finish typo corrections

Author: oscarlevin

source/exercises/counting-combine-outcomes.ptx

@@ -36,7 +36,7 @@
           <li>
             <p>
               For example, 63 is the number of choices you have if you will watch two movies,
-              first a comedy and then a horror.
+              first a comedy and then a horror movie.
             </p>
           </li>
         </ol>
@@ -55,7 +55,7 @@
     </statement>
     <hint>
       <p>
-        For a simpler example, there are 4 divisors of <m>6 = 2\cdot 3</m>.  They are <m>1 = 2^0\cdot 3^0</m>, <m>2 = 2^1\cdot 3^0</m>, <m>3 = 2^0\cdot 3^1</m> and <m>6 = 2^1\cdot 3^1</m>.
+        For a simpler example, there are 4 divisors of <m>6 = 2\cdot 3</m>.  They are <m>1 = 2^0\cdot 3^0</m>, <m>2 = 2^1\cdot 3^0</m>, <m>3 = 2^0\cdot 3^1</m>, and <m>6 = 2^1\cdot 3^1</m>.
       </p>
     </hint>
   </exercise>

source/exercises/counting-conc.ptx

@@ -8,8 +8,8 @@
   <exercise>
     <statement>
     <idx><h>bow ties</h></idx>
-    <idx><h>combination</h><h>vs permutation</h></idx>
-    <idx><h>permutation</h><h>vs combination</h></idx>
+    <idx><h>combination</h><h>vs. permutation</h></idx>
+    <idx><h>permutation</h><h>vs. combination</h></idx>
       <p>
         For each of the following counting problems,
         say whether the answer is <m>{10\choose 4}</m>,
@@ -380,7 +380,7 @@
       <idx><h>bow ties</h></idx>
       <p>
         You own 8 purple bow ties,
-        3 red bow ties, 3 blue bow ties and 5 green bow ties.
+        3 red bow ties, 3 blue bow ties, and 5 green bow ties.
         How many ways can you select one of each color bow tie to take with you on a trip? <m>8 \cdot 3 \cdot 3 \cdot 5</m> ways.
         How many choices do you have for a single bow tie to wear tomorrow? <m>8 + 3 + 3 + 5</m> choices.
       </p>
@@ -446,7 +446,7 @@
         <m>{n \choose 1}</m>, <m>{n \choose 2}</m>,
         <ellipsis/>, <m>{n \choose n}</m>
         (each time choosing which of the <m>n</m> steps to be to the right).
-        These two methods count the same quantity and so are equal.
+        These two methods count the same quantity, and so are equal.
       </p>
     </solution>
   </exercise>

source/exercises/counting-multisets.ptx

@@ -51,7 +51,7 @@
         <ol>
           <li>
             <p>
-              You take 3 strawberry, 1 lime, 0 licorice, 2 blueberry and 0 bubblegum.
+              You take 3 strawberry, 1 lime, 0 licorice, 2 blueberry, and 0 bubblegum.
             </p>
           </li>
 

source/exercises/counting-proofs.ptx

@@ -257,7 +257,7 @@
                 We have answered the question
                 (how many wedding parties can the bride choose from)
                 in two ways.
-                The first way gives the left-hand side of the identity and the second way gives the right-hand side of the identity.
+                The first way gives the left-hand side of the identity, and the second way gives the right-hand side of the identity.
                 Therefore the identity holds.
               </p>
             </li>
@@ -297,7 +297,7 @@
       </statement>
       <hint>
         <p>
-          Try <xref ref="exc-bridesmaids"/>
+          Try <xref ref="exc-bridesmaids"/>.
         </p>
       </hint>
     </exercise>
@@ -327,7 +327,7 @@
           <p>
             Answer 1: First select 2 of the <m>n</m> balls to put in the jar.
             Then select <m>k-2</m> of the remaining <m>n-2</m> balls to put in the box.
-            The first task can be completed in <m>{n \choose 2}</m> different ways,
+            The first task can be completed in <m>{n \choose 2}</m> different ways and
             the second task in <m>{n-2 \choose k-2}</m> ways.
             Thus there are <m>{n \choose 2}{n-2 \choose k-2}</m> ways to select the balls.
           </p>
@@ -336,14 +336,14 @@
             Answer 2: First select <m>k</m> balls from the <m>n</m> in the container.
             Then pick 2 of the <m>k</m> balls you picked to put in the jar,
             placing the remaining <m>k-2</m> in the box.
-            The first task can be completed in <m>{n \choose k}</m> ways,
+            The first task can be completed in <m>{n \choose k}</m> ways and
             the second task in <m>{k \choose 2}</m> ways.
             Thus there are <m>{n \choose k}{k \choose 2}</m> ways to select the balls.
           </p>
 
           <p>
             Since both answers count the same thing,
-            they must be equal and the identity is established.
+            they must be equal, and the identity is established.
           </p>
         </proof>
 
@@ -371,7 +371,7 @@
       </statement>
       <hint>
         <p>
-          For the combinatorial proof: what if you don't yet know how many bridesmaids you will have?
+          For the combinatorial proof: What if you don't yet know how many bridesmaids you will have?
         </p>
       </hint>
     </exercise>

source/exercises/gt-coloring.ptx

@@ -273,8 +273,8 @@
   </exercise>
 
   <exercise>
-    <idx><h>graph</h><h>coloring vertices vs edges</h></idx>
-    <idx><h>coloring</h><h>vertices vs edges</h></idx>
+    <idx><h>graph</h><h>coloring vertices vs. edges</h></idx>
+    <idx><h>coloring</h><h>vertices vs. edges</h></idx>
     <statement>
       <p>
         The two problems below can be solved using graph coloring.
@@ -374,8 +374,8 @@
     <solution>
       <p>
         The wheel graph below has this property.
-        The outside of the wheel forms an odd cycle, so requires 3 colors,
-        the center of the wheel must be different than all the outside vertices.
+        The outside of the wheel forms an odd cycle and so requires 3 colors;
+        the center of the wheel must be a different color from all the outside vertices.
       </p>
 
 
@@ -422,7 +422,7 @@
       </p>
 
       <p>
-        Note that you cannot use the 4-color theorem, or Brooke's theorem, or the clique number here.  In fact, this graph, called the <em>Grötzsch graph</em> is the smallest graph with chromatic number 4 that does not contain any triangles.
+        Note that you cannot use the 4-color theorem, or Brooke's theorem, or the clique number here.  In fact, this graph, called the <em>Grötzsch graph</em>, is the smallest graph with chromatic number 4 that does not contain any triangles.
       </p>
     </hint>
   </exercise>
@@ -450,13 +450,13 @@
     </statement>
     <solution>
       <p>
-        If we drew a graph with each letter representing a vertex,
+        If we drew a graph with each letter representing a vertex
         and each edge connecting two letters that were consecutive in the alphabet,
         we would have a graph containing two vertices of degree 1 (A and Z) and the remaining 24 vertices all of degree 2
         (for example,
         <m>D</m> would be adjacent to both <m>C</m> and <m>E</m>).
         By Brooks' theorem,
-        this graph has chromatic number at most 2, as that is the maximal degree in the graph and the graph is not a complete graph or odd cycle.
+        this graph has chromatic number at most 2, as that is the maximal degree in the graph, and the graph is not a complete graph or odd cycle.
         Thus only two boxes are needed.
       </p>
     </solution>
@@ -470,7 +470,7 @@
     </statement>
     <hint>
       <p>
-        You can color <m>K_5</m> in such a way that every vertex is adjacent to exactly two blue edges and two red edges.  However, there is a graph with only 5 edges that will result in a vertex incident to three edges of the same color no matter how they are colored.  What is it, and how can you generalize?
+        You can color <m>K_5</m> in such a way that every vertex is adjacent to exactly two blue edges and two red edges.  However, there is a graph with only 5 edges that will result in a vertex incident to three edges of the same color, no matter how they are colored.  What is it, and how can you generalize?
       </p>
     </hint>
     <solution>

source/exercises/gt-conc.ptx

@@ -228,7 +228,7 @@
           <li>
             <p>
               In general this should be possible:
-              the degree sequence does not determine the graph's isomorphism class.
+              The degree sequence does not determine the graph's isomorphism class.
               However, in this case, I was almost certain this was not possible.
               That is, until I stumbled up this:
             </p>
@@ -252,7 +252,7 @@
             <p>
               <m>G</m> is a tree
               (there are no cycles)
-              and as such also bipartite.
+              and as such is also bipartite.
             </p>
           </li>
 
@@ -430,7 +430,7 @@
       <p>
         Yes, as long as <m>n</m> is even.
         If <m>n</m> were odd,
-        then corresponding graph would have an odd number of odd degree vertices,
+        then the corresponding graph would have an odd number of odd degree vertices,
         which is impossible.
       </p>
     </solution>
@@ -503,7 +503,7 @@
               but we know from Euler's formula that there must be 18 vertices.
               We can write <m>64 = 3x + 4y</m> and solve for <m>x</m> and <m>y</m>
               (as integers).
-              We get that there must be 10 vertices with degree 4 and 8 with degree 3. (Note the number of faces joined at a vertex is equal to its degree in graph theoretic terms.)
+              We get that there must be 10 vertices with degree 4 and 8 with degree 3. (Note that the number of faces joined at a vertex is equal to its degree in graph theoretic terms.)
             </p>
           </li>
         </ol>
@@ -610,10 +610,10 @@
         We have that <m>K_{3,4}</m> has 7 vertices and 12 edges
         (each vertex in the group of 3 has degree 4).
         Then by Euler's formula we have that
-        <m>7 - 12 + f = 2</m> so if the graph were planar,
+        <m>7 - 12 + f = 2</m>, so if the graph were planar,
         it would have <m>f = 7</m> faces.
         However, since the girth of the graph is 4
-        (there are no cycles of length 3)
+        (there are no cycles of length 3),
         we get that <m>4f \le 2e</m>.
         But this would mean that <m>28 \le 24</m>, a contradiction.
       </p>
@@ -651,18 +651,18 @@
       <p>
         For all these questions, we are really coloring the vertices of a graph.
         You get the graph by first drawing a planar representation of the polyhedron and then taking its planar dual:
-        put a vertex in the center of each face
-        (including the outside)
+        Put a vertex in the center of each face
+        (including the outside),
         and connect two vertices if their faces share an edge.
 
         <ol>
           <li>
             <p>
               Since the planar dual of a dodecahedron contains a 5-wheel,
-              it's chromatic number is at least 4.
+              its chromatic number is at least 4.
               Alternatively,
               suppose you could color the faces using 3 colors without any two adjacent faces colored the same.
-              Take any face and color it blue.
+              Take any face, and color it blue.
               The 5 pentagons bordering this blue pentagon cannot be colored blue.
               Color the first one red.
               Its two neighbors
@@ -736,7 +736,7 @@
               To prove this,
               we can give an example of a pair of graphs with the same chromatic number that are not isomorphic.
               For example,
-              <m>K_{3,3}</m> and <m>K_{3,4}</m> both have chromatic number 2, but are not isomorphic.
+              <m>K_{3,3}</m> and <m>K_{3,4}</m> both have chromatic number 2 but are not isomorphic.
             </p>
           </li>
 
@@ -848,7 +848,7 @@
           <li>
             <p>
               The graph is planar.
-              Even though as it is drawn edges cross,
+              Even though as it is drawn with edges crossing,
               it is easy to redraw it without edges crossing.
             </p>
           </li>
@@ -924,14 +924,14 @@
           <li>
             <p>
               True.
-              The graph is bipartite so it is possible to divide the vertices into two groups with no edges between vertices in the same group.
+              The graph is bipartite, so it is possible to divide the vertices into two groups with no edges between vertices in the same group.
               Thus we can color all the vertices of one group red and the other group blue.
             </p>
           </li>
 
           <li>
             <p>
-              False. <m>K_{3,3}</m> has 6 vertices with degree 3, so contains no Euler trail.
+              False. <m>K_{3,3}</m> has 6 vertices with degree 3 and so contains no Euler trail.
             </p>
           </li>
 

source/exercises/gt-intro.ptx

@@ -94,7 +94,7 @@
   <solution>
     <p>
       The graphs are not equal.
-      For example, graph 1 has an edge <m>\{a,b\}</m> but graph 2 does not have that edge.
+      For example, graph 1 has an edge <m>\{a,b\}</m>, but graph 2 does not have that edge.
       They are isomorphic.
       One possible isomorphism is <m>f:G_1 \to G_2</m> defined by <m>f(a) = d</m>, <m>f(b) = c</m>, <m>f(c) = e</m>, <m>f(d) = b</m>, <m>f(e) = a</m>.
     </p>
@@ -631,7 +631,7 @@
 
   <hint>
     <p>
-      Try a small example first: any graph with 8 vertices must have two vertices of the same degree.
+      Try a small example first: Any graph with 8 vertices must have two vertices of the same degree.
       If not, what would the degree sequence be?
     </p>
   </hint>

source/exercises/gt-paths.ptx

@@ -35,7 +35,7 @@
         This is a question about finding Euler trails.
         Draw a graph with a vertex in each state,
         and connect vertices if their states share a border.
-        Exactly two vertices will have odd degree:
+        Exactly two vertices will have odd degree,
         the vertices for Nevada and Utah.
         Thus you must start your road trip at in one of those states and end it in the other.
       </p>
@@ -188,7 +188,7 @@
     </statement>
     <hint>
       <p>
-        This is harder than the previous three questions.  Think about which <q>side</q> of the graph the Hamilton path would need to be on every other step.
+        This is harder than the previous three questions.  Think about which <q>side</q> of the graph the Hamilton path would need to be on at every other step.
       </p>
     </hint>
 

source/exercises/gt-planar.ptx

@@ -91,7 +91,7 @@
 <!-- I should check that this is actually possible as a polyhedron -->
     <solution>
       <p>
-        Say the last polyhedron has <m>n</m> edges,
+        Say the last polyhedron has <m>n</m> edges
         and also <m>n</m> vertices.
         The total number of edges the polyhedron has then is <m>(7 \cdot 3 + 4 \cdot 4 + n)/2 = (37 + n)/2</m>.
         In particular,
@@ -310,13 +310,13 @@
           \end{tikzpicture}
         </latex-image>
         <description>
-          A graph with 11 vertices.  A single vertex in the center, then five vertices equally spaced around a ring around it, and five more equally spaced around a ring around those.  Edges form the sides of a pentagon for the outer ring of vertices and also the inner ring of vertices.  Each outer vertex is also adjacent to two inner vertices: the two on either side of the vertex closest to it.  Finally, every inner vertex is also adjacent to the center vertex. 
+          A graph with 11 vertices.  A single vertex in the center, then five vertices equally spaced around a ring around it, and five more equally spaced around a ring around those.  Edges form the sides of a pentagon for the outer ring of vertices and also the inner ring of vertices.  Each outer vertex is also adjacent to two inner vertices: the two on either side of the vertex closest to it.  Finally, every inner vertex is also adjacent to the center vertex.
         </description>
       </image>
     </statement>
     <hint>
       <p>
-        What has happened to the girth?  Careful: we have a different number of edges as well.  Better check Euler's formula.
+        What has happened to the girth?  Careful: We have a different number of edges as well.  Better check Euler's formula.
       </p>
     </hint>
   </exercise>

source/exercises/gt-trees.ptx

@@ -118,7 +118,7 @@
             This cannot be a tree.
             Each degree 3 vertex is adjacent to all but one of the vertices in the graph.
             Thus each must be adjacent to one of the degree 1 vertices (and not the other).
-            That means both degree 3 vertices are adjacent to the degree 2 vertex, and to each other, so that means there is a cycle.
+            That means both degree 3 vertices are adjacent to the degree 2 vertex and to each other, so that means there is a cycle.
           </p>
 
           <p>
@@ -129,7 +129,7 @@
         <li>
           <p>
             This might or might not be a tree.
-            The length 4 path has this degree sequence (this is a tree), but so does the union of a 3-cycle and a length 1 path (which is not connected, so not a tree).
+            The length 4 path has this degree sequence (this is a tree), but so does the union of a 3-cycle and a length 1 path (which is not connected, so this is not a tree).
           </p>
         </li>
 
@@ -209,7 +209,7 @@
 
   <hint>
     <p>
-      Try a proof by contradiction and consider a spanning tree of the graph.
+      Try a proof by contradiction, and consider a spanning tree of the graph.
     </p>
   </hint>
 </exercise>
@@ -226,7 +226,7 @@
       Yes.
       We will prove the contrapositive.
       Assume <m>G</m> does not contain a cycle.
-      Then <m>G</m> is a tree, so would have <m>v = e+1</m>, contrary to stipulation.
+      Then <m>G</m> is a tree, so this would have <m>v = e+1</m>, contrary to stipulation.
     </p>
   </solution>
 </exercise>
@@ -292,7 +292,7 @@
   <hint>
     <p>
       Minimality here should be in terms of the number of vertices.
-      If you had a minimum counterexample, and removed a leaf vertex, the resulting graph will be a smaller tree, so...
+      If you had a minimum counterexample and removed a leaf vertex, the resulting graph will be a smaller tree, so...
     </p>
   </hint>
 </exercise>
@@ -370,7 +370,7 @@
 
   <hint>
     <p>
-      If <m>e</m> is the root, then <m>b</m> will have three children (<m>a</m>, <m>c</m>, and <m>d</m>), all of which will be siblings, and have <m>b</m> as their parent.
+      If <m>e</m> is the root, then <m>b</m> will have three children (<m>a</m>, <m>c</m>, and <m>d</m>), all of which will be siblings and have <m>b</m> as their parent.
       <m>a</m> will not have any children.
     </p>
 
@@ -441,9 +441,9 @@
 
         <li>
           <p>
-            No, although there are graph for which this is true (note that if all spanning trees are isomorphic, then all spanning trees will have the same number of leaves).
+            No, although there are graphs for which this is true (note that if all spanning trees are isomorphic, then all spanning trees will have the same number of leaves).
             Again, <m>K_4</m> is a counterexample.
-            One spanning tree is a path, with only two leaves, another spanning tree is a star with 3 leaves.
+            One spanning tree is a path, with only two leaves, and another spanning tree is a star with 3 leaves.
           </p>
         </li>
       </ol>