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5.5.3 Rooted tree and binary tree. Definition 25: A directed graph is a directed tree if the graph is a tree in the underlying undirected graph.
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5.5.3 Rooted tree and binary tree • Definition 25: A directed graph is a directed tree if the graph is a tree in the underlying undirected graph. • Definition 26: A rooted tree is a directed tree if there are exactly a vertex that is 0 in-degree, and other vertices that are 1 in-degree. The vertex of 0 in-degree is called root. And the vertices of 0 out-degree are called leaves. The vertices that are not 0 out-degree are called internal vertices. • There is a unique path from the root to each vertex of the rooted tree by the definition 26
Definition 27: Let u be an internal vertex. If there is a directed edge (u,w) from u to w, then w is called child of u, and u is called the parent of w. If the vertices w1 and w2 are child of u, then w1 and w2 are called brothers. If there a directed path from u to z, then z is called descendant of u. and u is called ancestors of w. The level of a vertex v is the length of the unique path from the root to this vertex. The level of the root is defined to be zero. The height of a rooted tree is the maximum of the levels of all vertices. • Note: The parent of w is unique.
Root:1 • Leaf: 6, 8, 9, 10, 11, 12 • internal vertices: 1, 2, 3, 4, 5, 7 • the levels of 2, 3 are 1, the levels of 4, 5, 6, 7, 8 are 2 , • the levels of 9, 10, 11, 12 are 3 。 • The height of a rooted tree is 3
Definition 28: If v be a vertex of a rooted tree, then the subtree with v as its root is the subgraph of the tree consisting of v and its descendants and all edges incident to these descendants.
Definition 29: An ordered rooted tree is a rooted tree where the children of each internal vertex are ordered. Ordered rooted trees are drawn so that the children of each internal vertex and edges are shown in order from left to right, and these edges are marked on 1,2,,i,
Definition 30: An ordered rooted tree is called an m-ary tree if every vertex has no more than m children. The tree is called a full m-ary tree if every vertex has exactly m children. An m-ary tree with m=2 is called a binary tree. • Left subtree, right subtree
Theorem 5.19: A full binary tree with t leaves contains i=t-1 internal vertices. • Proof: Let the number vertices be n. • The sum of children of all internal vertices equals n-1, • internal vertices and leaves • the number vertices n=i+t, • i.e. 2i=i+t-1, • Thus i=t-1。 • A full m-ary tree with t leaves contains i=(t-1)/(m-1) internal vertices.
Theorem 5.20: Let T be a full binary tree. We denote the sum of length of simple paths from root to all internal vertices by I, and the sum of length of simple paths from root to all leaves by E. Then E=I+2i, where i is the number of internal vertices. • Proof: Let us apply induction on the number i of internal vertices. • E=2 and I=0 when i=1
Suppose that result holds for i=k-1 • For i=k, we chose internal vertex v so that its children are leaves. We have a new tree which is obtained by omitting edges of incident v and its children. • We denote the length of path from root to v by l. • E'=E- l-2, I'=I-l. • By the inductive hypothesis, E'=I'+2(k-1) • E= E'+l+2=I'+2(k-1)+l+2=I-l+2(k-1)+l+2=I+2k。 • Let T be a full m-ary tree. Then E=(m-1)I+mi, where i is the number of internal vertices.
5.6 Prefix codes and optimal tree • a b c d e • 00 110 010 10 01 • The set {00,110,010,10,01} is called code • 010010 • eador cc? • The string of e is prefix of string of c • c: 111 • The set {00,110,111,10,01} is called prefix code
Definition 31: Codes with this property which the bit string for a letter never occurs as the first part of the bit string for another letter are called prefix codes. • Theorem 5.21: We can construct a prefix code from any binary tree, and we can construct a binary tree from the prefix codes. • Proof: (1) We can construct a prefix code from any binary tree where the left edge at each internal vertex is labeled by 0 and the right edge by a 1 and where the leaves are labeled by characters • (2)We can construct a binary tree from the prefix codes
Huffman algorithm: • Let a1,a2,,an be n vertex with weight w1,w2,,wn and w1w2wn。 • F:=forest of n rooted tree each consisting of the single vertex ai with weight wi for i=1,2,...n. • While F is not a tree • Begin • Replace the rooted trees T and T’ of least weights from F with w(T)≥ w(T’) with a tree having a new root that has T as its left subtree and T’ as its right subtree. Assign w(T)+w(T’) as the weight of the new tree. • end
Theorem 5.22: Let T be built according to Huffman algorithm and leaves of T with weight w1,w2,,wn. Then T is an optimal tree. • Proof: Let us apply induction on the number n of vertices. • n=2, The results holds Suppose that result holds for n=k-1 For n=k, By the inductive hypothesis, Suppose that nodes in an optimal tree T have weights w1+w2,w3,,wk. Then This is an optimal tree with weight w1,w2,w3,,wk if T’s leaf with weight w1+w2 is replaced by subtree
Quiz: • 1.Let G be a tree with two or more vertices. Then G is a bipartite graph. • 2.Let G be a simple graph with n vertices. Show that ifδ(G) >[n/2]-1, then G is a tree or contains three spanning trees at least. • Theorem 5.22, • Transport Networks, 8.4 P307
Exercise: P259 23 • 1. Let T be a full m-ary tree. We denote the sum of length of simple paths from root to all internal vertices by I, and the sum of length of simple paths from root to all leaves by E. Then E=(m-1)I+2i, (m-1)i=t-1, where i is the number of internal vertices and t is the number of leaves • 2.(1)Find a optimal binary tree with weight 1,3,8,9,12, 15,16 • (2)Construct a algorithms for optimal 3-ary tree • (3)Find a optimal 3-ary tree with weight 1,2,3,4,5,6,7,8,9 • (4)Find a optimal 3-ary tree with weight 1,2,3,4,5,6,7,8,