Add "Cosa Nostra"

src/Advanced/Introduction to Quantum/src/parts/01 bits.tex

@@ -230,7 +230,7 @@ The \textit{tensor product} of two vectors is defined as follows:
 That is, we take our first vector, multiply the second
 vector by each of its components, and stack the result.
 You could think of this as a generalization of scalar
-mulitiplication, where scalar mulitiplication is a
+multiplication, where scalar multiplication is a
 tensor product with a vector in $\mathbb{R}^1$:
 \begin{equation*}
 	a

src/Advanced/Retrograde Analysis/parts/03 medium.tex

@@ -251,7 +251,7 @@ What is it, and what is its color? \par
 	\textbf{Part 4:}
 
 	The promoted black bishop on H2 must have been promoted on G1. The pawn which was promoted must have come from G7,
-	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accouted for).
+	since neither of the pawns from F6 or H6 could make a capture to get to the G-file (all six missing white pieces have been accounted for).
 	The Pawn from E7 has promoted to the bishop on A2.
 
 	What happened was this: the white pawn from G2 made its capture on H3 while the pawn on G3 was still on H2. This allowed the black pawn

src/Intermediate/Instant Insanity/main.tex

@@ -331,7 +331,7 @@
 	representing all four cubes. \\
 
 	\begin{center} \begin{small}
-	\begin{tikzpicture} \label{pic:II_comfiguration}
+	\begin{tikzpicture} \label{pic:II_configuration}
 		\filldraw [blue] (0,5) -- (1,5) -- (1,6) --
 		(0,6) -- (0,5);
 		\draw [line width = 1.5pt] (0,5) --

src/Warm-Ups/Cosa Nostra/main.typ

@@ -0,0 +1,50 @@
+#import "@local/handout:0.1.0": *
+
+#show: handout.with(
+  title: [Warm-Up: Cosa Nostra],
+  by: "Mark",
+)
+
+
+#problem()
+There are 36 gangsters in a certain district of Chicago.
+Some pairs of gangsters have feuds.
+
+- Each gangster is part of at least one outfit, and no two outfits share the same members.
+- If two gangsters are both in one outfit, there is no feud between them.
+- A gangster that is not in a certain outfit must have a feud with at least one if its members.
+
+What is the maximum number of outfits that can exist in this district?
+
+
+
+#solution[
+  *Definition:* Let the _authority_ of a gangster be the number of oufits they are a part of.
+
+  #v(5mm)
+
+  *Lemma:* Say two gangsters have a feud. Label them $x$ and $y$ so that $#text(`authority`) (x) > #text(`authority`) (y)$. \
+  Then, replacing $y$ with a clone of $x$ will strictly increase the number of outfits. \
+  If $#text(`authority`) (x) = #text(`authority`) (y)$, replacing $y$ with $x$ will not change the number of outfits.
+
+  #v(5mm)
+
+  *Proof:*
+  Let $A$ be the set of outfits that $y$ is a part of, and $B$ its complement (that is, all outfits that $a$ is _not_ a part of). If we delete $y$...
+  - all outfits in $B$ remain outfits.
+  - some outfits in $A$ cease to be outfits (as they are no longer maximal)
+
+  Also, no new outfits are formed. If a new outfit $o$ contains any enemies of $y$, it existed previously and is a member of $B$. If $o$ contains no enemies of $y$, it must have contained $y$ prior to deletion, and is thus a member of $A$. Therefore, the number of outfits is reduced by at most `authority(y)` when $y$ is deleted.
+
+  If we add a clone of $x$ after deleting $y$ (this clone has a feud with $x$), All previous outfits remain outfits, and `authority(x)` new outfits are created.
+  Therefore, replacing $y$ with a clone of $x$ strictly increases the number of outfits that exist.
+  We thus conclude that in the maximal case, all pairs of feuding gangsters have equal authority.
+
+  #v(5mm)
+
+  *Solution:* Consider an arbitrary gangster $g$. By the previous lemma, we can replace all gangsters $g$ has a feud with clones of itself. Repeat this for all gangsters, and we are left with groups of feuding gangsters who are friends with everyone outside their group. The total number of outfits is the product of the sizes of these groups.
+
+  #v(5mm)
+
+  This problem is now equivalent to the "Partition Products" warm-up. We want the list of numbers whose sum is 36 and whose product is maximal. The solution is to form 12 groups of three gangsters for a total of $3^12$ outfits.
+]

src/Warm-Ups/Cosa Nostra/meta.toml

@@ -0,0 +1,6 @@
+[metadata]
+title = "Cosa Nostra"
+
+[publish]
+handout = true
+solutions = true