Update on Overleaf.

rook/fitch-trick.tex

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+% !TEX TS-program = xelatex
+\documentclass{../leelavati}
+
+\usepackage{qrcode}
+
+\begin{document}
+
+\selectlanguage{english}
+\setstretch{1.1}
+
+\title{\LARGE The Fitch Trick and Variants \\
+\large ଫିଚ୍‌ଙ୍କ ତାସ ଖେଳ ଏବଂ ଏହାର ରହସ୍ୟ}
+\author{\AuthorDual{ନୀଲ୍‌ଧାରା ମିଶ୍ର}{Neeldhara Misra}}
+\date{}
+\maketitle
+
+\setstretch{1.2}
+
+\begin{abstract}
+ଯାଦୁକର ପ୍ରଥମେ ଦର୍ଶକଗଣଙ୍କୁ ଏକ ତାସ୍ ବିଡ଼ାକୁ ଭଲଭାବେ ମିଶାଇ ସେଥିରୁ ଯେକୌଣସି ପାଞ୍ଚଟି ତାସ୍ ପତ୍ର ବାଛିବା ପାଇଁ ନିର୍ଦ୍ଦେଶ ଦିଅନ୍ତି। ଏହାପରେ ଯାଦୁକର ସେହି ପାଞ୍ଚଟି ତାସ୍‌ଗୁଡ଼ିକୁ ନିରୀକ୍ଷଣ କରି ସେଥିରୁ ଗୋଟିଏ ତାସ୍‌କୁ ଲୁଚାଇ ଦିଅନ୍ତି ଏବଂ ବାକି ଚାରୋଟି ତାସ୍‌କୁ ଖୋଲାଭାବରେ (ମୁହଁ ଉପରକୁ କରି) ଗୋଟିଏ ଧାଡ଼ିରେ ସଜାଇ ରଖନ୍ତି। ତା'ପରେ ସେ ବାହାରେ ଅପେକ୍ଷା କରିଥିବା ନିଜର ଜଣେ ସହଯୋଗୀଙ୍କୁ ଡାକନ୍ତି। ସହଯୋଗୀ ଜଣକ ଆସି ସେହି ଚାରୋଟି ତାସ୍‌ଗୁଡ଼ିକ ଦେଖିବା ମାତ୍ରେ ହିଁ ଲୁଚାଯାଇଥିବା ତାସ୍‌ଟିକୁ ସଠିକ୍ ଭାବରେ ଚିହ୍ନଟ କରିଦିଅନ୍ତି। ଏହା କିପରି ସମ୍ଭବ?
+\end{abstract}
+
+\setstretch{1.05}
+
+Aalaya the magician and her assistant Badarika are about to perform an impossible feat of telepathy. To prove there is no foul play, Badarika leaves the room. She leaves behind her phone, smartwatch, and any other gadgets to ensure no one can communicate with her. 
+
+Then the audience picks out five cards from a shuffled deck and hands them to Aalaya. She studies the five cards, secretly removes one, and in a single fluid motion arranges the remaining four face-up neatly in a row.
+
+Once everything is set, Badarika is called back into the room. She carefully observes the four visible cards… and, to everyone’s amazement, correctly identifies the fifth card—the one she has never seen. How is this possible? 
+
+\section*{The Mathematics Behind the Magic}
+Notice that Badarika has to guess one out of $48$ possible cards, since the deck has $52$ cards of which four are face up. On the other hand, Aalaya can arrange the four cards in $4! = 24$ different orders. This feels like we are falling short by a factor of two! 
+
+But Aalaya actually has a point of leverage that we have not yet accounted for: she can \emph{choose} the card that she wants to set aside.
+
+\newpage 
+
+In other words, she is not just choosing an order for the four visible cards—she is also choosing which of the five cards to hide. This gives her $ 4! \times 5 = 120$ possible signals, more than enough to distinguish between the $48$ possibilities Badarika needs to resolve.
+
+{\par\centering
+\includegraphics[scale=0.21]{pic.jpg}
+\captionof*{figure}{In all these three cases, Aalaya shows the first four cards and Badarika guesses the fifth. What do you notice? Can you spot a pattern?}
+\par
+}
+
+So how do we think about a signaling strategy? Let's break down the numbers:
+\begin{itemize}
+    \item There are ${52 \choose 5} = 2,598,960$ possibilities for what the audience can hand Aalaya. Call this set $\color{magHighlight}{\mathbf{P}}$. 
+    \item The number of ordered sequences of four cards Aalaya can place on the table, say $\color{magHighlight}{\mathbf{Q}}$, is $52\cdot 51\cdot 50\cdot 49 =  6,497,400$.
+\end{itemize}
+
+Each of the $120$ possible signals corresponds to an element of $\color{magHighlight}{\mathbf{Q}}$.
+
+
+What we want is a rule that assigns each $\alpha \in \color{magHighlight}{\mathbf{P}}$ (what the audience does) to a unique and feasible choice of $\beta \in \color{magHighlight}{\mathbf{Q}}$ (what Aalaya would do). Badarika then comes in, sees $\beta$, works backward to $\alpha$, and identifies the missing card.
+
+In fact, since $\color{magHighlight}{\mathbf{Q}}$ has more elements that $\color{magHighlight}{\mathbf{P}}$, such a mapping definitely exists. But more interestingly, can you find a simple rule they could execute live, under pressure, with an audience watching?
+
+\begin{artprobox}
+We write cards by rank and suit, with $S, H, C,$ and $D$ denoting Spades ($\spadesuit$), Hearts ($\heartsuit$), Clubs ($\clubsuit$), and Diamonds ($\diamondsuit$). Let $\alpha = \langle 4S, 5H, 2C, 7C, 6D \rangle$ be a hand given to Aalaya. Which $\beta$ does $\alpha$ map to under your rule?
+\end{artprobox}
+
+\section*{Pushing the Limits}
+You may have noticed that there is a considerable gap between the number of signals ($6,497,400$) and the number of possibilities ($2,598,960$) we calculated earlier. This raises a natural question: can we turn this extra room into a more impressive trick? The effect remains the same: four cards are shown in a fixed order, and a fifth is guessed. But could the audience be choosing from a \emph{larger} deck?
+
+If the audience is choosing from a deck of size $n$ in general, then there are ${n \choose 5}$ possible sets of cards they could hand to Aalaya, while the number of ordered sequences of four cards she can display is $n\cdot (n-1)\cdot (n-2)\cdot (n-3)$. For the trick to work, the space of signals must be at least as large as the space of possibilities-otherwise, two different hands would have the same signal introducing ambiguity. 
+
+\begin{artprobox}
+From the inequality 
+\[
+n(n-1)(n-2)(n-3) \geq \binom{n}{5},
+\]
+conclude that $n \leqslant 124$.    
+\end{artprobox}
+
+This shows that $n \leq 124$ is a \emph{necessary} condition for the trick to work, but not necessarily a sufficient one. The next challenge is to show that the trick can, in fact, be made to work with a deck this large.
+
+So once you have figured out how to do it with a standard deck, try pushing further. Can you extend your method to larger decks, perhaps all the way up to $124$?
+
+\section*{Some Variations}
+A variation of this trick, performed by \textbf{Elwyn Berlekamp}, uses a deck of size $64$ and adds an extra twist: Badarika also correctly guesses the outcome of a coin flip made by an audience member. If you can figure out the 124-card version, you should start to see how to handle the version with the coin as well.
+
+You might also wonder whether the reveal really has to rely only on the \emph{ordering} of the four cards.  A variation suggested by \textbf{Colm Mulcahy} expands the signaling method: the cards are not only ordered but some are turned face up while others are face down. This allows the hidden card to be encoded using only \emph{three} cards instead of four! This makes for great theatrical flair, as discarding a card becomes part of the performance.
+
+We can frame the trick more generally as follows: the audience selects $p$ cards from a deck of size $n$, and the magician reveals $q$ of them. For which values of $n$, $p$, and $q$ is such a trick possible? And when it is possible, how many distinct signaling strategies are there?
+
+Please feel free to email your ideas to the author at the email below. And keep an eye out for future issues which will bring solutions and new puzzles!
+
+{\par\centering
+    \centering
+    \begin{minipage}{0.25\textwidth}
+        \centering \boxed{\color{magHighlight}\qrcode[height=1.1in]{https://short.neeldhara.website/fitch-video}}
+    \end{minipage}%
+    \begin{minipage}{0.25\textwidth}
+        \centering
+        \boxed{\qrcode[height=1.1in]{https://short.neeldhara.website/fitch-interactive}}
+    \end{minipage}
+    \captionof*{figure}{Left: link to a Youtube demonstration of the trick in its basic form. Right: link to an interactive where you can practice additional examples.}
+\par}
+
+{\small
+\vskip 0.05in
+\affiliation[1.6em]{Associate Professor, CSE; IIT Gandhinagar}
+\email{neeldhara.misra@gmail.com}
+}
+\end{document}

rook/rooks-fighting.tex

@@ -1,71 +0,0 @@
-% !TEX TS-program = xelatex
-\documentclass{../leelavati}
-\usepackage{framed}
-
-\begin{document}
-
-\selectlanguage{english}
-\setstretch{1.1}
-
-\title{\LARGE The Fitch Trick and Variants}
-\author{\AuthorDual{ନୀଲ୍‌ଧାରା ମିଶ୍ର}{Neeldhara Misra}}
-\date{}
-\maketitle
-
-\setstretch{1.2}
-
-\begin{abstract}
-The magician selects a volunteer to shuffle the deck and choose any five cards. The magician examines the cards, hides one of the five cards, sets the remaining four cards in a faceup row, and has the volunteer retrieve an accomplice from outside the room. The accomplice briefly examines the cards and identifies the missing card.
-\end{abstract}
-
-\setstretch{1.1}
-
-This is a trick that is to be performed by Aalaya (a magician) and Badarika (an accomplice) with a standard deck of $52$ cards. Badarika steps out of the room and leaves her phone, smartwatches, etc. behind, so nobody can communicate with her. Then the audience picks out five cards out of the deck and gives it to Aalaya. Aalaya then arranges four out of these five cards in a row\footnote{The cards can also be placed in a stack, if the audience worries that Aalaya is signaling by angularities in the layout of the cards.} and Badarika comes back and takes a good look at the four cards on the table, and then identifies the missing card.
-
-How is this possible? 
-
-\begin{framed}
-...image here.
-\vspace{140pt}
-\end{framed}
-
-\newpage 
-
-Notice that Badarika has to guess one out of 48 possible cards, since the deck has 52 cards of which four are face up. On the other hand, Aalaya can arrange the four cards in $4! = 24$ different orders, which is fewer than 48. This seems like we are falling short by a factor of two: but notice that Aalaya actually has a point of leverage that we have not accounted for: she can \emph{choose} the card that she wants to set aside.
-
-So what is the right way of thinking about a signaling strategy? On the one hand, there are: $${52 \choose 5} = 2,598,960$$ possibilities for what can happen at the trick. Call this set $\mathcal{P}$. Now the number of ordered sequences of four cards, say $\mathcal{Q}$, is $$52\cdot 51\cdot 50\cdot 49 =  6,497,400.$$
-
-As an example, let $\alpha = \langle 4S, 5H, 2C, 7C, 6D \rangle \in \mathcal{P}$ be the subset of cards handed to Aalaya by the audience. Now Aalaya has $120$ possibilities for what she can do with these cards. These possibilities are all recorded in $\mathcal{Q}$. What we want is a way to map every element $\alpha$ in $\mathcal{P}$ (what the audience does) to a distinct element $\beta$ in $\mathcal{Q}$ (what Aalaya would do). Now Badarika comes and sees $\beta$ and maps it back to $\alpha$ and figures out the missing card. 
-
-Can you figure out such a mapping that Aalaya and Badarika can use to perform and impress? You can use math to argue that such a mapping exists \emph{in principle}, but more usefully, you can also try to construct an explicit mapping that both Aalaya and Badarika can use in practice to impress!
-
-\newpage 
-
-Incidentally, if Aalaya is facing a tough crowd, they might not let her pick the hidden card, and might insisting on making the choice for her. Then indeed Aalaya has only $24$ signals: but turns out there is a way out even in this scenario. 
-
-One way out, for example is the following: Aalaya can attempt to use her 24 signals in a way that narrows down the possibilities for the hidden card to just 2. Once Badarika sees the 4 cards and figures out the two possibilities, she runs through the whole deck and finds the two cards and casually places one on the top of the deck and the other at the bottom. After this setup, she can distance herself from the deck and ask the audience what the card was.
-
-If the audience shouts out the card at the top, then Badarika can snap her fingers and pull up the top card, and if the audience indicates the other card, then Badarika can simply flip the whole deck over with a dramatic flourish instead. 
-
-You may have also noticed that there is a difference between the number of signals ($6,497,400$) and the number of possibilities ($2,598,960$) that we calculated earier. A natural question then is if we can make a more impressive trick with the same constraint on Aalaya and Badarika: that is the effect remains that four cards are shown in a fixed order and a fifth card is guessed, but can the audience make their pick from a larger deck?
-
-If the audience is choosing from a deck of size $n$ in general, then there are: ${n \choose 5}$ possibilities for what can happen at the trick and the number of ordered sequences of four cards is $n\cdot (n-1)\cdot (n-2)\cdot (n-3)$. We need, at the very least, that the space of the number of signals is at least as large as the space of possibilities. 
-
-In other words, we would need:
-
-$$\binom{n}{5} \leqslant n(n-1)(n-2)(n-3),$$
-
-which, upon simplification, works out to $n \leqslant 124$. This, of course, only tells us that $n \leqslant 124$ is a necessary condition for the card trick to work in the form that we have discussed --- it may or may not be sufficient. Our next challenge will be to show card trick does, in fact, work with a deck containing as many as 124 cards. So once you have figured out how to do it with a standard deck, either in principle or in practice, see if you can extend your arguments to larger and larger decks: perhaps all the way to 124?
-
-A variation of this trick, as performed by Elwyn Berlekamp, is to use a deck of size 64, with the additional feat of guessing the result of a coin flipped by an audience member. If you can figure out a way of doing the 124-variant, you will probably also have some ideas for the version with the coin. 
-
-You may also wonder if the reveal should necessarily involve an ordering of the four cards. A variation of the trick suggested by Colm Mulcahy involves revealing the cards in an order but also flipping some over, so that some cards are face up and some are face down. It turns out that with some clever observation, you can signal the hidden card using only three cards rather than four! This makes for great banter, as discarding a card is an excellent excuse for some drama.
-
-In its most general version, this trick has the audience pulling out $p$ cards out of a deck of $n$ and the magician revealing $q$ of the $p$ cards. For what values of $n$, $p$ and $q$ is this feasible? And for a feasible setup, how many distinct strategies are there?
-
-{\small
-\vskip 0.05in
-\affiliation[1.6em]{Associate Professor, CSE; IIT Gandhinagar}
-\email{neeldhara.misra@gmail.com}
-}
-\end{document}