Steven Clontzmathematician, professor, puzzler
http://clontz.org/
Fri, 09 Feb 2018 20:10:19 +0000Fri, 09 Feb 2018 20:10:19 +0000Jekyll v3.6.2pi-Base: A usable map of the forest<p>Here are my notes for today’s talk on <a href="http://pi-base.org">pi-Base</a> for
the University of South Alabama’s ACM and Math Clubs.</p>
<h2 id="clontzorg">Clontz.org</h2>
<ul>
<li>Notes/links from this talk are available on my <a href="http://clontz.org">website</a>.</li>
</ul>
<h2 id="relationship-of-math-and-comp-sci">Relationship of Math and Comp Sci</h2>
<ul>
<li>Math folks: knowing computer science/engineering can help you get a job.
<ul>
<li>Don’t have to take classes; just pick up a fun sideproject!</li>
</ul>
</li>
<li>CS folks: knowing math can help you get a job.
<ul>
<li>Having a math background makes people think you’re (a) smart,
(b) a problem-solver (in general).</li>
</ul>
</li>
<li>I hope to see future collaborations between ACM and Math Club, and I
encourage members of both groups to be involved with the other and
leadership to work together to support their memberships.</li>
</ul>
<h2 id="what-is-topology">What is Topology?</h2>
<ul>
<li>Topology is the a of mathematical structure that generalizes geometry
and calculus.
<ul>
<li>Classic example: <a href="https://en.wikipedia.org/wiki/Topology#/media/File:Mug_and_Torus_morph.gif">donut and coffee cup</a></li>
<li>My favorite example: <a href="http://mappmath.org/puzzles/">topology of fonts</a></li>
<li>Big data application: <a href="https://www.math.upenn.edu/~ghrist/preprints/barcodes.pdf">persistent homology</a></li>
</ul>
</li>
<li>One goal of topology is to identify the properties of topological spaces
that characterize them.
<ul>
<li>Compactness: \([0,1]\) is compact, but
\(\mathbb{R}\) is not.</li>
<li>Hausdorff: \([0,1]\) is Hausdorff, but an indiscrete space is not.</li>
<li>Normal: \([0,1]\) must be normal, because it is compact & Hausdorff.</li>
</ul>
</li>
</ul>
<h2 id="what-is-pi-base">What is pi-Base?</h2>
<ul>
<li>Topologists have studied thousands of topological spaces and thousands of
topological properties over the past century.
<ul>
<li><a href="https://en.wikipedia.org/wiki/Counterexamples_in_Topology">Counterexamples in Topology</a> surveyed about 160 of
these spaces and about 60 of these properties in 1970.</li>
<li>Paraphrasing the famous topologist Mary Ellen Rudin in her
review of Counterexamples,
<a href="http://www.jstor.org/stable/2318037?origin=crossref&seq=2#page_scan_tab_contents">topology is a dense forest of counterexamples, and a usable map of the forest is a fine thing</a>.</li>
<li><a href="http://pi-base.org">pi-Base</a> began as a sideproject of my colleague
<a href="http://jdabbs.com/">James Dabbs</a> to digitize this information into a user-friendly
web app.</li>
<li>Later pi-Base was opened up to community contributions.</li>
</ul>
</li>
<li>Features:
<ul>
<li><a href="https://topology.jdabbs.com/properties/P000013">Properties</a></li>
<li><a href="https://topology.jdabbs.com/spaces/S000026">Spaces</a></li>
<li><a href="https://topology.jdabbs.com/spaces/S000026/properties/P000013">Theorems</a></li>
<li>Canonical naming as in <a href="http://oeis.org/A000045">OEIS</a>.</li>
</ul>
</li>
</ul>
<h2 id="where-is-pi-base-going">Where is pi-Base going?</h2>
<ul>
<li>Currently I have a small grant to convert pi-Base into a modern tool for
mathematical researchers.
<ul>
<li>Major flaw in current version: lack of citations and peer-review.</li>
<li>ACM officer Cody Martin worked for me last summer to add citations
from pi-Base to Counterexamples.</li>
<li>Once this audit is complete, all contributions will require references
to a peer-reviewed manuscript to be marked as verified.</li>
</ul>
</li>
<li>pi-Base will become a treasure trove for undergraduate research.
<ul>
<li>There are still many unknowns in topology.</li>
<li>pi-Base can automatically catalog space/property pairs that are missing.</li>
<li>Three possibilities:
<ul>
<li>The question has been answered in literature not in pi-Base.</li>
<li>The question hasn’t been answered because it’s hard.</li>
<li>The question hasn’t been answered because no one’s tried: perfect
for undergrads!</li>
</ul>
</li>
</ul>
</li>
</ul>
Tue, 12 Sep 2017 10:16:00 +0000
http://clontz.org/blog/2017/09/12/acm-pibase-talk/
http://clontz.org/blog/2017/09/12/acm-pibase-talk/Alternate Standard Deviation Formula<p>While teaching MA 110 (Finite Mathematics), I casually introduced
standard deviation using the formula provided by our textbook,
where \(\sum x\) represents the sum of a dataset and
\(\overline{x}=\frac{\sum x}{n}\) is its mean:</p>
<p>\[
s
=
\sqrt{
\frac{
\sum (x^2) - n(\overline{x})^2
}{
n-1
}
}
\]</p>
<p>But, this isn’t my favorite formula for standard deviation, because it’s
not really clear what it’s trying to measure from its formula. The formula
I was more accustomed to makes it clear that standard deviation measures
the differences between datapoints and the mean, and then sums up the
squares of those differences. That is to say, it gives a measure of how
different the datapoints are from the average.</p>
<p>\[
s
=
\sqrt{
\frac{
\sum (x-\overline{x})^2
}{
n-1
}
}
\]</p>
<p>But while the textbook uses that first formula, some of our online homework
uses the second formula when working out example solutions. So, for the
record, here is a proof that both formulas do in fact measure the same thing.</p>
<p>Note that we need only prove that the numerators of each fraction are equal.
So…</p>
<p>\[
\sum (x-\overline{x})^2
\]</p>
<p>Applying \((a-b)^2=a^2-2ab+b^2\), we get:</p>
<p>\[
= \sum (x^2-2x\overline{x}+(\overline{x})^2)
\]</p>
<p>We’re allowed to add up each term separately.</p>
<p>\[
= \sum (x^2)-2\sum (x\overline{x})+\sum((\overline{x})^2)
\]</p>
<p>Now, since the last term is just adding the same number
\((\overline{x})^2\) once for each of the \(n\) members of the dataset,
we may simplify it as follows.</p>
<p>\[
= \sum (x^2)-2\sum (x\overline{x})+n(\overline{x})^2
\]</p>
<p>We may also factor out the constant \(\overline{x}\) from the middle sum.</p>
<p>\[
= \sum (x^2)-2\overline{x}\sum x+n(\overline{x})^2
\]</p>
<p>It follows from \(\overline{x}=\frac{\sum x}{n}\) that
\(n\overline{x}=\sum x\), giving the following.</p>
<p>\[
= \sum (x^2)-2\overline{x}(n\overline{x})+n(\overline{x})^2
\]</p>
<p>The solution is revealed by rearranging and combining like terms.</p>
<p>\[
= \sum (x^2)-2n(\overline{x})^2+n(\overline{x})^2
\]</p>
<p>\[
= \sum (x^2)-n(\overline{x})^2 \hspace{1em}\Box
\]</p>
Fri, 21 Apr 2017 11:52:00 +0000
http://clontz.org/blog/2017/04/21/alternate-standard-deviation/
http://clontz.org/blog/2017/04/21/alternate-standard-deviation/Job Hunting Workshop Notes<p>First things first, here are the relevant links.</p>
<ul>
<li><a href="http://overleaf.com">Overleaf.com</a> and <a href="http://sharelatex.com">ShareLaTeX.com</a>
<ul>
<li>Online LaTeX editors</li>
<li>Typeset professional documents and save/share them in the cloud</li>
</ul>
</li>
<li><a href="http://researchgate.net">ResearchGate.net</a>
<ul>
<li>Professional network for academics</li>
<li>Connect with colleagues at other institutions (useful
for potential name recommendation, but cannot replace
networking in real life)</li>
</ul>
</li>
<li><a href="http://mathjobs.org">MathJobs.org</a>
<ul>
<li>One of the main job posting websites for academic jobs
in mathematics</li>
</ul>
</li>
</ul>
<p>Some more detailed notes follow.</p>
<hr />
<h2 id="typesetting-professional-documents-with-overleafsharelatex">Typesetting Professional Documents with Overleaf/ShareLaTeX</h2>
<p>Something you can start doing now, no matter how far off your job hunt is,
is learn to typeset professional documents. In addition to writing research
papers, it’s good practice to use professional-looking templates based on the
LaTeX typesetting system. Learning LaTeX isn’t hard, especially if you use
cloud LaTeX services like <a href="http://overleaf.com">Overleaf</a> and <a href="http://sharelatex.com">ShareLaTeX</a>.</p>
<p>Both websites allow you to select pre-made templates for CVs, cover letters,
and so on. You can then open up the LaTeX code within your web browser,
make your edits to customize its contents to yourself, and then save/share the
result online or download a PDF. First impressions are important, and using
LaTeX to typeset documents is the industry standard, so it’s wise to use it when
presenting yourself to prospective future collegues.</p>
<h2 id="networking-in-real-life-and-online">Networking in real life and online</h2>
<p>Your main goal as an applicant is to get your foot in the door. In a sea of
qualified applicants, you only have a few opporunities to catch a search committee’s
attention. One important way is to have a colleague at the school you’re applying to.
That means that as soon as you have research to share, you need to start going to
conferences and meeting people in your field.</p>
<p>There’s a very small chance that you’ll
know someone on the search committee. Howeer, many search committees will ask their
colleagues for input. Those collegues will probably (at most) look through the list of
applicant names, their schools, and their fields. So, you want your name to be familiar
to that colleague, at least enough that they will take the extra step to read your cover
letter.</p>
<p>There’s no substitute for meeting colleagues in real life. However, another tool you
can use to follow up on a real life connection is <a href="http://researchgate.net">ResearchGate</a>. While
search committees should read every cover letter they receive, your colleague may not
be serving on that committee. In that case, they may only have a list of names,
schools, and fields of research; thus, you want to increase your chances of name
recognition.</p>
<h2 id="finding-jobs-and-submitting-applications">Finding jobs and submitting applications</h2>
<p>While there are other sites for academic mathematics jobs, the only one I used during
my search was <a href="http://mathjobs.org">MathJobs</a>. You should begin looking on this site 13 months
before you want to start; for example, if you’re looking for a job to start in August
2018, you should begin looking in July 2017.</p>
<p>Here’s my recommended workflow. Start by making an account. Once you’ve logged in,
go through all the job postings by clicking All on the View Jobs page. You’ll
probably notice that 80% of the postings don’t apply to you, depending on your
field of research and geographical preferences. Mark those jobs with an X so they
will be hidden by default when logged in. Other jobs you will know that you want
to apply. Mark those with a checkmark.</p>
<p>Now once you’ve filtered all the postings, you can click the yellow jagged cloud
on future logins to see what’s been posted since the last time you visited. I
recommend doing this once a week so you don’t get overwhelmed.</p>
<p>There are other places for advice on how to write CVs, cover letters, etc.,
beyond my general advice above on using professional LaTeX typesetting.
One particular thing I’d like to mention: make sure you make the case in your
cover letter for why you want that job particularly. A connection with
a faculty member or the local community is a great way to catch the attention
of search committees you have a particular interest in.</p>
<hr />
<p>Best of luck!</p>
Fri, 31 Mar 2017 18:16:00 +0000
http://clontz.org/blog/2017/03/31/job-hunting-workshop/
http://clontz.org/blog/2017/03/31/job-hunting-workshop/Katamari on the Rocks<p>So here’s an arrangement by myself (Steven Clontz, alias Steben on the Overclocked ReMix forums) of the Katamari Damacy theme, Katamari on the Rocks, for marching band. The percussion was arranged by Christopher Nelson (alias Rainman DX on the Overclocked ReMix forums). This was written somewhere around 2006, when I was a student at Auburn University (hence why I stuck a bit of “War Eagle” in the arrangement).</p>
<p>Over the last decade I’d get requests for the arrangement every couple years or so, but somewhere down the road I had misplaced the hard drive that contained them. Fortunately, I recently had the presence of mind to track down Chris, who had kept up with his copies better than I.</p>
<p>The arrangement’s Finale source files and a burned MP3 of the resulting MIDI have been stored in the following GitHub repo for posterity.</p>
<p><a href="https://github.com/StevenClontz/katamari-on-the-rocks-marching-band">https://github.com/StevenClontz/katamari-on-the-rocks-marching-band</a></p>
<p>Here’s a direct link to the MP3 if you want to give it a listen.</p>
<p><a href="https://github.com/StevenClontz/katamari-on-the-rocks-marching-band/blob/master/katamari-on-the-rocks.mp3?raw=true">https://github.com/StevenClontz/katamari-on-the-rocks-marching-band/blob/master/katamari-on-the-rocks.mp3?raw=true</a></p>
Sat, 18 Mar 2017 16:28:00 +0000
http://clontz.org/blog/2017/03/18/katamari-on-the-rocks/
http://clontz.org/blog/2017/03/18/katamari-on-the-rocks/k-Markov Strategies in Selection Games<p>Slides for my talk may be found at <a href="https://docs.google.com/presentation/d/1arkW_nLv8ph8fjn-_zqgCcSJAfdGgxIB2d7mDxPPe_Q/edit?usp=sharing">Google Slides</a>.</p>
<p>I’ve also put my notes on a more in-depth talk on my website
<a href="/blog/2017/02/05/auburn-chalk-talk/">here</a>.</p>
Sat, 04 Mar 2017 14:30:00 +0000
http://clontz.org/blog/2017/03/04/stdc2017/
http://clontz.org/blog/2017/03/04/stdc2017/Markov Strategies in Selection Games<p>Here is an outline of the talk I gave at the AU set-theoretic topology
seminar on 2017 Feb 06.</p>
<ul>
<li>Definition of the <a href="https://en.wikipedia.org/wiki/Banach%E2%80%93Mazur_game">Banach-Mazur Game</a>:
<ul>
<li>Introduced by Mazur as Problem 43 of the <a href="https://en.wikipedia.org/wiki/Scottish_Book">Scottish Book</a>
<a href="http://kielich.amu.edu.pl/Stefan_Banach/pdf/ks-szkocka/ks-szkocka3ang.pdf">[pdf]</a>, a notebook kept by mathematicians frequenting
the Scottish Cafe in Lwów, Poland in the 1930s and 1940s.</li>
<li>Let \(S\subseteq\mathbb R\). Two players alternate choosing
decreasing intervals
\(N_0\supseteq E_0\supseteq N_1\supseteq E_1\supseteq\dots\).
The first player wins if \(\bigcap_{n<\omega}N_n\) contains a point of
\(S\); the first player wins otherwise.</li>
<li>Banach (1935) showed that the second player has a winning strategy for
this game if and only if \(S\) is <em>meager</em>, a countable union
of nowhere-dense sets. The prize was a bottle of wine, awarded by Mazur.</li>
<li>Our variant of interest: two players choose<br />
decreasing non-empty open subsets
\(E_0\supseteq N_0\supseteq E_1\supseteq N_1\supseteq\dots\)
of a topological space \(X\), and let the <em>second</em> player win if and
only if \(\bigcap_{n<\omega}N_n\not=\emptyset\).
Call this game \(BM(X)\).</li>
</ul>
</li>
<li>Limited information strategies
<ul>
<li>Let \(M\) be a set of possible moves for a game.</li>
<li>A (perfect information) strategy is a function
\(\sigma:M^{<\omega}\to M\) that determines the next move for a player
based upon all the previous moves of her opponent.</li>
<li>A \(k\)-tactical strategy is a function
\(\sigma:M^{\leq k}\to M\) that determines the next move for a player
based upon the most recent \(k\) previous moves of her opponent.</li>
<li>A \(k\)-Markov strategy is a function
\(\sigma:M^{\leq k}\times\omega\to M\)
that determines the next move for a player
based upon the most recent \(k\) moves of her opponent and the current
round number.</li>
<li>A coding strategy is a function
\(\sigma:M^{\leq 2}\to M\)
that determines the next move for a player
based upon the most recent moves of both players.</li>
<li>A strategy is <em>winning</em> if a player that follows the strategy is
guaranteed to win the game regardless of the moves of the opponent.</li>
<li>A winning (perfect information) strategy for the second player in
\(BM(X)\) may always be improved to a winning coding strategy.
(Debs 1985; Galvin,Telgarsky 1986)</li>
<li>A winning Markov (that is, \(1\)-Markov)
strategy for the second player in
\(BM(X)\) may always be improved to a winning tactical
(that is, \(1\)-tactical, also known as stationary) strategy.
(Galvin,Telgarsky 1986)</li>
<li>Debs (1985) found examples of spaces for which the second player
in \(BM(X)\) has winning
(\(2\)-tactical; see Bartoszynski,Scheepers,Just 1993)
strategies, but no winning tactical strategies.</li>
<li><a href="http://www.telgarsky.com/1987-RMJM-Telgarsky-Topological-Games.pdf">Telgarksy conjectures (1987)</a>
that there exist spaces \(X_k\) for \(k<\omega\)
such that the second player in
\(BM(X_k)\) has a winning \((k+1)\)-tactical
strategy, but no winning \(k\)-tactical strategy.</li>
</ul>
</li>
<li>Countable/Finite Games
<ul>
<li>Scheepers (1992) published the first of many papers under the
title of <em>Meager-Nowhere Dense Games: \(n\)-Tactics</em>.</li>
<li>Let \(MG(X)\) be a game where the first player
chooses meager subsets \(M_{n+1}\supseteq M_n\) during round \(n+1\),
followed by the second player choosing a nowhere dense subset
\(N_{n+1}\). The second player wins if
\(\bigcup_{n<\omega}N_n\supseteq\bigcup_{n<\omega}M_n\).</li>
<li>The special case where \(X=\kappa\) has the co-finite topology
(and therefore the first player chooses countable sets and the second
player chooses finite sets) was studied by Scheepers in
<em>Concerning \(n\)-tactics in the countable-finite game</em>.
Call this game \(Sch^{\cup,\supsetneq}(\kappa)\).</li>
<li>Scheepers showed that any \((k+3)\)-tactical strategy for the second
player in \(Sch^{\cup,\supsetneq}(\kappa)\)
may be improved to a \(k\)-tactical strategy.</li>
<li>Modification: \(Sch^{\cap}(\kappa)\): the first player chooses
any countable set \(C_n\), and the second player chooses any finite set
\(F_n\). The second player wins if
\(\bigcup_{n<\omega}F_n\supseteq\bigcap_{n<\omega}C_n\).</li>
<li>The axiom \(\mathcal A’(\kappa)\) implies that the second player
has a winning \(2\)-tactical strategy in
\(Sch^{\cup,\supsetneq}(\kappa)\) (Scheepers) and
a winning \(2\)-Markov strategy in
\(Sch^{\cap}(\kappa)\) (Clontz).</li>
<li>\(A’(\aleph_n)\) holds in ZFC for \(n<\omega\) (Clontz,Dow to appear);
\(A’(\kappa)\) may be forced to hold for arbitrarily large
\(\kappa\leq\mathfrak c\) (Scheepers).</li>
</ul>
</li>
<li>Selection Games
<ul>
<li>Let \(G_{fin}(\mathcal A,\mathcal B)\)
be the game where the first player chooses
\(A_n\in \mathcal A\) and the second player chooses a finite subset
\(B_n\in[A_n]^{<\omega}\). The second player wins if
\(\bigcup_{n<\omega}B_n\in\mathcal B\).</li>
<li>For each cardinal \(\kappa\), let \(L(\kappa)=\kappa\cup\{\infty\}\)
be the space with \(\kappa\) discrete and \(\infty\) having
co-countable neighborhoods.</li>
<li>Let the Menger game be
\(Men(X)=G_{fin}(\mathcal O,\mathcal O)\) where \(\mathcal O\)
is the set of open covers of \(X\).</li>
<li>In regards to \(k\)-Markov strategies for the second player,
\(Men(L(\kappa))\) and \(Sch^{\cap}(\kappa)\) are equivalent
(Clontz).</li>
</ul>
</li>
<li>Markov strategies in Selection Games
<ul>
<li>Any \((k+2)\)-Markov strategy for the second player in
\(G_{fin}(\mathcal A,\mathcal B)\) may be improved to a
\(2\)-Markov strategy.
<ul>
<li>Source: <a href="https://www.researchgate.net/publication/282155672_Applications_of_limited_information_strategies_in_Menger%27s_game">Applications of limited information strategies in Menger’s game (to appear, CUMC)</a></li>
</ul>
</li>
<li>Assume \(|\bigcup\mathcal A|\leq\aleph_0\) and \(\mathcal B\)
is closed under supersets. Then any winning (perfect-information)
strategy for the second player in \(G_{fin}(\mathcal A,\mathcal B)\)
may be improved to a Markov strategy.
<ul>
<li>Source: <a href="https://www.researchgate.net/publication/309202868_Relating_games_of_Menger_countable_fan_tightness_and_selective_separability">Relating games of Menger, countable fan tightness, and selective separability (preprint)</a></li>
</ul>
</li>
<li>Corollaries:
<ul>
<li>Let \(\mathcal D\) give the dense subsets of a space \(X\).
The second player having a winning strategy in
\(G_{fin}(\mathcal D,\mathcal D)\) characterizes
strategic selective separability, \(SS^+\). Thus all countable
\(SS^+\) spaces are Markov selectively separable, \(SS^{+mark}\).
(Barman,Dow 2012).</li>
<li>The second player having a winning strategy in
\(G_{fin}(\mathcal O,\mathcal O)\) characterizes the
strategic Menger property.
All second-countable strategic Menger spaces are Markov Menger.
Markov Menger spaces are exactly the
\(\sigma\)-relatively-compact spaces
(equivalent to \(\sigma\)-compact for regular spaces). (Clontz)</li>
</ul>
</li>
<li>Question (Gruenhage): Does there exist an \(SS^+\) space that is not
\(SS^{+mark}\)?</li>
</ul>
</li>
</ul>
Sun, 05 Feb 2017 19:54:00 +0000
http://clontz.org/blog/2017/02/05/auburn-chalk-talk/
http://clontz.org/blog/2017/02/05/auburn-chalk-talk/Fall 2016 Classes<p>Another semester of <a href="/teaching/">teaching</a> has begun! I’m teaching
two sections of Calculus 2 at my new university: The University of
South Alabama.</p>
<p><a href="https://stevenclontz.github.io/teaching-2016-08-17-fall/">https://stevenclontz.github.io/teaching-2016-08-17-fall/</a></p>
<p>I’ve flipped the classroom this semester, so students will watch
YouTube lectures posted to that site, and then work on problems in
the classroom with me and their classmates.</p>
Fri, 19 Aug 2016 08:31:00 +0000
http://clontz.org/blog/2016/08/19/fall-2016-classes/
http://clontz.org/blog/2016/08/19/fall-2016-classes/31st Summer Conference on Topology and its Applications<p>Slides for my upcoming talk at the summer topology conference in
Leicester, England on Wednesday August 3rd:</p>
<p><a href="https://stevenclontz.github.io/scta-presentation-20160803/#/">https://stevenclontz.github.io/scta-presentation-20160803/#/</a></p>
Sun, 31 Jul 2016 15:43:00 +0000
http://clontz.org/blog/2016/07/31/summer-topology/
http://clontz.org/blog/2016/07/31/summer-topology/Zero Time Dilemma<p>The <a href="https://en.wikipedia.org/wiki/Zero_Escape">Zero Escape trilogy</a>
is my favorite video game series. To be clear, I’m not claiming it’s the
<em>best</em>, but merely my favorite. Its escape rooms can be hit-or-miss, but
the stories in each game have kept me up to 3am many nights as I tried
to unravel their intricate storylines.</p>
<p>What’s so great about their plots? Well, unfortunately, telling you would
be a huge spoiler! I can say this: each game pushes the boundaries of
story-telling within the visual novel format just a little further.</p>
<p>Unfortunately, there are a few questionable design
decisions with the latest game, Zero Time Dilemma, which can block players’
progress through the game.
Most of them could have been
solved with a better hinting system, not only within the escape rooms,
but also in how to progress the story itself. Normally such problems can
be solved by checking Google for answers, but in a game like ZTD, it’s
very easy to run across spoilers which will detract from the player’s
experience for a first playthrough.</p>
<p>My goal for this post is to
provide first-time players a guide to push them in the correct direction,
without any risk of spoilers. It should be referred to sparingly, and
only when the player feels she does not know how to proceed in furthering
the story.</p>
<p><img src="/img/20160705/fragment.jpg" alt="Fragments" /><br />
<em>To emulate the loss of the characters’ memories, gameplay is fragmented into
several sections.</em></p>
<p><img src="/img/20160705/global.jpg" alt="Global" /><br />
<em>As more of the story is revealed, the fragmented storylines are positioned
within a branching timeline, based upon the choices of the characters.</em></p>
<p>When it is unclear how to progress the story, the player either has to make
a new <strong>Decision</strong> (marked with a balance scale on the flowchart),
or needs to complete a <strong>Quest</strong> (often, escape a room; marked with a door
on the flowchart).
Searching online for Quest walkthroughs is easy enough, and has little risk
for spoilers, so I’ll be focusing on the several Decision games. Here’s an
example:</p>
<ul>
<li><strong>Read Spoiler?</strong> (example)
<ul>
<li>Outcome 1
<ul>
<li>Prerequisite:
<span class="spoiler">CQD-END:2</span></li>
<li>Hint 1:
<span class="spoiler">Think about this Decision’s title.</span></li>
<li>Hint 2:
<span class="spoiler">Well, you aren’t going to WRITE it.</span></li>
<li>Choice:
<span class="spoiler">“READ”</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisite: none</li>
<li>Choice:
<span class="spoiler">“DONT”, or let time expire</span></li>
</ul>
</li>
</ul>
</li>
</ul>
<p>The outcomes are numbered as they appear left-to-right on the fragment and
global flowcharts, not necessarily in the order they should be played.
The given prerequisites are not enforced by the game; however, unless you’ve
fulfilled them, then you shouldn’t expect to know how to make that choice
at that point of the game (and should try doing something else for now).
Here’s my suggestions before using this guide:</p>
<ol>
<li>Play any Fragments marked New.</li>
<li>Play any nodes of the Global Flowchart marked with an exclamation point.</li>
<li>Read my <em>Note on Execution Voting</em> for an explanation
on how to get all the Execution Results.</li>
<li>Look for a Decision you are missing outcomes for on the Global Flowchart
in the below list. If you have satisfied its prerequisites, then peek
at its hints. If you give up, read its choice and move on.</li>
</ol>
<p>Shoot me feedback on the Reddit post <a href="https://www.reddit.com/r/ZeroEscape/comments/4rmz6d/ztd_decision_guide_for_firsttime_players/">here</a>.
Thanks to /u/mythriz for pointing out a potential spoiler in that thread.</p>
<h2 id="time-to-decide">Time to Decide!</h2>
<p>Decisions are ordered alphabetically in this list.</p>
<ul>
<li><strong>3-Way Standoff</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">What two people are involved in the
standoff with the child?</span></li>
<li>Choice: <span class="spoiler">“ERIC”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">What two people are involved in the
standoff with the child?</span></li>
<li>Choice: <span class="spoiler">“MIRA”.</span></li>
</ul>
</li>
<li>Outcome 3
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">Violence isn’t always the answer.</span></li>
<li>Choice: <span class="spoiler">“NO ONE”.</span></li>
</ul>
</li>
<li>Outcome 4
<ul>
<li>Prerequisites: <span class="spoiler">C-END:1</span>,
<span class="spoiler">D-END:2</span>, or
<span class="spoiler">Q-END:2</span>.</li>
<li>Hint 1: <span class="spoiler">Start by trying any names you
can think of from elsewhere in the game.</span></li>
<li>Hint 2: <span class="spoiler">“MYSELF” fails - you cannot commit
suicide. Who else is there?</span></li>
<li>Hint 3: <span class="spoiler">If you’ve seen C-END:1, try to
complete the anagram.</span></li>
<li>Hint 4: <span class="spoiler">Consider attacking the character named
in D-END:2 or Q-END:2.</span></li>
<li>Choice: <span class="spoiler">“DELTA”.</span></li>
</ul>
</li>
</ul>
</li>
<li><span class="spoiler">AB</span> <strong>Game</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“ALLY”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: <span class="spoiler">“Ally” and “Betray” gameovers.</span></li>
<li>Choice: <span class="spoiler">Automatic once prerequisites are
satisfied.</span></li>
</ul>
</li>
<li>Outcome 3
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“BETRAY”.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>“Don’t Press” Button</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Let time expire.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Press the button.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Door of Truth</strong>
<ul>
<li>Outcome
<ul>
<li>Prerequisites:
<span class="spoiler">D-END:2 and Yellow Door (Q-Team Cinema)</span></li>
<li>Hint 1: <span class="spoiler">You’ve seen a hint on the other
side of the door during Yellow Door (Q-Team Cinema)</span></li>
<li>Hint 2: <span class="spoiler">You saw the birth of two babies
leading up to D-END:2.</span></li>
<li>Choice: <span class="spoiler">“11162029” (for 11/16/2029).</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Dream or Reality</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Let time expire.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Press the button.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Execution Vote: C/Q/D</strong>
<ul>
<li>See <em>Note on Execution Voting</em> below.</li>
</ul>
</li>
<li><strong>Final Decision</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“DON’T SHIFT” or let time expire.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“SHIFT”.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Find the Mask</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Hint 1: <span class="spoiler">Think about the Monty Hall problem.</span></li>
<li>Hint 2: <span class="spoiler">You have only a 10% chance of choosing
the correct locker without any added information.</span></li>
<li>Choice: <span class="spoiler">The initial choice is irrelevant, but
you have a 90% chance of getting this Correct outcome if you
switch.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Hint 1: <span class="spoiler">Think about the Monty Hall problem.</span></li>
<li>Hint 2: <span class="spoiler">You have only a 10% chance of choosing
the correct locker without any added information.</span></li>
<li>Choice: <span class="spoiler">The initial choice is irrelevant, but
you have a 90% chance of getting this Incorrect outcome if you
don’t switch.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Force Quit Box: C</strong>
<ul>
<li>Outcome
<ul>
<li>Prerequisites: <span class="spoiler">C-END:1</span></li>
<li>Hint 1: <span class="spoiler">Recall the “spell to open Pandora’s
box”.</span></li>
<li>Hint 2: <span class="spoiler">Zero tells you this Latin phrase, which
means “live for today”.</span></li>
<li>Choice: <span class="spoiler">“VIVEHODIE”</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Force Quit Box: D</strong>
<ul>
<li>Outcome
<ul>
<li>Prerequisites:
<span class="spoiler">strong recollection of Virtue’s Last
Reward</span> or
<span class="spoiler">Gun vs Incinerator Outcome 3 plus some
recollection of VLR</span> or
<span class="spoiler">D-END:2</span></li>
<li>Hint 1: <span class="spoiler">If you remember the events of Virtue’s
Last Reward, you may be able to figure this out. Neither
item belongs to Sigma, so…</span></li>
<li>Hint 2: <span class="spoiler">For the first item,
recall Diana’s memory of Phi from
D-COM, a few days before the events of the game. For the second
you’ll need a certain ending from this game, or Luna’s ending
from VLR.</span></li>
<li>Hint 3: <span class="spoiler">What items were sent along in a
transporter?</span></li>
<li>Choice: <span class="spoiler">“BROOCH” & “BLUE BIRD”/”MUSIC BOX”</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Force Quit Box: Q</strong>
<ul>
<li>Outcome
<ul>
<li>Prerequisites: <span class="spoiler">Q-END:2</span></li>
<li>Hint: <span class="spoiler">Zero reveals the code to this panel
in another timeline.</span></li>
<li>Choice: <span class="spoiler">“38080832”</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Gift from Gab</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“DRINK”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“DON’T DRINK”.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Gun vs Incinerator</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Shoot the gun. This outcome occurs
50% of the time.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Shoot the gun. This outcome occurs
50% of the time.</span></li>
</ul>
</li>
<li>Outcome 3
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Let time expire.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Heads or Tails</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: <span class="spoiler">CDQ-END:1</span></li>
<li>Choice: <span class="spoiler">Irrelevant. Always triggers after
fulfilling prerequisite, then 50% afterwards.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Irrelevant. Always triggers before
Outcome 1’s prerequisite, then 50% afterwards.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Helmet Code</strong>
<ul>
<li>Outcome
<ul>
<li>Prerequisites: <span class="spoiler">Reality (Bad end)</span></li>
<li>Hint: <span class="spoiler">Zero reveals the code to the helmet
in another timeline.</span></li>
<li>Choice: <span class="spoiler">“61404091”</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Inject</strong> <span class="spoiler">Radical-6</span><strong>?</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“INJECT RADICAL-6”</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“DON’T INJECT”</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Kill Button: C/Q/D</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Let time expire.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Press the button.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>The</strong> <span class="spoiler">Antidote</span> <strong>Is?</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Hint 1: <span class="spoiler">The antidote can be uniquely
determined by considering which characters’ tounges are numb.</span></li>
<li>Hint 2: <span class="spoiler">Akane numbered the eight vials
A=0 through H=7, and converted them to binary, 000 through 111.
Carlos tested the vial if the first binary digit was 1, Junpei tested
the vial if the second binary digit was 1, and Akane tested the vial
if the last binary digit was 1.</span></li>
<li>Hint 3: <span class="spoiler">Match the tasting results with
the background image in the Decision. Carlos took the top
row, Junpei took the second row, and Akane took the final row.
Or, read the Log carefully and use logic to deduce the solution.</span></li>
<li>Choice: <span class="spoiler">Choose correctly using the
above hints.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Choose incorrectly, which occurs
with 87.5% likelihood when selecting randomly anyway.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>The</strong> <span class="spoiler">Bomb</span>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“RUN AWAY”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“DON’T RUN”.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>The Three Dice</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">There’s nothing you can do/learn to
improve your odds, but the game is gracious enough to cheat in
your favor.</span></li>
<li>Choice: <span class="spoiler">Keep rolling the dice. The game will
always give it to you on your third attempt.
(If it didn’t, you would have to roll the dice 150 times just
for a 50% chance at victory!!)</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Just roll the dice. It is the 99.537%
reality, after all.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>Transporter</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“DON’T TRANSPORT”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">“TRANSPORT”.</span></li>
</ul>
</li>
</ul>
</li>
<li><span class="spoiler">Who Killed</span> <strong>Junpei?</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">Who is the biggest danger to Carlos
right now?</span></li>
<li>Choice: <span class="spoiler">“AKANE”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">In this timeline,
Carlos seemed shocked that he
voted for Q-Team instead of D-Team.</span></li>
<li>Choice: <span class="spoiler">“CARLOS”.</span></li>
</ul>
</li>
</ul>
</li>
<li><span class="spoiler">Who Killed</span> <strong>Mira?</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Hint: <span class="spoiler">Even if you knew, Eric’s not
exactly interested in hearing it.</span></li>
<li>Choice: <span class="spoiler">Let time expire, “DONNO”,
or “I DONT KNOW”.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Almost anything, including
gibberish. However, you’ll get some interesting unique reactions if
you try various names from the game.</span></li>
</ul>
</li>
</ul>
</li>
<li><strong>You</strong> <span class="spoiler">are Zero</span><strong>!</strong>
<ul>
<li>Outcome 1
<ul>
<li>Prerequisites: none</li>
<li>Choice: <span class="spoiler">Almost anything, including
gibberish. However, you’ll get some unique reactions if
you try names of characters you can see in the room with you.</span></li>
</ul>
</li>
<li>Outcome 2
<ul>
<li>Prerequisites: <span class="spoiler">C-END:1</span> or
<span class="spoiler">D-END:2</span>.</li>
<li>Hint 1: <span class="spoiler">If you’ve seen C-END:1, try to
complete the anagram.</span>.</li>
<li>Hint 2: <span class="spoiler">If you’ve seen D-END:2, you’ve
seen this person in this room before.</span>.</li>
<li>Choice: <span class="spoiler">“DELTA”.</span></li>
</ul>
</li>
</ul>
</li>
</ul>
<h2 id="note-on-execution-voting">Note on Execution Voting</h2>
<p>My biggest peeve with Zero Time Dilemma
is the lack of clarity in how to trigger the different
Execution Results. You may reasonably think that as soon as your flowchart
contains the appropriate Decisions to execute another team, the game should
recognize that execution.</p>
<p>But nope. Instead, the game secretly stores the latest Decisions you made
for each team in Execution Vote: C/Q/D. So, you need to be careful of the
order you make your Decisions, or go back and redo all three each time.
For your convenience,
here’s one of the quickest ways you can trigger all four Election Results:</p>
<ul>
<li>C votes for D</li>
<li>D votes for Q</li>
<li>Q votes for C
<ul>
<li>All Teams Survived</li>
</ul>
</li>
<li>C votes for Q
<ul>
<li>Q-Team Executed</li>
</ul>
</li>
<li>D votes for C
<ul>
<li>C-Team Executed</li>
</ul>
</li>
<li>Q votes for D</li>
<li>C votes for D
<ul>
<li>D-Team Executed</li>
</ul>
</li>
</ul>
Tue, 05 Jul 2016 22:21:00 +0000
http://clontz.org/blog/2016/07/05/zero-time-dilemma/
http://clontz.org/blog/2016/07/05/zero-time-dilemma/Infinite Blue Eyes Puzzle<p>I was asked by my friend and colleague
<a href="http://stephengware.com/">Dr. Stephen “Danger” Ware</a> on a Facebook post about another
<a href="https://www.reddit.com/r/AskReddit/comments/4kz3di/whats_your_favourite_maths_fact/d3jj88j">interesting puzzle concerning infinity</a> if the
<a href="http://xkcd.com/blue_eyes.html">Blue Eyes Puzzle</a> could be extended to an infinite number
of islanders. Intuitively, it seems reasonable that the
<a href="https://xkcd.com/solution.html">solution</a> should generalize nicely: if there are infintely
many pairs of blue eyes, it will still take “infinitely many days”. But, these
things can be subtle once infinity is thrown into the mix,
so I thought it’d be fun to attempt a rigorous treatment
of the question.</p>
<p>First things first, I’m going to reformulate the puzzle to avoid the words
“everyone is a perfect logician”, which for some reason bugs me more than
dealing with infinity or islanders who cannot communicate with each other.</p>
<p>So, suppose there is an island with several inhabitants. These islanders cannot
communicate with each other, but they share a few truths:</p>
<ul>
<li>All islanders are immortal, except as explained below.</li>
<li>All islanders know the color of every other islander’s eyes, but not their own.</li>
<li>If an islander has enough information to logically deduce the color of their
own eyes, they will perish at midnight of the following day, which will be
noticed by the other islanders.</li>
<li>Any message washed ashore in a bottle can be trusted.</li>
</ul>
<p>Now as it happens, at 12:01am on Day 0, a message washes ashore in a bottle,
and is read by all the islanders:</p>
<blockquote>
<p>There is at least one islander with blue eyes.</p>
</blockquote>
<p>What happens?</p>
<h2 id="finite-solution">Finite solution</h2>
<p>Of course, since it wasn’t specified, you would reasonably assume that there
are only finitely many islanders. But, in fact, the following solution works
for infinitely many islanders, as long as only finitely many have blue eyes.</p>
<p>The easy case is when there is only one islander with blue eyes. She reads
the message, looks around, and sees no one else with blue eyes. Therefore,
she correctly deduces she must be the islander with blue eyes. So, at midnight
on Day 1, she’s dead as a doornail.</p>
<p>What about the rest of them? Well, suppose you are another islander. You
may be right to fear that you have blue eyes when that message arrives.
However, you saw at least one other person on the island who did have blue eyes,
which means that the message told you absolutely nothing new about yourself.
Importantly, you have no other way of gaining more information
<em>except by observing when other islanders perish or survive on following days</em>.
Thus, you will not die on Day 1.</p>
<p>Now, when you see the blue-eyed islander pass away at noon on Day 1, her
passing is itself new information. Putting yourself in her shoes, you realize
that if you had blue eyes, she’d have been in the same position as you were
the previous days, and would have lived just as you did. Alas, she did not,
so you conclude she saw that your eyes were not blue, and since they could
be any other color, you rest easy with the lack of knowledge. Thus you,
and every other islander, live on.</p>
<p>So, we have proven the following fact.</p>
<p><strong>Proposition</strong> If there is only one islander with blue eyes, she will
perish at midnight on Day 1, and all others will survive.</p>
<p>However, we are more interested in the lemmas which led us to this
conclusion.</p>
<p><strong>Lemma 1’</strong> If an islander sees at least one islander with blue eyes, she cannot
deduce her eye color by Day 1.</p>
<p><strong>Lemma 2’</strong> If an islander sees no other islanders with blue eyes, she can
deduce her eye color by Day 1.</p>
<p>We’d like to prove two generalizations. We will always assume
\(N\in\omega=\{0,1,2,\dots\}\).</p>
<p><strong>Lemma 1</strong> If an islander sees at least \(N\) other islanders with blue
eyes, she knows on Day \(0\) that she will not be able to deduce her eye
color before Day \(N\).</p>
<p><em>Proof.</em> This is trivially true for \(N=0\), so by induction we may assume
it is true for
\(N\) and aim to show it for \(N+1\). Let \(Alice\) be such
an islander who sees at least
\(N+1\) blue-eyed islanders. By induction, she cannot deduce her eye color
by noon on Day \(N\) as she sees at least \(N\) blue-eyed islanders.
Importantly, Alice realizes that no other islander can deduce their eye
color by then either! Any blue-eyed islander besides \(Alice\) can see
the other \(N\) blue-eyed islanders \(Alice\) observed, so they also are
subject to the induction hypothesis. And of course the non-blue-eyed islanders
can see the same \(N+1\) blue-eyed islanders \(Alice\) noticed, so they
too see at least \(N\) pairs of blue eyes.</p>
<p>Since \(Alice\) knows from Day \(0\) that no other islander can deduce his
own eye color during the first \(N\) days, so she gains no information she
cannot deduce herself from the beginning when noon on day \(N\) rolls around
and everyone is still alive. Therefore,
she still cannot deduce her eye color by noon on day \(N+1\). \(\square\)</p>
<p><strong>Lemma 2</strong> If an islander sees \(N\) other islanders with blue eyes who
are alive on Day \(N\), then she can deduce her eye color by Day \(N+1\).</p>
<p><em>Proof.</em> We again proceed by induction as this holds for \(N=0\). Let
\(Alice\) be an islander who sees \(N+1\) other islanders with blue eyes
who are alive on Day \(N+1\). Let \(Bob\) be one of the islanders she
observes with blue eyes. She proceeds by contradiction. If she lacked blue
eyes, then on Day \(N\) Bob would have seen the \(N\) other islanders with
blue eyes that Alice observed. Thus, by induction, he would have perished
at midnight on Day \(N+1\), a contradiction. Therefore, Alice may deduce
that she has blue eyes. \(\square\)</p>
<p><small><i>
It’s interesting that I used contradiction above. It would seem that Alice must
assume the <a href="https://en.wikipedia.org/wiki/Law_of_excluded_middle">
Law of the Excluded Middle</a> here, but I am not a logician, so perhaps
I missed something.
</i></small></p>
<p><strong>Theorem 3</strong> If there are \(N+1\) islanders with blue eyes, they will
perish at midnight on Day \(N+1\), and all others will survive.</p>
<p><em>Proof.</em> Each blue-eyed islander sees \(N\) other islanders with blue
eyes, so by the above lemmas they will be alive on Day \(N\), but must
perish on Day \(N+1\). The other islanders see \(N+1\) other islanders
with blue eyes, and thus survive on Day \(N+1\). Since the blue-eyed islanders
perished that morning, the others deduce that they cannot have blue eyes,
but cannot specify their eye color further and thus survive.
\(\square\)</p>
<h2 id="whats-the-wait-for">What’s the wait for?</h2>
<p>It’s worth pointing out that when there is more than one islander with blue eyes,
no one gains information from the message on the bottle. So, what’s the hold up?
It wasn’t obvious to me from the above proof (even though I wrote it!),
so here’s my intuition.</p>
<p>The important thing to note is that if you see \(N+2\) people with blue eyes,
you are waiting until they give you more information. You cannot tell if your
eyes are blue or not. And if they aren’t, a blue-eyed islander sees only
\(N+1\) people with blue eyes. In their shoes, they have the same uncertainty,
and they should consider the possibility they lack blue eyes, in which case
the other blue-eyed islanders see only \(N\) people with blue eyes.
And so on, and so forth. The idea is that you know it will take at
least \(N+1\) days to remove the uncertainty from the other islanders with blue
eyes, because they lack information you have about their own eyes.
That’s why the wait is required.</p>
<h2 id="the-infinite-case">The infinite case</h2>
<p>So, we’ve handled the case where there are an arbitrary number of islanders
(even infinite!), but only finitely many islanders with blue eyes. So,
what if there are infinitely many islanders with blue eyes?</p>
<p>Well, we already have a very nice lemma above, <strong>Lemma 1</strong>. That allows us
to prove that, since all islanders see infinitely many pairs of blue eyes,
no islander can deduce their eye color by Day \(N\) for all
\(N\in\{0,1,2,\dots\}\).</p>
<p>So, assuming that our days are limited to the finite ordinals, the case is
closed. But why should we stop there? What if there exists a day for every
<a href="https://en.wikipedia.org/wiki/Ordinal_number">ordinal number</a>?</p>
<p>We should consider what happens on Day \(\omega\). Will anyone wake up dead?
The answer is of course, no. All islanders were able to deduce from the
beginning that no one could perish after a finite number of days, so
by the time Day \(\omega\) begins, no one has gained information they didn’t
already have. We shall be able to extend this argument by
<a href="https://en.wikipedia.org/wiki/Transfinite_induction">transfinite induction</a> to handle every ordinal
(regardless of the cardinality of the ordinal or the set of blue-eyed islanders).</p>
<p><strong>Theorem 4</strong> If there infinitely many islanders with blue eyes, then no islanders
can deduce their eye color by Day \(\alpha\), where \(\alpha\) is any ordinal.</p>
<p><em>Proof.</em> If \(\alpha\) is finite, we are done by Lemma 1. By transfinite induction,
we may assume the theorem holds for all \(\beta\lt\alpha\). In order to
deduce her eye color by Day \(\alpha\), \(Alice\) requires new information on
a previous day. However, by the induction hypothesis, \(Alice\) knows from
the beginning that every islander cannot deduce their eye color by a previous
day. So, if \(\alpha\) is a limit ordinal, we are done. The other case
is when \(\alpha=\beta+1\) for some \(\beta\lt\alpha\). Alice wakes up
on Day \(\beta\), and is immediately bored. She already knew that everyone
sees infinitely many islanders with blue eyes, so she already knew that
everyone would be alive. Indeed, they are. So, she has gained no new information
that morning, and therefore cannot deduce anything new by the end of the day,
the beginning of Day \(\alpha\). \(\square\)</p>
<h2 id="in-conclusion">In conclusion</h2>
<p>In working on this article, I was particularly amused by the following:</p>
<ul>
<li>If the cardinality \(\kappa\) of islanders with blue eyes is <strong>finite</strong>,
then all blue-eyed islanders will <strong>perish</strong> after finitely-many days.
(Specifically, \(\kappa\)-many.)</li>
<li>If the cardinality \(\kappa\) of islanders with blue eyes is <strong>infinite</strong>,
then all blue-eyed islanders will <strong>remain alive</strong> no matter how many days
pass, even after \(\kappa\)-many days.</li>
</ul>
<p>I have some intuition here: in any case, the unknown datum for each
islander (their own eye color) is finite. Thus, if there are finitely many
blue-eyed islanders, this datum is important! But, if there are infinitely many
blue-eyed islanders, then the status of an islander’s own eyes is irrelevant
to the cardinality of that set. Therefore, the bottled message, which says
that the cardinality of blue-eyed islanders is positive, gives no new information
to the islanders, <strong>because they already have exactly the same information about
the cardinality of blue-eyed islanders</strong>.</p>
<p>I hope you found all that interesting. If you have any comments or questions,
shoot them to me at <a href="http://twitter.com/StevenXClontz">@StevenXClontz</a>.
Thanks for reading! :-)</p>
Thu, 26 May 2016 10:18:00 +0000
http://clontz.org/blog/2016/05/26/infinite-blue-eyes/
http://clontz.org/blog/2016/05/26/infinite-blue-eyes/