Update Fitch trick solutions.

Co-authored-by: Cursor <cursoragent@cursor.com>

fitch/fitch-trick-solutions-part-one.tex

@@ -0,0 +1,522 @@
+% !TEX TS-program = xelatex
+\documentclass{../leelavati}
+
+\usepackage{qrcode}
+
+% -------------------------------------------------------------------
+% Playing-card drawing helpers (TikZ)
+% -------------------------------------------------------------------
+\definecolor{cardred}{HTML}{C0392B}
+\definecolor{cardblack}{HTML}{1A1A1A}
+
+% Map a suit letter S/H/C/D to a symbol (\cursym) and a colour (\curcol)
+\newcommand{\cursym}{}
+\newcommand{\curcol}{cardblack}
+\newcommand{\setsuit}[1]{%
+  \ifstrequal{#1}{S}{\def\cursym{\spadesuit}\def\curcol{cardblack}}{}%
+  \ifstrequal{#1}{H}{\def\cursym{\heartsuit}\def\curcol{cardred}}{}%
+  \ifstrequal{#1}{C}{\def\cursym{\clubsuit}\def\curcol{cardblack}}{}%
+  \ifstrequal{#1}{D}{\def\cursym{\diamondsuit}\def\curcol{cardred}}{}%
+}
+
+% Inline coloured card token, e.g. \cc{4}{S} -> 4 of spades
+\newcommand{\cc}[2]{\setsuit{#2}\textcolor{\curcol}{\textsf{\textbf{#1}}\,\ensuremath{\cursym}}}
+
+\tikzset{cardframe/.style={draw=black!55, line width=0.4pt, rounded corners=1pt, fill=white}}
+
+% Face-up card centred at (#1,#2) with rank #3 and suit letter #4
+\newcommand{\drawcard}[4]{%
+  \setsuit{#4}%
+  \draw[cardframe] (#1,#2) +(-0.375,-0.55) rectangle +(0.375,0.55);
+  \node[anchor=north west, font=\sffamily\bfseries\footnotesize, text=\curcol]
+        at (#1-0.32,#2+0.5) {#3};
+  \node[font=\Large, text=\curcol] at (#1,#2-0.04) {$\cursym$};
+}
+
+% Face-down card centred at (#1,#2)
+\newcommand{\drawback}[2]{%
+  \draw[cardframe, fill=magHighlight!80, draw=magHighlight!55] (#1,#2) +(-0.375,-0.55) rectangle +(0.375,0.55);
+  \draw[white, line width=0.5pt] (#1,#2) +(-0.26,-0.42) rectangle +(0.26,0.42);
+}
+
+% Numeric card (for the 124-card deck), centred at (#1,#2) showing number #3
+\newcommand{\drawnum}[3]{%
+  \draw[cardframe, fill=magHighlight!12] (#1,#2) +(-0.42,-0.52) rectangle +(0.42,0.52);
+  \node[font=\sffamily\bfseries\small] at (#1,#2) {#3};
+}
+
+% Small face-up / face-down tiles for the "Ups and Downs" figure
+\newcommand{\udU}[2]{%
+  \draw[cardframe, fill=white] (#1,#2) +(-0.12,-0.2) rectangle +(0.12,0.2);
+  \node[font=\tiny, text=cardblack] at (#1,#2) {$\spadesuit$};}
+\newcommand{\udD}[2]{%
+  \draw[cardframe, fill=magHighlight!75, draw=magHighlight!55] (#1,#2) +(-0.12,-0.2) rectangle +(0.12,0.2);}
+
+\begin{document}
+
+\selectlanguage{english}
+\setstretch{1.1}
+
+\title{\LARGE The Fitch Trick: Solutions \& Variations \\
+\large ଫିଚ୍‌ଙ୍କ ତାସ ଖେଳ ଏବଂ ଏହାର ରହସ୍ୟ}
+\author{\AuthorDual{ନୀଲ୍‌ଧାରା ମିଶ୍ର}{Neeldhara Misra}}
+\date{}
+\maketitle
+
+\setstretch{1.2}
+
+\begin{abstract}
+In last issue's puzzle, five cards from the audience went to Badarika, who
+tucked one away and laid the remaining four face-up in a row; Aalaya, absent
+throughout, returned and named the hidden card. How? This article delivers the
+complete answer. We give the full strategy for a standard $52$-card deck, then
+ask how far the trick can be pushed: it turns out $124$ is the largest deck size
+for which a winning strategy can exist, and, perhaps more surprisingly, one
+always does. We prove this bound, construct an explicit scheme that realizes it,
+and offer a gentler warm-up that reaches $124$ by bending one small rule. The
+article closes with a tour of variations, including a version where Aalaya
+predicts the outcome of a coin flip and one where the four cards may be placed
+face-up or face-down.
+\end{abstract}
+
+\setstretch{1.05}
+
+In the trick we presented in the last issue, Badarika the assistant studied
+five cards an audience handed her, slipped one away, and laid the other four
+face-up in a row. The magician Aalaya—who had been out of the room
+throughout—walked in, glanced at the four cards, and named the fifth. 
+
+Here we revisit this magic trick and discuss how it works: we begin with the standard deck and then show how to perform the same effect with a massively extended deck of size $124$, and conclude with a discussion of several variations.
+
+\section*{The Two Levers}
+
+Aalaya seems to face long odds. Badarika pockets one of five cards, lays the
+other four face-up in a row, and Aalaya—standing across the room, seeing
+only those four—must name the hidden card from the remaining $48$
+possibilities. But look more carefully at what \emph{Badarika} controls. She
+chooses which of the five cards to hide, picking from five candidates; and she
+chooses the order in which the remaining four cards are placed, one of
+$4! = 24$ arrangements. These are two separate levers, and the trick turns on
+using both together.
+
+The first lever comes from a simple counting observation. Five cards drawn
+from a $52$-card deck must include at least two of the same suit, because there
+are only four suits and five cards—this is the so-called the pigeonhole principle at work. Indeed, if no suit repeated, we could have at most four cards, which contradicts the situation that there are, in fact, five. Note that there may be more than one pair of cards of the same suit and there may also be more than two cards that belong to the same suit. The claim is only that at \emph{any} subset of five cards from a standard deck of 52 cards \emph{is guaranteed} to contain a pair of cards that have the same suit.
+
+Badarika
+finds that matching pair and hides one of them. The other she lays face-up in
+the first position. The moment Aalaya sees that first card, she knows the
+hidden card's suit. A problem with $48$ unknowns suddenly becomes a problem
+with at most $12$: the hidden card is one of the twelve remaining cards of that suit.
+
+The second lever handles the rank. Picture the thirteen ranks—ace, two,
+three, up to ten, jack, queen, king—arranged around a circle, the ace at the
+top and each rank one step clockwise from the last (see Figure~\ref{fig:clock}).
+Any two positions on this clock are at most six steps apart the short way
+round. To see this, say the numbers are $x$ and $y$. Try going clockwise from $x$ to $y$. If you consume at most six steps, you are done, if not, then you consumed at least seven, so going counter clockwise from $x$ to $y$ (or equivalently, going clockwise from $y$ to $x$) will consume at most six steps, since there are only thirteen steps available altogether. 
+
+
+So among Badarika's two same-suit cards, one always sits at most six
+clockwise steps ahead of the other. She reveals the card she counts \emph{from}
+and keeps the one that is $k$ steps \emph{ahead} hidden, where $k$ lies between
+$1$ and $6$. By choosing which of the pair to hide, she guarantees the hidden
+card is always ahead and never more than six steps away.
+
+\begin{figure}[t]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm]
+  \def\R{2.3}\def\rc{0.45}
+  \pgfmathsetmacro{\Rout}{\R+\rc+0.16}
+  \foreach \i/\lab in {1/A,2/2,3/3,4/4,5/5,6/6,7/7,8/8,9/9,10/10,11/J,12/Q,13/K}{
+    \pgfmathtruncatemacro{\rem}{mod(\i,13)}
+    \coordinate (C) at ({90-(\i-1)*360/13}:\R);
+    \ifnum\i=3
+      \colorlet{cf}{magHighlight!22}\colorlet{cd}{magHighlight}
+    \else\ifnum\i=10
+      \colorlet{cf}{magHighlight!22}\colorlet{cd}{magHighlight}
+    \else
+      \colorlet{cf}{white}\colorlet{cd}{black!35}
+    \fi\fi
+    \draw[draw=cd, fill=cf, line width=0.5pt] (C) circle (\rc);
+    \draw[black!25, line width=0.4pt] ($(C)+(20:\rc)$) -- ($(C)+(200:\rc)$);
+    \node[font=\scriptsize\bfseries, text=black!85] at ($(C)+(0,0.2)$) {\lab};
+    \node[font=\scriptsize, text=magHighlight] at ($(C)+(0,-0.2)$) {\rem};
+  }
+  % short-way arc: six steps clockwise from 10 to 3, crossing the K
+  \draw[magHighlight, line width=1pt, -{Stealth[length=4.5pt]}]
+    ({90 - 9*360/13 - 6}:\Rout) arc[start angle={90 - 9*360/13 - 6},
+       end angle={90 - 2*360/13 - 360 + 6}, radius=\Rout];
+  \node[font=\scriptsize\bfseries, text=magHighlight] at ({-242.3}:{\Rout+0.4}) {$k=6$};
+\end{tikzpicture}
+\caption{The thirteen ranks on a clock, each split to show its remainder
+mod $13$ underneath (so $K$ is $0$, $Q$ is $12$, and so on).
+In this example, the cards we have have ranks $3$ and $10$. Note that $10$ is seven steps ahead of $3$, and $3$ lies six
+steps past the $10$: so we reveal the $10$, hide the $3$, and $k=6$. In terms of remainders, $10 + 6 = 16 \equiv 3 \pmod{13}$.}
+\label{fig:clock}
+\end{figure}
+
+Now the whole problem collapses to a single number: Aalaya needs to receive
+$k$, somewhere from $1$ to $6$. The three remaining cards are exactly the tool.
+Beforehand the partners fix an ordering of the entire deck—by rank first, and
+among equal ranks by suit in the order clubs, hearts, spades, diamonds\footnote{This is sometimes called the CHaSeD order, which makes it easy to remember. Of course, you can pick any order you like, as long as you and your partner both swear by the same one!}. Under
+this ordering any three distinct cards fall into one Lowest, one Middle, one
+Highest. There are $3! = 6$ ways to arrange three distinct things, and the
+partners agree on the table below.
+
+\begin{center}\small
+\setlength{\extrarowheight}{1pt}
+\begin{tabular}{cccccc}
+\textsf{LMH} & \textsf{LHM} & \textsf{MLH} & \textsf{MHL} & \textsf{HLM} & \textsf{HML}\\
+$1$ & $2$ & $3$ & $4$ & $5$ & $6$
+\end{tabular}
+\end{center}
+
+Putting it together, here is Badarika's recipe. Find the two same-suit cards.
+Decide which to hide—the one that is $1$ to $6$ clockwise steps ahead of the
+other—and lay the other face-up first. Then arrange the remaining three in the
+order that encodes $k$. Aalaya decodes in reverse: read the suit from the
+first card; read $k$ from the Low-Middle-High pattern of the other three;
+count $k$ steps clockwise from the first card's rank; and name that card in the
+suit she already knows.
+
+Let's go over a concrete example. Suppose the five cards are
+the \cc{4}{S}, \cc{5}{H}, \cc{2}{C}, \cc{7}{C}, \cc{6}{D}—exactly the hand
+posed as a problem in the companion article. The two clubs are the matching
+pair. Going clockwise from the $2$ to the $7$ is exactly five steps, so $k=5$:
+Badarika hides the \cc{7}{C} and sets down the \cc{2}{C} first. The three left
+over are \cc{4}{S}, \cc{5}{H}, \cc{6}{D}; in the global order their ranks
+$4,5,6$ are distinct, so $\textsf{L}=\cc{4}{S}$, $\textsf{M}=\cc{5}{H}$,
+$\textsf{H}=\cc{6}{D}$. To send $k=5$ she needs \textsf{HLM}, so she lays the
+\cc{6}{D}, then the \cc{4}{S}, then the \cc{5}{H}. Aalaya sees the row in
+Figure~\ref{fig:basicrow}. She reads clubs from the first card, recognises
+\textsf{HLM} $=5$ in the rest, counts five steps from rank $2$ to rank $7$, and
+names the \cc{7}{C}. In the notation of the companion piece, the hand
+$\alpha=\langle \cc{4}{S},\cc{5}{H},\cc{2}{C},\cc{7}{C},\cc{6}{D}\rangle$ maps
+to $\beta=\langle \cc{2}{C},\cc{6}{D},\cc{4}{S},\cc{5}{H}\rangle$.
+
+\begin{figure}[t]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm]
+  \drawcard{0}{0}{2}{C}
+  \drawcard{1}{0}{6}{D}
+  \drawcard{2}{0}{4}{S}
+  \drawcard{3}{0}{5}{H}
+  \drawback{4.25}{0}
+  \node[font=\scriptsize] at (0,-0.85) {suit};
+  \node[font=\scriptsize] at (2,-0.85) {\textsf{HLM} $=5$};
+  \node[font=\scriptsize, text=magHighlight] at (4.25,-0.85) {\cc{7}{C}?};
+\end{tikzpicture}
+\caption{The four cards Aalaya sees, in order, and the card she must name.
+First card $\Rightarrow$ suit (clubs); the rest spell \textsf{HLM} $=5$;
+$2+5=7$, so the hidden card is the \cc{7}{C}.}
+\label{fig:basicrow}
+\end{figure}
+
+Rehearsing until this is automatic matters more than performers like to admit. The good news is that the mechanics do
+become second nature with practice. However, if you are going to repeat the trick though, there is one structural tell: the
+suit-revealing card always sits first, and an alert audience that sees several
+rounds will start to wonder why position one seems special. There is a way to workaround this, and
+it is worth keeping in mind—--we will return to it among the variations.
+
+\begin{artprobox}
+Warm up as the magician. Using the basic scheme above (deck order
+\textsf{A,2,\dots,K}, suits \cc{}{C}\,\cc{}{D}\,\cc{}{H}\,\cc{}{S}), name the
+hidden card for each row:
+
+\smallskip
+\noindent
+(a)~$\langle\cc{2}{C},\cc{6}{S},\cc{6}{H},\cc{K}{C}\rangle$\\[2pt]
+(b)~$\langle\cc{4}{D},\cc{K}{S},\cc{3}{C},\cc{10}{S}\rangle$.
+% \footnote{Answers:
+% (a) the first card \cc{2}{C} fixes clubs; \cc{6}{S},\cc{6}{H},\cc{K}{C} read as
+% \textsf{MLH} $=3$, and $2+3=5$, so the \cc{5}{C}. (b) \cc{4}{D} fixes diamonds;
+% \cc{K}{S},\cc{3}{C},\cc{10}{S} read as \textsf{HLM} $=5$, and $4+5=9$, so the
+% \cc{9}{D}.}
+\end{artprobox}
+
+\section*{How Big Can the Deck Be?}
+
+The trick works beautifully with a standard $52$-card deck, but a natural
+question lingers: how large could the deck be? Could Badarika and Aalaya
+perform the same feat with $60$ cards? A hundred?
+
+Let us count. Suppose the deck has $n$ cards. When the audience draws five and
+hands them to Badarika, the number of possible five-card hands is $\binom{n}{5}$.
+Badarika's message is an ordered row of four cards, and the number of such rows is
+$n(n-1)(n-2)(n-3)$. For the trick to be possible at all, there must be at least
+as many messages as hands: if two different hands forced Badarika to send the
+same message, Aalaya would see that message and have no way of telling them
+apart.
+
+The inequality resolves to $n \leqslant 124$. That is a striking answer. The hard ceiling from information theory is $124$
+cards. A standard deck of $52$ sits comfortably below it; in principle the
+trick could work for a deck more than twice as large. Most people who meet the
+trick for the first time assume the numbers must be far tighter than this.
+
+We should be careful, though. The condition $n \leqslant 124$ is necessary but
+not obviously sufficient. Having enough messages to go around does not
+automatically mean a valid strategy exists, because Badarika can only use the
+five cards she was dealt. She cannot lay down a message built from a card she
+never received. \emph{Every hand must be matched to a message whose four cards
+all belong to that hand.} Showing that such a matching always exists takes a
+little more work.
+
+The right framework is a bipartite graph (Figure~\ref{fig:bipartite}). Place
+all possible five-card hands on one side, and all possible ordered four-card
+messages on the other. Draw an edge between a hand and a message whenever the
+four cards of the message all appear in that hand. A workable strategy for
+Badarika and Aalaya is exactly a \emph{perfect matching} in this graph: each
+hand paired with a distinct message it contains. Does such a matching exist? It turns out that an important idea in graph theory which is known as the Hall's Marriage Theorem gives us a criterion: a
+perfect matching exists if and only if every set of $k$ hands collectively
+reaches at least $k$ messages. 
+
+\begin{figure*}[t]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm,
+   hand/.style={circle, draw=black!45, fill=magHighlight!12, inner sep=1.6pt, font=\scriptsize},
+   msg/.style={circle, draw=black!45, fill=black!4, inner sep=1.6pt, font=\scriptsize}, scale=0.8]
+  \foreach \i/\y in {1/2,2/1,3/0,4/-1,5/-2}{ \node[hand] (H\i) at (0,\y) {$H_{\i}$}; }
+  \foreach \i/\y in {1/2,2/1,3/0,4/-1,5/-2}{ \node[msg]  (M\i) at (5,\y) {$M_{\i}$}; }
+  \node[font=\small\bfseries] at (0,2.7) {hands};
+  \node[font=\small\bfseries] at (5,2.7) {messages};
+  % a few "eligible" edges (faint)
+  \foreach \a/\b in {1/1,1/2,2/1,2/3,3/2,3/4,4/3,4/5,5/4,5/5,1/5,2/4}{
+     \draw[black!22, line width=0.4pt] (H\a) -- (M\b);
+  }
+  % a highlighted perfect matching
+  \foreach \a/\b in {1/2,2/1,3/4,4/3,5/5}{
+     \draw[magHighlight, line width=1.4pt] (H\a) -- (M\b);
+  }
+  \node[anchor=west, font=\footnotesize, text=black!70, align=left] at (6.4,1)
+    {An edge means ``the four cards of $M_j$\\
+     all lie inside hand $H_i$.''\\
+     \vspace{0.2em}~\\
+     A \textcolor{magHighlight}{perfect matching}  (thick edges)\\
+    is a complete strategy. \\
+    \vspace{0.2em}~\\    
+    At $n=124$ every vertex has degree exactly $120$, \\
+    which forces
+     such a matching to exist.};
+\end{tikzpicture}
+\caption{A toy picture of the real bipartite graph. At $n=124$ the two sides are
+equal in size: $\binom{124}{5} = 124\cdot 123\cdot 122\cdot 121 = 225{,}150{,}024$
+vertices each, and the graph is $120$-regular.}
+\label{fig:bipartite}
+\end{figure*}
+
+
+To see if we meet this condition, checking each subset by hand would be hopeless, but fortunately the graph has a regularity that makes the condition automatic. Every
+five-card hand contains exactly five four-card subsets, and each of those can
+be arranged in $24$ orders, so every hand-vertex has degree $5\times 24 = 120$.
+On the other side, every ordered four-card message can be extended to a hand by
+appending any one of the remaining $120$ cards, so every message-vertex also
+has degree $120$. In this $120$-regular bipartite graph, any collection of $k$
+hands sends out $120k$ edges in total; each message can absorb at most $120$ of
+them, so those hands must touch at least $k$ distinct messages. Hall's condition
+holds for every subset at once, and a perfect matching exists.
+
+The regularity carries a bonus. A finite $d$-regular bipartite graph splits into
+exactly $d$ edge-disjoint perfect matchings—this is König's edge-colouring
+theorem, equivalently the integer form of the Birkhoff--von Neumann theorem (a
+nonnegative-integer matrix whose every row and column sums to $d$ is a sum of
+$d$ permutation matrices). Our graph is $120$-regular, so it yields $120$
+edge-disjoint perfect matchings, and Badarika and Aalaya do not have just one
+valid strategy—they have at least $120$ to choose from.
+
+The catch is that the proof only confirms existence. At $n=124$, the bipartite
+graph has $225{,}150{,}024$ vertices on each side. The argument tells us a
+strategy exists; it gives no recipe a person could actually memorise and use.
+We give two methods that are performable: first a relaxed \emph{warm-up} that
+reaches $124$ by extending the ideas we had with the standard-deck and introducing one small
+physical gimmick, and then a more elegant scheme that works with $124$ using nothing but the order of four cards, so it can be, in principle, performed over a phone call!
+
+\section*{A Warm-Up: Four Suits and a Little Cheating}
+
+Begin by re-suiting the deck. Number the $124$ cards $0$ through $123$. Give
+each card $c$ a ``suit'' $s = c \bmod 4$—four suits, just as in a real deck—and
+a ``rank'' $r = \lfloor c/4 \rfloor$, the whole-number part of $c/4$, which runs
+from $0$ to $30$. Then $c = 4r + s$, and each suit holds exactly $124/4 = 31$
+cards. The architecture is identical to a standard deck; only the number of
+ranks per suit has changed, from $13$ to $31$ (Figure~\ref{fig:resuit}).
+
+\begin{figure}[t]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm]
+  \foreach \s in {0,1,2,3}{
+    \node[font=\scriptsize\bfseries, text=magHighlight] at (\s,0.46) {$\equiv\s$};
+  }
+  \node[font=\scriptsize, text=black!55] at (-0.82,0.46) {$r$};
+  \foreach \r in {0,...,30}{
+    \node[font=\scriptsize, text=black!55] at (-0.82,{-\r*0.44}) {\r};
+    \foreach \s/\tint/\tc in {0/0/black,1/20/black,2/42/black,3/64/white}{
+      \pgfmathtruncatemacro{\num}{4*\r+\s}
+      \fill[magHighlight!\tint] (\s-0.46,{-\r*0.44-0.2}) rectangle (\s+0.46,{-\r*0.44+0.2});
+      \draw[black!15, line width=0.2pt] (\s-0.46,{-\r*0.44-0.2}) rectangle (\s+0.46,{-\r*0.44+0.2});
+      \node[font=\scriptsize, text=\tc] at (\s,{-\r*0.44}) {\num};
+    }
+  }
+\end{tikzpicture}
+\caption{All $124$ numbers re-suited as $4\times 31$. Each row is one rank $r$
+(the four consecutive numbers $\{4r,\dots,4r+3\}$); each column is one suit
+$s = c \bmod 4$, shaded from white ($\equiv 0$) to blue ($\equiv 3$). Two cards
+share a suit exactly when they lie in the same column.}
+\label{fig:resuit}
+\end{figure}
+
+The pigeonhole argument needs no adjustment. Five cards arrive from four suits,
+so at least two share a suit. Badarika picks any such pair: one will be hidden,
+the other revealed first, announcing the suit to Aalaya—exactly as before.
+
+Now the clock. The $31$ ranks sit on a clock face with $31$ marks
+(Figure~\ref{fig:warmclock}). Because $31$ is odd, no two distinct marks sit
+exactly opposite each other, so any two different ranks have a well-defined
+shorter arc between them, always $1$ to $15$ steps. Badarika reveals the card she
+counts from, call it $Y$, and hides the card $X$ that is $k$ steps clockwise
+ahead, for some $k$ between $1$ and $15$. The card $Y$ announces the suit; the
+remaining job is to transmit $k$.
+
+\begin{figure}[t]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm]
+  \def\R{1.65}
+  \foreach \i in {0,...,30}{ \fill[black!35] ({90 - \i*360/31}:\R) circle (0.7pt); }
+  \node[circle, draw=magHighlight, fill=magHighlight!20, inner sep=0.8pt, font=\tiny\bfseries]
+     at ({90 - 0*360/31}:\R) {$Y$};
+  \node[circle, draw=magHighlight, fill=magHighlight!20, inner sep=0.8pt, font=\tiny\bfseries]
+     at ({90 - 8*360/31}:\R) {$X$};
+  \draw[magHighlight, line width=1pt, -{Stealth[length=4pt]}]
+     ({90 - 0*360/31}:\R+0.33) arc[start angle={90 - 0*360/31}, end angle={90 - 8*360/31}, radius=\R+0.33];
+  \node[font=\scriptsize, text=magHighlight] at (2.25,1.5) {$k \leqslant 15$};
+\end{tikzpicture}
+\caption{The $31$ ranks of one suit, on a clock. Reveal $Y$ and hide the card
+$X$ that is $k$ short steps ahead; since $31$ is odd, $k$ is always one of
+$1,\dots,15$.}
+\label{fig:warmclock}
+\end{figure}
+
+The three leftover cards carry this job partway. In the
+agreed deck ordering they fall into Low, Medium, and High, and their six
+arrangements code a number from $1$ to $6$: \textsf{LMH} means $1$, \textsf{LHM}
+$2$, \textsf{MLH} $3$, \textsf{MHL} $4$, \textsf{HLM} $5$, \textsf{HML} $6$. That
+was the complete mechanism from the basic trick. The trouble here is that $k$ can
+reach $15$, and six arrangements reach only $6$.
+
+So here we are going to cheat a bit. Lay the three cards along a straight line: say the near edge of the table. Now add a second signal --- either leave
+all three flush against the baseline, or nudge exactly one of them slightly
+forward (Figure~\ref{fig:warmnudge}). So we have four physical states in all—no nudge, or
+the first, second, or third card nudged—carry addends of $0$, $6$, $7$, and
+$12$. Aalaya reads the arrangement for a number from $1$ to $6$, reads the
+nudge for an addend, and adds them.
+
+\begin{figure}[H]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm,scale=1.2]
+  % +0 : all flush
+  \draw[black!35,dashed,line width=0.5pt] (-0.35,0) -- (1.55,0);
+  \node[font=\scriptsize,text=magHighlight] at (-0.72,0.24) {$+0$};
+  \draw[cardframe,fill=white] (-0.18,0) rectangle (0.18,0.48);
+  \draw[cardframe,fill=white] (0.42,0) rectangle (0.78,0.48);
+  \draw[cardframe,fill=white] (1.02,0) rectangle (1.38,0.48);
+  % +6 : first raised
+  \draw[black!35,dashed,line width=0.5pt] (-0.35,-0.95) -- (1.55,-0.95);
+  \node[font=\scriptsize,text=magHighlight] at (-0.72,-0.71) {$+6$};
+  \draw[cardframe,fill=magHighlight!18] (-0.18,-0.77) rectangle (0.18,-0.29);
+  \draw[cardframe,fill=white] (0.42,-0.95) rectangle (0.78,-0.47);
+  \draw[cardframe,fill=white] (1.02,-0.95) rectangle (1.38,-0.47);
+  % +7 : second raised
+  \draw[black!35,dashed,line width=0.5pt] (-0.35,-1.9) -- (1.55,-1.9);
+  \node[font=\scriptsize,text=magHighlight] at (-0.72,-1.66) {$+7$};
+  \draw[cardframe,fill=white] (-0.18,-1.9) rectangle (0.18,-1.42);
+  \draw[cardframe,fill=magHighlight!18] (0.42,-1.72) rectangle (0.78,-1.24);
+  \draw[cardframe,fill=white] (1.02,-1.9) rectangle (1.38,-1.42);
+  % +12 : third raised
+  \draw[black!35,dashed,line width=0.5pt] (-0.35,-2.85) -- (1.55,-2.85);
+  \node[font=\scriptsize,text=magHighlight] at (-0.72,-2.61) {$+12$};
+  \draw[cardframe,fill=white] (-0.18,-2.85) rectangle (0.18,-2.37);
+  \draw[cardframe,fill=white] (0.42,-2.85) rectangle (0.78,-2.37);
+  \draw[cardframe,fill=magHighlight!18] (1.02,-2.67) rectangle (1.38,-2.19);
+\end{tikzpicture}
+\caption{The nudge channel. All three cards resting on the baseline means add
+$0$; lifting the first, second, or third card slightly means add $6$, $7$, or
+$12$.}
+\label{fig:warmnudge}
+\end{figure}
+
+The arrangement picks a row of the table below and the nudge picks a column; the
+two add to give $k$.
+
+\begin{center}\small
+\setlength{\extrarowheight}{1pt}
+\begin{tabular}{r|cccc}
+ & $+0$ & $+6$ & $+7$ & $+12$\\\hline
+\textsf{LMH} $=1$ & $1$ & $7$  & $8$  & $13$\\
+\textsf{LHM} $=2$ & $2$ & $8$  & $9$  & $14$\\
+\textsf{MLH} $=3$ & $3$ & $9$  & $10$ & $15$\\
+\textsf{MHL} $=4$ & $4$ & $10$ & $11$ & \textcolor{black!40}{$16$}\\
+\textsf{HLM} $=5$ & $5$ & $11$ & $12$ & \textcolor{black!40}{$17$}\\
+\textsf{HML} $=6$ & $6$ & $12$ & $13$ & \textcolor{black!40}{$18$}
+\end{tabular}
+\end{center}
+
+The four addends do exactly what you want: $0$, $6$, $7$, and $12$, one per
+nudge option. Adding $0$ covers arrangement values $1$ through $6$; adding $6$
+covers $7$ through $12$; adding $7$ covers $8$ through $13$; adding $12$ covers
+$13$ through $18$. Together these ranges blanket every $k$ from $1$ to $15$.
+Values $8$ through $13$ are each reachable two ways, a harmless overlap, and the
+few totals above $15$ (greyed in the table) are quiet ghosts, since $k$ never
+reaches them. 
+
+If the choice of $12$ seems a bit abrupt, you could also instead use the addends $0$, $6$, $7$, $8$. This will top out at $14$, but you can still handle
+$k=15$ with a flourish: lay the suit card face up and turn the other three face down, with an agreement beforehand that this special arrangement conveys a gap of $15$. Make some banter about how you don't even need all the four cards, and your magic senses feel especially powerful today.
+
+Let's work out a couple of examples. Suppose the five cards are $7,19,50,88,103$. Their
+remainders mod $4$ are $3,3,2,0,3$, so Badarika works in suit $3$ with the pair
+$7$ and $19$. Their ranks are $1$ and $4$ (since $7 = 4\cdot 1 + 3$ and
+$19 = 4\cdot 4 + 3$), so $k=3$—comfortably within the honest range. She reveals
+$7$ and places the leftover cards $50,88,103$ in \textsf{MLH} order, that is
+$88,50,103$, all flush to the baseline. Aalaya sees $7$ followed by
+$88,50,103$, reads suit $3$, rank $1$, no nudge, \textsf{MLH} $=3$, so $k=3$, and
+the hidden card sits at rank $1+3=4$ in suit $3$: that is $4\cdot 4 + 3 = 19$.
+
+Now let's try a scenario involving a nudge. Take the hand $2,34,11,60,99$. Cards $2$ and $34$ share suit $2$;
+their ranks are $0$ and $8$, giving $k=8$. The arrangement alone reaches only
+$6$, so Badarika combines: \textsf{LHM} encodes $2$, and nudging the first leftover
+card adds $6$, giving $2+6=8$. The leftover cards $11,60,99$ in \textsf{LHM}
+order are $11,99,60$, with the $11$ nudged forward. Aalaya reads suit $2$,
+rank $0$, a first-card nudge ($+6$), \textsf{LHM} $=2$, and concludes $k=8$: the
+hidden card is rank $8$ in suit $2$, which is $4\cdot 8 + 2 = 34$.
+
+Note that performing this does not even need a deck of cards! Five numbers from $0$ to $123$ are called out, and
+Badarika jots them in a notebook. She writes four of
+them, in order, on a board or a sheet of paper that everyone can see, and writes
+the hidden one on a small folded slip set aside as a \emph{prediction}. 
+physical nudge becomes a faint written mark instead—an underline, a dot, a
+casual scrawl—under exactly one of the three arrangement numbers (never the
+suit-giver), or under none. Aalaya walks in, reads the four numbers and the
+single mark (or notes the absence of it), names the hidden value, and the slip is opened to confirm.
+
+If you work with the scheme of adding $6, 7, 8$ then to communicate a 15, you can write the four numbers in any order, then make a cheesy remark along the lines of the prediction sharing a
+``very special connection'' with one of these numbers, followed by erasing the three irrelevant numbers, and again you are done.
+\begin{artprobox}
+    Suppose you are the assistant and the five numbers handed to you are $10,\ 2,\ 99,\ 42,\ 115$. What will you hide? What will you write?
+\end{artprobox}
+\begin{artprobox}
+Suppose you are the magician and you find that the four numbers written in a row are
+$46,\ 17,\ 5,\ 88$, and the $5$ is underlined. You are the magician: name the
+hidden number.
+% \footnote{$46$ has suit $46 \bmod 4 = 2$ and rank
+% $\lfloor 46/4\rfloor = 11$. The underline sits under the \emph{second} of the
+% three numbers, so add $7$; and $17,5,88$ read as \textsf{MLH} $=3$. Thus
+% $k = 3+7 = 10$, the hidden rank is $11+10 = 21$, and the card is
+% $4\cdot 21 + 2 = 86$.}
+\end{artprobox}
+
+Now: can we avoid the cheating? Imagine you had to perform this over a phone call to the magician, who's on speaker. Then you only have the order to work with, since using tonality of voice to communicate nudges is too unreliable, and coughing is suspicious. In a follow up article, we'll see how to work this magic with no cheating at all. Stay tuned!
+
+
+{\small
+\vskip 0.05in
+\affiliation[1.6em]{Associate Professor, CSE; IIT Gandhinagar}
+\email{neeldhara.misra@gmail.com}
+}
+\end{document}

fitch/fitch-trick-solutions-part-two.tex

@@ -0,0 +1,381 @@
+% !TEX TS-program = xelatex
+\documentclass{../leelavati}
+
+\usepackage{qrcode}
+
+% -------------------------------------------------------------------
+% Playing-card drawing helpers (TikZ)
+% -------------------------------------------------------------------
+\definecolor{cardred}{HTML}{C0392B}
+\definecolor{cardblack}{HTML}{1A1A1A}
+
+% Map a suit letter S/H/C/D to a symbol (\cursym) and a colour (\curcol)
+\newcommand{\cursym}{}
+\newcommand{\curcol}{cardblack}
+\newcommand{\setsuit}[1]{%
+  \ifstrequal{#1}{S}{\def\cursym{\spadesuit}\def\curcol{cardblack}}{}%
+  \ifstrequal{#1}{H}{\def\cursym{\heartsuit}\def\curcol{cardred}}{}%
+  \ifstrequal{#1}{C}{\def\cursym{\clubsuit}\def\curcol{cardblack}}{}%
+  \ifstrequal{#1}{D}{\def\cursym{\diamondsuit}\def\curcol{cardred}}{}%
+}
+
+% Inline coloured card token, e.g. \cc{4}{S} -> 4 of spades
+\newcommand{\cc}[2]{\setsuit{#2}\textcolor{\curcol}{\textsf{\textbf{#1}}\,\ensuremath{\cursym}}}
+
+\tikzset{cardframe/.style={draw=black!55, line width=0.4pt, rounded corners=1pt, fill=white}}
+
+% Face-up card centred at (#1,#2) with rank #3 and suit letter #4
+\newcommand{\drawcard}[4]{%
+  \setsuit{#4}%
+  \draw[cardframe] (#1,#2) +(-0.375,-0.55) rectangle +(0.375,0.55);
+  \node[anchor=north west, font=\sffamily\bfseries\footnotesize, text=\curcol]
+        at (#1-0.32,#2+0.5) {#3};
+  \node[font=\Large, text=\curcol] at (#1,#2-0.04) {$\cursym$};
+}
+
+% Face-down card centred at (#1,#2)
+\newcommand{\drawback}[2]{%
+  \draw[cardframe, fill=magHighlight!80, draw=magHighlight!55] (#1,#2) +(-0.375,-0.55) rectangle +(0.375,0.55);
+  \draw[white, line width=0.5pt] (#1,#2) +(-0.26,-0.42) rectangle +(0.26,0.42);
+}
+
+% Numeric card (for the 124-card deck), centred at (#1,#2) showing number #3
+\newcommand{\drawnum}[3]{%
+  \draw[cardframe, fill=magHighlight!12] (#1,#2) +(-0.42,-0.52) rectangle +(0.42,0.52);
+  \node[font=\sffamily\bfseries\small] at (#1,#2) {#3};
+}
+
+% Small face-up / face-down tiles for the "Ups and Downs" figure
+\newcommand{\udU}[2]{%
+  \draw[cardframe, fill=white] (#1,#2) +(-0.12,-0.2) rectangle +(0.12,0.2);
+  \node[font=\tiny, text=cardblack] at (#1,#2) {$\spadesuit$};}
+\newcommand{\udD}[2]{%
+  \draw[cardframe, fill=magHighlight!75, draw=magHighlight!55] (#1,#2) +(-0.12,-0.2) rectangle +(0.12,0.2);}
+
+\begin{document}
+
+\selectlanguage{english}
+\setstretch{1.1}
+
+\title{\LARGE The Fitch Trick: Solutions \& Variations \\
+\large ଫିଚ୍‌ଙ୍କ ତାସ ଖେଳ ଏବଂ ଏହାର ରହସ୍ୟ}
+\author{\AuthorDual{ନୀଲ୍‌ଧାରା ମିଶ୍ର}{Neeldhara Misra}}
+\date{}
+\maketitle
+
+\setstretch{1.2}
+
+\begin{abstract}
+TBD.
+\vspace{14em}
+\end{abstract}
+\setstretch{1.05}
+
+In the trick we presented in the last issue, Badarika the assistant studied
+five cards an audience handed her, slipped one away, and laid the other four
+face-up in a row. The magician Aalaya—who had been out of the room
+throughout—walked in, glanced at the four cards, and named the fifth. 
+
+These five cards could be cards from a standard deck of 52 cards, or cards from a deck of 124 cards. Last time, we saw how to work with 124 cards if we were allowed a physical gimmick. This time we'll see how to pull off the trick fair and square: so it's doable even over a phone call, where the assistant has to rely purely on card order and no extra signals. 
+
+\section*{A Recipe for 124 Cards}
+
+So here is our upgrade to the pro version of this trick. Let's discuss the encoding first. When Badarika receives five cards, she sorts them mentally into increasing order
+and assigns them positions $0,1,2,3,4$ from smallest to largest. She adds all
+five card-numbers together, divides by $5$, and notes the remainder; call it
+$i$. \emph{She hides the card in position $i$} and lays the other four face-up
+in a carefully chosen order. That is the complete hiding rule.
+
+The reason this choice is clever comes down to a small bookkeeping trick. Write
+the sorted hand as
+\[
+c_0<c_1<c_2<c_3<c_4,
+\]
+and suppose Badarika hides $c_i$. Aalaya sees the other four cards and calls
+their sum $s$. Now imagine setting those four visible cards aside and
+renumbering the remaining $120$ cards in increasing order, from $0$ up to
+$119$.
+
+What is the new number of the hidden card? Deleting cards only affects a card's
+label by removing the deleted cards below it. Since $c_i$ was the $i$-th
+smallest card in the original five-card hand, exactly $i$ of the visible cards
+lie below it. So, in the renumbered deck, the hidden card has new label
+\[
+c_i-i.
+\]
+But $i$ was chosen to be the total hand-sum modulo $5$. Since that total is
+$s+c_i$, we have $i \equiv s+c_i \pmod 5$, and therefore
+\[
+c_i-i \equiv c_i-(s+c_i) \equiv -s \pmod 5.
+\]
+This is the key fact. The choice of which position to hide exactly compensates
+for the leftward shift caused by deleting the visible cards.
+
+{\color{red}Potential to add a problem here.}
+
+Among $120$ candidates, exactly $120/5 = 24$ share any given remainder, so the
+forced remainder cuts Aalaya's search to $24$ cards
+(Figure~\ref{fig:residue}). The arrangement of the four revealed cards then says
+\emph{which} of those $24$ it is—and since four distinct cards admit exactly
+$24$ orderings, the match is exact. Reading the arrangement as a number $p$ from
+$0$ to $23$, Aalaya computes the hidden card's new number as
+$5p + ((-s)\bmod 5)$. Then she converts that new number to an actual card number
+by a simple counting-over step: she runs through the revealed cards in order,
+and each time a revealed card sits at or below her running answer, she bumps the
+answer up by one.
+
+\begin{figure}[H]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm,
+  shown/.style={draw=black!35, fill=black!5, rounded corners=1pt, inner sep=2.5pt, font=\scriptsize},
+  hidden/.style={draw=magHighlight, fill=magHighlight!18, rounded corners=1pt, inner sep=2.5pt, font=\scriptsize\bfseries},
+  cand/.style={draw=black!25, fill=white, rounded corners=1pt, inner sep=2pt, font=\scriptsize},
+  hit/.style={draw=magHighlight, fill=magHighlight!18, rounded corners=1pt, inner sep=2pt, font=\scriptsize\bfseries}, scale=1.2
+]
+  \node[font=\scriptsize\bfseries, text=black!70] at (1.25,0.45) {old labels};
+  \node[font=\scriptsize\bfseries, text=black!70] at (4.95,0.45) {new labels $\equiv 1$};
+
+  % Old labels from the worked example.
+  \draw[black!35, line width=0.6pt, -{Stealth[length=3.5pt]}] (1.25,0.15) -- (1.25,-3.75);
+  \foreach \y/\num/\sty in {0/7/shown,-0.95/37/shown,-1.45/38/hidden,-2.45/61/shown,-3.35/94/shown}{
+    \draw[black!35] (1.14,\y) -- (1.36,\y);
+    \node[\sty] at (1.25,\y) {\num};
+  }
+  \node[font=\scriptsize, text=black!60] at (1.25,-0.48) {$\vdots$};
+  \node[font=\scriptsize, text=black!60] at (1.25,-1.95) {$\vdots$};
+  \node[font=\scriptsize, text=black!60] at (1.25,-2.90) {$\vdots$};
+
+  % Candidate new labels with the forced residue.
+  \draw[black!35, line width=0.6pt, -{Stealth[length=3.5pt]}] (4.95,0.15) -- (4.95,-3.75);
+  \foreach \y/\num/\sty in {0/1/cand,-0.45/6/cand,-0.90/11/cand,-1.35/31/cand,-1.80/36/hit,-2.25/41/cand,-3.35/116/cand}{
+    \draw[black!35] (4.84,\y) -- (5.06,\y);
+    \node[\sty] at (4.95,\y) {\num};
+  }
+  \node[font=\scriptsize, text=black!60] at (4.95,-1.15) {$\vdots$};
+  \node[font=\scriptsize, text=black!60] at (4.95,-2.80) {$\vdots$};
+
+  \draw[magHighlight, line width=0.9pt, -{Stealth[length=3.8pt]}]
+    (1.70,-1.45) -- (4.48,-1.80);
+  \node[font=\scriptsize, text=magHighlight, align=center] at (3.05,-1.18)
+    {$38-2=36$\\[-1pt]{\tiny delete $7$ and $37$ below}};
+  \node[font=\scriptsize, text=magHighlight] at (5.75,-1.80) {$p=7$};
+\end{tikzpicture}
+\caption{A concrete picture of the renumbering in the worked example. The
+shown cards below $38$ are $7$ and $37$, so deleting the shown cards changes
+the hidden card's label from $38$ to $38-2=36$. Since $36\equiv 1 \pmod 5$, it
+appears as the $p=7$ entry among the $24$ possible new labels
+$1,6,11,\dots,116$.}
+\label{fig:residue}
+\end{figure}
+
+
+{\color{red}Need some better pictures here.}
+
+A concrete run-through helps. Suppose Badarika lays down four cards in this order:
+$37,\,7,\,94,\,61$. Their sum is $s=199$. Since $199$ leaves remainder $4$ on
+division by $5$, the forced remainder is $(-4)\bmod 5 = 1$. The arrangement of
+the four cards encodes the value $p=7$ (the mechanism comes next). So the
+hidden card's new number is $5\cdot 7 + 1 = 36$. To recover the actual card,
+start from $36$: the revealed card $7$ is at most $36$, so the answer becomes
+$37$; the revealed card $37$ is now at most $37$, so the answer becomes $38$;
+the cards $61$ and $94$ sit above $38$, so no more bumps. \textbf{The hidden
+card is $38$.}
+
+Now for the only step that might seem fiddly: turning an arrangement of four
+cards into a number from $0$ to $23$, and back, without any factorial
+arithmetic. Write $p = 6a + b$, where $a$ runs over $0,1,2,3$ and $b$ over
+$0,1,2,3,4,5$; every $p$ in $0,\dots,23$ factors this way uniquely. Badarika
+places the \emph{lowest} of her four cards into slot number $a$, counting slots
+$0,1,2,3$ from left to right. She arranges the other three cards in the
+remaining slots using the very same Low-Medium-High table from the basic
+trick: the patterns \textsf{LMH}, \textsf{LHM}, \textsf{MLH}, \textsf{MHL},
+\textsf{HLM}, \textsf{HML} encode $b=0,1,2,3,4,5$. To decode, Aalaya notes
+which slot holds the lowest card (giving $a$), reads the three-card pattern in
+the remaining slots (giving $b$), and computes $p = 6a + b$. 
+
+The worked example closes neatly on itself (Figure~\ref{fig:slots}). The four
+revealed cards are $37,7,94,61$. The lowest, $7$, sits in slot $1$, so $a=1$.
+The other three, read left to right, are $37,94,61$—that is Low, High, Medium,
+the \textsf{LHM} pattern, so $b=1$. Hence $p = 6\cdot 1 + 1 = 7$, exactly the
+value the calculation required.
+
+{\color{red}Add the factorial method for encoding a permutation as well.}
+
+\begin{figure}[t]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm]
+  \drawnum{0}{0}{37}
+  \drawnum{1}{0}{7}
+  \drawnum{2}{0}{94}
+  \drawnum{3}{0}{61}
+  % mark the lowest card's slot
+  \draw[magHighlight, line width=1pt, -{Stealth[length=4pt]}] (1,0.95) -- (1,0.58);
+  \node[font=\scriptsize, text=magHighlight] at (1,1.2) {lowest $\Rightarrow a=1$};
+  \foreach \x/\lab in {0/0,1/1,2/2,3/3}{ \node[font=\scriptsize, text=black!60] at (\x,-0.8) {slot \lab}; }
+  \node[font=\scriptsize, anchor=west] at (-0.5,-1.45)
+     {others $37,94,61=\textsf{LHM}\Rightarrow b=1$, \ so $p=6a+b=7$};
+\end{tikzpicture}
+\caption{The encoding of the permutation: the slot of the lowest card gives $a$; the
+\textsf{LMH} pattern of the other three gives $b$; and $p=6a+b$.}
+\label{fig:slots}
+\end{figure}
+
+% \AuthorNote{%
+% \textbf{A friendlier engine (a small proposal).} The classical exposition asks
+% the performer to compute a permutation's rank in ``factorial base''—the most
+% error-prone step at the table. The $p = 6a + b$ split above replaces that with
+% two moves the duo already use in the $52$-card trick: \emph{spot the lowest
+% card} and \emph{read an \textsf{LMH} pattern}. It is a genuine bijection
+% between the $24$ orderings and $\{0,\dots,23\}$ (four slot-choices times six
+% patterns), so it is just as valid as the textbook version, and it happens to
+% reproduce the classic $37,7,94,61\mapsto 38$ example exactly. I have found
+% nothing that removes the renumbering step entirely; whether a truly
+% relabel-free decoding exists strikes me as a nice open question.
+% }
+
+One last observation. At $124$ cards the arithmetic comes out perfectly: the
+number of five-card hands, $\binom{124}{5}$, equals the number of ordered
+four-card messages, $124\cdot 123\cdot 122\cdot 121$. There is no slack at all: the correspondence between hands and messages is
+a bijection, and not one of the $24$ orderings is spare.
+
+\begin{artprobox}
+    You are given the numbers:
+    $10,\ 3,\ 16,\ 121,\ 42$. What will you hide? What will you show?
+    % \footnote{Sort
+    % the hand as $3,10,16,42,121$. The total is $192\equiv 2\pmod 5$, so hide the
+    % card in position $2$, namely $16$. The visible cards are $3,10,42,121$, whose
+    % sum is $176\equiv 1\pmod 5$, so the forced remainder is $4$. In the renumbered
+    % deck the hidden card has label $16-2=14$, hence $p=(14-4)/5=2$. With
+    % $p=6a+b$, this gives $a=0,b=2$: put the lowest visible card $3$ in slot $0$,
+    % and arrange the remaining three in the pattern $\textsf{MLH}$. Thus one valid
+    % display is $3,\ 42,\ 10,\ 121$.}
+\end{artprobox}
+    
+\begin{artprobox}
+The four numbers laid in a row are
+$9,\ 40,\ 2,\ 88$. Name the hidden number.
+% \footnote{The sum is
+% $s = 139 \equiv 4$, so $(-s)\bmod 5 = 1$. The lowest number $2$ sits in slot
+% $a=2$; the others $9,40,88$ are already in \textsf{LMH} order, so $b=0$ and
+% $p = 6\cdot 2 + 0 = 12$. The new label is $5\cdot 12 + 1 = 61$; stepping past
+% the four shown numbers gives the hidden card $64$.}
+\end{artprobox}
+
+\section*{Variations on the Theme}
+
+The scheme we built has two moving parts: Badarika's choice of which card to
+hide, and her arrangement of the four she shows. Both levers can be adjusted,
+and each adjustment opens a different room.
+
+\textbf{The coin flip (Elwyn Berlekamp).} Stretch the deck to $64$ cards. After
+Badarika lays out four cards from a hand of five, $60$ cards remain unseen; the
+hiding rule leaves the hidden card in a residue class of exactly
+$60/5 = 12$. The four shown cards still arrange in $24$ ways, and $24$ is
+exactly twice $12$. That spare factor of two carries one extra bit: Aalaya
+names the hidden card \emph{and} announces the result of a coin an audience
+member flipped in secret before the show. The same four-card layout does double
+duty, and the audience sees nothing unusual.
+
+\begin{artprobox}
+Explain the magic number $64$. Show four cards of a deck of $D$ and ask why the
+$24$ orderings can carry both the hidden card \emph{and} a coin bit only when
+$D \leqslant 64$.\footnote{After four cards are shown, $D-4$ remain, and the
+hiding rule pins the hidden card to a residue class of $(D-4)/5$ candidates.
+Naming it \emph{and} a coin flip needs $24 \geqslant 2\cdot (D-4)/5$, that is
+$D-4 \leqslant 60$, so $D \leqslant 64$—with equality exactly at $64$, where
+$(D-4)/5 = 12$ and $2\cdot 12 = 24$.}
+\end{artprobox}
+
+\textbf{Face up, face down (Colm Mulcahy's ``Ups and Downs'').} Keep the
+suit-and-pair logic, so again only a number from $1$ to $6$ must be sent.
+Instead of encoding it through the ordering of three cards, Badarika encodes it
+through their \emph{orientation}: \textsf{D} for face-down, \textsf{U} for
+face-up. The six patterns \textsf{DDU, DUD, DUU, UDD, UDU, UUD} represent $1$
+through $6$ (Figure~\ref{fig:updown}), and two patterns are left over. Since
+\textsf{UUU} never appears, Aalaya can treat it as a private signal to switch
+back to the original trick; the performer shifts modes mid-show with a casual
+``Shall we make it harder and only show some of the cards?'' and nobody notices
+the mechanism changed. Since \textsf{DDD} never appears either, at least one
+card always lies face up, and that first face-up card can itself announce the
+suit. The dedicated suit card becomes redundant: only \emph{three} cards need be
+shown at all, and the fourth can be set aside with a theatrical flourish.
+
+\begin{figure}[H]
+\centering
+\begin{tikzpicture}[x=1cm,y=1cm]
+  % DDU = 1
+  \udD{0.00}{0}\udD{0.30}{0}\udU{0.60}{0} \node[font=\scriptsize] at (0.30,-0.42) {$\textsf{DDU}=1$};
+  % DUD = 2
+  \udD{1.55}{0}\udU{1.85}{0}\udD{2.15}{0} \node[font=\scriptsize] at (1.85,-0.42) {$\textsf{DUD}=2$};
+  % DUU = 3
+  \udD{3.10}{0}\udU{3.40}{0}\udU{3.70}{0} \node[font=\scriptsize] at (3.40,-0.42) {$\textsf{DUU}=3$};
+  % UDD = 4
+  \udU{0.00}{-1}\udD{0.30}{-1}\udD{0.60}{-1} \node[font=\scriptsize] at (0.30,-1.42) {$\textsf{UDD}=4$};
+  % UDU = 5
+  \udU{1.55}{-1}\udD{1.85}{-1}\udU{2.15}{-1} \node[font=\scriptsize] at (1.85,-1.42) {$\textsf{UDU}=5$};
+  % UUD = 6
+  \udU{3.10}{-1}\udU{3.40}{-1}\udD{3.70}{-1} \node[font=\scriptsize] at (3.40,-1.42) {$\textsf{UUD}=6$};
+\end{tikzpicture}
+\caption{``Ups and Downs.'' Three cards, each face-up (\textsf{U}) or face-down
+(\textsf{D}), send a number from $1$ to $6$. The patterns \textsf{UUU} and
+\textsf{DDD} are never used—each becomes a hidden bonus.}
+\label{fig:updown}
+\end{figure}
+
+\textbf{Hiding the tell (Mulcahy's ``Suit Alterations'').} A careful watcher of
+the basic trick notices, after a few rounds, that the suit-giving card always
+sits first. Here is the promised fix. Once Badarika has her four cards, she adds
+their values and reduces modulo $4$ (using $4$ in place of $0$); the result
+names \emph{which position} carries the suit, and the other three carry the
+number as usual. For example, the displayed cards \cc{4}{S}, \cc{Q}{C},
+\cc{5}{H}, \cc{10}{D} sum to $4+12+5+10 = 31 \equiv 3 \pmod 4$, so the third
+card, the \cc{5}{H}, names the suit; the remaining three in \textsf{HLM} order
+communicate $6$, and the hidden card is the \cc{8}{H}.
+
+\textbf{When the volunteer chooses (after Gardner's ``Eigen's Value'').} Now the
+volunteer, not Badarika, keeps one card back, so Badarika's best lever is gone. She
+has only the ordering of four cards—$24$ arrangements—to point at one of $48$
+unknowns, and $24$ can narrow $48$ down to at most two. The resolution is
+stagecraft. With a fixed total order agreed in advance, the four displayed
+cards single out two candidates (say the $15$th and $39$th in the residual
+ordered list of $48$); Aalaya quietly brings one to the top of the deck and
+the other to the bottom. Once the chosen card is named aloud, she turns over
+either the top or the bottom card to reveal it. The deciding bit is smuggled in
+by \emph{which end} she turns over, while every eye is on the volunteer.
+
+\textbf{Pushing to the extreme (Tom Edgar).} With a ``one-way'' deck—backs with
+no rotational symmetry—each card can point in four directions rather than two.
+Two face-down cards of this kind, where one may also sit slightly atop the
+other, produce $2\times 4\times 4 = 32$ distinct signals; combined with modular
+arithmetic on the single revealed card's value, even \emph{two} cards suffice
+for the $52$-card trick. There is even a one-card version using two
+differently-coloured decks, where the relative placement and orientation of a
+single card carry the whole message. These feel more like combinatorial puzzles
+than performances, but they show how far the core idea stretches.
+
+\textbf{The general pattern.} Behind every variation sits the same inequality.
+Suppose the audience draws a hand of $n$ cards, Aalaya is shown $r$ of them,
+and she must name the rest. A strategy exists exactly when
+\[
+\frac{(N-r)!}{(N-n)!} \;\geqslant\; n! ,
+\]
+where $N$ is the deck size. When $r = n-1$ (all but one card shown) the largest
+workable deck satisfies $N = n! + n - 1$; for $n=5$ this gives $124$, and at
+that maximum every hand of five matches exactly one arrangement of four—no
+slack, no redundancy. Such a ``perfect'' trick is the card-magic cousin of a
+perfect code.
+
+Several questions sit at the edges of this picture. For which pairs $(n,r)$ does
+a perfect deck size exist—one where hands and messages match with no remainder,
+a clean Diophantine condition $\tfrac{(N-r)!}{(N-n)!} = n!$? How many
+essentially different strategies are there for given $(n,r,N)$, once one strips
+away relabellings of the deck? And whenever a strategy exists, is there always
+one simple enough to carry in one's head, or does existence sometimes outrun any
+hope of performance?
+
+{\small
+\vskip 0.05in
+\affiliation[1.6em]{Associate Professor, CSE; IIT Gandhinagar}
+\email{neeldhara.misra@gmail.com}
+}
+\end{document}

fitch/fitch-trick.tex

@@ -0,0 +1,106 @@
+% !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}