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	<title>Comments on: Is the &#8220;Hardest Logic Puzzle Ever&#8221; too Easy?</title>
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	<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/</link>
	<description>Some things in mathematical logic that I find interesting</description>
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		<title>By: David</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-150</link>
		<dc:creator>David</dc:creator>
		<pubDate>Fri, 13 Feb 2009 03:44:22 +0000</pubDate>
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		<description>TJ, that has the same effect as the first solution given - you can only ascertain the answer to one of the three options at a time. Thus, you may need to ask two questions to work out how many fingers are being held up.

But with the exploding heads approach you only ever need to ask *one* question.</description>
		<content:encoded><![CDATA[<p>TJ, that has the same effect as the first solution given &#8211; you can only ascertain the answer to one of the three options at a time. Thus, you may need to ask two questions to work out how many fingers are being held up.</p>
<p>But with the exploding heads approach you only ever need to ask *one* question.</p>
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		<title>By: TJ Shannon</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-146</link>
		<dc:creator>TJ Shannon</dc:creator>
		<pubDate>Wed, 14 Jan 2009 14:29:57 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=61#comment-146</guid>
		<description>Just stumbled on this webpage.

I thought the answer was to ask two questions to the Knight (or Knave) as follows:
&quot;If I were to ask a member of the other tribe whether you were holding up one finger, what would his answer be?&quot;

If the inhabitant is actually holding up one finger, whether he is a Knight or a Knave the answer will be &quot;No&quot;, and you will know that he is holding  up one finger. If the inhabitant is holding up instead two (or three) fingers, the answer would be &quot;Yes&quot;.
If he does answer &quot;Yes&#039; then you must ask a second question, &quot;If I were to ask a member of the other tribe whether you were holding up two fingers, what would his answer be&quot;
Once again, if the inhabitant answers &quot;No&quot;, he actually is holding up two fingers; if instead he answers &quot;Yes&quot; again, you can then deduce that he is holding up three fingers.

The point being that a Knight will always give the hypothetical Knave&#039;s lying response, and the Knave will always reverse the hypothetical Knight&#039;s truthful response.

No need for exploding heads.</description>
		<content:encoded><![CDATA[<p>Just stumbled on this webpage.</p>
<p>I thought the answer was to ask two questions to the Knight (or Knave) as follows:<br />
&#8220;If I were to ask a member of the other tribe whether you were holding up one finger, what would his answer be?&#8221;</p>
<p>If the inhabitant is actually holding up one finger, whether he is a Knight or a Knave the answer will be &#8220;No&#8221;, and you will know that he is holding  up one finger. If the inhabitant is holding up instead two (or three) fingers, the answer would be &#8220;Yes&#8221;.<br />
If he does answer &#8220;Yes&#8217; then you must ask a second question, &#8220;If I were to ask a member of the other tribe whether you were holding up two fingers, what would his answer be&#8221;<br />
Once again, if the inhabitant answers &#8220;No&#8221;, he actually is holding up two fingers; if instead he answers &#8220;Yes&#8221; again, you can then deduce that he is holding up three fingers.</p>
<p>The point being that a Knight will always give the hypothetical Knave&#8217;s lying response, and the Knave will always reverse the hypothetical Knight&#8217;s truthful response.</p>
<p>No need for exploding heads.</p>
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		<title>By: mkoconnor</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-29</link>
		<dc:creator>mkoconnor</dc:creator>
		<pubDate>Fri, 22 Aug 2008 15:08:17 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=61#comment-29</guid>
		<description>It occurred to me that the fact that $latex \neg\square P \wedge (\square P \vee \square\neg P)$ is a sentence whose decidability is undecidable in PA is actually not at all remarkable: every sentence which is undecidable in PA is such that its decidability is undecidable in PA, since for PA to prove a sentence is undecidable in PA, it would have to prove that it can&#039;t prove something, which is equivalent to it proving its consistency.   So I will have to work a little harder to find a provability model of this.</description>
		<content:encoded><![CDATA[<p>It occurred to me that the fact that <img src='http://l.wordpress.com/latex.php?latex=%5Cneg%5Csquare+P+%5Cwedge+%28%5Csquare+P+%5Cvee+%5Csquare%5Cneg+P%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg\square P \wedge (\square P \vee \square\neg P)' title='\neg\square P \wedge (\square P \vee \square\neg P)' class='latex' /> is a sentence whose decidability is undecidable in PA is actually not at all remarkable: every sentence which is undecidable in PA is such that its decidability is undecidable in PA, since for PA to prove a sentence is undecidable in PA, it would have to prove that it can&#8217;t prove something, which is equivalent to it proving its consistency.   So I will have to work a little harder to find a provability model of this.</p>
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		<title>By: mkoconnor</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-28</link>
		<dc:creator>mkoconnor</dc:creator>
		<pubDate>Wed, 20 Aug 2008 14:28:42 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=61#comment-28</guid>
		<description>As another note, hopefully it will be possible to adapt the hierarchy-of-gods interpretation (or one similar to it) to provability logic.

For example, you can adapt what I called the Second-Order Liar to a Second-Order Gödel Sentence: one whose decidability is undecidable:  Given an arithmetical sentence $latex P$, let $latex \square P$ be the sentence asserting that $latex P$ is provable in PA (Peano Arithmetic).

Then, by diagonalization or fixed point or some such proposition, we know that there is a sentence $latex P$ such that Peano Arithmetic proves that $latex P \leftrightarrow (\neg\square P \wedge (\square P \vee \square \neg P))$.

Now we would like to show that $latex \square P \vee \square \neg P$ is undecidable in PA. 

First suppose that PA proves $latex \square P \vee \square \neg P$.  Then PA proves $latex P \leftrightarrow (\neg\square P \wedge \top)$ and thus $latex P \leftrightarrow \neg\square P$.  By the $latex \omega$-consistency of PA and the proof of Gödel&#039;s first incompleteness theorem, PA does not prove $latex P$ and does not prove $latex \neg P$.  Now, using the fact that all statements provable in PA are actually true and the assumption that PA proves $latex \square P \vee \square \neg P$, we have a contradiction.

Now suppose that PA proves $latex \neg(\square P\vee \square \neg P)$.   Then PA proves $latex P\leftrightarrow (\neg\square P \wedge \bot)$.  So PA proves $latex P \leftrightarrow \bot$.  So PA proves $latex \neg P$.  So PA proves $latex \square \neg P$.  On the other hand, since PA proves $latex \neg(\square P \vee \square \neg P)$, PA proves $latex \neg \square \neg P$.  So PA is inconsistent.</description>
		<content:encoded><![CDATA[<p>As another note, hopefully it will be possible to adapt the hierarchy-of-gods interpretation (or one similar to it) to provability logic.</p>
<p>For example, you can adapt what I called the Second-Order Liar to a Second-Order Gödel Sentence: one whose decidability is undecidable:  Given an arithmetical sentence <img src='http://l.wordpress.com/latex.php?latex=P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P' title='P' class='latex' />, let <img src='http://l.wordpress.com/latex.php?latex=%5Csquare+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\square P' title='\square P' class='latex' /> be the sentence asserting that <img src='http://l.wordpress.com/latex.php?latex=P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P' title='P' class='latex' /> is provable in PA (Peano Arithmetic).</p>
<p>Then, by diagonalization or fixed point or some such proposition, we know that there is a sentence <img src='http://l.wordpress.com/latex.php?latex=P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P' title='P' class='latex' /> such that Peano Arithmetic proves that <img src='http://l.wordpress.com/latex.php?latex=P+%5Cleftrightarrow+%28%5Cneg%5Csquare+P+%5Cwedge+%28%5Csquare+P+%5Cvee+%5Csquare+%5Cneg+P%29%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P \leftrightarrow (\neg\square P \wedge (\square P \vee \square \neg P))' title='P \leftrightarrow (\neg\square P \wedge (\square P \vee \square \neg P))' class='latex' />.</p>
<p>Now we would like to show that <img src='http://l.wordpress.com/latex.php?latex=%5Csquare+P+%5Cvee+%5Csquare+%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\square P \vee \square \neg P' title='\square P \vee \square \neg P' class='latex' /> is undecidable in PA. </p>
<p>First suppose that PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Csquare+P+%5Cvee+%5Csquare+%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\square P \vee \square \neg P' title='\square P \vee \square \neg P' class='latex' />.  Then PA proves <img src='http://l.wordpress.com/latex.php?latex=P+%5Cleftrightarrow+%28%5Cneg%5Csquare+P+%5Cwedge+%5Ctop%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P \leftrightarrow (\neg\square P \wedge \top)' title='P \leftrightarrow (\neg\square P \wedge \top)' class='latex' /> and thus <img src='http://l.wordpress.com/latex.php?latex=P+%5Cleftrightarrow+%5Cneg%5Csquare+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P \leftrightarrow \neg\square P' title='P \leftrightarrow \neg\square P' class='latex' />.  By the <img src='http://l.wordpress.com/latex.php?latex=%5Comega&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\omega' title='\omega' class='latex' />-consistency of PA and the proof of Gödel&#8217;s first incompleteness theorem, PA does not prove <img src='http://l.wordpress.com/latex.php?latex=P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P' title='P' class='latex' /> and does not prove <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg P' title='\neg P' class='latex' />.  Now, using the fact that all statements provable in PA are actually true and the assumption that PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Csquare+P+%5Cvee+%5Csquare+%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\square P \vee \square \neg P' title='\square P \vee \square \neg P' class='latex' />, we have a contradiction.</p>
<p>Now suppose that PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Cneg%28%5Csquare+P%5Cvee+%5Csquare+%5Cneg+P%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg(\square P\vee \square \neg P)' title='\neg(\square P\vee \square \neg P)' class='latex' />.   Then PA proves <img src='http://l.wordpress.com/latex.php?latex=P%5Cleftrightarrow+%28%5Cneg%5Csquare+P+%5Cwedge+%5Cbot%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P\leftrightarrow (\neg\square P \wedge \bot)' title='P\leftrightarrow (\neg\square P \wedge \bot)' class='latex' />.  So PA proves <img src='http://l.wordpress.com/latex.php?latex=P+%5Cleftrightarrow+%5Cbot&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='P \leftrightarrow \bot' title='P \leftrightarrow \bot' class='latex' />.  So PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg P' title='\neg P' class='latex' />.  So PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Csquare+%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\square \neg P' title='\square \neg P' class='latex' />.  On the other hand, since PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Cneg%28%5Csquare+P+%5Cvee+%5Csquare+%5Cneg+P%29&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg(\square P \vee \square \neg P)' title='\neg(\square P \vee \square \neg P)' class='latex' />, PA proves <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+%5Csquare+%5Cneg+P&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg \square \neg P' title='\neg \square \neg P' class='latex' />.  So PA is inconsistent.</p>
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		<title>By: mkoconnor</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-27</link>
		<dc:creator>mkoconnor</dc:creator>
		<pubDate>Wed, 20 Aug 2008 07:07:41 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=61#comment-27</guid>
		<description>Thanks for the detailed reply Brian! I should have made it more clear that the &quot;hierarchy-of-gods&quot; that I was proposing wasn&#039;t the only interpretation of the circumstances under which one&#039;s head might explode, and that it was certainly different from the one you used in your paper.

Here&#039;s a stab at a formulation of the &quot;hierarchy-of-gods&quot; interpretation:  There are a chain of gods, god_0, god_1, ... .  The zeroth god, god_0, is asked a question (I&#039;ll get to how to formalize the question in a moment.)  For each god, there is a decision that that god might have to make.  For god_0, the decision is whether to answer &quot;yes&quot; or &quot;no&quot; to the question.  For i &gt; 0, god_i&#039;s decision is whether or not to absolve each god_j for j &lt; i of the responsibility of making their decision.  We also stipulate that each god is stingy with their absolutions, which is to say that, given what all the other gods have done, they only absolve the lower gods if there was nothing they could do that is, ultimately, consistent with giving a correct answer to the question.  

Note that it may be the case that a god is absolved of the responsibility of making a decision, but makes one anyway.

Now to flesh it out a bit.  Let $latex Y_0, N_0, Y_1, N_1, \ldots$ be propositional variables.  The variable $latex Y_0$ is intended to mean that god_0 replies &quot;yes.&quot;  The variable $latex N_0$ is intended to mean that god_0 replies &quot;no.&quot;  For $latex i &gt; 0$, the variable $latex Y_i$ is intended to mean that god_i does not absolve the lesser gods of their responsibilities.  The variable $latex N_i$ is intended to mean that god_i does absolve the lesser gods of their responsibilities.

A question $latex Q$ will then be a finite propositional formula whose propositional variables are among the $latex Y_0, N_0, Y_1, N_1, \ldots$. 

We call an assignment of truth values to the propositional variables a world.  

In a given world, we say that the gods are law-abiding if for all $latex j$, if $latex \neg Y_j \wedge \neg N_j$ holds then for some $latex i &gt; j$, $latex N_i$ holds.

In a given world, we say that the gods act singly if for all $latex i$, $latex \neg Y_i \vee \neg N_i$ holds.

In a given world, we say that god_0 acts acceptably if $latex Y_0 \rightarrow Q$ and $latex N_0 \rightarrow \neg Q$ hold. 

In a given world, we say that god_1 is stingy with his absolutions if the following holds: If $latex N_1$ holds, then in the world that is the same as the current one except that $latex Y_0\wedge \neg N_0$ is true, god_0 does not act acceptably (i.e., $latex Q$ does not hold) and in the world that is the same as the current one except that $latex \neg Y_0\wedge N_0$ is true, god_0 does not act acceptably (i.e., $latex Q$ does hold).

For $latex i &gt; 1$, we say that god_i is stingy with absolutions if the following holds: If $latex N_i$ holds, then in the world that is the same as the current one except that $latex Y_{i-1}\wedge N_{i-1}$ is true, god_{i-1} is not stingy with absolutions and in the world that is the same as the current one except that $latex \neg Y_{i-1}\wedge N_{i-1}$ is true, god_{i-1} is not stingy with absolutions.

Then, for a given question, we say that a solution to the question is one where the gods are law-abiding, act singly, where god_0 acts acceptably and the rest are all stingy with their absolutions.

I believe you can then prove existence of a solution for all questions.  (I may have to tweak the definitions a bit; I&#039;m not sure yet as I haven&#039;t written this out formally.)  The existence should follow from the naive algorithm: Start with no god being absolved of their responsibility, and work your way up.  Each god tries his best and if he fails, the next god absolves him of his responsibility.  Because the question might refer to that eventuality however, some lower god might now be able to succeed, in which case there may need to be more absolutions.  If all else fails, the first god not mentioned in the question, which involves only finitely many gods, simply absolves all the gods mentioned of responsibility.

Of course, there is no uniqueness for solutions, since a question like &quot;Will you answer &#039;yes&#039; to this question?&quot; may be asked.</description>
		<content:encoded><![CDATA[<p>Thanks for the detailed reply Brian! I should have made it more clear that the &#8220;hierarchy-of-gods&#8221; that I was proposing wasn&#8217;t the only interpretation of the circumstances under which one&#8217;s head might explode, and that it was certainly different from the one you used in your paper.</p>
<p>Here&#8217;s a stab at a formulation of the &#8220;hierarchy-of-gods&#8221; interpretation:  There are a chain of gods, god_0, god_1, &#8230; .  The zeroth god, god_0, is asked a question (I&#8217;ll get to how to formalize the question in a moment.)  For each god, there is a decision that that god might have to make.  For god_0, the decision is whether to answer &#8220;yes&#8221; or &#8220;no&#8221; to the question.  For i &gt; 0, god_i&#8217;s decision is whether or not to absolve each god_j for j &lt; i of the responsibility of making their decision.  We also stipulate that each god is stingy with their absolutions, which is to say that, given what all the other gods have done, they only absolve the lower gods if there was nothing they could do that is, ultimately, consistent with giving a correct answer to the question.  </p>
<p>Note that it may be the case that a god is absolved of the responsibility of making a decision, but makes one anyway.</p>
<p>Now to flesh it out a bit.  Let <img src='http://l.wordpress.com/latex.php?latex=Y_0%2C+N_0%2C+Y_1%2C+N_1%2C+%5Cldots&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_0, N_0, Y_1, N_1, \ldots' title='Y_0, N_0, Y_1, N_1, \ldots' class='latex' /> be propositional variables.  The variable <img src='http://l.wordpress.com/latex.php?latex=Y_0&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_0' title='Y_0' class='latex' /> is intended to mean that god_0 replies &#8220;yes.&#8221;  The variable <img src='http://l.wordpress.com/latex.php?latex=N_0&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='N_0' title='N_0' class='latex' /> is intended to mean that god_0 replies &#8220;no.&#8221;  For <img src='http://l.wordpress.com/latex.php?latex=i+%3E+0&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='i &gt; 0' title='i &gt; 0' class='latex' />, the variable <img src='http://l.wordpress.com/latex.php?latex=Y_i&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_i' title='Y_i' class='latex' /> is intended to mean that god_i does not absolve the lesser gods of their responsibilities.  The variable <img src='http://l.wordpress.com/latex.php?latex=N_i&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='N_i' title='N_i' class='latex' /> is intended to mean that god_i does absolve the lesser gods of their responsibilities.</p>
<p>A question <img src='http://l.wordpress.com/latex.php?latex=Q&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Q' title='Q' class='latex' /> will then be a finite propositional formula whose propositional variables are among the <img src='http://l.wordpress.com/latex.php?latex=Y_0%2C+N_0%2C+Y_1%2C+N_1%2C+%5Cldots&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_0, N_0, Y_1, N_1, \ldots' title='Y_0, N_0, Y_1, N_1, \ldots' class='latex' />. </p>
<p>We call an assignment of truth values to the propositional variables a world.  </p>
<p>In a given world, we say that the gods are law-abiding if for all <img src='http://l.wordpress.com/latex.php?latex=j&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='j' title='j' class='latex' />, if <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+Y_j+%5Cwedge+%5Cneg+N_j&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg Y_j \wedge \neg N_j' title='\neg Y_j \wedge \neg N_j' class='latex' /> holds then for some <img src='http://l.wordpress.com/latex.php?latex=i+%3E+j&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='i &gt; j' title='i &gt; j' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=N_i&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='N_i' title='N_i' class='latex' /> holds.</p>
<p>In a given world, we say that the gods act singly if for all <img src='http://l.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='i' title='i' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+Y_i+%5Cvee+%5Cneg+N_i&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg Y_i \vee \neg N_i' title='\neg Y_i \vee \neg N_i' class='latex' /> holds.</p>
<p>In a given world, we say that god_0 acts acceptably if <img src='http://l.wordpress.com/latex.php?latex=Y_0+%5Crightarrow+Q&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_0 \rightarrow Q' title='Y_0 \rightarrow Q' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=N_0+%5Crightarrow+%5Cneg+Q&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='N_0 \rightarrow \neg Q' title='N_0 \rightarrow \neg Q' class='latex' /> hold. </p>
<p>In a given world, we say that god_1 is stingy with his absolutions if the following holds: If <img src='http://l.wordpress.com/latex.php?latex=N_1&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='N_1' title='N_1' class='latex' /> holds, then in the world that is the same as the current one except that <img src='http://l.wordpress.com/latex.php?latex=Y_0%5Cwedge+%5Cneg+N_0&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_0\wedge \neg N_0' title='Y_0\wedge \neg N_0' class='latex' /> is true, god_0 does not act acceptably (i.e., <img src='http://l.wordpress.com/latex.php?latex=Q&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Q' title='Q' class='latex' /> does not hold) and in the world that is the same as the current one except that <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+Y_0%5Cwedge+N_0&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg Y_0\wedge N_0' title='\neg Y_0\wedge N_0' class='latex' /> is true, god_0 does not act acceptably (i.e., <img src='http://l.wordpress.com/latex.php?latex=Q&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Q' title='Q' class='latex' /> does hold).</p>
<p>For <img src='http://l.wordpress.com/latex.php?latex=i+%3E+1&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='i &gt; 1' title='i &gt; 1' class='latex' />, we say that god_i is stingy with absolutions if the following holds: If <img src='http://l.wordpress.com/latex.php?latex=N_i&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='N_i' title='N_i' class='latex' /> holds, then in the world that is the same as the current one except that <img src='http://l.wordpress.com/latex.php?latex=Y_%7Bi-1%7D%5Cwedge+N_%7Bi-1%7D&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='Y_{i-1}\wedge N_{i-1}' title='Y_{i-1}\wedge N_{i-1}' class='latex' /> is true, god_{i-1} is not stingy with absolutions and in the world that is the same as the current one except that <img src='http://l.wordpress.com/latex.php?latex=%5Cneg+Y_%7Bi-1%7D%5Cwedge+N_%7Bi-1%7D&#038;bg=ffffff&#038;fg=444444&#038;s=0' alt='\neg Y_{i-1}\wedge N_{i-1}' title='\neg Y_{i-1}\wedge N_{i-1}' class='latex' /> is true, god_{i-1} is not stingy with absolutions.</p>
<p>Then, for a given question, we say that a solution to the question is one where the gods are law-abiding, act singly, where god_0 acts acceptably and the rest are all stingy with their absolutions.</p>
<p>I believe you can then prove existence of a solution for all questions.  (I may have to tweak the definitions a bit; I&#8217;m not sure yet as I haven&#8217;t written this out formally.)  The existence should follow from the naive algorithm: Start with no god being absolved of their responsibility, and work your way up.  Each god tries his best and if he fails, the next god absolves him of his responsibility.  Because the question might refer to that eventuality however, some lower god might now be able to succeed, in which case there may need to be more absolutions.  If all else fails, the first god not mentioned in the question, which involves only finitely many gods, simply absolves all the gods mentioned of responsibility.</p>
<p>Of course, there is no uniqueness for solutions, since a question like &#8220;Will you answer &#8216;yes&#8217; to this question?&#8221; may be asked.</p>
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		<title>By: brian rabern</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-26</link>
		<dc:creator>brian rabern</dc:creator>
		<pubDate>Wed, 20 Aug 2008 05:02:37 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=61#comment-26</guid>
		<description>Nice! It looks like you are attempting to construct a question that has some kind of analogous relationship to a revenge liar. The key here is where you say &quot;it is inconsistent that his head explodes&quot;.  You must be thinking, &quot;Look, if he explodes, then the correct answer to the question is &#039;no&#039; (it is not the case that his head failed to explode) and since he is a truth-teller, then he should answer &#039;no&#039;. So exploding is &quot;inconsistent&quot; with telling the truth.

The problem is that head explosion is not meant to be like a third truth-value. The head explodes when there is no other recourse.  The way we were thinking of the procedure is something like this: If you ask him a yes-no question, then (i) he will say &#039;yes&#039;, if he can provide a truthful answer without contradicting himself and said truthful answer is &#039;yes&#039;, (ii) he will say &#039;no&#039; if he can provide a truthful answer without contradicting himself and said truthful answer is &#039;no&#039; (iii) and he will explode otherwise.

Thus, all insurmountable problems lead to head explosion; the knight reasons &quot;if my head explodes, then I could have answered &#039;no&#039;&quot;, and he continues &quot;but if I answer &#039;no&#039;, then I lie, so that doesn&#039;t work, I can&#039;t answer &#039;yes&#039;, so I must explode, but then I could answer &#039;no&#039;, but nope that&#039;s no good, ...&quot;, hence his head explodes.  So if he gets stuck in a loop he cannot resolve, then he explodes regardless of how he got in the loop, by definition.

The important point is that the knight explodes instead of trying to save face by responding with &#039;neuter&#039; or &#039;gappy&#039;. If &#039;neuter&#039; is taken to mean &quot;neither &#039;yes&#039; or &#039;no&#039; are the truthful answer&quot;, then there is a revenge problem (just as you outline). Ask: &quot;Will you answer this question by responding either &#039;neuter&#039; or &#039;no&#039;?&quot; He can&#039;t respond &#039;yes&#039;, he can&#039;t respond &#039;no&#039; and he can&#039;t respond &#039;neuter&#039;.

But exploding is understood to be different. It is like the knight replying &quot;I cannot provide a truthful yes-no response without contradicting myself&quot;. If we ask: &quot;Will you answer this question by responding either &#039;I cannot provide a truthful yes-no response&#039; or &#039;no&#039;?&quot;. The knight will just say &quot;I cannot provide a truthful yes-no response&quot; (or he could just say &quot;donkey!&quot;). This doesn&#039;t get him into trouble because he is not attempting to answer the question. He is just throwing up his hands, shrugging his shoulders, etc. We imagined the knight (or god in the original case) being very reluctant to give up; he would try so hard to provide a true yes-no answer that his head would explode. So if you ask him, &quot;Are you going to answer &#039;no&#039; to this question or explode when I pose it to you?&quot;, he will just explode. But there is no revenge problem here (he couldn&#039;t answer without contradicting himself, so boom!). When he explodes we might infer that the correct answer was &#039;yes&#039;. But just because that was the correct answer it doesn&#039;t mean that the god could answer &#039;yes&#039; truthfully.

I like the idea of adding super gods to the story such that we can get an unbounded amount of information from a single yes-or-no question by watching the world explode all around us. I am not sure exactly how it is supposed to work, though.  You imagine an infinite chain reaction of explosions up the hierarchy. But why is it that the lower gods explode given that they are just waiting on word from the higher gods? Each god_i is waiting for god_{i+1} to instruct them; we will hear a great silence instead of a big boom (a clever way to distract the gods for eternity). In any case, the details need to be fleshed out.

It is an interesting question what the relationship between these self-referential questions, truth-telling gods and the semantic paradoxes are. Perhaps trying to solve the Liar paradox is like trying to figure out what the god should answer without being inconsistent. What answer should the god give? It seems that any attempt is open to a revenge problem. So the god just throws up his hands (or explodes). Does this suggest that if we are asked if the Liar sentence is true the best we can do is throw up our hands? I am not sure. Maybe this does suggest a kind of quietism. But I am not very satisfied with that [...would a dialetheist god would respond &quot;yes &amp; no&quot;?].</description>
		<content:encoded><![CDATA[<p>Nice! It looks like you are attempting to construct a question that has some kind of analogous relationship to a revenge liar. The key here is where you say &#8220;it is inconsistent that his head explodes&#8221;.  You must be thinking, &#8220;Look, if he explodes, then the correct answer to the question is &#8216;no&#8217; (it is not the case that his head failed to explode) and since he is a truth-teller, then he should answer &#8216;no&#8217;. So exploding is &#8220;inconsistent&#8221; with telling the truth.</p>
<p>The problem is that head explosion is not meant to be like a third truth-value. The head explodes when there is no other recourse.  The way we were thinking of the procedure is something like this: If you ask him a yes-no question, then (i) he will say &#8216;yes&#8217;, if he can provide a truthful answer without contradicting himself and said truthful answer is &#8216;yes&#8217;, (ii) he will say &#8216;no&#8217; if he can provide a truthful answer without contradicting himself and said truthful answer is &#8216;no&#8217; (iii) and he will explode otherwise.</p>
<p>Thus, all insurmountable problems lead to head explosion; the knight reasons &#8220;if my head explodes, then I could have answered &#8216;no&#8217;&#8221;, and he continues &#8220;but if I answer &#8216;no&#8217;, then I lie, so that doesn&#8217;t work, I can&#8217;t answer &#8216;yes&#8217;, so I must explode, but then I could answer &#8216;no&#8217;, but nope that&#8217;s no good, &#8230;&#8221;, hence his head explodes.  So if he gets stuck in a loop he cannot resolve, then he explodes regardless of how he got in the loop, by definition.</p>
<p>The important point is that the knight explodes instead of trying to save face by responding with &#8216;neuter&#8217; or &#8216;gappy&#8217;. If &#8216;neuter&#8217; is taken to mean &#8220;neither &#8216;yes&#8217; or &#8216;no&#8217; are the truthful answer&#8221;, then there is a revenge problem (just as you outline). Ask: &#8220;Will you answer this question by responding either &#8216;neuter&#8217; or &#8216;no&#8217;?&#8221; He can&#8217;t respond &#8216;yes&#8217;, he can&#8217;t respond &#8216;no&#8217; and he can&#8217;t respond &#8216;neuter&#8217;.</p>
<p>But exploding is understood to be different. It is like the knight replying &#8220;I cannot provide a truthful yes-no response without contradicting myself&#8221;. If we ask: &#8220;Will you answer this question by responding either &#8216;I cannot provide a truthful yes-no response&#8217; or &#8216;no&#8217;?&#8221;. The knight will just say &#8220;I cannot provide a truthful yes-no response&#8221; (or he could just say &#8220;donkey!&#8221;). This doesn&#8217;t get him into trouble because he is not attempting to answer the question. He is just throwing up his hands, shrugging his shoulders, etc. We imagined the knight (or god in the original case) being very reluctant to give up; he would try so hard to provide a true yes-no answer that his head would explode. So if you ask him, &#8220;Are you going to answer &#8216;no&#8217; to this question or explode when I pose it to you?&#8221;, he will just explode. But there is no revenge problem here (he couldn&#8217;t answer without contradicting himself, so boom!). When he explodes we might infer that the correct answer was &#8216;yes&#8217;. But just because that was the correct answer it doesn&#8217;t mean that the god could answer &#8216;yes&#8217; truthfully.</p>
<p>I like the idea of adding super gods to the story such that we can get an unbounded amount of information from a single yes-or-no question by watching the world explode all around us. I am not sure exactly how it is supposed to work, though.  You imagine an infinite chain reaction of explosions up the hierarchy. But why is it that the lower gods explode given that they are just waiting on word from the higher gods? Each god_i is waiting for god_{i+1} to instruct them; we will hear a great silence instead of a big boom (a clever way to distract the gods for eternity). In any case, the details need to be fleshed out.</p>
<p>It is an interesting question what the relationship between these self-referential questions, truth-telling gods and the semantic paradoxes are. Perhaps trying to solve the Liar paradox is like trying to figure out what the god should answer without being inconsistent. What answer should the god give? It seems that any attempt is open to a revenge problem. So the god just throws up his hands (or explodes). Does this suggest that if we are asked if the Liar sentence is true the best we can do is throw up our hands? I am not sure. Maybe this does suggest a kind of quietism. But I am not very satisfied with that [...would a dialetheist god would respond "yes &amp; no"?].</p>
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		<title>By: Tim</title>
		<link>http://xorshammer.com/2008/08/18/hardest-logic-puzzle-ever/#comment-23</link>
		<dc:creator>Tim</dc:creator>
		<pubDate>Tue, 19 Aug 2008 01:54:37 +0000</pubDate>
		<guid isPermaLink="false">http://xorshammer.wordpress.com/?p=61#comment-23</guid>
		<description>This is fantastic. My favorite sentence is &quot;If we can observe inhabitants’ heads exploding and reason based on it, we should be able to ask inhabitants about it.&quot; and my favorite sentence fragment is &quot;...by choosing the question carefully and then observing how much of the universe is destroyed by our asking it.&quot;
Deduction by exploding heads -- brilliant!</description>
		<content:encoded><![CDATA[<p>This is fantastic. My favorite sentence is &#8220;If we can observe inhabitants’ heads exploding and reason based on it, we should be able to ask inhabitants about it.&#8221; and my favorite sentence fragment is &#8220;&#8230;by choosing the question carefully and then observing how much of the universe is destroyed by our asking it.&#8221;<br />
Deduction by exploding heads &#8212; brilliant!</p>
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