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	<title>Comments on: Proving Without Explaining, and Verifying Without Understanding</title>
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	<link>http://www.scottaaronson.com/blog/?p=1170</link>
	<description>The Blog of Scott Aaronson</description>
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		<title>By: Laura</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57992</link>
		<dc:creator>Laura</dc:creator>
		<pubDate>Mon, 17 Dec 2012 03:00:23 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57992</guid>
		<description><![CDATA[Slipper #34: I don&#039;t think the paper you refer to claims that there are no known graphs with quantum chromatic number 3 and chromatic number greater than 3. In fact, Imai, Le Gall, and Fukawa describe such a graph. To achieve this (3 vs &gt;3) kind of separations, quantum provers *must* use measurements with projectors of rank higher than one (i.e. full basis measurements are not sufficient).

Unfortunately, the paper &quot;Quantum Coloring Games via Symmetric SAT Games&quot; by Imai, Le Gall, Fukawa that I refer to above, can only be found in the conference proceedings of AQIS 11.]]></description>
		<content:encoded><![CDATA[<p>Slipper #34: I don&#8217;t think the paper you refer to claims that there are no known graphs with quantum chromatic number 3 and chromatic number greater than 3. In fact, Imai, Le Gall, and Fukawa describe such a graph. To achieve this (3 vs &gt;3) kind of separations, quantum provers *must* use measurements with projectors of rank higher than one (i.e. full basis measurements are not sufficient).</p>
<p>Unfortunately, the paper &#8220;Quantum Coloring Games via Symmetric SAT Games&#8221; by Imai, Le Gall, Fukawa that I refer to above, can only be found in the conference proceedings of AQIS 11.</p>
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		<title>By: Slipper.Mystery</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57815</link>
		<dc:creator>Slipper.Mystery</dc:creator>
		<pubDate>Tue, 11 Dec 2012 05:02:48 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57815</guid>
		<description><![CDATA[&lt;i&gt;Slipper.Mystery December 5th, 2012 at 9:53 am Says: Your comment is awaiting moderation.&lt;/i&gt;

(The AI underlying the blog moderation appears to be running on an Atari 2600 these days)

If we&#039;re to believe that the refs in http://arxiv.org/abs/1212.1724 (Graph Homomorphisms for Quantum Players) are up to date (it just appeared today), then (a) the object in question is called the quantum chromatic number of a graph (the minimum c such that there exists a perfect quantum strategy, i.e., winning every time despite the lack of a classical c-coloring), and (b) currently there are no known graphs with c=3.

If this interpretation is correct, this small part of the talk may need to be modified next time around.]]></description>
		<content:encoded><![CDATA[<p><i>Slipper.Mystery December 5th, 2012 at 9:53 am Says: Your comment is awaiting moderation.</i></p>
<p>(The AI underlying the blog moderation appears to be running on an Atari 2600 these days)</p>
<p>If we&#8217;re to believe that the refs in <a href="http://arxiv.org/abs/1212.1724" rel="nofollow">http://arxiv.org/abs/1212.1724</a> (Graph Homomorphisms for Quantum Players) are up to date (it just appeared today), then (a) the object in question is called the quantum chromatic number of a graph (the minimum c such that there exists a perfect quantum strategy, i.e., winning every time despite the lack of a classical c-coloring), and (b) currently there are no known graphs with c=3.</p>
<p>If this interpretation is correct, this small part of the talk may need to be modified next time around.</p>
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		<title>By: Slipper.Mystery</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57622</link>
		<dc:creator>Slipper.Mystery</dc:creator>
		<pubDate>Wed, 05 Dec 2012 14:53:25 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57622</guid>
		<description><![CDATA[Scott #30: &lt;i&gt;Would be interesting to know whether it can be done for 3-coloring (or maybe it’s already known).&lt;/i&gt;

Yes, the Magic Square and Pentagram games are described in Mermin&#039;s &#039;93 article http://link.aps.org/doi/10.1103/RevModPhys.65.803 , but the 3-color case still requires more thought. It is conceivable that Arkhipov the author of http://arxiv.org/abs/1209.3819 would already know the answer (though the problem is slightly different from his generalization of Mermin-Peres).]]></description>
		<content:encoded><![CDATA[<p>Scott #30: <i>Would be interesting to know whether it can be done for 3-coloring (or maybe it’s already known).</i></p>
<p>Yes, the Magic Square and Pentagram games are described in Mermin&#8217;s &#8217;93 article <a href="http://link.aps.org/doi/10.1103/RevModPhys.65.803" rel="nofollow">http://link.aps.org/doi/10.1103/RevModPhys.65.803</a> , but the 3-color case still requires more thought. It is conceivable that Arkhipov the author of <a href="http://arxiv.org/abs/1209.3819" rel="nofollow">http://arxiv.org/abs/1209.3819</a> would already know the answer (though the problem is slightly different from his generalization of Mermin-Peres).</p>
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		<title>By: ngm</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57583</link>
		<dc:creator>ngm</dc:creator>
		<pubDate>Tue, 04 Dec 2012 15:02:23 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57583</guid>
		<description><![CDATA[Great explanation! Thanks!]]></description>
		<content:encoded><![CDATA[<p>Great explanation! Thanks!</p>
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		<title>By: Scott</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57561</link>
		<dc:creator>Scott</dc:creator>
		<pubDate>Tue, 04 Dec 2012 04:43:05 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57561</guid>
		<description><![CDATA[ngm #29: To make it work, you need an encryption scheme where every message has a unique decryption (also known as a commitment scheme).  But those exist under pretty minimal cryptographic assumptions.  As the simplest example, Merlin could encrypt a 0 with a product of 2 huge primes and a 1 with a product of 3 huge primes, then factor the products to reveal the bits he commited to.  (Note that Arthur can check in polynomial time that a given number is prime.)]]></description>
		<content:encoded><![CDATA[<p>ngm #29: To make it work, you need an encryption scheme where every message has a unique decryption (also known as a commitment scheme).  But those exist under pretty minimal cryptographic assumptions.  As the simplest example, Merlin could encrypt a 0 with a product of 2 huge primes and a 1 with a product of 3 huge primes, then factor the products to reveal the bits he commited to.  (Note that Arthur can check in polynomial time that a given number is prime.)</p>
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		<title>By: Scott</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57560</link>
		<dc:creator>Scott</dc:creator>
		<pubDate>Tue, 04 Dec 2012 04:38:03 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57560</guid>
		<description><![CDATA[Slipper.Mystery #28: Yes, sorry, I&#039;d forgotten that their examples with perfect entangled cheating strategies are for k-coloring rather than 3-coloring specifically.  Would be interesting to know whether it can be done for 3-coloring (or maybe it&#039;s already known).

For other similar examples, see the Mermin Magic Square Game and Pentagram Game.]]></description>
		<content:encoded><![CDATA[<p>Slipper.Mystery #28: Yes, sorry, I&#8217;d forgotten that their examples with perfect entangled cheating strategies are for k-coloring rather than 3-coloring specifically.  Would be interesting to know whether it can be done for 3-coloring (or maybe it&#8217;s already known).</p>
<p>For other similar examples, see the Mermin Magic Square Game and Pentagram Game.</p>
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		<title>By: ngm</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57553</link>
		<dc:creator>ngm</dc:creator>
		<pubDate>Tue, 04 Dec 2012 00:01:01 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57553</guid>
		<description><![CDATA[Re: Hamiltonian path proof: Sorry, there&#039;s one part of this presentation that I&#039;m stuck on. Could a bad Merlin fake knowledge of a Hamiltonian path? For example, could he always send back the Arthur graph with &quot;clever&quot; edge encryption such that: (1) If Arthur asks Merlin to reveal the whole graph, then Merlin could do so (decrypt all) and (2) If Arthur asks Merlin to show the Hamiltonian path, Merlin could just reveal &#124;V&#124; fake edges?

In other words, how does Arthur know that the boxes can only be decrypted one way and Merlin isn&#039;t just picking a decryption key based on Arthur&#039;s question? Do you need some kind of signature scheme to keep Merlin honest (in addition to encryption)? Or does your encryption scheme make it hard to fake the true contents of a message?

[I don&#039;t think the question changes much if Merlin is only allowed to send back one key for all &#124;V&#124; Hamiltonian edge boxes. Maybe that just means he needs each box to have 2^&#124;V&#124; hidden compartments, rather than 2].]]></description>
		<content:encoded><![CDATA[<p>Re: Hamiltonian path proof: Sorry, there&#8217;s one part of this presentation that I&#8217;m stuck on. Could a bad Merlin fake knowledge of a Hamiltonian path? For example, could he always send back the Arthur graph with &#8220;clever&#8221; edge encryption such that: (1) If Arthur asks Merlin to reveal the whole graph, then Merlin could do so (decrypt all) and (2) If Arthur asks Merlin to show the Hamiltonian path, Merlin could just reveal |V| fake edges?</p>
<p>In other words, how does Arthur know that the boxes can only be decrypted one way and Merlin isn&#8217;t just picking a decryption key based on Arthur&#8217;s question? Do you need some kind of signature scheme to keep Merlin honest (in addition to encryption)? Or does your encryption scheme make it hard to fake the true contents of a message?</p>
<p>[I don't think the question changes much if Merlin is only allowed to send back one key for all |V| Hamiltonian edge boxes. Maybe that just means he needs each box to have 2^|V| hidden compartments, rather than 2].</p>
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		<title>By: Slipper.Mystery</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57195</link>
		<dc:creator>Slipper.Mystery</dc:creator>
		<pubDate>Wed, 28 Nov 2012 01:15:40 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57195</guid>
		<description><![CDATA[p.s. scott #25: &lt;i&gt;An excellent starting point, however, is this paper [arXiv:quant-ph/0404076] by Cleve, Hoyer, Toner, and Watrous&lt;/i&gt;

OK I have looked. In 3.2 they discuss convincing a ref that an odd cycle of length n is 2-colorable, using a single Bell pair.  The optimal strategy is correct with probability cos^2(pi/4n): better than the classical 1-1/2n, but not a perfect quantum strategy. (So now it&#039;s not clear if your two Merlin&#039;s were just beating the classical odds, or always successful.)

In 4.1, they discuss the graph coloring proof system, with k colors, where there is a perfect quantum strategy for a class of graphs G_n, but the smallest number of colors for which one of these is uncolorable (and hence would require the quantum strategy) is k=n=16, so inapplicable to the k=3 case.

So still slightly unclear as to whether there&#039;s a perfect quantum strategy for two Merlins purporting to 3-color, or just one that beats the classical odds (i.e., does better than 1-1/(V+E) ), and in either case how may qubits are used.

(Sorry if this takes your talk too literally)]]></description>
		<content:encoded><![CDATA[<p>p.s. scott #25: <i>An excellent starting point, however, is this paper [arXiv:quant-ph/0404076] by Cleve, Hoyer, Toner, and Watrous</i></p>
<p>OK I have looked. In 3.2 they discuss convincing a ref that an odd cycle of length n is 2-colorable, using a single Bell pair.  The optimal strategy is correct with probability cos^2(pi/4n): better than the classical 1-1/2n, but not a perfect quantum strategy. (So now it&#8217;s not clear if your two Merlin&#8217;s were just beating the classical odds, or always successful.)</p>
<p>In 4.1, they discuss the graph coloring proof system, with k colors, where there is a perfect quantum strategy for a class of graphs G_n, but the smallest number of colors for which one of these is uncolorable (and hence would require the quantum strategy) is k=n=16, so inapplicable to the k=3 case.</p>
<p>So still slightly unclear as to whether there&#8217;s a perfect quantum strategy for two Merlins purporting to 3-color, or just one that beats the classical odds (i.e., does better than 1-1/(V+E) ), and in either case how may qubits are used.</p>
<p>(Sorry if this takes your talk too literally)</p>
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		<title>By: Slipper.Mystery</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57172</link>
		<dc:creator>Slipper.Mystery</dc:creator>
		<pubDate>Tue, 27 Nov 2012 15:27:19 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57172</guid>
		<description><![CDATA[Scott #25: re the two Merlin&#039;s cheating, for a graph with N vertices I&#039;m willing to believe this can be done with an entangled state of 2N qubits, each Merlin holding N of them, one for each vertex, with the entanglement such that the qubits corresponding to the same vertex always measure the same, and qubits corresponding to neighboring vertices always measure different.
I even suspect this can be done with 2logN qubits, such that each vertex is associated to applying a specific binary pattern of Hadamards before measurement, again such that same and neighboring patterns measure same and different.
But do you happen to know the minimum number of qubits necessary for this case of cheating, specifically whether constant or how it scales with N?]]></description>
		<content:encoded><![CDATA[<p>Scott #25: re the two Merlin&#8217;s cheating, for a graph with N vertices I&#8217;m willing to believe this can be done with an entangled state of 2N qubits, each Merlin holding N of them, one for each vertex, with the entanglement such that the qubits corresponding to the same vertex always measure the same, and qubits corresponding to neighboring vertices always measure different.<br />
I even suspect this can be done with 2logN qubits, such that each vertex is associated to applying a specific binary pattern of Hadamards before measurement, again such that same and neighboring patterns measure same and different.<br />
But do you happen to know the minimum number of qubits necessary for this case of cheating, specifically whether constant or how it scales with N?</p>
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		<title>By: Scott</title>
		<link>http://www.scottaaronson.com/blog/?p=1170#comment-57147</link>
		<dc:creator>Scott</dc:creator>
		<pubDate>Tue, 27 Nov 2012 04:45:14 +0000</pubDate>
		<guid isPermaLink="false">http://www.scottaaronson.com/blog/?p=1170#comment-57147</guid>
		<description><![CDATA[rrtucci #23: You can have pretty much any combination you want!  People have defined and studied quantum interactive proofs, quantum zero-knowledge proofs, etc.  However, I don&#039;t think there&#039;s any sense in which &quot;quantum&quot; is some sort of umbrella notion that captures everything.  For one thing, you can certainly have &lt;i&gt;non&lt;/i&gt;-interactive and &lt;i&gt;non&lt;/i&gt;-zero-knowledge quantum proofs.  Even more important, no matter how you chose to define &quot;quantum proofs,&quot; I bet I could find a classical notion of proof that wasn&#039;t encompassed by your notion!  (By simply adding more Merlins, making their messages longer, giving Arthur more time to verify them, etc.)

Adding quantum seems to make many kinds of proof more powerful than they would be otherwise: for example, it&#039;s conjectured that QMA strictly contains MA (where MA = Merlin-Arthur), and that QSZK strictly contains SZK (where SZK = Statistical Zero Knowledge).  (Though on the other hand, we now know that QIP=IP=PSPACE, meaning that quantum does &lt;i&gt;not&lt;/i&gt; increase the power of poly(n)-round interactive proofs!)  On the other hand, quantum is &lt;i&gt;not&lt;/i&gt; a panacea whose addition makes a given proof notion &quot;leapfrog&quot; over every other proof notion that could ever be defined: to give one example, no one knows whether QSZK contains classical MA, and they might well be incomparable.

In other words: as far as complexity theory knows or cares, quantumness is basically &quot;just another interesting resource to add to the mix,&quot; similar to nondeterminism, multiple provers, etc.  Of course, the fact that the actual physical universe seems to &lt;i&gt;provide&lt;/i&gt; quantumness makes that resource extremely special!  But then we&#039;re talking physics, not complexity theory. :-)

(Incidentally, I think your comment stopped making sense at &quot;interactive depends on what observables one is allowed to measure.&quot;  I don&#039;t know how to define interactive proofs by talking only about &quot;measuring observables,&quot; without making reference to a hypothetical entity that implements a &lt;i&gt;strategy&lt;/i&gt; with the &lt;i&gt;goal&lt;/i&gt; of &lt;i&gt;maximizing&lt;/i&gt; the verifier&#039;s acceptance probability.)]]></description>
		<content:encoded><![CDATA[<p>rrtucci #23: You can have pretty much any combination you want!  People have defined and studied quantum interactive proofs, quantum zero-knowledge proofs, etc.  However, I don&#8217;t think there&#8217;s any sense in which &#8220;quantum&#8221; is some sort of umbrella notion that captures everything.  For one thing, you can certainly have <i>non</i>-interactive and <i>non</i>-zero-knowledge quantum proofs.  Even more important, no matter how you chose to define &#8220;quantum proofs,&#8221; I bet I could find a classical notion of proof that wasn&#8217;t encompassed by your notion!  (By simply adding more Merlins, making their messages longer, giving Arthur more time to verify them, etc.)</p>
<p>Adding quantum seems to make many kinds of proof more powerful than they would be otherwise: for example, it&#8217;s conjectured that QMA strictly contains MA (where MA = Merlin-Arthur), and that QSZK strictly contains SZK (where SZK = Statistical Zero Knowledge).  (Though on the other hand, we now know that QIP=IP=PSPACE, meaning that quantum does <i>not</i> increase the power of poly(n)-round interactive proofs!)  On the other hand, quantum is <i>not</i> a panacea whose addition makes a given proof notion &#8220;leapfrog&#8221; over every other proof notion that could ever be defined: to give one example, no one knows whether QSZK contains classical MA, and they might well be incomparable.</p>
<p>In other words: as far as complexity theory knows or cares, quantumness is basically &#8220;just another interesting resource to add to the mix,&#8221; similar to nondeterminism, multiple provers, etc.  Of course, the fact that the actual physical universe seems to <i>provide</i> quantumness makes that resource extremely special!  But then we&#8217;re talking physics, not complexity theory. <img src='http://www.scottaaronson.com/blog/wp-includes/images/smilies/icon_smile.gif' alt=':-)' class='wp-smiley' /> </p>
<p>(Incidentally, I think your comment stopped making sense at &#8220;interactive depends on what observables one is allowed to measure.&#8221;  I don&#8217;t know how to define interactive proofs by talking only about &#8220;measuring observables,&#8221; without making reference to a hypothetical entity that implements a <i>strategy</i> with the <i>goal</i> of <i>maximizing</i> the verifier&#8217;s acceptance probability.)</p>
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