Shawn Zhong

Shawn Zhong

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Shawn Zhong

 钟万祥
  • Tutorials
  • Mathematics
    • Math 240
    • Math 375
    • Math 431
    • Math 514
    • Math 521
    • Math 541
    • Math 632
    • Abstract Algebra
    • Linear Algebra
    • Category Theory
  • Computer Sciences
    • CS/ECE 252
    • CS/ECE 352
    • Learn Haskell
  • AP Notes
    • AP Microecon
    • AP Macroecon
    • AP Statistics
    • AP Chemistry
    • AP Physics E&M
    • AP Physics Mech
    • CLEP Psycho

Home / 2018 / May / Page 2

Category Theory – Video 5

  • May 23, 2018
  • Shawn
  • Category Theory
  • No comments yet
Calculating The Number of Maps Between Sets • Consider A={a_0,a_1 }, B={b_0,b_1,b_2 } • There are 3×3=9 maps from A to B • There are 2×2×2=8 maps from B to A General Formula for The Number of Maps A→B • In general, the number of maps A→B is |B|^|A| because each a_i has |B| independent choices • The number of maps A→B≡|B^A |=|B|^|A| • In Video 2, we have defined ○ X={x_0,x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_10,x_11 } ○ C={C♮,(C♯/D♭),D♮,(D♯\/E♭),E♮,F♮,(F♯/G♭),G♮,(G♯/A♭),A♮,(A♯\/B♭),B♮} • So |X^C |=|〖12〗^12 |=8,916,100,448,256 Universal Mapping Properties (UMPs) • A Universal Mapping Property asserts something about one or more objects with zero or more maps between them in relation to all other objects X in the Category • A UMP is an object P, that may or may not have a relationship to a diagram of some sort • The principal is that there is only one map between P and all other objects X in that Category • With X as either the Domain or Codomain of that map, such that P still obeys its restriction Initial Object • In any Category C, an object 0 is said to be an Initial Object of C, if ∀X in C, there is a unique C-arrow 0→X • In S, a set 0 is said to be an Initial Object of S, if for all sets X, there is a unique map 0→X • Number of maps A→B is represented by |B^A |=|B|^|A| , so |X|^|0| =1 must hold for all X • So the Initial Set is the Empty Set, because any number raised to the power of 0 is 1 • There are no maps X→0, unless X≅0, since 0 raised to the power of any number (except 0) is 0 Duality • Whenever you define a UMP in S, to get another UMP just "do the opposite of what the definition says" • Usually just by reversing the arrows, and you ll get another UMP for free • Just add the prefix "co-" to the original definition and you’re done Terminal Object • In any Category C, an object T is said to be a Terminal Object of C, if ∀X in C, there is a unique C-arrow X→T • In S, a set T is said to be a Terminal Object of S, if for all sets X, there is a unique map X→T • Number of maps A→B is represented by |B^A |=|B|^|A| , so |T|^|X| =1 must hold for all X • The Terminal Set is 1, because 1 raised to the power of any number is 1 The Opposite Category and Duality • Duality ○ What if for every map f:A→B, we define a map f^op:B→A that makes the same associations between A and B, just from B^′ s perspective? ○ The map f^op:B→A assigns to each b_i∈B a subset A_i of A called the Splitting of b_i, defined as {a∈A_i⇔f^op (b_i )=a}, such that for all a∈A, there is exactly one A_i such that a∈A_i • Opposite Category ○ We can form a new category S^op that is the "mirror", or Opposite Category of S ○ For every object in S, there is a correspondent object in S^op ○ For every arrow f:A→B in S, there is a correspondent arrow f^op:B→A in S^op Calculating the Number of Maps 3→2 in S^op • The number of maps 3→2 in S^op must be 9, because there are 9 maps 2→3 in S • Ideally each element of the Domain in S^op has 3 choices in a map into the Codomain 2 ○ 0: Do nothing ○ 1: Connect ○ 2: Split into 2 • If we want to partition 2 into 3 parts, our options are 2+0+0, and 1+1+0 • We will represent the size of the subset that f^op assigns to b_0,b_1,b_2 as ⟨|A_0 |,|A_1 |,|A_2 |⟩ The Terminal Object in S^op • The Terminal Object in S^op is the Initial Object in S • The Initial Object in S^op is the Terminal Object in S
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Category Theory – Video 6

  • May 23, 2018
  • Shawn
  • Category Theory
  • No comments yet
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11.1 Introduction to Trees

  • May 07, 2018
  • Shawn
  • Math 240
  • No comments yet
Trees • Definition ○ A tree is a connected undirected graph with no simple circuits. • Example ○ Which of these graphs are trees? • Solution ○ G_1 and G_2 are trees - both are connected and have no simple circuits ○ Because e, b, a, d, e is a simple circuit, G_3 is not a tree. ○ G_4 is not a tree because it is not connected. • Definition ○ A forest is a graph that has no simple circuit, but is not connected. ○ Each of the connected components in a forest is a tree. • Theorem ○ An undirected graph is a tree if and only if there is a unique simple path between any two of its vertices. • Proof ○ (⟹)Assume that T is a tree. § Then T is connected with no simple circuits. § Hence, if x and y are distinct vertices of T, there is a simple path between them (by Theorem 1 of Section 10.4). § This path must be unique - for if there were a second path, there would be a simple circuit in T (by Exercise 59 of Section 10.4). § Hence, there is a unique simple path between any two vertices of a tree. ○ (⟸)Now assume that there is a unique simple path between any two vertices of a graph T. § Then T is connected because there is a path between any two of its vertices. § Furthermore, T can have no simple circuits since if there were a simple circuit, there would be two paths between some two vertices. ○ Hence, a graph with a unique simple path between any two vertices is a tree. Rooted Trees • Definition ○ A rooted tree is a tree in which one vertex has been designated as the root and every edge is directed away from the root. • An unrooted tree is converted into different rooted trees when different vertices are chosen as the root. Rooted Tree Terminology • Terminology for rooted trees is a mix from botany and genealogy (such as this family tree of the Bernoulli family of mathematicians). • If v is a vertex of a rooted tree other than the root, the parent of v is the unique vertex u such that there is a directed edge from u to v. • When u is a parent of v, v is called a child of u. • Vertices with the same parent are called siblings. • The ancestors of a vertex are the vertices in the path from the root to this vertex, excluding the vertex itself and including the root. • The descendants of a vertex v are those vertices that have v as an ancestor. • A vertex of a rooted tree with no children is called a leaf. • Vertices that have children are called internal vertices. • If a is a vertex in a tree, the subtree with a as its root is the subgraph of the tree consisting of a and its descendants and all edges incident to these descendants. Examples of Rooted Trees • Example: In the rooted tree T (with root a): ○ Find the parent of c, the children of g, the siblings of h, the ancestors of e, and the descendants of b. ○ Find all internal vertices and all leaves. ○ What is the subtree rooted at G? • Solution: ○ The parent of c is b. The children of g are h, i, and j. The siblings of h are i and j. The ancestors of e are c, b, and a. The descendants of b are c, d, and e. ○ The internal vertices are a, b, c, g, h, and j. The leaves are d, e, f, i, k, l, and m. ○ We display the subtree rooted at g. Properties of Trees • Theorem 2 ○ A tree with n vertices has n − 1 edges. • Proof (by mathematical induction): ○ Basis Step § When n = 1, a tree with one vertex has no edges. § Hence, the theorem holds when n = 1. ○ Inductive Step § Assume that every tree with k vertices has k − 1 edges. § Suppose that a tree T has k + 1 vertices and that v is a leaf of T. § Let w be the parent of v. § Removing the vertex v and the edge connecting w to v produces a tree T′ with k vertices. § By the inductive hypothesis, T′ has k − 1 edges. § Because T has one more edge than T′, we see that T has k edges. § This completes the inductive step. m-ary Rooted Trees • Definition ○ A rooted tree is called an m-ary tree if every internal vertex has no more than m children. ○ The tree is called a full m-ary tree if every internal vertex has exactly m children. ○ An m-ary tree with m = 2 is called a binary tree. • Example ○ Are the following rooted trees full m-ary trees for some positive integer m? • Solution ○ T_1 is a full binary tree because each of its internal vertices has two children. ○ T_2 is a full 3-ary tree because each of its internal vertices has three children. ○ In T_3 each internal vertex has five children, so T3 is a full 5-ary tree. ○ T_4 is not a full m-ary tree for any m because some of its internal vertices have two children and others have three children. Counting Vertices in Full m-Ary Trees • Theorem 3 ○ A full m-ary tree with i internal vertices has n = mi + 1 vertices. • Proof ○ Every vertex, except the root, is the child of an internal vertex. ○ Because each of the i internal vertices has m children, ○ there are mi vertices in the tree other than the root. ○ Hence, the tree contains n = mi + 1 vertices. • Theorem 4 ○ A full m-ary tree with § n vertices has i=(n−1)/m internal vertices and l=((m−1)n+1)/m leaves, § i internal vertices has n = mi + 1 vertices and l=(m−1)i+1 leaves, § l leaves has n=(ml−1)/(m−1) vertices and i=(l−1)/(m−1) internal vertices. ○ Proof (of part i) § Solving for i in n = mi + 1 (from Theorem 3) gives i=(n−1)/m. § Since each vertex is either a leaf or an internal vertex, n=l+i. § By solving for l and using the formula for i, we see that § l=n−i=n−(n−1)/m=((m−1)n+1)/m Level of vertices and height of trees • When working with trees, we often want to have rooted trees where the subtrees at each vertex contain paths of approximately the same length. • To make this idea precise we need some definitions: ○ The level of a vertex v in a rooted tree is the length of the unique path from the root to this vertex. ○ The height of a rooted tree is the maximum of the levels of the vertices. • Example: ○ Find the level of each vertex in the tree to the right. ○ What is the height of the tree? • Solution: ○ The root a is at level 0. ○ Vertices b, j, and k are at level 1. ○ Vertices c, e, f, and l are at level 2. ○ Vertices d, g, i, m, and n are at level 3. ○ Vertex h is at level 4. ○ The height is 4, since 4 is the largest level of any vertex. Balanced m-Ary Trees • Definition ○ A rooted m-ary tree of height h is balanced if all leaves are at levels h or h − 1. • Example ○ Which of the rooted trees shown below is balanced? • Solution ○ T_1 and T_3 are balanced, but T_2 is not because it has leaves at levels 2, 3, and 4. The Bound for the Number of Leaves in an m-Ary Tree • Theorem 5 ○ There are at most mh leaves in an m-ary tree of height h. • Proof (by mathematical induction on height): ○ Basis Step § Consider an m-ary trees of height 1. § The tree consists of a root and no more than m children, all leaves. § Hence, there are no more than m^1=m leaves in an m-ary tree of height 1. ○ Inductive Step § Assume the result is true for all m-ary trees of height h. § Let T be an m-ary tree of height h. § The leaves of T are the leaves of the subtrees of T we get when we delete the edges from the root to each of the vertices of level 1. § Each of these subtrees has height ≤ h−1. § By the inductive hypothesis, each of these subtrees has at most mh− 1 leaves. § Since there are at most m such subtrees, there are at most m⋅m^(h1) = mh leaves in the tree. • Corollary 1 ○ If an m-ary tree of height h has l leaves, then h ≥ ⌈log_m⁡l⌉. ○ If the m-ary tree is full and balanced, then h = ⌈log_m⁡l⌉. (see text for the proof)
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10.4 Connectivity

  • May 07, 2018
  • Shawn
  • Math 240
  • No comments yet
Paths • Informal Definition ○ A path is a sequence of edges that begins at a vertex of a graph and travels from vertex to vertex along edges of the graph. ○ As the path travels along its edges, it visits the vertices along this path, that is, the endpoints of these. • Applications ○ Numerous problems can be modeled with paths formed by traveling along edges of graphs such as ○ determining whether a message can be sent between two computers. ○ efficiently planning routes for mail delivery. • Definition ○ Let n be a nonnegative integer and G an undirected graph ○ A path of length n from u to v in G is a sequence of n edges e_1,…,e_n of G for which § there exists a sequence x_0=u,x_1,…,x_(n−1),x_n=v of vertices such that § e_i has, for i=1,…,n, the endpoints x_(i−1) and x_i. ○ When the graph is simple, we denote this path by its vertex sequence § x_0,x_1,…,x_n (since listing the vertices uniquely determines the path). ○ The path is a circuit if it begins and ends at the same vertex and has length greater than zero. ○ The path or circuit is said to pass through the vertices x_1,x_2,…,x_(n−1) and traverse the edges e_1,…,e_n. ○ A path or circuit is simple if it does not contain the same edge more than once. • Example ○ In the simple graph here: ○ a, d, c, f, e is a simple path of length 4. ○ d, e, c, a is not a path because e is not connected to c. ○ b, c, f, e, b is a circuit of length 4. ○ a, b, e, d, a, b is a path of length 5, but it is not a simple path. Degrees of Separation • Paths in Acquaintanceship Graphs ○ In an acquaintanceship graph there is a path between two people if there is a chain of people linking these people, where two people adjacent in the chain know one another. ○ In this graph there is a chain of six people linking Kamini and Ching. • Note: ○ Some have speculated that almost every pair of people in the world are linked by a small chain of no more than six, or maybe even, five people. ○ The play Six Degrees of Separation by John Guare is based on this notion. Connectedness in Undirected Graphs • Definition ○ An undirected graph is called connected if there is a path between every pair of vertices. ○ An undirected graph that is not connected is called disconnected. ○ We say that we disconnect a graph when we remove vertices or edges, or both, to produce a disconnected subgraph. • Example ○ G_1 is connected because there is a path between any pair of its vertices, as can be easily seen. ○ However G_2 is not connected because there is no path between vertices a and f, for example. • Theorem ○ Every pair of distinct vertices in a connect graph is connected by a simple path • Proof ○ Let u and v be two distinct vertices of the connected undirected graph G=(V,E) ○ Because G is connected, there is at least one path between u and v ○ Let x_0,x_1,…,x_n, where x_0=u and x_n=v, be the vertex sequence of a path of least length. ○ This path of least length is simple. ○ To see this, suppose is not simple, then x_i=x_j for some i and j with 0≤i j ○ This means that there is a path from u to v of shorter length with vertex sequence § x_0,x_1,…,x_(i−1),x_j,…,x_n ○ Obtained by deleting the edges corresponding to the vertex sequence x_i,…,x_(j−1) Connected Components Definition A connected component of a graph G is a connected subgraph of G that is not a proper subgraph of another connected subgraph of G A graph G that is not connected has two or more connected components that are disjoint and have G as their union. Example The graph H is the union of three disjoint subgraphs H_1, H_2, and H_3, none of which are proper subgraphs of a larger connected subgraph of G These three subgraphs are the connected components of H. Connectedness in Directed Graphs • Definition ○ A directed graph is strongly connected if there is a path from a to b and a path from b to a whenever a and b are vertices in the graph. • Definition ○ A directed graph is weakly connected if there is a path between every two vertices in the underlying undirected graph, which is the undirected graph obtained by ignoring the directions of the edges of the directed graph. Counting Paths between Vertices • We can use the adjacency matrix of a graph to find the number of paths between two vertices in the graph. • Theorem ○ Let G be a graph with adjacency matrix A with respect to the ordering v_1,…,v_n of vertices ○ (with directed or undirected edges, multiple edges and loops allowed). ○ The number of different paths of length r from v_i to v_j, where r 0 is a positive integer, equals the (i,j)th entry of A^r. • Proof by mathematical induction: ○ Basis Step § By definition of the adjacency matrix, the number of paths from v_i to v_j of length 1 is the (i,j)th entry of A. ○ Inductive Step § For the inductive hypothesis, we assume that that the (i,j)th entry of A^r is the number of different paths of length r from v_i to v_j. § Because A^(r+1)=A^r A, the (i,j)th entry of A^(r+1) equals b_i1 a_1j+b_i2 a_2j+…+b_in a_nj, where b_ik is the (i,k)th entry of A^r. § By the inductive hypothesis, b_ik is the number of paths of length r from v_i to v_k. § A path of length r + 1 from v_i to v_j is made up of a path of length r from v_i to some v_k, and an edge from v_k to v_j. § By the product rule for counting, the number of such paths is the product of the number of paths of length r from v_i to v_k (i.e., b_ik ) and the number of edges from from v_k to v_j (i.e, akj). § The sum over all possible intermediate vertices v_k is b_i1 a_1j+b_i2 a_2j+…+b_in a_nj.
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10.3 Representing Graphs and Graph Isomorphism

  • May 07, 2018
  • Shawn
  • Math 240
  • No comments yet
Representing Graphs: Adjacency Lists • Definition ○ An adjacency list can be used to represent a graph with no multiple edges by specifying the vertices that are adjacent to each vertex of the graph. • Example 1 • Example 2 Representation of Graphs: Adjacency Matrices • Definition ○ Suppose that G=(V,E) is a simple graph where |V|=n. ○ Arbitrarily list the vertices of G as v_1,v_2,…,v_n. ○ The adjacency matrix A_G of G, with respect to the listing of vertices, is the n×n zero-one matrix with 1 as its (i, j)th entry when v_i and v_j are adjacent, and 0 as its (i, j)th entry when they are not adjacent. ○ In other words, if the graphs adjacency matrix is A_G=[a_ij ], then ○ a_ij={■8(1&if {v_i,v_j } is an edge of G@0&otherwise)┤ • Example 1 ○ The ordering of vertices is a, b, c, d. • Example 2 ○ The ordering of vertices is a, b, c, d. • Note ○ The adjacency matrix of a simple graph is symmetric, i.e., a_ij=a_ji ○ Also, since there are no loops, each diagonal entry a_ii for i = 1, 2, 3, …, n, is 0. • Note ○ When a graph is sparse, that is, it has few edges relatively to the total number of possible edges, it is much more efficient to represent the graph using an adjacency list than an adjacency matrix. ○ But for a dense graph, which includes a high percentage of possible edges, an adjacency matrix is preferable. Adjacency Matrices: Graphs with Loops and Multiple Edges • Adjacency matrices can also be used to represent graphs with loops and multiple edges. • A loop at the vertex v_i is represented by a 1 at the (i, j)th position of the matrix. • When multiple edges connect the same pair of vertices v_i and v_j, (or if multiple loops are present at the same vertex), the (i, j)th entry equals the number of edges connecting the pair of vertices. • Example ○ We give the adjacency matrix of the pseudograph shown here using the ordering of vertices a, b, c, d. Adjacency Matrices: Directed graphs • Adjacency matrices can also be used to represent directed graphs. • The matrix for a directed graph G=(V, E) has a 1 in its (i, j)th position if there is an edge from v_i to v_j, where v_1,v_2,…,v_n is a list of the vertices. • In other words, if the graphs adjacency matrix is A_G=[a_ij ], then ○ a_ij={■8(1&if (v_i,v_j ) is an edge of G@0&otherwise)┤ • The adjacency matrix for a directed graph does not have to be symmetric, because there may not be an edge from v_i to v_j, when there is an edge from v_j to v_i. • To represent directed multigraphs, the value of a_ij is the number of edges connecting v_i to v_j. Representation of Graphs: Incidence Matrices • Definition ○ Let G=(V, E) be an undirected graph with vertices v_1,v_2,…,v_n and edges e_1,e_2,…,e_m. ○ The incidence matrix with respect to the ordering of V and E is the n×m matrix ○ M=[m_ij ],where m_ij={■8(1&when edge e_j is incident with v_i@0&otherwise)┤ • Example ○ Simple Graph and Incidence Matrix ○ ○ The rows going from top to bottom represent v_1 through v_5 ○ The columns going from left to right represent e_1 through e_6. • Example ○ The rows going from top to bottom represent v_1 through v_5 ○ The columns going from left to right represent e_1 through e_8. Isomorphism of Graphs • Definition ○ The simple graphs G_1 = (V_1, E_1) and G_2 = (V_2, E_2) are isomorphic if § there is a one-to-one and onto function f from V_1 to V_2 with the property that § a and b are adjacent in G_1 if and only if f(a) and f(b) are adjacent in G_2, for all a and b in V_1. ○ Such a function f is called an isomorphism. ○ Two simple graphs that are not isomorphic are called nonisomorphic. • Example ○ Show that the graphs G=(V, E) and H=(W, F) are isomorphic. • Solution ○ The function f with f(u_1) = v_1, f(u_2)=v_4, f(u_3) = v_3, and f(u_4) = v_2 is a one-to-one correspondence between V and W. ○ Note that adjacent vertices in G are u_1 and u_2, u_1 and u_3, u_2 and u_4, and u_3 and u_4. ○ Each of the pairs f(u_1) = v_1 and f(u_2) = v_4, f(u_1)=v_1 and f(u_3)=v_3 , f(u_2)=v_4 f(u_4)=v_2, f(u_3) = v_3 and f(u_4) = v_2 consists of two adjacent vertices in H • Note ○ It is difficult to determine whether two simple graphs are isomorphic using brute force because there are n! possible one-to-one correspondences between the vertex sets of two simple graphs with n vertices. ○ The best algorithms for determining whether two graphs are isomorphic have exponential worst case complexity in terms of the number of vertices of the graphs. ○ Sometimes it is not hard to show that two graphs are not isomorphic. ○ We can do so by finding a property, preserved by isomorphism, that only one of the two graphs has. ○ Such a property is called graph invariant. ○ There are many different useful graph invariants that can be used to distinguish nonisomorphic graphs, such as the number of vertices, number of edges, and degree sequence (list of the degrees of the vertices in nonincreasing order). ○ We will encounter others in later sections of this chapter. • Example ○ Determine whether these two graphs are isomorphic • Solution ○ Both graphs have eight vertices and ten edges. ○ They also both have four vertices of degree two and four of degree three. ○ However, G and H are not isomorphic. ○ Note that since deg(a) = 2 in G, a must correspond to t, u, x, or y in H, ○ because these are the vertices of degree 2. ○ But each of these vertices is adjacent to another vertex of degree two in H ○ which is not true for a in G. ○ Alternatively, note that the subgraphs of G and H made up of vertices of ○ degree three and the edges connecting them must be isomorphic. ○ But the subgraphs, as shown at the right, are not isomorphic. • Example ○ Determine whether these two graphs are isomorphic. • Solution ○ Both graphs have six vertices and seven edges. ○ They also both have four vertices of degree two and two of degree three. ○ The subgraphs of G and H consisting of all the vertices of degree two and the edges connecting them are isomorphic ○ So, it is reasonable to try to find an isomorphism f. ○ We define an injection f from the vertices of G to the vertices of H that preserves the degree of vertices. ○ We will determine whether it is an isomorphism. ○ The function f with f(u_1)=v_6,f(u_2)=v_3,f(u_3)=v_4,f(u_4)=v_5,f(u_5)=v_1,f(u_6)=v_2 is a one-to-one correspondence between G and H ○ Showing that this correspondence preserves edges is straightforward, so we will omit the details here. ○ Because f is an isomorphism, it follows that G and H are isomorphic graphs. ○ See the text for an illustration of how adjacency matrices can be used for this verification.
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