).

This is why I don't understand your logic, Fred. You think that Joy Christian has proven that QM correlations can be generated by a local realistic process. Hence it follows logically that a computer implementation of Joy's model can win my challenge, or win the QRC. Yet now you're saying that nothing can.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Fri Feb 14, 2014 12:37 am

OK, here is the QRC with the Minkwe programming showing anti-correlation OK, Bell violated and CHSH violated.

"Anti-correlation at equal angles OK."
(199, 28, 167),
The Bell inequality predicts that the first number is smaller than the sum of the second and third numbers.
So to violate Bell, the first number must be larger than the sum of the second two."
"Bell's inequality is violated! Please play again."
2.0205, "CHSH inequality is violated!"

A good start but as you can see in the PDF, I need to fix the Sign function in the A and B Tables. It is returning the wrong values since we are working with 0 and 1 as the values for A and B.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Fri Feb 14, 2014 1:35 am

Good start. But do remember that to win the QRC, you have to violate the Bell inequality often. There are easy local realistic models which violate it with probability 50%. This is totally uninteresting. But at sample size N=800, QM can violate it with probability 95% (or something like that). You need to write code which gives a violation in much more than 50% of the attempts.

BTW I posted a new topic on https://groups.google.com/forum/#!forum ... oundations but it seems that the moderator has not approved it yet... I have a new simulation of Christian's model which I think is much better than minkwe's (faster, shorter, more accurate).
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sat Feb 15, 2014 11:17 am

The flaw in the QRC is the same as the flaw in Gill's version. It has been pointed in various ways by many people but they are simply ignored.

In the simplest of terms: You can not refute that a relationship exists between the head-size and length of snakes in the amazon, if you measure head-size in one set of snakes and length in a completely different disjoint set of snakes. (I'm sure as a statistician, Richard is familiar with Simpson's paradox)

If I derive an inequality based on the assumed relationship between snake head-size and snake length, no experiment which measures head-size in one set of snakes and length in a completely different set, is a valid test of my inequality and should be expected to violate it. It does not matter how randomly the sets of snakes were chosen, so long as you don't JOINTLY measure head-size and length, you can say absolutely nothing about my inequality. Worse still, if you do not know anything about the variables in my inequalities, especially if they are HIDDEN. Then you could not seriously claim that you have truely randomly selected the snakes, notwithstanding the earlier difficulty of disjointedness. (I'm sure as a statistician, Richard is familiar with Bertrand paradox)

Similarly, if you write a simulation which jointly considers head-size and length, you will never violate my simulation which was based on joint consideration of head-size and length. But this simulation will say nothing about the meaning of real experiments where head-size and length were not JOINTLY measured. Worse still, if it were not possible to JOINTLY measure head-size and length, then my inequality which considers head-size and length JOINTLY, will NEVER be testable in any REAL experiment.

Translation:
Bell and other CHSH type inequalities (including all statistical variations) are derived based on JOINT consideration of mutually EXCLUSIVE experiments. Although they are mathematically valid, they are untestable in any REAL experiment since it is IMPOSSIBLE to JOINTLY measure MUTUALLY EXCLUSIVE outcomes.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby FrediFizzx » Sat Feb 15, 2014 12:47 pm

Hmm... Well, unlike Gill's version of CHSH that NOTHING can violate, not even QM, the QRC can show violation if there is a connection between the a and b vectors. Do you have Mathematica? I have programmed the QRC with your parameters and the notebook file is here.

I am just trying to find what the exact mathematical flaw might be but I hear what you are saying.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sat Feb 15, 2014 1:19 pm

I am highly amused by the non-statisticians' problems in grasping the simple fact that a sample average tells us about a population mean. And since the difference between two means equals the mean of the differences, it follows that the difference between two sample averages tells us about the mean of the difference in the population.

It doesn't matter a hoot whether the samples are the same or different (at least, not to first order; it does make a difference to the accuracy of the estimate, but that is another mater.

minkwe wrote:The flaw in the QRC is the same as the flaw in Gill's version. It has been pointed in various ways by many people but they are simply ignored.

In the simplest of terms: You can not refute that a relationship exists between the head-size and length of snakes in the amazon, if you measure head-size in one set of snakes and length in a completely different disjoint set of snakes. (I'm sure as a statistician, Richard is familiar with Simpson's paradox)


This analogy is quite completely inappropriate.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby gill1109 » Sat Feb 15, 2014 1:24 pm

FrediFizzx wrote:Hmm... Well, unlike Gill's version of CHSH that NOTHING can violate, not even QM, the QRC can show violation if there is a connection between the a and b vectors. Do you have Mathematica? I have programmed the QRC with your parameters and the notebook file is here.

I am just trying to find what the exact mathematical flaw might be but I hear what you are saying.


Fred, you too are still so completely in the dark, it is getting quite amusing.

I have given you a data set that my CHSH does violate. The point is, that it is only violated by a lot very rarely, if the data set is generated in the way I describe in my paper.

It is easy to arrange that both Bell's original inequality and the CHSH inequality are violated by sample averages about half the time. Just arrange that the mean values satisfy Bell/CHSH at equality. The sample averages will be sometimes larger, sometimes smaller.

Wake up, guys. Start thinking. And try some reading, as well.
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Re: Is the Quantum Randi Challenge (QRC) valid?

Postby minkwe » Sat Feb 15, 2014 1:44 pm

The analogy is very appropriate. Any statistician should recognize it immediately. You cannot use disjoint samples to disprove a claim about JOINT properties. If you disagree about this crucial point say so. You have repeatedly avoided this question.

Do you agree or disagree that You can not use disjoint samples to prove or disprove a claim about JOINT properties.This is a simple question, a simple yes/no should be appropriate here, or are you afraid to take a stand on this question. This is what the Simpson's paradox is all about.

The CHSH is a claim about JOINT properties in the following way. We have 4 properties A,B,A',B' (each +/- 1)
AB + AB' + A'B - A'B' = +/-2
A(B+B') +A'(B-B')

Note that not only are the pairs joint to each other in each paired term, but each paired term shares a property with two of the other 3 terms. This is the combinatorial-topological cyclicity that Hans de Raedt and Karl Hess often talk about such as in (http://arxiv.org/pdf/1108.3583.pdf). The CHSH is a statement about JOINT properties.

No experiment which measures the properties independently on separate sets of particles can prove or disprove the CHSH (all versions, including statistical variants). You can measure pairs jointly and still not meet the cyclicity requirement which is required in the crucial factorization step in order to derive the inequalities (all variants, including statistical ones). This is so simple it is astounding it is simply being ignored.
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