diff --git a/src/Advanced/De Bruijn/parts/0 intro.tex b/src/Advanced/De Bruijn/parts/0 intro.tex index 10ed2c7..d669ce7 100644 --- a/src/Advanced/De Bruijn/parts/0 intro.tex +++ b/src/Advanced/De Bruijn/parts/0 intro.tex @@ -21,7 +21,7 @@ Unlock this lock with only 5 keypresses. \end{solution} \vfill -Now, consider the same lock, now set with a three-digit binary code. +Now consider the same lock, but configured with a three-digit binary code. \problem{} How many codes are possible? \vfill diff --git a/src/Advanced/De Bruijn/parts/1 words.tex b/src/Advanced/De Bruijn/parts/1 words.tex index 527cd10..8e56ec1 100644 --- a/src/Advanced/De Bruijn/parts/1 words.tex +++ b/src/Advanced/De Bruijn/parts/1 words.tex @@ -20,7 +20,11 @@ We say $v$ is a \textit{subword} of $w$ if $v$ is contained in $w$. \par For example, \texttt{11} is a subword of \texttt{011}, but \texttt{00} is not. \definition{} -Recall \ref{lockproblem}. Let's generalize this to the \textit{$n$-subword problem}: \par +Recall the lock problem from the previous page. +Let's generalize this to the \textit{$n$-subword problem}: + +\vspace{1mm} + Given an alphabet $A$ and a positive integer $n$, we want a word over $A$ that contains all possible length-$n$ subwords. The shortest word that solves a given $n$-subword problem is called the \textit{optimal solution}. @@ -67,7 +71,7 @@ Find the following: \problem{} Let $w$ be a word over an alphabet of size $k$. \par -Prove the following: +Show that all of the following are true: \begin{itemize} \item $\mathcal{S}_n(w) \leq k^n$ \item $\mathcal{S}_n(w) \geq \mathcal{S}_{n-1}(w) - 1$ @@ -103,7 +107,7 @@ Prove the following: \definition{} Let $v$ and $w$ be words over the same alphabet. \par -The word $vw$ is the word formed by writing $v$ after $w$. \par +The word $vw$ is the word formed by writing $w$ after $v$. \par For example, if $v = \texttt{1001}$ and $w = \texttt{10}$, $vw$ is $\texttt{100110}$. \problem{} @@ -116,7 +120,6 @@ We'll call this the \textit{Fibonacci word} of order $k$. \item What are $F_3$, $F_4$, and $F_5$? \item Compute $\mathcal{S}_0$ through $\mathcal{S}_5$ for $F_5$. \item Show that the length of $F_k$ is the $(k + 2)^\text{th}$ Fibonacci number. \par - \hint{Induction.} \end{itemize} \begin{solution} diff --git a/src/Advanced/Retrograde Analysis/parts/02 easy.tex b/src/Advanced/Retrograde Analysis/parts/02 easy.tex index c51140e..839be1c 100644 --- a/src/Advanced/Retrograde Analysis/parts/02 easy.tex +++ b/src/Advanced/Retrograde Analysis/parts/02 easy.tex @@ -127,10 +127,6 @@ Mate the king in one move. \par \pagebreak - - - - % Sherlock, a question of survival \problem{An empty board} \difficulty{2}{5} @@ -161,42 +157,6 @@ There is one more piece on the board, which isn't shown. What color square does \pagebreak -% Sherlock, another monochromatic -\problem{The knight's grave} -\difficulty{3}{5} - -In the game below, no pieces have moved from a black square to a white square, or from a white square to a black square. -The white king has made less than fourteen moves. \par -Use this information to show that a pawn was promoted. \par - -% spell:off -\manyboards{ - ke8, - Pb2,Pd2, - Ke1 -} -% spell:on - -\begin{solution} - Knights always move to a different colored square, so all four missing knights must have been captured on their home square. - What pieces captured them? - - \vspace{2mm} - - We can easily account for the white knights and the black knight on G8, but who could've captured the knight from B8? - The only white pieces that can move to black squares are pawns, the Bishop (which is trapped on C1), the rook (which is stuck on column A and row 1), or the king (which would need at least 14 moves to do so). - - \vspace{2mm} - - If this knight was captured by a pawn, that pawn would be immediately promoted. If it was captured by a piece that wasn't a pawn, that piece must be a promoted pawn. -\end{solution} - -\vfill -\pagebreak - - - - % Arabian Knights, intro (given with solution) @@ -372,4 +332,122 @@ Which bishop was it, and what did it capture? \par \end{solution} \vfill -\pagebreak \ No newline at end of file +\pagebreak + + +% Sherlock, appendix +\problem{Moriarty's first} +\difficulty{3}{5} + + +No captures have been made in the last four moves. \par +It is White's move. What was the previous move? + +% spell:off +\manyboards{ + Bc8, + pg6, + Pg5,kh5, + Pd4,Qg4,Bh4, + pd3, + Pd2,Be2,Bg2, + Nc1,rd1,Ne1,Kf1,Qg1,Rh1 +} +% spell:on + +\begin{solution} + To see what the position was four moves ago, + move the Black queen to E4, the knight on E1 to F3, + the Black bishop to E1, and the White bishop on C8 to H3. + + The following sequence of moves brought the game to the present position: + + \begin{itemize} + \item bishop to c8, check + \item bishop to h4, check + \item knight to e1, check + \item queen to g4. + \end{itemize} + + This is the only way the present position could have arisen, + so Black's last move was with the queen from E4 to G4. + + + Try any other last move, and you will find it impossible to play back three more moves. +\end{solution} + +\vfill +\pagebreak + + +% Sherlock, appendix +\problem{Moriarty's second} +\difficulty{3}{5} + + +Neither the White king nor queen has moved +during the last five moves, nor has any piece +been captured during that time. +What was the last move? + +% spell:off +\manyboards{ + kh8, + Kg6,Bh6, + pa4, + Qa2 +} +% spell:on + +\begin{solution} + Put the Black pawn on A7, the Black king on G8, remove the + White bishop, and put a White pawn on d5; this was the position + five moves ago. The following sequence of moves brought the + game to its present position: + + \begin{itemize} + \item White: P-d6 + \item Black: K-h8 + \item White: P-d7 + \item Black: P-a6 + \item White: P-d8 = B + \item Black: P-a5 + \item White: B-g5 + \item Black: P-a4 + \item White: B-h6 + \end{itemize} +\end{solution} + +\vfill +\pagebreak + + +% Sherlock, appendix +\problem{Moriarty's third} +\difficulty{3}{5} + + +No pawn has moved, nor has any piece been +captured in the last five moves. \par +The Black king has been accidentally +knocked off the board. \par +On what square should he stand? + +% spell:off +\manyboards{ + rh8, + pa7,pb7,pc7,pd7,pe7,Kf7,pg7,Ph7, + Pg6, + na2 +} +% spell:on + +\begin{solution} + The only way to avoid a retrograde stalemate for White is by + placing the Black king on C8. Black's last move was with + the rook from D8, White's move before that was with his + king from G8, and Black's move before that was to castle. +\end{solution} + +\vfill +\pagebreak diff --git a/src/Advanced/Retrograde Analysis/parts/03 medium.tex b/src/Advanced/Retrograde Analysis/parts/03 medium.tex index b49a2d2..c3f8f65 100644 --- a/src/Advanced/Retrograde Analysis/parts/03 medium.tex +++ b/src/Advanced/Retrograde Analysis/parts/03 medium.tex @@ -169,16 +169,156 @@ White to move. Which side of the board did each color start on? \par +% Sherlock, another monochromatic +\problem{Monochromatic} +\difficulty{4}{5} + +In the game below, no pieces have moved from a black square to a white square or from a white square to a black square. +The white king has made fewer than fourteen moves. \par +Use this information to show that a pawn was promoted. \par + +% spell:off +\manyboards{ + ke8, + Pb2,Pd2, + Ke1 +} +% spell:on + +\begin{solution} + Knights always move to a different colored square, so all four missing knights must have been captured on their home square. + What pieces captured them? + + \vspace{2mm} + + We can easily account for the white knights and the black knight on G8, but who could've captured the knight from B8? + The only white pieces that can move to black squares are pawns, the Bishop (which is trapped on C1), the rook (which is stuck on column A and row 1), or the king (which would need at least 14 moves to do so). + + \vspace{2mm} + + If this knight was captured by a pawn, that pawn would be immediately promoted. If it was captured by a piece that wasn't a pawn, that piece must be a promoted pawn. +\end{solution} + +\vfill +\pagebreak +% Sherlock, another question of location +\problem{Superposition} +\difficulty{4}{5} + +A white pawn is missing; it is either on F2 or G2. \par +Where is it? + +% spell:off +\manyboards{ + ke8,rh8, + pa7,pf7,pg7, + pa6,pb6, + pb5, + Pa4,Pb4,Pc4, + pa3, + Pa2,Pb2, + Ke1 +} +% spell:on + +\vfill +\pagebreak + +% Sherlock, another question of location +\problem{Possibility} +\difficulty{4}{5} + +Show that black can castle to either side. \par +We know the following: + +\begin{itemize} + \item White started the game missing one rook. + \item White has not moved either knight + \item No promotions have been made + \item White's last move was from E2 to E4. +\end{itemize} + +% spell:off +\manyboards{ + ra8,ke8,rh8, + pa7,bb7,pc7,pd7,pf7,pg7,ph7, + nc6,nh6, + pe5,qg5, + bb4,Pe4, + Pb2,Pc2,Pd2,Pf2,Pg2,Ph2, + Nb1,Bc1,Qd1,Ke1,Bf1,Ng1,Rh1 +} +% spell:on + + +\vfill +\pagebreak + + + + + +% Sherlock, little exercise 2 +\problem{Kidnapping} +\difficulty{4}{5} + +On which square was the White queen captured?. \par + +% spell:off +\manyboards{ + ra8,qd8,ke8,ng8,rh8, + pa7,pb7,pc7,pe7,pf7,ph7, + nc6,pe6,ph6, + Pb3, + Na2,Pb2,Pc2,Pd2,Pe2,Pf2,Pg2,Ph2, + Ra1,Ke1,Rh1 +} +% spell:on + +\begin{solution} + White is missing a queen, both bishops, and one knight. \par + The black pawns on E6 and H6 account for two captures. + + \vspace{2mm} + + Neither white bishop could've been captured by these pawns, + since both are trapped by their pawns. Thus, these black pawns must have captured a queen and a knight. + + \vspace{4mm} + + The white pawn on B3 must have captured a black bishop. \par + The white queen got onto the board through A2. \par + Therefore, the pawn on B3 made its capture before the queen escaped, + and the black bishop was captured before the white queen. + + \vspace{4mm} + + Similarly, the bishop from C8 must have been + captured on B3 after the capture on E6, since it + got on the board through D7. + + + \vspace{4mm} + + The capture on E6 was made before the capture on B3 (black bishop), + which was made before the white queen was captured. + Therefore, the white queen was not captured on E6, and must + have been lost on H6. +\end{solution} + +\vfill +\pagebreak + % Arabian Knights 4 \problem{A missing piece} -\difficulty{4}{5} +\difficulty{6}{8} There is a piece at G4, marked with a $\odot$. \par diff --git a/src/Advanced/Retrograde Analysis/parts/04 hard.tex b/src/Advanced/Retrograde Analysis/parts/04 hard.tex index 4ca8eb1..c29a464 100644 --- a/src/Advanced/Retrograde Analysis/parts/04 hard.tex +++ b/src/Advanced/Retrograde Analysis/parts/04 hard.tex @@ -2,7 +2,7 @@ % Arabian Knights 5 \problem{The hidden castle} -\difficulty{7}{7} +\difficulty{8}{8} There is a white castle hidden on this board. Where is it? \par None of the royalty has moved or been under attack. \par @@ -30,7 +30,7 @@ None of the royalty has moved or been under attack. \par % Arabian Knights 6 \problem{Who moved last?} -\difficulty{7}{7} +\difficulty{8}{8} After many moves of chess, the board looks as follows. \par Who moved last? \par @@ -58,7 +58,7 @@ Who moved last? \par % Arabian Knights 3 \problem{The king in disguise} -\difficulty{7}{7} +\difficulty{8}{8} The white king is exploring his kingdom under a disguise. He could look like any piece of any color.\par Show that he must be on C7. @@ -119,7 +119,7 @@ Show that he must be on C7. % Arabian Knights 3 \problem{The king in disguise once more} -\difficulty{5}{7} +\difficulty{5}{8} The white king is again exploring his kingdom, now under a different disguise. Where is he? \par \hint{\say{different disguise} implies that the white king looks like a different piece!} diff --git a/src/Warm-Ups/Painting/main.typ b/src/Warm-Ups/Painting/main.typ index 1426b35..400e0f2 100644 --- a/src/Warm-Ups/Painting/main.typ +++ b/src/Warm-Ups/Painting/main.typ @@ -2,9 +2,8 @@ #import "@preview/cetz:0.4.2" #show: handout.with( - title: [Warm-Up: What's an AST?], + title: [Warm-Up: The Painting], by: "Mark", - subtitle: "Based on a true story.", ) #problem() @@ -13,7 +12,8 @@ Hang the painting on two nails so that if either is removed, the painting falls. #v(2mm) You may detach the string as you hang the painting, but it must be re-attached once you're done. \ -#hint[The solution to this problem isn't a "think outside the box" trick, it's a clever wrapping of the string.] + +The solution to this problem isn't a trick, it's a clever wrapping of the string. #v(2mm)