Update Fitch trick solutions.
Co-authored-by: Cursor <cursoragent@cursor.com>
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% !TEX TS-program = xelatex
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\documentclass{../leelavati}
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\begin{document}
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\selectlanguage{odia}
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\title{\LARGE ମର୍କେଟର୍ ମାନଚିତ୍ରର କାହାଣୀ - ଭାଗ ୧ \vskip 0.04in \large The Story of the Mercator Map - Part 1}
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\author{\AuthorDual{ସୌମ୍ୟଶାନ୍ତ ନାୟକ}{Soumyashant Nayak}}
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\date{}
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\maketitle
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\setstretch{1.2}
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\begin{abstract}
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For centuries, navigating the open ocean was a high-stakes challenge of geometry and survival. This article is the first of a two-part series which follows the transition from the usage of environmental cues to the mathematical precision of modern cartography. We look at the geometric challenges of flattening a round Earth onto a flat grid and why simple maps can lead sailors astray. Ultimately, it is a story of how the vastness of the sea forced us to invent the clever coordinate systems that organized our world.
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\end{abstract}
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\setstretch{1.35}
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\section{ନୌଚାଳନାରେ ସଠିକ୍ ଅବସ୍ଥିତି ନିର୍ଣ୍ଣୟର ଜଟିଳତା}
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ଆଜିକାଲି ପକେଟ୍ରେ ଧରି ବୁଲୁଥିବା ଗୋଟିଏ ଛୋଟ ସ୍ମାର୍ଟଫୋନ୍ ଭିତରେ ହିଁ ଜିପିଏସ୍ (GPS) ଏବଂ ଗୁଗୁଲ୍ ମ୍ୟାପ୍ସ (Google Maps) ପରି ଡିଜିଟାଲ୍ ମାନଚିତ୍ର ରହୁଛି। ତେଣୁ ନୌଚାଳନା ଯେ ଦିନେ 'ସଠିକ୍ ଅବସ୍ଥିତି' ଜାଣିବା ଉପରେ ଆଧାରିତ ନଥିଲା, ସେକଥା ଭାବି ହୁଏ ନାହିଁ। ଅଥଳ ସମୁଦ୍ର ମଝିରେ ଥରେ କୂଳ ଆଖି ଆଗରୁ ହଜିଗଲା ପରେ, ନିଜର ସଠିକ୍ ଅବସ୍ଥିତି ଜାଣିବା ଇତିହାସର ଅଧିକାଂଶ ନାବିକଙ୍କ ପାଇଁ ପ୍ରାୟ ଅସମ୍ଭବ ଥିଲା।
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ଯଦିଓ ଦ୍ୱିତୀୟ ଶତାବ୍ଦୀରେ ଗ୍ରୀକ୍ ପଣ୍ଡିତ ଟଲେମି (Ptolemy) ତାଙ୍କ ପୁସ୍ତକ 'ଜିଓଗ୍ରାଫିଆ' (Geographia) ରେ ପୃଥିବୀକୁ ଅକ୍ଷାଂଶ (latitude) ଓ ଦ୍ରାଘିମାର (longitude) ଏକ ଗ୍ରିଡ୍ ଭାବେ ବର୍ଣ୍ଣନା କରିସାରିଥିଲେ, ତଥାପି ସେହି ଦୁଇଟି ସ୍ଥାନାଙ୍କ (coordinate) ମାଧ୍ୟମରେ ବାସ୍ତବ ଜଗତରେ ପଥ ଖୋଜିବା ପାଇଁ ମଣିଷକୁ ଆହୁରି ଅନେକ ଶତାବ୍ଦୀ ସଂଘର୍ଷ କରିବାକୁ ପଡ଼ିଥିଲା।
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ଏହା ବୁଦ୍ଧିମତାର ବିଫଳତା ନଥିଲା; ବରଂ ଏହା ପୃଥିବୀର ପ୍ରାକୃତିକ ସଂରଚନାର ଏକ ପରିଣାମ ଥିଲା। ପୃଥିବୀ ନିଜ ଅକ୍ଷ ଚାରିପଟେ ଆବର୍ତ୍ତନ କରୁଥିବାରୁ ଏହା ସ୍ୱାଭାବିକ ଭାବେ ଦୁଇଟି ମେରୁ (pole) ସୃଷ୍ଟି କରେ ଏବଂ ଏହି ଦୁଇ ମେରୁଠାରୁ ସମାନ ଦୂରତାରେ ରହିଥାଏ ବିଷୁବବୃତ୍ତ (equator)। ଏହି ଭୌଗୋଳିକ ତଥ୍ୟ ହିଁ ଅକ୍ଷାଂଶକୁ ବ୍ୟାଖ୍ୟା କରିବା ଏବଂ ମାପିବା ପାଇଁ ଏକ ସ୍ୱାଭାବିକ ଆଧାର ପ୍ରଦାନ କରେ। ସେଥିପାଇଁ ସୂର୍ଯ୍ୟ କିମ୍ବା ତାରାମାନଙ୍କ ସାହାଯ୍ୟରେ ଅକ୍ଷାଂଶ ଆକଳନ ସହଜରେ କରାଯାଇପାରୁଥିଲା (ସମସ୍ୟା \ref{artprob:altitude} ଦେଖ)।
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\begin{figure}[H]
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\centering
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\begin{tikzpicture}
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\node[anchor=center, inner sep=0] (image) at (0,0) {
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\includegraphics[width=.77\linewidth, angle=0]{grid_Orthographic.pdf}
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};
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\def\r{2.64}
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\coordinate (O) at (0,0);
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\fill (O) circle (1.2pt) node[below] {$\textbf{O}$};
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\coordinate (N) at (0,\r);
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\fill (N) circle (1.2pt) node[above left] {$N$};
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\draw[thick,dashed,->] (N) -- ++(0,1.0)
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node[above] {ଧ୍ରୁବତାରା};
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\coordinate (S) at (0,-\r);
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\fill (S) circle (1.2pt) node[below] {$S$};
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\coordinate (P) at ($(O)+(25:\r)$);
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\draw[dashed] (O) -- (P);
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\coordinate (H) at ($(P)+(115:2cm)$);
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\coordinate (PN) at ($(P) + (0,2.4)$);
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\fill (P) circle (1.2pt) node[right] {$\textbf{P}$};
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\draw[thick,dashed,->] (P) -- (PN) node[above] {ଧ୍ରୁବତାରା};
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\draw[thick, ->] (P) -- (H) node[shift={(130:0.3)}] {ଉତ୍ତରଦିଗ};
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\draw[thick] (P) -- ++(-60:1cm);
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\foreach \a/\lbl in {
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-60/{60^\circ S},
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-40/{40^\circ S},
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-20/{20^\circ S},
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0/{0^\circ},
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20/{20^\circ N},
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40/{40^\circ N},
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60/{60^\circ N}
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}{
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\node[left] at ($(O)+(180-\a:\r)$) {$\lbl$};
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}
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\coordinate (E) at (\r,0);
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\tkzMarkAngle[size=0.8, mark=none, arc=l](E,O,P)
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\tkzLabelAngle[pos=1.2](E,O,P){$\phi$}
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\tkzMarkAngle[size=0.8, mark=none, arc=l](PN,P,H)
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\tkzLabelAngle[pos=1.2](PN,P,H){$\psi$}
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\end{tikzpicture}
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\caption{ଅବସ୍ଥିତି $P$ ଠାରେ ଉତ୍ତରଦିଗ ଓ ଧ୍ରୁବତାରାର ଦିଗ}
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\label{fig:alt_lat}
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\end{figure}
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\begin{artprobox}
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ପ୍ରତିଛବି \ref{fig:alt_lat} ରେ, ଅକ୍ଷାଂଶ $\phi$ ର ସହାୟତାରେ $\psi$ କୋଣର ମାପ ନିର୍ଣ୍ଣୟ କର।
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\end{artprobox}
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\begin{artprobox}[altitude]
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ଧର, $P$ ପୃଥିବୀ ପୃଷ୍ଠରେ ତୁମ ଅବସ୍ଥିତିକୁ ସୂଚାଉଛି। ରାତିର ଆକାଶକୁ ନିରୀକ୍ଷଣ କରି ଏବଂ ସେକ୍ସଟାଣ୍ଟ (sextant) ପରି ଏକ କୋଣମାପକ ଯନ୍ତ୍ର ସାହାଯ୍ୟରେ କୋଣ $\psi$ ନିର୍ଣ୍ଣୟ କରିବା ପାଇଁ ଏକ କୌଶଳ ପ୍ରସ୍ତୁତ କର। ଉପରୋକ୍ତ ସମସ୍ୟାର ସମାଧାନକୁ ଭିତ୍ତି କରି ଓ $\psi$ ବ୍ୟବହାର କରି, ନିଜର ଅକ୍ଷାଂଶ ନିର୍ଣ୍ଣୟ କର।
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\end{artprobox}
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\vspace{-.1cm}
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\begin{figure}[H]
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\centering
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\begin{tikzpicture}[scale=.7]
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\coordinate (p) at (0,.7);
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\node at (p) {\faStar};
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\draw[black, dashed, thick] (p) -- (0,-2);
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\node[left] at (0,-.3) {$\psi$};
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\node[above] at (0,.9) {ଧ୍ରୁବତାରା};
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\node at (3.75,2.5) {ତାରାର};
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\node at (3.75,2) {ଅହୋରାତ୍ର ବୃତ୍ତ};
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\tkzDrawArc[R, color=magHighlight, dashed, thick, ->](p, 1.5cm)(0, 330)
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\coordinate (s1) at ($(p) + (-15:1.5cm)$);
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\node[scale=.5] at (s1) {\faStar};
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\tkzDrawArc[R, color=magHighlight, dashed, thick, ->](p, 3cm)(20, 240)
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\tkzDrawArc[R, color=magHighlight, dashed, thick, ->](p, 3cm)(300, 370)
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\coordinate (s1) at ($(p) + (15:3cm)$);
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\node[scale=.5] at (s1) {\faStar};
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\draw[magHighlight, line width=1pt]
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(-4,-2.5) .. controls (0,-1.9) .. (4,-2.5);
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\node[scale=3.5, text=black!80] at (0,-2.2) {\faUser};
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\end{tikzpicture}
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\caption{ଧ୍ରୁବତାରାର କୌଣିକ ଉଚ୍ଚତା $\psi$}
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\label{fig:pole_star_horizon}
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\end{figure}
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\vspace{-.2cm}
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ବିଷୁବବୃତ୍ତ ଯେପରି ଅକ୍ଷାଂଶ ପାଇଁ ଏକ ପ୍ରାକୃତିକ ଆଧାର, ଦ୍ରାଘିମା ପାଇଁ ସେପରି କୌଣସି ସ୍ଵାଭାବିକ ଆରମ୍ଭ ବିନ୍ଦୁ (origin) ନାହିଁ। ଆଜି ଆମେ ବ୍ୟବହାର କରୁଥିବା 'ଗ୍ରୀନଉଇଚ୍ ମେରିଡିଆନ୍' (Greenwich Meridian) କେବଳ ଏକ ସର୍ବସମ୍ମତ ଇଚ୍ଛାଧୀନ ନିଷ୍ପତ୍ତି ମାତ୍ର। ଦ୍ରାଘିମା ନିର୍ଣ୍ଣୟ କରିବା ପାଇଁ ନିଜର ସ୍ଥାନୀୟ ସମୟକୁ (local time) ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଆଧାର ଦ୍ରାଘିମାର (reference longitude) ସମୟ ସହିତ ତୁଳନା କରିବାକୁ ପଡ଼ିଥାଏ। ସାମୁଦ୍ରିକ କାଳମାପକର (marine chronometer) ଉଦ୍ଭାବନ ପୂର୍ବରୁ, ଦୋଳାୟମାନ ଜାହାଜରେ ସଠିକ୍ ସମୟ ରଖିବା ସମ୍ଭବ ନଥିଲା, ଯାହା ଫଳରେ ଦ୍ରାଘିମା ନିର୍ଣ୍ଣୟ କରିବା ଅସମ୍ଭବ ହୋଇପଡ଼ୁଥିଲା।
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\begin{transtable}
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ଅକ୍ଷାଂଶ & latitude & ମେରୁ & pole \\
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ଦ୍ରାଘିମା & longitude & ମାନଚିତ୍ର & map \\
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ବିଷୁବବୃତ୍ତ & equator & ଅକ୍ଷ & axis \\
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କାଳମାପକ & chronometer & ଆବର୍ତ୍ତନ & rotation\\
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ନୌଚାଳନା & navigation & ସ୍ଥାନାଙ୍କ & coordinate
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\end{transtable}
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\section{ମାନଚିତ୍ର ବିହୀନ ପାରମ୍ପରିକ ନୌଚାଳନା}
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ଜଣେ ନାବିକ ନିଜର ଉତ୍ତର-ଦକ୍ଷିଣ ଅବସ୍ଥିତି ଜାଣିପାରୁଥିଲେ ମଧ୍ୟ, ପୂର୍ବ-ପଶ୍ଚିମ ଅବସ୍ଥିତିରେ ସବୁବେଳେ ଅନିଶ୍ଚିତତା ରହୁଥିଲା।
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\begin{questionbox}
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ଏହି ସମସ୍ୟା ସତ୍ତ୍ୱେ ଜାହାଜଗୁଡ଼ିକ ନିୟମିତ ଭାବେ ସମୁଦ୍ର ପାର ହେଉଥିଲେ; ଏହା କିପରି ସମ୍ଭବ ହେଉଥିଲା?
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\end{questionbox}
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ମହାନଦୀର ମୁହାଣରୁ ଦକ୍ଷିଣ-ପୂର୍ବ ଏସିଆ ଅଭିମୁଖେ ଯାତ୍ରା କରୁଥିବା କଳିଙ୍ଗର ନାବିକମାନଙ୍କ କଥା ଚିନ୍ତା କରନ୍ତୁ। ସେମାନଙ୍କର ନୌଚାଳନା କୌଣସି ମାନଚିତ୍ର କିମ୍ବା ଅକ୍ଷାଂଶ-ଦ୍ରାଘିମାର ସ୍ଥାନାଙ୍କ ଗ୍ରିଡ୍ ଉପରେ ନିଜର ଅବସ୍ଥିତି ଚିହ୍ନଟ କରିବା ଉପରେ ନିର୍ଭର କରୁନଥିଲା। ବରଂ, ଏହା ପରିବେଶର ଏକ ଗଭୀର ବୁଝାମଣା ଉପରେ ଆଧାରିତ ଥିଲା।
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ସେମାନେ ମୌସୁମୀ ବାୟୁର ଗତିପଥକୁ ଅନୁସରଣ କରୁଥିଲେ, ଯାହା ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସମୟ-ଚକ୍ର ଅନୁସାରେ ନିଜର ଦିଗ ପରିବର୍ତ୍ତନ କରିଥାଏ। ସମ୍ଭବ ହେଲେ ସେମାନେ ଉପକୂଳ ଅନୁସରଣ କରୁଥିଲେ, ଗୋଟିଏ ବନ୍ଦରରୁ ଅନ୍ୟ ବନ୍ଦରକୁ ଯାଉଥିଲେ ଏବଂ ଦ୍ୱୀପପୁଞ୍ଜଗୁଡ଼ିକୁ ପାହାଚ ଭାବରେ ବ୍ୟବହାର କରୁଥିଲେ। ସେମାନଙ୍କୁ ସମୁଦ୍ରରୁ ବିଭିନ୍ନ ସଙ୍କେତ ପଢ଼ିବାକୁ ପଡ଼ୁଥିଲା: ଯଥା ଲହରୀର ଦିଗ ବା ଗତି, ପକ୍ଷୀ ଓ ଜଳଜ ଉଦ୍ଭିଦମାନଙ୍କର ଉପସ୍ଥିତି, ସମୁଦ୍ର ପାଣିର ରଙ୍ଗରେ ପରିବର୍ତ୍ତନ ଏବଂ ପବନରେ ଭାସି ଆସୁଥିବା ସ୍ଥଳଭାଗର ବାସ୍ନା।
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କିନ୍ତୁ, ଯେତେବେଳେ ସମୁଦ୍ର ଯାତ୍ରାର ପରିସର ବୃଦ୍ଧି ପାଇଲା, ସେତେବେଳେ ପରିବେଶର ଏହି ସୂକ୍ଷ୍ମ ଅନୁଭବ ଏବଂ ପରୋକ୍ଷ ସଙ୍କେତ ସୀମିତ ହୋଇପଡ଼ିଲା। ଆନ୍ତଃ-ମହାସାଗରୀୟ ଯାତ୍ରାର ଅର୍ଥ ଥିଲା ସ୍ଥଳଭାଗ ନ ଦେଖି ସପ୍ତାହ ସପ୍ତାହ ଧରି ସମୁଦ୍ର ମଧ୍ୟରେ ରହିବା। ଏଠାରେ ପରିଚିତ ଚିହ୍ନ ଅପେକ୍ଷା ପଥ ନିର୍ଣ୍ଣୟ କରିବାର ଗାଣିତିକ ପଦ୍ଧତି ଅତ୍ୟାବଶ୍ୟକ ହୋଇପଡ଼ିଲା। କିନ୍ତୁ ଗଣିତର ପ୍ରୟୋଗ ପୂର୍ବରୁ ଆମକୁ ଏହା ବୁଝିବାକୁ ହେବ ଯେ ମାନଚିତ୍ର ପ୍ରକୃତରେ କିପରି ବିବର୍ତ୍ତିତ ହୋଇଥିଲା।
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\section{ନୌଚାଳନା ପାଇଁ ବିଶ୍ୱ ମାନଚିତ୍ରର ସୃଷ୍ଟି}
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ମାନଚିତ୍ରର ଇତିହାସ ବହୁ ପୁରୁଣା ହେଲେ ମଧ୍ୟ, ଦୀର୍ଘ ସମୟ ପର୍ଯ୍ୟନ୍ତ ଏହାର ନୌଚାଳନା ସହିତ କୌଣସି ସିଧାସଳଖ ସମ୍ପର୍କ ନଥିଲା। ପ୍ରାଥମିକ ସମୟର ମାନଚିତ୍ରଗୁଡ଼ିକ, ଯେପରିକି ବାବିଲୋନୀୟ ମୃଣ୍ମୟ ଫଳକ (Babylonian clay tablets) କିମ୍ବା ମଧ୍ୟଯୁଗୀୟ 'ମାପ୍ପା ମୁଣ୍ଡି' (Mappa Mundi), ବାସ୍ତବିକ ପଥ ପ୍ରଦର୍ଶନ ପାଇଁ ଉଦ୍ଦିଷ୍ଟ ନ ହୋଇ ମୁଖ୍ୟତଃ ପ୍ରତୀକାତ୍ମକ ଓ ସୂଚନାତ୍ମକ ଥିଲେ। ମୌଳିକ ଭାବେ, ସେଗୁଡ଼ିକ କେବଳ ସ୍ଥାନୀୟ ସମ୍ପର୍କକୁ ଦର୍ଶାଉଥିଲେ - ଯେପରିକି ପାହାଡ଼ରୁ ବାହାରୁଥିବା ନଦୀ, ରାସ୍ତା ଦ୍ୱାରା ସଂଯୁକ୍ତ ସହର କିମ୍ବା ଧାର୍ମିକ ସ୍ଥଳ। ଏଥିରେ ଦିଗ ବା ଦୂରତାର ସଠିକତା ଅପେକ୍ଷା ସାଙ୍କେତିକ ସୂଚନାକୁ ଅଧିକ ଗୁରୁତ୍ୱ ଦିଆଯାଉଥିଲା।
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କାଳକ୍ରମେ ମାନଚିତ୍ରଗୁଡ଼ିକ ଅଧିକ ବିସ୍ତୃତ ହେଲେ। ସେତେବେଳେ ମଧ୍ୟ ସେଗୁଡ଼ିକ ମୁଖ୍ୟତଃ ସ୍ଥାନୀୟ ଅଞ୍ଚଳ ମଧ୍ୟରେ ସୀମିତ ଥିଲା: ଯେପରିକି ବନ୍ଦରର ନକ୍ସା, ନଦୀ ଅବବାହିକା (river basin) କିମ୍ବା ସ୍ଥଳପଥ। ସେତେବେଳେ ବିଶ୍ୱସ୍ତରୀୟ ଜ୍ୟାମିତିର ବିଶେଷ ଆବଶ୍ୟକତା ନଥିଲା। ବାରମ୍ବାର ନୌଚାଳନା କରି ନାବିକମାନେ ଏକ ଉପକୂଳର ସଠିକ୍ ମାନଚିତ୍ର ପ୍ରସ୍ତୁତ କରିପାରୁଥିଲେ। ମାତ୍ର ମହାସାଗର ପରି ବିଶାଳ ପରିସର ପାଇଁ ଏତିକି ଜ୍ଞାନ ଯଥେଷ୍ଟ ନଥିଲା।
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୧୪୯୨ ମସିହାରେ, ସ୍ପେନ୍ର (Spain) ରାଜକୀୟ ଅନୁଦାନରେ ଖ୍ରୀଷ୍ଟୋଫର୍ କଲମ୍ବସ୍ (Christopher Columbus) ଏକ ଦୁଃସାହସିକ ଆବିଷ୍କାର ଅଭିଯାନ ଆରମ୍ଭ କରିଥିଲେ। ପୃଥିବୀର ପରିଧିକୁ ନେଇ ପ୍ରଚଳିତ ବିଭିନ୍ନ ମତବାଦ ମଧ୍ୟରେ କଲମ୍ବସ୍ଙ୍କ ଗଣନା ତ୍ରୁଟିପୂର୍ଣ୍ଣ ଥିଲା; ସେ କେବଳ ପୃଥିବୀର ଆକାରକୁ ଛୋଟ ମାପିନଥିଲେ, ବରଂ ଏସିଆର ପରିସରକୁ ମଧ୍ୟ ଅତ୍ୟଧିକ ବଡ଼ ବୋଲି ଅନୁମାନ କରିଥିଲେ। ନିଜର ଏହି ଭୁଲ୍ ଗଣନା ଯୋଗୁଁ, କଲମ୍ବସ୍ ୟୁରୋପରୁ ପଶ୍ଚିମ ଦିଗ ଦେଇ ଏସିଆ ପହଞ୍ଚିବାର ପ୍ରକୃତ ଦୂରତାକୁ ଯଥେଷ୍ଟ କମ୍ କରି ଆକଳନ କରିଥିଲେ। ସେ ବିଶ୍ୱାସ କରୁଥିଲେ ଯେ ଆଟଲାଣ୍ଟିକ୍ ମହାସାଗରରେ ପଶ୍ଚିମାଭିମୁଖୀ ଯାତ୍ରାର ଅଳ୍ପ କିଛି ସପ୍ତାହ ମଧ୍ୟରେ ସେ ସିଧାସଳଖ ଏସିଆର ଉପକୂଳରେ ପହଞ୍ଚିଯିବେ। କୌଣସି ସଠିକ୍ ମାନଚିତ୍ର ବିନା, ସେ କେବଳ କମ୍ପାସ୍ ଏବଂ ଜାହାଜର ଗତି ଓ ଦିଗର ଏକ ଆନୁମାନିକ ହିସାବ (dead reckoning) ରଖି ସୌଭାଗ୍ୟବଶତଃ ଏକ ଭୂଖଣ୍ଡରେ ପହଞ୍ଚିଥିଲେ, ଯାହାକୁ ଆଜି ଆମେ ଆମେରିକା ଭାବେ ଜାଣୁ। କିନ୍ତୁ ମୃତ୍ୟୁ ପର୍ଯ୍ୟନ୍ତ ସେ ଏହି ଭୂଖଣ୍ଡକୁ ଏସିଆ ବୋଲି ଭାବି ଭ୍ରମରେ ରହିଲେ।
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ପଞ୍ଚଦଶ ଶତାବ୍ଦୀର ଶେଷ ଭାଗର ଏହି ଅନ୍ୱେଷଣ, ଷୋଡ଼ଶ ଶତାବ୍ଦୀ ବେଳକୁ ଏକ ବିଶାଳ ବାଣିଜ୍ୟିକ ସମ୍ଭାବନାରେ ପରିଣତ ହେଲା। ଆଟଲାଣ୍ଟିକ୍ ମହାସାଗରର ଅଜ୍ଞାତ ଜଳରାଶି ମଧ୍ୟକୁ ପ୍ରବେଶ କରୁଥିବା ୟୁରୋପୀୟ ନାବିକମାନଙ୍କ ନିମନ୍ତେ ପୂର୍ବରୁ ବ୍ୟବହୃତ ଉପକୂଳବର୍ତ୍ତୀ ମାନଚିତ୍ରଗୁଡ଼ିକ ଖୋଲା ସମୁଦ୍ରରେ ନିରର୍ଥକ ପ୍ରମାଣିତ ହେଲା। ସେଠାରେ ହଜାର ହଜାର ମାଇଲ୍ ପର୍ଯ୍ୟନ୍ତ କୌଣସି ସ୍ଥଳଭାଗର ସଙ୍କେତ ନଥିଲା କିମ୍ବା ପଥ ପ୍ରଦର୍ଶନ ପାଇଁ ପରିଚିତ ପକ୍ଷୀଙ୍କ ଦେଖା ମଧ୍ୟ ମିଳୁନଥିଲା।
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ଫଳସ୍ୱରୂପ, 'ଟ୍ରାନ୍ସ-ଆଟଲାଣ୍ଟିକ୍' ଯାତ୍ରାକୁ ନିରାପଦ, ନିର୍ଭରଯୋଗ୍ୟ ଓ ପୁନରାବୃତ୍ତି ଯୋଗ୍ୟ କରିବା ପାଇଁ ମାନଚିତ୍ରର ପରିକଳ୍ପନାକୁ ସ୍ଥାନୀୟ ବର୍ଣ୍ଣନା ପରିସରରୁ ବାହାର କରି ଏକ ଜ୍ୟାମିତି ଆଧାରିତ ବିଶ୍ୱ-ମାନଚିତ୍ରରେ (global map) ରୂପାନ୍ତରିତ କରିବା ଅପରିହାର୍ଯ୍ୟ ହୋଇପଡ଼ିଲା। ନାବିକମାନଙ୍କୁ ଏବେ ଏମିତି ଏକ ମାନଚିତ୍ର ଦରକାର ଥିଲା, ଯେଉଁଠାରେ ପ୍ରତ୍ୟେକ ବିନ୍ଦୁର ଏକ ସୁନିଶ୍ଚିତ ସ୍ଥାନାଙ୍କ ରହିବା ସହ ପ୍ରତ୍ୟେକ ରେଖା ସମୁଦ୍ରର ବାସ୍ତବିକ ଦିଗ ସହିତ ମେଳ ଖାଇବ।
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ତେବେ, ଏହି ଗାଣିତିକ ଆବଶ୍ୟକତା ପୂରଣ କରିବାକୁ ହେଲେ ଆମକୁ ସର୍ବପ୍ରଥମେ ପୃଥିବୀର ଗୋଲାକାର ଜ୍ୟାମିତି ଏବଂ ସମତଳ କାଗଜ ମଧ୍ୟରେ ଥିବା ଏକ ମୌଳିକ ଦ୍ୱନ୍ଦ୍ୱକୁ ବୁଝିବାକୁ ହେବ।
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\section{ସମାନ ବ୍ୟବଧାନର ଗ୍ରିଡ୍ରେ ଜ୍ୟାମିତିକ ବିକୃତି}
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ଧର, ଆମେ ଏକ ସମତଳ କାଗଜରେ ସମଗ୍ର ପୃଥିବୀର ମାନଚିତ୍ର ଆଙ୍କିବା ପାଇଁ ଅକ୍ଷାଂଶ ଓ ଦ୍ରାଘିମାକୁ ସମାନ ବ୍ୟବଧାନରେ କିଛି ଭୂସମାନ୍ତର (horizontal) ଓ ଭୂଲମ୍ବ (vertical) ରେଖା ଦ୍ୱାରା ଚିହ୍ନିତ କଲୁ (ପ୍ରତିଛବି \ref{fig:uniform_grid_map} ଦେଖ)।
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\begin{figure}[H]
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\centering \includegraphics[width=1.0\linewidth]{world_map_PlateCarree.pdf}
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\caption{ସମାନ ବ୍ୟବଧାନର ଗ୍ରିଡ୍ }
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\label{fig:uniform_grid_map}
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\end{figure}
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ପ୍ରଥମ ଦେଖାରେ ଏହା ଯୁକ୍ତିଯୁକ୍ତ ମନେହୁଏ, କାରଣ ଅକ୍ଷାଂଶ ଓ ଦ୍ରାଘିମା ହେଉଛି କେବଳ କୋଣ (angle) ଏବଂ କୋଣୀୟ ମାପ ସବୁ ସ୍ଥାନରେ ସମାନ ମନେହୁଏ। କିନ୍ତୁ ଏହି ସରଳ ଗ୍ରିଡ୍ ଚିତ୍ରଣ, ବାସ୍ତବିକ ଦୂରତା ଓ ଆକୃତିକୁ ନେଇ କିଛି ଭ୍ରାନ୍ତ ଧାରଣା ସୃଷ୍ଟି କରେ।
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|
||||
ବର୍ତ୍ତମାନ ପୃଥିବୀକୁ $R$ ବ୍ୟାସାର୍ଦ୍ଧ ବିଶିଷ୍ଟ ଏକ ଗୋଲକ ଭାବରେ ବିଚାର କର। ବିଷୁବବୃତ୍ତରୁ ଉତ୍ତରକୁ $\phi$ କୋଣରେ ଥିବା ଏକ ଅକ୍ଷାଂଶ-ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ହେଉଛି, $R \cos \phi$। ତେଣୁ ସେହି ଅକ୍ଷାଂଶ-ବୃତ୍ତର ଲମ୍ବ ହେଉଛି, $2 \pi R \cos \phi $ (ପ୍ରତିଛବି \ref{fig:rad_change} ଦେଖ)।
|
||||
|
||||
\begin{figure}
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=1]
|
||||
|
||||
% Parameters
|
||||
\def\R{2.5} % Earth radius
|
||||
\def\phi{40} % Latitude angle (for drawing only)
|
||||
|
||||
% Center
|
||||
\coordinate (O) at (0,0);
|
||||
\coordinate (N) at (0,\R);
|
||||
\coordinate (S) at (0,-\R);
|
||||
|
||||
\fill (N) circle (1.2pt);
|
||||
\fill (S) circle (1.2pt);
|
||||
|
||||
% Earth
|
||||
\draw[thick] (O) circle (\R);
|
||||
\draw[dashed] (N)--(S);
|
||||
|
||||
% Equator
|
||||
\draw (-\R,0) -- (\R,0);
|
||||
\node[below left] at (0,0) {ବିଷୁବବୃତ୍ତ};
|
||||
|
||||
% Latitude line
|
||||
\draw
|
||||
(0,{\R*sin(\phi)}) --
|
||||
({\R*cos(\phi)},{\R*sin(\phi)});
|
||||
|
||||
% Radius to latitude
|
||||
\draw (O) -- ({\R*cos(\phi)},{\R*sin(\phi)});
|
||||
|
||||
% Horizontal radius (R cos phi)
|
||||
\draw
|
||||
(0,{ \R*sin(\phi) }) -- ({\R*cos(\phi)},{\R*sin(\phi)});
|
||||
\node[above] at ({0.4*\R*cos(\phi)},{\R*sin(\phi)})
|
||||
{$R\cos φ$};
|
||||
|
||||
% Radius R
|
||||
\draw[dashed] (O) -- ({\R*cos(\phi)},{\R*sin(\phi)});
|
||||
\node at (1.2,1.2) {$R$};
|
||||
|
||||
% Angle phi
|
||||
\draw (0,0) ++(0:0.6) arc (0:\phi:0.6);
|
||||
\node[right] at (0.5,0.25) {$φ$};
|
||||
|
||||
% Poles
|
||||
\node[above] at (0,\R) {ଉତ୍ତର ମେରୁ};
|
||||
\node[below] at (0,-\R) {ଦକ୍ଷିଣ ମେରୁ};
|
||||
|
||||
\end{tikzpicture}
|
||||
\caption{ଅକ୍ଷାଂଶ-ବୃତ୍ତର ବ୍ୟାସାର୍ଦ୍ଧ ପରିବର୍ତ୍ତନ}
|
||||
\label{fig:rad_change}
|
||||
\end{figure}
|
||||
|
||||
\begin{table}[h]
|
||||
\centering
|
||||
\renewcommand{\arraystretch}{2}
|
||||
\begin{tabular}{|c|c|c|}
|
||||
\hline
|
||||
\textbf{ଅକ୍ଷାଂଶ ($\phi$) } & $\bm{\cos} (\phi)$ & \textbf{ଅକ୍ଷାଂଶ-ବୃତ୍ତର ଲମ୍ବ}\\
|
||||
\hline
|
||||
$0^\circ$ & $1$ & $6.28 \times R$ \\
|
||||
\hline
|
||||
$30^\circ$ & $\frac{\sqrt{3}}{2}$ & $5.44 \times R$ \\
|
||||
\hline
|
||||
$45^\circ$ & $\frac{1}{\sqrt{2}}$ & $4.44 \times R$ \\
|
||||
\hline
|
||||
$60^\circ$ & $\frac{1}{2}$ & $3.14 \times R$ \\
|
||||
\hline
|
||||
$90^\circ$ & 0 & $0$ \\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\caption{କିଛି ସାଧାରଣ କୋଣ ପାଇଁ ଅକ୍ଷାଂଶ-ବୃତ୍ତର ଲମ୍ବ}
|
||||
\label{tab:cos_val}
|
||||
\end{table}
|
||||
|
||||
|
||||
ଟେବୁଲ୍ \ref{tab:cos_val} ରୁ ସ୍ପଷ୍ଟ ହେଉଛି ଯେ, ପୃଥିବୀର ମେରୁ ଆଡ଼କୁ ଗଲେ ଅକ୍ଷାଂଶ-ବୃତ୍ତଗୁଡ଼ିକର ପରିଧି କ୍ରମଶଃ ହ୍ରାସ ପାଇଥାଏ। ଫଳସ୍ୱରୂପ, ସମାନ ଅକ୍ଷାଂଶରେ ଅବସ୍ଥିତ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଦ୍ରାଘିମା-ପାର୍ଥକ୍ୟ (longitude difference) ବିଶିଷ୍ଟ ଦୁଇଟି ସ୍ଥାନର ବାସ୍ତବ ପୂର୍ବ-ପଶ୍ଚିମ ଦୂରତା ମଧ୍ୟ ସେହି ଅନୁପାତରେ ସଙ୍କୁଚିତ ହୋଇଯାଏ। ଉଦାହରଣ ସ୍ୱରୂପ, ବିଷୁବବୃତ୍ତ ଉପରେ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ଦ୍ରାଘିମା-ପାର୍ଥକ୍ୟ ଥିବା ଦୁଇଟି ବିନ୍ଦୁ ମଧ୍ୟରେ ଯେତିକି ବାସ୍ତବ ଦୂରତା ଥାଏ, $45^{\circ}$ ଅକ୍ଷାଂଶରେ ତାହାର ମାତ୍ର $71 \%$ ହୁଏ (ଯେହେତୁ $\cos 45^{\circ} \approx 0.71$)। କିନ୍ତୁ ଆମ ସମତଳ ଗ୍ରିଡ୍ରେ ଏହି ଦୂରତାକୁ ସବୁ ସ୍ଥାନରେ ସମାନ ଦେଖାଯାଉଥିବାରୁ, ବାସ୍ତବିକତା ଏବଂ ମାନଚିତ୍ର ମଧ୍ୟରେ ଏକ ବଡ଼ ଧରଣର ଅସଙ୍ଗତି ସୃଷ୍ଟି ହୁଏ।
|
||||
|
||||
ଏହାକୁ ବୁଝିବା ପାଇଁ କଳ୍ପନା କର ଯେ ତୁମେ ଅକ୍ଷାଂଶ $\phi$ ରେ ପୃଥିବୀର ଏକ ଅତ୍ୟଣୁ (infinitesimal) ଅଞ୍ଚଳକୁ ବଡ଼ କରି ଦେଖୁଛ, ଯାହାର ବିସ୍ତାର ହେଉଛି Δϕ ଅକ୍ଷାଂଶ-ପାର୍ଥକ୍ୟ ଏବଂ Δλ ଦ୍ରାଘିମା-ପାର୍ଥକ୍ୟ (ଏଠାରେ ରେଡିଆନ୍ ଏକକ ବ୍ୟବହୃତ)। ତେବେ ଗୋଲକ ଉପରେ ଏହାର ଉତ୍ତର-ଦକ୍ଷିଣ ପାର୍ଶ୍ୱର ପ୍ରକୃତ ଦୂରତା $R \Delta \phi$ ହୋଇଥିବା ବେଳେ ପୂର୍ବ-ପଶ୍ଚିମ ପାର୍ଶ୍ୱର ପ୍ରକୃତ ଦୂରତା ହେଉଛି $R \cos \phi \Delta \lambda$ (ପ୍ରତିଛବି \ref{fig:dist_change} ଦେଖ)।
|
||||
|
||||
\begin{figure}[H]
|
||||
\centering
|
||||
\begin{tikzpicture}[scale=1]
|
||||
|
||||
\begin{scope}
|
||||
\node at (-2,2.3) {\textbf{Earth}};
|
||||
\coordinate (O) at (0,0);
|
||||
\draw[thick] (O) ++(20:2.2)
|
||||
arc (20:160:2.2);
|
||||
% Latitude line
|
||||
\draw[dashed] (-1.8,1.0) -- (1.8,1.0);
|
||||
\node[left] at (-2,1.0) {$\phi$};
|
||||
|
||||
% Local patch
|
||||
\draw[fill=magHighlight!40] (-0.6,0.8) rectangle (0.6,1.3);
|
||||
|
||||
% North–South arrow
|
||||
\draw[<->] (-.8,0.65) -- (-.8,1.5);
|
||||
\node[left] at (-.8,.8) {$R\; \Delta \phi$};
|
||||
|
||||
\draw[<->] (.8,0.65) -- (.8,1.5);
|
||||
\node[right] at (.8,.8) {N-S};
|
||||
|
||||
\draw[<->] (-0.6,0.5) -- (0.6,0.5);
|
||||
\node[below] at (0,0.5) {$R\cos\phi \; \Delta \lambda$};
|
||||
|
||||
\draw[<->] (-0.6,1.5) -- (0.6,1.5);
|
||||
\node[above] at (0,1.5) {E-W};
|
||||
|
||||
\end{scope}
|
||||
|
||||
\begin{scope}[yshift=-3cm]
|
||||
|
||||
\node at (-2,2.5) {\textbf{Map}};
|
||||
|
||||
\draw[thick] (-2.2,.75) -- (-2.2,2.2)--(2.2,2.2) -- (2.2,.75);
|
||||
|
||||
% Latitude line
|
||||
\draw[dashed] (-2.2,1.0) -- (2.2,1.0);
|
||||
\node[left] at (-2.2,1.0) {$\phi$};
|
||||
|
||||
% Distorted patch
|
||||
\draw[fill=magHighlight!50] (-.75,.8) rectangle (.75,1.3);
|
||||
|
||||
% North–South arrow
|
||||
\draw[<->] (-.95,0.65) -- (-.95,1.5);
|
||||
\node[left] at (-.95,.8) {$\Delta \phi$};
|
||||
|
||||
\draw[<->] (.95,0.65) -- (.95,1.5);
|
||||
\node[right] at (.95,.8) {N-S};
|
||||
|
||||
\draw[<->] (-0.75,0.5) -- (0.75,0.5);
|
||||
\node[below] at (0,0.5) {$\Delta \lambda$};
|
||||
|
||||
\draw[<->] (-0.75,1.5) -- (0.75,1.5);
|
||||
\node[above] at (0,1.5) {E-W};
|
||||
|
||||
\end{scope}
|
||||
|
||||
\end{tikzpicture}
|
||||
\caption{ମାନଚିତ୍ରରେ ପୂର୍ବ-ପଶ୍ଚିମ ଦୂରତାର ପ୍ରସାରଣ}
|
||||
\label{fig:dist_change}
|
||||
\end{figure}
|
||||
|
||||
ଆମ ମାନଚିତ୍ରରେ ଆୟତକ୍ଷେତ୍ରର ପାର୍ଶ୍ୱଗୁଡ଼ିକ $\Delta \phi$ ଏବଂ $\Delta \lambda$ ସହିତ ସମାନୁପାତିକ ଅଟେ। ଫଳସ୍ୱରୂପ, ଭୂପୃଷ୍ଠର ପ୍ରକୃତ ଦୂରତା ଅନୁପାତ ତୁଳନାରେ ଏହି ଆୟତକ୍ଷେତ୍ରଟି ମାନଚିତ୍ରରେ ଭୂସମାନ୍ତର ଭାବେ $\frac{1}{\cos \phi} = \sec \phi$ ଗୁଣ ପ୍ରସାରିତ ହୋଇ ଦେଖାଯାଏ।
|
||||
|
||||
|
||||
|
||||
ଏହି ଅସମାନ ପ୍ରସାରଣ ଯୋଗୁଁ ମାନଚିତ୍ରର ଉପଯୋଗିତା କ୍ଷେତ୍ରରେ କିଛି ପ୍ରତିବନ୍ଧକ ସୃଷ୍ଟି ହୁଏ। ଯଦି ପୃଥିବୀ ପୃଷ୍ଠରେ ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ପଥ ଉତ୍ତର ଦିଗ ସହିତ $\theta$ କୋଣ ସୃଷ୍ଟି କରେ, ତେବେ ଆମ ମାନଚିତ୍ରରେ ସେହି ପଥଟି ଏକ ଭିନ୍ନ କୋଣ $\theta '$ ସୃଷ୍ଟି କରିବ। ଏହି ଦୁଇ କୋଣ ମଧ୍ୟରେ ସମ୍ପର୍କ ହେଉଛି,
|
||||
\begin{equation}
|
||||
\label{eqn:non_conform}
|
||||
\tan \theta '= \sec \phi \tan \theta \quad \text{।}
|
||||
\end{equation}
|
||||
|
||||
\begin{figure}[H]
|
||||
\centering
|
||||
\begin{tikzpicture}
|
||||
\begin{scope}
|
||||
\coordinate (P) at (0,0);
|
||||
\fill (P) circle (2pt) node[left] {$P$};
|
||||
% North direction
|
||||
\draw[thick,->] (P) -- (0,1);
|
||||
\node[above] at (0,1) {$N$};
|
||||
|
||||
% Arbitrary direction
|
||||
\draw[thick,->] (P) -- (1.3,1);
|
||||
\node[above] at (1.2,1) {$D$};
|
||||
|
||||
\draw[thick,dashed] (0,1) -- (1.2,1);
|
||||
|
||||
% Angle marking
|
||||
\draw (0,0) ++(90:0.5) arc (90:35:0.5);
|
||||
\node[above] at (0.35,0.4) {$\theta$};
|
||||
|
||||
\draw[<->] (-.5,0) -- (-.5,1);
|
||||
\node[left] at (-.5,.5) {$N-S$};
|
||||
|
||||
\draw[<->] (0,-.5) -- (1.3,-0.5);
|
||||
\node[below] at (.65,-0.5) {$R\cos\phi \; \Delta \lambda$};
|
||||
\end{scope}
|
||||
|
||||
\begin{scope}[xshift = 2.5cm]
|
||||
\coordinate (P) at (0,0);
|
||||
\fill (P) circle (2pt) node[left] {$P$};
|
||||
% North direction
|
||||
\draw[thick,->] (P) -- (0,1);
|
||||
\node[above] at (0,1) {$N$};
|
||||
|
||||
% Arbitrary direction
|
||||
\draw[thick,->] (P) -- (1.5,1);
|
||||
\node[above] at (1.4,1) {$D'$};
|
||||
|
||||
\draw[thick,dashed] (0,1) -- (1.5,1);
|
||||
|
||||
% Angle marking
|
||||
\draw (0,0) ++(90:0.5) arc (90:35:0.5);
|
||||
\node[above] at (0.35,0.4) {$\theta'$};
|
||||
|
||||
\draw[<->] (0,-.5) -- (1.5,-0.5);
|
||||
\node[below] at (.65,-0.5) {$R \; \Delta \lambda$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\caption{ମାନଚିତ୍ରରେ ଅନୁକୋଣତାର କ୍ଷୟ}
|
||||
\label{fig:non_conform}
|
||||
\end{figure}
|
||||
|
||||
ଏଠାରେ ଉଲ୍ଲେଖଯୋଗ୍ୟ ଯେ, ଯଦିଓ ଏହି ମାନଚିତ୍ରରେ ଯେକୌଣସି ଦୁଇଟି ରେଖା ମଧ୍ୟରେ ଥିବା କୋଣର ବିକୃତି ଘଟିଥାଏ, ଏହି ସୂତ୍ରଟି କେବଳ ଉତ୍ତର ଦିଗ ସହିତ ସମ୍ପର୍କିତ କୋଣର ପରିବର୍ତ୍ତନକୁ ପ୍ରକାଶ କରେ।
|
||||
|
||||
|
||||
\begin{artprobox}
|
||||
ପ୍ରତିଛବି \ref{fig:non_conform} ସାହାଯ୍ୟରେ ସମୀକରଣ \ref{eqn:non_conform} କୁ ପ୍ରମାଣ କର।
|
||||
\end{artprobox}
|
||||
|
||||
ବିଷୁବବୃତ୍ତ ($\phi = 0^{\circ}$, $\cos 0^{\circ} = 1$) ବ୍ୟତୀତ ଅନ୍ୟ ସବୁ ସ୍ଥାନରେ କୋଣଗୁଡ଼ିକରେ ଏହି ପରିବର୍ତ୍ତନ ଘଟିଥାଏ। ଆମ ମାନଚିତ୍ରର ଗ୍ରିଡ୍ ଦେଖିବାକୁ ପରିଷ୍କାର ମନେହୁଏ ସତ, କିନ୍ତୁ ଏହାର ଅନୁକୋଣତା (conformality) ନ ଥାଏ - ଅର୍ଥାତ୍ ମାନଚିତ୍ରରେ ଦର୍ଶାଯାଇଥିବା ଦୁଇଟି ରେଖା ମଧ୍ୟରେ ଥିବା କୋଣ ପ୍ରାୟତଃ ବାସ୍ତବିକ କୋଣ ସହ ସମାନ ରହେ ନାହିଁ।
|
||||
|
||||
\begin{transtable}
|
||||
ଗୋଲକ & sphere & ଅତ୍ୟଣୁ & infinitesimal\\
|
||||
ବିକୃତି & distortion & ପରିଧି & circumference \\
|
||||
ବ୍ୟାସାର୍ଦ୍ଧ & radius & ବର୍ଗକ୍ଷେତ୍ର & square \\
|
||||
ବୃତ୍ତ & circle & ଆୟତକ୍ଷେତ୍ର & rectangle \\
|
||||
ଉପବୃତ୍ତ & ellipse & ଅନୁକୋଣତା & conformality
|
||||
\end{transtable}
|
||||
|
||||
କୋଣଗୁଡ଼ିକୁ ଅକ୍ଷୁଣ୍ଣ ରଖିବା ଏବଂ ଅତ୍ୟଣୁ ସ୍ତରର ଆକୃତିଗୁଡ଼ିକୁ ବଜାୟ ରଖିବା\textemdash ଏ ଉଭୟ ଓତପ୍ରୋତ ଭାବେ ଜଡ଼ିତ। ଯଦି ମାନଚିତ୍ରରେ ପରସ୍ପରକୁ ଛେଦ କରୁଥିବା ରେଖାଗୁଡ଼ିକ ମଧ୍ୟରେ ଥିବା କୋଣ ଅପରିବର୍ତ୍ତିତ ରହେ, ତେବେ ପ୍ରତ୍ୟେକ ଅତ୍ୟଣୁ ସ୍ତରର ଆକୃତି ଏକ ସମାନୁପାତିକ ପରିମାପରେ ବଜାୟ ରହିଥାଏ। ଏହାର ବିପରୀତ ପକ୍ଷରେ, ଯଦି ଅନୁକୋଣତା (conformality) ବଜାୟ ନରହେ, ତେବେ ଅତ୍ୟଣୁ ସ୍ତରରେ ବାସ୍ତବିକ ଆକୃତିଗୁଡ଼ିକ ବିକୃତ ହୋଇଯାଏ।
|
||||
|
||||
ଏହାକୁ ଅଧିକ ସ୍ପଷ୍ଟ ଭାବେ କହିଲେ: ଏକ ଉପାଦେୟ ମାନଚିତ୍ରରେ ଛୋଟ ବର୍ଗକ୍ଷେତ୍ରଗୁଡ଼ିକ (square) ଆକାରରେ ବଡ଼ ବା ଛୋଟ ହୋଇପାରନ୍ତି, କିନ୍ତୁ ସେଗୁଡ଼ିକ ପ୍ରସାରିତ ହୋଇ ଆୟତକ୍ଷେତ୍ର (rectangle) ରେ ପରିଣତ ନ ହୋଇ ବର୍ଗକ୍ଷେତ୍ର ହୋଇ ହିଁ ରହନ୍ତି। ସେହିପରି, ଛୋଟ ବୃତ୍ତଗୁଡ଼ିକ (circle) ବିକୃତ ହୋଇ ଉପବୃତ୍ତର (ellipse) ରୂପ ନେବା ପରିବର୍ତ୍ତେ ନିଜର ବୃତ୍ତାକାର ବଜାୟ ରଖନ୍ତି।
|
||||
|
||||
ଆମର ଏହି ସରଳ ଗ୍ରିଡ୍ରେ ଅନୁକୋଣତା ନଷ୍ଟ ହେଉଥିବାରୁ, ଏହା କେବଳ ଦୂରତାକୁ ନୁହେଁ, ବରଂ ସ୍ଥାନୀୟ ସ୍ତରରେ ମାନଚିତ୍ରର ପ୍ରତିଟି ଆକୃତିକୁ ମଧ୍ୟ ବିଗାଡ଼ି ଦିଏ।
|
||||
|
||||
\section{ମର୍କେଟର୍ଙ୍କ ଦୃଷ୍ଟିକୋଣ}
|
||||
|
||||
ଦ୍ୱାଦଶ-ତ୍ରୟୋଦଶ ଶତାବ୍ଦୀ ବେଳକୁ ସାମୁଦ୍ରିକ ଯାତ୍ରାରେ ଚୁମ୍ବକୀୟ କମ୍ପାସର (magnetic compass) ବ୍ୟବହାର ନାବିକମାନଙ୍କୁ ଏକ ନୂତନ ସାମର୍ଥ୍ୟ ପ୍ରଦାନ କଲା। ଆକାଶ ମେଘାଚ୍ଛନ୍ନ ହେଉ କିମ୍ବା ଜାହାଜ କୂଳରୁ ବହୁ ଦୂରକୁ ଚାଲିଯାଉ—ଉତ୍ତର ଦିଗର ଏହି ଅଚ୍ୟୁତ ସୂଚନା ନାବିକମାନଙ୍କୁ ନିଜ ଯାତ୍ରାପଥ ସମ୍ପର୍କରେ ଏକ ସ୍ପଷ୍ଟ ଧାରଣା ଦେଇପାରିଲା।
|
||||
|
||||
ସେ ସମୟରେ ନାବିକମାନେ ମୁଖ୍ୟତଃ 'ଡେଡ୍ ରେକନିଂ' (dead reckoning) ପଦ୍ଧତି ଉପରେ ନିର୍ଭର କରୁଥିଲେ। ପ୍ରତି ମୁହୂର୍ତ୍ତରେ ନିଜର ସଠିକ୍ ଭୌଗୋଳିକ ଅବସ୍ଥିତି ଗଣନା କରିବା ପରିବର୍ତ୍ତେ, ସେମାନେ କେବଳ କମ୍ପାସର ଦିଗକୁ ଅକ୍ଷୁଣ୍ଣ ରଖିବା ଉପରେ ଗୁରୁତ୍ୱ ଦେଉଥିଲେ। ଯାତ୍ରା ଆରମ୍ଭ କରିବା ପୂର୍ବରୁ ସାମୁଦ୍ରିକ ମାନଚିତ୍ର ବା ନିର୍ଦ୍ଦେଶାବଳୀ ସହାୟତାରେ ଦୁଇଟି ବନ୍ଦର ମଧ୍ୟରେ ଏକ ସ୍ଥିର କୋଣ (constant bearing) ନିର୍ଦ୍ଧାରଣ କରାଯାଉଥିଲା ଏବଂ ସେହି ଅନୁସାରେ ଜାହାଜ ଚଲାଯାଉଥିଲା।
|
||||
|
||||
ତେବେ ଏଠାରେ ଏକ ଗାଣିତିକ ଜଟିଳତା ଥିଲା। ଯଦି ଏକ ଜାହାଜ କମ୍ପାସର କଣ୍ଟା ସହିତ ଏକ ସ୍ଥିର କୋଣ ବଜାୟ ରଖି ଅଗ୍ରସର ହୁଏ, ତେବେ ତାହା ପୃଥିବୀ ପୃଷ୍ଠରେ ଏକ କୁଣ୍ଡଳୀ (spiral) ସଦୃଶ ପଥ ଅନୁସରଣ କରେ (ପ୍ରତିଛବି \ref{fig:straight_vs_rhumb} ଦେଖ) । ଏହି ବକ୍ର ପଥକୁ 'ରମ୍ବ ଲାଇନ' (Rhumb line) ବା 'ଲକ୍ସୋଡ୍ରୋମ୍' (Loxodrome) କୁହାଯାଏ, ଯାହା ପ୍ରତ୍ୟେକ ଦ୍ରାଘିମା ରେଖାକୁ ଏକ ସମାନ କୋଣରେ ଛେଦ କରିଥାଏ। ସେ ସମୟର ପାରମ୍ପରିକ ସମତଳ ମାନଚିତ୍ରଗୁଡ଼ିକରେ ଏହି ବକ୍ର ଗତିପଥକୁ ସଠିକ୍ ଭାବେ ଚିତ୍ରଣ କରିବା କଷ୍ଟକର ଥିଲା।
|
||||
|
||||
ଷୋଡ଼ଶ ଶତାବ୍ଦୀର ବିଶିଷ୍ଟ ମାନଚିତ୍ରକାର ଜେରାର୍ଡସ୍ ମର୍କେଟର୍ (Gerardus Mercator) ଏହି ସମସ୍ୟାକୁ 'ଯାତ୍ରା-ପୂର୍ବ ଯୋଜନା' ଦୃଷ୍ଟିକୋଣରୁ ବିଚାର କଲେ। ୧୫୬୯ ମସିହାରେ ସେ ଏକ ଅଭିନବ ‘ମାନଚିତ୍ର ଅଭିକ୍ଷେପ’ (map projection) ଉଦ୍ଭାବନ କଲେ, ଯାହାର ମୁଖ୍ୟ ଲକ୍ଷ୍ୟ ଥିଲା ମାନଚିତ୍ର ଓ କମ୍ପାସ ମଧ୍ୟରେ ଥିବା ଏହି ଗାଣିତିକ ଅସଙ୍ଗତିକୁ ଦୂର କରିବା।
|
||||
|
||||
ମର୍କେଟର୍ ଚାହୁଁଥିଲେ ଯେ କମ୍ପାସର ସ୍ଥିର ଦିଗଗୁଡ଼ିକ ମାନଚିତ୍ରରେ ବକ୍ର ରେଖା ପରିବର୍ତ୍ତେ ସରଳ ରେଖା ଭାବରେ ଦେଖାଯାଉ। ଏହାଦ୍ୱାରା ଜଣେ ନାବିକ କେବଳ ଏକ ମାପକାଠି (ruler) ସାହାଯ୍ୟରେ ଦୁଇଟି ବିନ୍ଦୁ ମଧ୍ୟରେ ସରଳ ରେଖା ଟାଣି, ଏକ କୋଣମାପକ (protractor) ସାହାଯ୍ୟରେ ଉତ୍ତର ଦିଗ ସହିତ ଏହାର କୋଣ ମାପି, ନିଜର ପଥ ନିରୂପଣ କରିପାରିବେ ଏବଂ ଯାତ୍ରା ସମୟରେ କୌଣସି ଜଟିଳ ଗଣନା ନକରି ସେହି ସ୍ଥିର ଦିଗରେ ଗତି କରି ଗନ୍ତବ୍ୟ ସ୍ଥଳରେ ପହଞ୍ଚି ପାରିବେ।
|
||||
|
||||
\begin{figure}[H]
|
||||
\centering
|
||||
\begin{tikzpicture}
|
||||
% 1. Place the image as a node
|
||||
\node[inner sep=0pt, anchor=south west] (map) at (0,0)
|
||||
{\includegraphics[width=1\linewidth]{rhumb_lines_PlateCarree.pdf}};
|
||||
|
||||
\node[below, font=\odiafont] at (3.8, 2.9) {ରମ୍ବ ଲାଇନ};
|
||||
|
||||
\node[font=\odiafont] at (5.5, 3.25) {\textmb{ଆଫ୍ରିକା}};
|
||||
|
||||
\node[font=\odiafont] at (2.22, 1.9) {\textmb{ଦକ୍ଷିଣ}};
|
||||
\node[font=\odiafont] at (2.22, 1.5) {\textmb{ଆମେରିକା}};
|
||||
|
||||
\end{tikzpicture}
|
||||
\caption{ଆମ ମାନଚିତ୍ରରେ ରମ୍ବ ଲାଇନ ବନାମ ସରଳ ରେଖା}
|
||||
\label{fig:straight_vs_rhumb}
|
||||
\end{figure}
|
||||
|
||||
\begin{questionbox}
|
||||
କିନ୍ତୁ ମର୍କେଟର୍ ଏହି ଅସମ୍ଭବକୁ ସମ୍ଭବ କଲେ କିପରି? କେଉଁ ଗାଣିତିକ ଯାଦୁ ବଳରେ ସେ ମାନଚିତ୍ରକୁ ଏପରି ଭାବେ ବଦଳାଇଲେ, ଯାହା ଫଳରେ ବକ୍ର ପଥଗୁଡ଼ିକ ସରଳ ରେଖାରେ ପରିଣତ ହୋଇଗଲେ?
|
||||
\end{questionbox}
|
||||
|
||||
ଏହି ଜ୍ୟାମିତିକ ରହସ୍ୟ ଏବଂ ମର୍କେଟର୍ ମାନଚିତ୍ରର ଗାଣିତିକ ସୂତ୍ର ଜାଣିବା ପାଇଁ ପଢ଼ନ୍ତୁ ଏହି ପ୍ରବନ୍ଧର ପରବର୍ତ୍ତୀ ଭାଗ (ଆସନ୍ତା ମାର୍ଚ୍ଚ ୨୦୨୬ ସଂଖ୍ୟାରେ ପ୍ରକାଶ ପାଇବ)।
|
||||
|
||||
|
||||
\begin{transtable}
|
||||
ଅଭିକ୍ଷେପ & projection & ମାପକାଠି & ruler \\
|
||||
ସରଳ ରେଖା & straight line & କୋଣମାପକ & protractor \\
|
||||
ବକ୍ର ରେଖା & curve & କୁଣ୍ଡଳୀ & spiral
|
||||
\end{transtable}
|
||||
|
||||
|
||||
{\small
|
||||
\vskip 0.05in
|
||||
\affiliation{Assistant Professor, Statistics \& Mathematics Unit, Indian Statistical Institute, Bangalore}
|
||||
\email{soumyashant@leelavati.in}
|
||||
}
|
||||
|
||||
\vskip 0.2in
|
||||
|
||||
\begin{leelabox}[leelaAccent, title=Kaprekar's Curious Procedure]
|
||||
\noindent
|
||||
Begin with any four-digit number whose digits are not all equal.
|
||||
|
||||
\medskip
|
||||
Rearrange its digits to form the largest possible number and the smallest possible number, and subtract the smaller from the larger.
|
||||
|
||||
\medskip
|
||||
Apply the same procedure repeatedly to the result you obtain.
|
||||
|
||||
\medskip
|
||||
No matter which number you start with, something unexpected happens in the end.
|
||||
Try to discover what it is.
|
||||
|
||||
\medskip
|
||||
This is called \textsb{Kaprekar's routine}, named after Dattatreya Ramchandra Kaprekar (ଦତ୍ତାତ୍ରେୟ ରାମଚନ୍ଦ୍ର କାପ୍ରେକର). Despite having no formal postgraduate training and working as a schoolteacher, he published extensively and became well known in recreational mathematics circles.
|
||||
\end{leelabox}
|
||||
|
||||
|
||||
\end{document}
|
||||
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88
clean-latex-junk.sh
Executable file
88
clean-latex-junk.sh
Executable file
|
|
@ -0,0 +1,88 @@
|
|||
#!/usr/bin/env bash
|
||||
set -euo pipefail
|
||||
|
||||
ROOT="$(cd "$(dirname "${BASH_SOURCE[0]}")" && pwd)"
|
||||
DRY_RUN=0
|
||||
|
||||
usage() {
|
||||
cat <<'USAGE'
|
||||
Usage: ./clean-latex-junk.sh [--dry-run]
|
||||
|
||||
Removes common LaTeX build byproducts from this repository, recursively.
|
||||
Source files and PDFs are not removed.
|
||||
|
||||
Options:
|
||||
--dry-run Print files that would be removed without deleting them.
|
||||
-h, --help Show this help.
|
||||
USAGE
|
||||
}
|
||||
|
||||
for arg in "$@"; do
|
||||
case "$arg" in
|
||||
--dry-run) DRY_RUN=1 ;;
|
||||
-h|--help) usage; exit 0 ;;
|
||||
*) echo "Unknown option: $arg" >&2; usage; exit 2 ;;
|
||||
esac
|
||||
done
|
||||
|
||||
patterns=(
|
||||
"*.aux"
|
||||
"*.bcf"
|
||||
"*.blg"
|
||||
"*.fdb_latexmk"
|
||||
"*.fls"
|
||||
"*.lof"
|
||||
"*.log"
|
||||
"*.lot"
|
||||
"*.nav"
|
||||
"*.out"
|
||||
"*.run.xml"
|
||||
"*.snm"
|
||||
"*.synctex.gz"
|
||||
"*.toc"
|
||||
"*.vrb"
|
||||
"*.xdv"
|
||||
)
|
||||
|
||||
files=()
|
||||
while IFS= read -r -d '' file; do
|
||||
files+=("$file")
|
||||
done < <(
|
||||
find "$ROOT" \
|
||||
\( -path "$ROOT/.git" -o -path "$ROOT/.cursor" \) -prune \
|
||||
-o -type f \
|
||||
\( \
|
||||
-name "${patterns[0]}" \
|
||||
-o -name "${patterns[1]}" \
|
||||
-o -name "${patterns[2]}" \
|
||||
-o -name "${patterns[3]}" \
|
||||
-o -name "${patterns[4]}" \
|
||||
-o -name "${patterns[5]}" \
|
||||
-o -name "${patterns[6]}" \
|
||||
-o -name "${patterns[7]}" \
|
||||
-o -name "${patterns[8]}" \
|
||||
-o -name "${patterns[9]}" \
|
||||
-o -name "${patterns[10]}" \
|
||||
-o -name "${patterns[11]}" \
|
||||
-o -name "${patterns[12]}" \
|
||||
-o -name "${patterns[13]}" \
|
||||
-o -name "${patterns[14]}" \
|
||||
-o -name "${patterns[15]}" \
|
||||
\) -print0
|
||||
)
|
||||
|
||||
if ((${#files[@]} == 0)); then
|
||||
echo "No LaTeX build byproducts found."
|
||||
exit 0
|
||||
fi
|
||||
|
||||
if ((DRY_RUN)); then
|
||||
echo "Would remove ${#files[@]} LaTeX build byproduct(s):"
|
||||
printf ' %s\n' "${files[@]#$ROOT/}"
|
||||
exit 0
|
||||
fi
|
||||
|
||||
printf 'Removing %s LaTeX build byproduct(s):\n' "${#files[@]}"
|
||||
printf ' %s\n' "${files[@]#$ROOT/}"
|
||||
rm -f -- "${files[@]}"
|
||||
echo "Done."
|
||||
BIN
counting/two-mysteries.pdf
Normal file
BIN
counting/two-mysteries.pdf
Normal file
Binary file not shown.
|
Before Width: | Height: | Size: 2.1 KiB After Width: | Height: | Size: 2.1 KiB |
BIN
fitch/fitch-trick-solutions-part-one.pdf
Normal file
BIN
fitch/fitch-trick-solutions-part-one.pdf
Normal file
Binary file not shown.
522
fitch/fitch-trick-solutions-part-one.tex
Normal file
522
fitch/fitch-trick-solutions-part-one.tex
Normal file
|
|
@ -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}
|
||||
BIN
fitch/fitch-trick-solutions-part-two.pdf
Normal file
BIN
fitch/fitch-trick-solutions-part-two.pdf
Normal file
Binary file not shown.
381
fitch/fitch-trick-solutions-part-two.tex
Normal file
381
fitch/fitch-trick-solutions-part-two.tex
Normal file
|
|
@ -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}
|
||||
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Add table
Add a link
Reference in a new issue