typeset chapters and appendices by scanning through the sources

appendixa/problems/1/a.tex

@@ -1,4 +1,3 @@
-\subproblem
 Asymptotic upper bound:
 \begin{align*}
     \sum_{k=1}^n k^r &\le \sum_{k=1}^n n^r \\

appendixa/problems/1/b.tex

@@ -1,4 +1,3 @@
-\subproblem
 Asymptotic upper bound:
 \begin{align*}
     \sum_{k=1}^n \lg^s k &\le \sum_{k=1}^n \lg^s n \\

appendixa/problems/1/c.tex

@@ -1,4 +1,3 @@
-\subproblem
 Asymptotic upper bound:
 \begin{align*}
     \sum_{k=1}^n k^r \lg^s k &\le \sum_{k=1}^n n^r \lg^s n \\

appendixa/problems/1/main.tex

@@ -1,7 +1,2 @@
-\problem{Bounding summations}
 In order to determine the asymptotic tight bounds on each summation, we will find their asymptotic upper bound and asymptotic lower bound by substituting each term in the summations by appropriate values.
 Since the bounds on the summations don't depend on the parity of $n$, for simplicity we will assume that $n$ is even.
-
-\subimport{./}{a}
-\subimport{./}{b}
-\subimport{./}{c}

appendixa/sections/1/1.tex

@@ -1,4 +1,3 @@
-\exercise
 Let $g_1$, $g_2$, \dots, $g_n$ be functions, so that $g_k(i)=O(f_k(i))$ for each $k=1$, 2, \dots, $n$.
 From the definition of $O$-notation, there exist positive constants $c_1$, $c_2$, \dots, $c_n$ and $i_0$, such that $g_k(i)\le c_k f_k(i)$ for all $i\ge i_0$.
 Let $c=\max\,\{c_k: 1\le k\le n\}$.

appendixa/sections/1/10.tex

@@ -1,4 +1,3 @@
-\exercise
 By formulas (A.7) and (A.11), we have:
 \begin{align*}
     \sum_{k=1}^\infty(2k+1)x^{2k} &= 2\sum_{k=0}^\infty k(x^2)^k + \sum_{k=0}^\infty(x^2)^k-1 \\

appendixa/sections/1/11.tex

@@ -1,4 +1,3 @@
-\exercise
 We have
 \begin{align*}
     \prod_{k=2}^n\biggl(1-\frac{1}{k^2}\biggr) &= \prod_{k=2}^n\frac{k^2-1}{k^2} \\[3pt]

appendixa/sections/1/2.tex

@@ -1,4 +1,3 @@
-\exercise
 We apply the linearity property and formula (A.1):
 \begin{align*}
     \sum_{k=1}^n(2k-1) &= 2\sum_{k=1}^n k - \sum_{k=1}^n 1 \\

appendixa/sections/1/3.tex

@@ -1,4 +1,3 @@
-\exercise
 If we let $x=10$ in equation (A.6), we have:
 \begin{align*}
     111{,}111{,}111 &= \sum_{k=0}^8 10^k \\

appendixa/sections/1/4.tex

@@ -1,4 +1,3 @@
-\exercise
 Each subsequent term is the previous term multiplied by $-1/2$, so letting $x=-1/2$ in equation (A.7) gives us the value of the series:
 \begin{align*}
     \sum_{k=0}^\infty(-1/2)^k &= \frac{1}{1-(-1/2)} \\

appendixa/sections/1/5.tex

@@ -1,4 +1,3 @@
-\exercise
 For the upper bound, we bound $k$ by $n$:
 \begin{align*}
     \sum_{k=1}^n k^c &\le \sum_{k=1}^n n^c \\

appendixa/sections/1/6.tex

@@ -1,4 +1,3 @@
-\exercise
 Let's differentiate both sides of equation (A.11):
 \begin{align*}
     \sum_{k=0}^\infty k\cdot kx^{k-1} &= \frac{(1-x)^2+2x(1-x)}{(1-x)^4} \\

appendixa/sections/1/7.tex

@@ -1,4 +1,3 @@
-\exercise
 The asymptotic upper bound:
 \begin{align*}
     \sum_{k=1}^n \sqrt{k\lg k} &\le \sum_{k=1}^n \sqrt{n\lg n} \\

appendixa/sections/1/8.tex

@@ -1,4 +1,3 @@
-\exercise
 By the upper bound on $H_n$ (A.9), we have:
 \begin{align*}
     \sum_{k=1}^n\frac{1}{2k-1} &< 1+\sum_{k=2}^{n+1}\frac{1}{2k-2} \\

appendixa/sections/1/9.tex

@@ -1,4 +1,3 @@
-\exercise
 By formulas (A.7) and (A.11), we have:
 \begin{align*}
     \sum_{k=0}^\infty\frac{k-1}{2^k} &= \sum_{k=0}^\infty\frac{k}{2^k} - \sum_{k=0}^\infty\frac{1}{2^k} \\

appendixa/sections/2/1.tex

@@ -1,4 +1,3 @@
-\exercise
 Using the properties of telescoping series, we get
 \begin{align*}
     \sum_{k=1}^n\frac{1}{k^2} &\le 1 + \sum_{k=2}^n\frac{1}{k(k-1)} \\

appendixa/sections/2/2.tex

@@ -1,4 +1,3 @@
-\exercise
 Thanks to inequality $\lceil n\rceil < n+1$ (3.2) and identity (A.7), we can write:
 \begin{align*}
 	\sum_{k=0}^{\lfloor\lg n\rfloor} \left\lceil\frac{n}{2^k}\right\rceil &< \sum_{k=0}^{\lfloor\lg n\rfloor} \left(\frac{n}{2^k}+1\right) \\

appendixa/sections/2/3.tex

@@ -1,4 +1,3 @@
-\exercise
 Proceeding similarly as when searching for an upper bound on the $n$th harmonic number, we split the range 1 to $n$ into $\lfloor\lg n\rfloor$ pieces and lower-bound the contribution of each piece by $1/2$:
 \begin{align*}
     \sum_{k=1}^n\frac{1}{k} &\ge \sum_{i=0}^{\lfloor\lg n\rfloor-1} \sum_{j=0}^{2^i-1} \frac{1}{2^i+j} \\

appendixa/sections/2/4.tex

@@ -1,4 +1,3 @@
-\exercise
 The function $f(k)=k^3$ is monotonically increasing, so we approximate the summation using inequality (A.18):
 \[
 	\int_0^n x^3\,dx \le \sum_{k=1}^n k^3 \le \int_1^{n+1} x^3\,dx.

appendixa/sections/2/5.tex

@@ -1,4 +1,3 @@
-\exercise
 Applying inequality (A.19) to $\sum_{k=1}^n 1/k$ leads to an improper integral:
 \begin{align*}
     \sum_{k=1}^n \frac{1}{k} &\le \int_0^n \frac{dx}{x} \\

appendixb/problems/1/a.tex

@@ -1,3 +1,2 @@
-\subproblem
 For a given tree let us distinguish any of its nodes and view the tree as a rooted tree in that node.
 Edges in such a tree are incident on nodes from adjacent levels, so nodes can be colored according to the parity of their depth in the tree (e.g., those on odd levels get color 1, and those on even levels get color 2).

appendixb/problems/1/b.tex

@@ -1,4 +1,3 @@
-\subproblem
 $(1)\Rightarrow(2)$: Since $G=(V,E)$ is a bipartite graph, the vertex set $V$ can be partitioned into two subsets $V_1$ and $V_2$, such that vertices from one subset aren't adjacent to vertices from the other subset.
 Thanks to this property it is possible to assign one color to each vertex from $V_1$ and a different color to each vertex from $V_2$, obtaining a 2-coloring of graph $G$.
 

appendixb/problems/1/c.tex

@@ -1,4 +1,3 @@
-\subproblem
 We carry out the proof by induction on the number of vertices in graph $G=(V,E)$.
 If the graph has only one vertex (with degree 0), then of course one color is enough.
 

appendixb/problems/1/d.tex

@@ -1,4 +1,3 @@
-\subproblem
 For any $k\ge2$, if graph $G$ is $k$-colorable, but is not $(k-1)$-colorable, then for every two different colors used in a $k$-coloring of $G$ there are two adjacent vertices, one with the first color and the other with the second color.
 If that weren't true, there would be two different colors that we could not distinguish, and so we could reduce the number of colors required, obtaining a $(k-1)$-coloring of $G$.
 

appendixb/problems/2/a.tex

@@ -1,4 +1,3 @@
-\subproblem
 \begin{theorem}
     In any undirected graph $G=(V,E)$, where $|V|\ge2$, there are two vertices with the same degree.
 \end{theorem}

appendixb/problems/2/b.tex

@@ -1,4 +1,3 @@
-\subproblem
 \begin{theorem}
     An undirected graph $G=(V,E)$ of 6 vertices, or its complement\footnote{See page 1085 in the book.} $\overline{G}$, contain a clique\footnote{See page 1081 in the book.} of size 3.
 \end{theorem}

appendixb/problems/2/c.tex

@@ -1,4 +1,3 @@
-\subproblem
 \begin{theorem}
     For any undirected graph $G=(V,E)$ the set $V$ can be partitioned into two subsets such that for any vertex $v\in V$ at least half of its neighbors do not belong to the subset that $v$ belongs to.
 \end{theorem}

appendixb/problems/2/d.tex

@@ -1,4 +1,3 @@
-\subproblem
 Before we formulate the main theorem, we will need the following lemma.
 \begin{lemma}
     Let $G=(V,E)$ be an undirected graph, where $|V|\ge3$ and

appendixb/problems/3/a.tex

@@ -1,4 +1,3 @@
-\subproblem
 The fact trivially holds for binary trees with at most three nodes, so in the rest of the proof we will assume that $n\ge4$.
 We define a \concept{branch} of a binary tree as a path $\langle x_0$, $x_1$, \dots, $x_k\rangle$, where $x_0$ is the root, $x_k$ is a leaf, and for $i=1$, 2, \dots, $k$, $x_i$ is the root of the subtree of $x_{i-1}$ with more nodes.
 Note that a tree may have more than one branch.

appendixb/problems/3/b.tex

@@ -1,7 +1,6 @@
-\subproblem
 The constant $3/4$ is sufficient to make balanced partitions, as we have shown in part (a).
 The example of the binary tree in \refFigure{B-3b} shows that the constant cannot be lower.
 \begin{figure}[htb]
-    \subimport{./}{b.tikz}
+    \input{b.tikz}
     \caption{The binary tree whose most evenly balanced partition upon removal of a single edge has a subset with 3 vertices.} \label{fig:B-3b}
 \end{figure}

appendixb/problems/3/c.tex

@@ -1,4 +1,3 @@
-\subproblem
 Assume that $n\ge6$.
 We'll remove one edge from the original tree to cut off a subtree of at most $\lfloor n/2\rfloor$ vertices.
 Consider any of the tree's branches $\langle x_0$, $x_1$, \dots, $x_k\rangle$, as well as the function $s$, both defined in the solution to part (a).

appendixb/sections/1/1.tex

@@ -1,6 +1,5 @@
-\exercise
 See \refFigure{B.1-1}.
 \begin{figure}[htb]
-    \subimport{./}{1.tikz}
+    \input{1.tikz}
     \caption{A Venn diagram illustrating the first of the distributive laws (B.1).} \label{fig:B.1-1}
 \end{figure}

appendixb/sections/1/2.tex

@@ -1,4 +1,3 @@
-\exercise
 We prove the first formula by induction on $n$.
 For $n=1$ the proof is trivial, and for $n=2$ the formula is the first of DeMorgan's laws.
 Let's assume that $n>2$, and that the formula holds for a family of $n-1$ sets.

appendixb/sections/1/3.tex

@@ -1,4 +1,3 @@
-\exercise
 We prove the principle of inclusion and exclusion by induction on $n$.
 For $n=1$ the proof is trivial, and for $n=2$ the formula is identical to (B.3).
 If $n>2$, then by (B.3) and the generalized first distributive law, we have:

appendixb/sections/1/4.tex

@@ -1,3 +1,2 @@
-\exercise
 An example of a one-to-one correspondence between $\mathbb{N}$ and the set of odd natural numbers is: $n \to 2n+1$.
 Thus, the set of odd natural numbers is countable.

appendixb/sections/1/5.tex

@@ -1,4 +1,3 @@
-\exercise
 We prove by induction on the cardinality of the set $S$.
 
 When $|S|=0$, it holds that $|2^S|=2^{|S|}=1$, because $S$ has exactly one subset\dash the empty set.

appendixb/sections/1/6.tex

@@ -1,4 +1,3 @@
-\exercise
 An $n$-tuple of elements $a_1$, $a_2$, \dots, $a_n$ is defined as:
 \[
     (a_1,a_2,\dots,a_n) =

appendixb/sections/2/1.tex

@@ -1,4 +1,3 @@
-\exercise
 To show that the relation ``$\subseteq$'' on the power set $2^\mathbb{Z}$ is a partial order, we will show that ``$\subseteq$'' is reflexive, antisymmetric and transitive.
 Let $A$, $B$, $C\in2^\mathbb{Z}$.
 Of course, $x\in A$ implies $x\in A$, so reflexivity trivially holds.

appendixb/sections/2/2.tex

@@ -1,4 +1,3 @@
-\exercise
 Let us denote the relation ``equivalent modulo $n$'', for $n\in\mathbb{N}-\{0\}$, by
 \[
     R_n = \{(a,b)\in\mathbb{Z}\times\mathbb{Z}: a=b\!\!\!\pmod{n}\}.

appendixb/sections/2/3.tex

@@ -1,4 +1,3 @@
-\exercise
 \subexercise
 The relation $\{(a,a),(b,b),(c,c),(a,b),(b,a),(a,c),(c,a)\}$ on the set $\{a,b,c\}$.
 

appendixb/sections/2/4.tex

@@ -1,4 +1,3 @@
-\exercise
 If $R$ is an equivalence relation, then $s\in[s]$ for all $s\in S$.
 By antisymmetry of $R$, if $s'\,R\,s$ and $s\,R\,s'$, then $s=s'$, so there are no such $s'$ that $s'\in[s]-\{s\}$.
 This means that for all $s\in S$, $[s]=\{s\}$.

appendixb/sections/2/5.tex

@@ -1,4 +1,3 @@
-\exercise
 Symmetry and transitivity are defined using implications.
 For an implication ``$p$ implies $q$'' to hold true, it isn’t required for $p$ to be true.
 A relation on a set remains symmetric and transitive, if there is an element in the set, that is not related to any element in that set.

appendixb/sections/3/1.tex

@@ -1,4 +1,3 @@
-\exercise
 \subexercise
 For any function $f:A\to B$, $f(A)\subseteq B$, which results directly from the definition of the range of $f$\!.
 Additionally, if $f$ is injective, then each argument $a\in A$ has a distinct image $f(a)\in f(A)$, so $|A|=|f(A)|$.

appendixb/sections/3/2.tex

@@ -1,3 +1,2 @@
-\exercise
 The function $f(x)=x+1$, with $\mathbb{N}$ as both the domain and the codomain, is not bijective because there is no $x\in\mathbb{N}$ such that $f(x)=0$.
 If we consider $\mathbb{Z}$ as the domain and the codomain, then $f$ becomes bijective, since every integer $y$ in the codomain is an image under $f$ of a uniquely determined integer in the domain: $y=f(y-1)$.

appendixb/sections/3/3.tex

@@ -1,4 +1,3 @@
-\exercise
 The definition comes as a natural generalization of the functional inverse.
 Let $R$ be a binary relation on sets $A$ an $B$.
 The binary relation $R^{-1}$ on sets $B$ and $A$, such that

appendixb/sections/3/4.tex

@@ -1,4 +1,3 @@
-\exercise
 Let $h:\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$ be the function we are looking for.
 Since the inverse of a bijection is also a bijection, we will focus on finding the inverse $h^{-1}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$, which is equivalent to finding a way to sequentially number each ordered pair with integer elements, so that each integer is used as the number of some pair.
 We will now describe the construction of one of such mappings.
@@ -10,7 +9,7 @@
 Consider the numbering of ordered pairs with integer elements shown in \refFigure{B.3-4} in the form of a spiral.
 Since each such pair $(x,y)$ is assigned a unique natural number, we can treat this spiral as a description of the function $g$.
 \begin{figure}[htb]
-    \subimport{./}{4.tikz}
+    \input{4.tikz}
     \caption{A bijection from $\mathbb{Z}\times\mathbb{Z}$ to $\mathbb{Z}$.
     Individual natural numbers denote the values of this bijection for points with integer coordinates in the Cartesian coordinate system.
     } \label{fig:B.3-4}

appendixb/sections/4/1.tex

@@ -1,3 +1,2 @@
-\exercise
 We can represent the relation of ``shaking hands'' on the faculty party as an undirected graph $G=(V,E)$, where $V$ is the set of professors attending the party and $E$ is the set of unordered pairs of professors who shook hands.
 Summing up the degrees of all vertices of $G$, we get twice the number of its edges, because each edge is incident on exactly two vertices.

appendixb/sections/4/2.tex

@@ -1,4 +1,3 @@
-\exercise
 Let $\langle v_0$, $v_1$, \dots, $v_n\rangle$ be a path from vertex $u$ to vertex $v$ in a directed or undirected graph.
 If the path is simple, the vertices in the path are distinct.
 Otherwise, the path contains a cycle $\langle v_i$, $v_{i+1}$, \dots, $v_{i+k}$, $v_i\rangle$ for some $i$ and $k$, and by eliminating the subpath $\langle v_i$, $v_{i+1}$, \dots, $v_{i+k}\rangle$ from the path, we discard an extra repetition of $v_i$.

appendixb/sections/4/3.tex

@@ -1,4 +1,3 @@
-\exercise
 We proceed by induction on the number of vertices in the graph $G=(V,E)$.
 When $|V|=1$, the graph has no edges and the inequality trivially holds.
 For the inductive step, pick a vertex $v\in V$.

appendixb/sections/4/4.tex

@@ -1,4 +1,3 @@
-\exercise
 Every vertex of a directed or undirected graph is reachable from itself because there is a path of length 1 containing only that vertex, therefore the ``is reachable from'' relation is reflexive.
 
 Let $u$, $v$, $w$ be vertices of a directed or undirected graph such that $u\overset{p}{\leadsto}v$ and $v\overset{q}{\leadsto}w$.

appendixb/sections/4/5.tex

@@ -1,8 +1,7 @@
-\exercise
 See \refFigure{B.4-5}.
 \begin{figure}[htb]
-    \subcaptionbox{\label{fig:B.4-5a}}[0.45\textwidth]{\subimport{./}{5a.tikz}}
-    \subcaptionbox{\label{fig:B.4-5b}}[0.45\textwidth]{\subimport{./}{5b.tikz}}
+    \subcaptionbox{\label{fig:B.4-5a}}[0.45\textwidth]{\input{5a.tikz}}
+    \subcaptionbox{\label{fig:B.4-5b}}[0.45\textwidth]{\input{5b.tikz}}
     \caption{\textbf{(a)}\, The undirected version of the directed graph in Figure B.2(a).\,
     \textbf{(b)}\, The directed version of the undirected graph in Figure B.2(b).} \label{fig:B.4-5}
 \end{figure}

appendixb/sections/4/6.tex

@@ -1,4 +1,3 @@
-\exercise
 A hypergraph $H=(V_H,E_H)$ can be represented as a bipartite graph $G=(V_1\cup V_2,E)$, where $V_1=V_H$ and $V_2=E_H$.
 A pair $\{v_1,v_2\}$, such that $v_1\in V_1$ and $v_2\in V_2$, is an edge of the graph $G$, if and only if the hyperedge $v_2$ is incident on vertex $v_1$ in the hypergraph $H$.
 In the graph $G$ there are no edges between elements of $V_1$ or between elements of $V_2$, so $G$ is indeed bipartite.

appendixb/sections/5/1.tex

@@ -1,10 +1,9 @@
-\exercise
 See \refFigure{B.5-1}.
 \begin{figure}[htb]
-    \subcaptionbox{\label{fig:B.5-1a}}[0.15\textwidth]{\subimport{./}{1a.tikz}}
-    \subcaptionbox{\label{fig:B.5-1b}}[0.25\textwidth]{\subimport{./}{1b.tikz}}
-    \subcaptionbox{\label{fig:B.5-1c}}[0.35\textwidth]{\subimport{./}{1c.tikz}}
-    \subcaptionbox{\label{fig:B.5-1d}}[0.8\textwidth]{\subimport{./}{1d.tikz}}
+    \subcaptionbox{\label{fig:B.5-1a}}[0.15\textwidth]{\input{1a.tikz}}
+    \subcaptionbox{\label{fig:B.5-1b}}[0.25\textwidth]{\input{1b.tikz}}
+    \subcaptionbox{\label{fig:B.5-1c}}[0.35\textwidth]{\input{1c.tikz}}
+    \subcaptionbox{\label{fig:B.5-1d}}[0.8\textwidth]{\input{1d.tikz}}
     \caption{\textbf{(a)}\, All the free trees with vertices $x$, $y$, and $z$.\,
     \textbf{(b)}\, All the rooted trees with nodes $x$, $y$, and $z$ with $x$ as the root.\,
     \textbf{(c)}\, All the ordered trees with nodes $x$, $y$, and $z$ with $x$ as the root.\,

appendixb/sections/5/2.tex

@@ -1,4 +1,3 @@
-\exercise
 Let $G'=(V,E')$ be the undirected version of the graph $G$.
 Note that $G'$ is connected, since every vertex $v\in V$ is reachable from vertex $v_0$ in $G'$.
 

appendixb/sections/5/3.tex

@@ -1,4 +1,3 @@
-\exercise
 The proof is by induction on the number $n$ of nodes in the tree.
 The base case is when $n=1$.
 A binary tree with only one node has one leaf and no degree-2 nodes, thus the statement holds.

appendixb/sections/5/4.tex

@@ -1,4 +1,3 @@
-\exercise
 We construct a sequence of full binary trees $T_1$, $T_2$, \dots, as follows.
 Let $T_1$ be a binary tree containing a single node, which obviously is a single leaf.
 For any $k\ge2$, let $T_k$ be a binary tree whose left subtree is $T_{k-1}$ and whose right subtree is a one-node binary tree.

appendixb/sections/5/5.tex

@@ -1,4 +1,3 @@
-\exercise
 The proof is by induction on the height $h$ of the binary tree.
 
 The only nonempty binary tree with $h=0$ is the one with a single node, so $n=1$ and the statement $h\ge\lfloor\lg n\rfloor$ holds.