Trees and Distance

1
Chapter 2

• Trees and Distance

2.1 Basic properties 2.2 Spanning tree and
enumeration 2.3 Optimization and trees
2
Acyclic, tree2.1.1

• A graph with no cycle is acyclic
• A forest is an acyclic graph
• A tree is a connected acyclic graph
• A leaf (or pendant vertex) is a vertex of degree 1

leaf
tree
tree
forest
3
Spanning Subgraph 2.1.1

• A spanning subgraph of G is a subgraph with
  vertex set V(G)
• A spanning tree is a spanning subgraph that is a
  tree

Spanning tree
Spanning subgraph
4
Lemma. Every tree with at least two vertices has
at least two leaves. Deleting a leave from a
n-vertex tree produces a tree with n-1 vertices.
2.1.3

• Proof (1/2)
• A connected graph with at least two vertices has
  an edge.
• In an acyclic graph, an endpoint of a maximal
  nontrivial path has no neighbor other than its
  neighbor on the path.
• Hence the endpoints of such a path are leaves.

Impossible! Cycle occurs
Impossible! It is Maximal.
5
Lemma. Every tree with at least two vertices has
at least two leaves. Deleting a leave from a
n-vertex tree produces a tree with n-1 vertices.
2.1.3

• Proof (2/2)
• Let v be a leave of a tree G, and that GG-v.
• A vertex of degree 1 belongs to no path
  connecting two other vertices.
• Therefore, for u, w ? V(G), every u, w-path in G
  is also in G.
• Hence G is connected.
• Since deleting a vertex cannot create a cycle, G
  also is acyclic.
• Thus G is a tree with n-1 vertices.

6
Theorem 2.1.4. For an n-vertex graph G (with
n?1), the following are equivalent (and
characterize the trees with n vertices)A) G is
connected and has no cyclesB) G is connected and
has n-1 edgesC) G has n-1 edges and no cyclesD)
For, u, v?V(G), G has exactly one u, v-path 2.1.4

• Proof We first demonstrate the equivalence of A,
  B, and C by proving that any two of connected,
  acyclic, n-1 edges together imply the third

7
Theorem 2.1.4 Continue

• A?B,C. connected, acyclic ? n-1 edges
• We use induction on n.
• For n1, an acyclic 1-vertex graph has no edge.
• For ngt1, we suppose that implication holds for
  graphs with fewer than n vertices.
• Given an acyclic connected graph G, Lemma 2.1.3
  provides a leaf v and states that GG-v also is
  acyclic and connected (see figure above).
• Applying the induction hypothesis to G yields
  e(G)n-2.
• Since only one edge is incident to v, we have
  e(G)n-1.

8
Theorem 2.1.4 continue

• B?A, C. connected and n-1 edges ? acyclic
• Delete edges from cycles of G one by one until
  the resulting graph G is acyclic.
• Since no edge of a cycle is a cut-edge(Theorem
  1.2.14), G is connected.
• Now the preceding paragraph implies that
  e(G)n-1.
• Since we are given e(G)n-1, no edges were
  deleted.
• Thus GG, and G is acyclic.

9
Theorem 2.1.4 continue

• C?A, B. n-1 edges and no cycles ? connected
• Let G1,,Gk be the components of G.
• Since every vertex appears in one component,
  ?in(Gi)n.
• Since G has no cycles, each component satisfies
  property A.
• Thus e(Gi)n(Gi)-1.
• Summing over i yields e(G)?In(Gi)-1n-k.
• We are given e(G)n-1,so k1, and G is connected.

10
Theorem 2.1.4

• A?D. Connected and no cycles ? For u, v?V(G),
  one and only one u, v-path exists.
• Since G is connected, each pair of vertices is
  connected by a path.
• If some pair is connected by more than one , we
  choose a shortest (total length) pair P, Q of
  distinct paths with the same endpoints.
• By this extremal choice, no internal vertex of P
  or Q can belong to the other path.
• This implies that P?Q is a cycle, which
  contradicts the hypothesis A.

p
v
u
q
11
Theorem 2.1.4

• D?A.
• For u, v?V(G), one and only one u, v-path
  exists ? connected and no cycles.
• If there is a u,v-path for every u,v?V(G), then G
  is connected.
• If G has a cycle C, then G has two u,v-paths for
  u,v ?V(G) which contradicts the hypothesis D
• Hence G is acyclic (this also forbids loops).

12
Corollarya) Every edge of a tree is a cut-edge
b) Adding one edge to a tree forms exactly one
cyclec) Every connected graph contains a
spanning tree 2.1.5

• Proof
• A tree has no cycles, so Theorem 1.2.14 implies
  that every edge is a cut-edge.
• A tree has a unique path linking each pair of
  vertices (Theorem2.1.4D), so joining two vertices
  by an edge creates exactly one cycle.
• As in the proof of B?A, C in Theorem 2.1.4,
  iteratively deleting edges from cycles in a
  connected graph yields a connected acyclic
  subgraph.

13
Proposition If T, T are spanning trees of a
connected graph G and e?E(T)-E(T), then there is
an edge e?E(T)- E(T) such that T-ee is a
spanning tree of G. 2.1.6

• Proof By Corollary 2.1.5a, every edge of T is a
  cut-edge of T. Let U and U be the two components
  of T-e. Since T is connected, T has an edge e
  with endpoints in U and U. Now T-ee is
  connected, has n(G)-1 edges, and is a spanning
  tree of G.

e
e
G
T
T
T-ee
14
Distance in trees and Graphs

• If G has a u, v-path, then the distance from u to
  v, written dG(u,v) or simply d(u,v), is the least
  length of a u,v-path. If G has no such path, then
  d(u,v) ?
• The diameter (diam G) is maxu,v?V(G) d(u,v).
• Upper bound of distance between every pair.
• The eccentricity of a vertex u, written ?(u), is
  maxv?V(G) d(u,v).
• Upper bound of the distance from u to the others.
• The radius of a graph G, written rad G, is
  minu?V(G) ?(u).
• Lower bound of the eccentricity.

15
Distance, Diameter, Eccentricity, and Radius
a
b
f
e
c
g
d
Distance(f,c) 2 Distance(g,c) 2 Distance(a,c)
3
radius 2
diameter 3
eccentricity(f)2 eccentricity(a) 3
16
If G is a simple graph, then diam G?3 ? diam
?3 2.1.11

• Proof 1/2
• Since diam Ggt2, there exist nonadjacent vertices
  u, v?V(G) with no common neighbor.
• If any pair of nonadjacent vertices has a common
  neighbor, the distance of every pair is less than
  or equal to 2 and diam G2

17
If G is a simple graph, then diam G?3 ? diam
?3 2.1.11

• Proof 2/2
• Hence every x?V(G)-u,v has at least one of
  u,v as a nonneighbor.
• Equivalently, this makes x adjacent to at least
  one of u,v in .
• Since also uv?E( ), for every pair x, y there
  is an x, y-path of length at most 3 in
  through u,v. Hence diam ?3

u
v
18
Center 2.1.12

• Definition The center of a graph G is the
  subgraph induced by the vertices of minimum
  eccentricity.
• The center of a graph is the full graph if and
  only if the radius and diameter are equal.

Center
19
Theorem The center of a tree is a vertex or an
edge 2.1.13

• Proof We use induction on the number of vertices
  in a tree T.
• Basis step n(T)?2. With at most two vertices,
  the center is the entire tree.

20
Theorem. 2.1.13 Continue

• Induction step n(T)gt2.
• Let T T- leaves. By Lemma 2.1.3, T is a
  tree.
• Since the internal vertices on the paths between
  leaves of T remain, T has at least one vertex.
• Every vertex at maximum distance in T from a
  vertex u?V(T) is a leaf (otherwise, the path
  reaching it from u can be extended farther).
• Since all the leaves have been removed and no
  path between two other vertices uses a leave,
  ?T(u) ?T(u)-1 for every u?V(T).
• Also, the eccentricity of a leaf in T is greater
  than the eccentricity of its neighbor in T.
• Hence the vertices minimizing ?T(u) are the same
  as the vertices minimizing ?T(u).
• It is shown T and T have the same center. By the
  induction hypothesis, the center of T is a
  vertex or an edge.

21
Spanning Trees and Enumeration 2.2

• There are 2c(n,2) simple graphs with vertex set
  n1,,n.
• since each pair may or may not form an edge.
• How many of these are trees?

22
Enumeration of Trees 2.2

• One or two vertices, then there is only one tree.
• Three vertices, three trees.
• Four vertices, then four stars and 12 paths,
  total 16.
• Five vertices, then there are 125 trees.

23
Algorithm for tree generation by using Prufer
code 2.2.1

• Algorithm. (Prufer code) Production of
  f(T)(a1,,an-2)
• Input A tree T with vertex set S?N
• Iteration At the ith step, delete the least
  remaining leaf, and say that ai is the neighbor
  of the deleted leaf

1. Delete 2 a1 7 2. Delete 3 a2 4 3. Delete 5
a3 4 4. Delete 4 a4 1 5. Delete 6 a5 7 6.
Delete 7 a6 1
4
7
1
3
2
5
6
8
24
Theorem For a set S?N of size n, there are nn-2
trees with vertex set S 2.2.3

• Proof (sketch)
• Trees with vertex set S ?Sn-2 of lists of length
  n-2
• One prufer code ? one tree.
• There are nn-2 codes.
• Proved by induction

25
Spanning Trees in Graphs 2.2.6

• Example. A kite. To count the spanning trees
• Four are path around the outside cycle in the
  drawing
• The remaining spanning trees use the diagonal
  edge
• Since we must include an edge to each vertex of
  degree 2, we obtain four more spanning trees.
• The total is eight.

26
Contraction 2.2.7

• In a graph G, contraction of edge e with
  endpoints u, v is the replacement of u and v with
  a single vertex whose incident edges are the
  edges other than e that were incident to u or v.
  The resulting graph Ge has one less edge than G.

u
e
G
v
27
Proposition. Let ?(G) denote the number of
spanning trees of a graph G. If e?E(G) is not a
loop, then ?(G) ?(G-e) ?(Ge) 2.2.8

• Proof 1/2
• The spanning trees of G that omit e are precisely
  the spanning trees of G-e.
• We need show that G has ?(Ge) spanning trees
  containing e
• It must be shown that contraction of e defines a
  bijection from the set of spanning trees of G
  containing e to the set of spanning trees of Ge.

28
Proposition. Let ?(G) denote the number of
spanning trees of a graph G. If e?E(G) is not a
loop, then ?(G) ?(G-e) ?(Ge) 2.2.8

• Proof2/2
• When we contract e in a spanning tree that
  contains e, we obtain a spanning tree of Ge
  because
• the resulting subgraph of Ge is spanning and
  connected and
• It has the right number of edges.
• The other edges maintain their identity under
  contraction,
• So no two trees are mapped to the same spanning
  tree of Ge by this operation.

29
Proposition. Let ?(G) denote the number of
spanning trees of a graph G. If e?E(G) is not a
loop, then ?(G) ?(G-e) ?(Ge) 2.2.8

• ?(G-e) The number of trees without e
• ?(Ge) The number of trees with e
• A spanning tree in G.e ? A spanning tree having
  e in G

G
G-e
u
e
v

Ge
30
A Matrix Tree computation. 2.2.12

• Given a loopless graph G with vertex set v1, .,
  vn, let aij be the number of edges with endpoints
  vi and vj. Let Q be the matrix in which entry (i,
  j) is ai,j when i? j and is d(vi) when ij. If
  Q is a matrix obtained by deleting row s and
  column t of Q, then
• ?(G) (-1)st detQ

31
A Matrix Tree computation. 2.2.11

• Theorem 2.2.12 instructs us to form a matrix by
  putting the vertex degrees on the diagonal and
  subtracting the adjacency matrix. We then delete
  a row and a column and take the determinant.
• When G is the kite of Example 2.2.9, the vertex
  degrees are 3,3,2,2. We form the matrix on the
  left below and take the determinant of the matrix
  in the middle. The result is the number of
  spanning trees.

32
Example 2.2.11
v1
v3
v4
v2
v1 v2 v3 v4
v1 v2 v3 v4
8
33
Minimum Spanning Tree 2.3

• In a connected weighted graph of possible
  communication links, all spanning trees have n-1
  edges we seek one that minimizes or maximizes
  the sum of the edge weights.
• Kruskals Algorithm
• Prims Algorithm

34
Kruskals Algorithm for Minimum Spanning Tree
2.3.1

• Input A weighted connected graph
• Idea
• Maintain an acyclic spanning subgraph H.
• Enlarging it by edges with low weight to form a
  spanning tree.
• Consider edges in nondecreasing order of weight.

35
Kruskals Algorithm for Minimum Spanning Tree
2.3.1

• Initialization Set E(H)?.
• Iteration If the next cheapest edge joins two
  components of H, then include it otherwise,
  discard it. Terminate when H is connected.

H
Join Two vertices in one component. Cycle
occurs. Not Allowed!
Join two components. It works
36
Example of using Kruskals Algorithm
9
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9
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8
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37
Theorem In a connected weighted graph G,
Kruskals Algorithm constructs a minimum-weight
spanning tree. 2.3.3

• Proof 1/3
• We show first that the algorithm produces a tree.
  
• It never chooses an edge that completes a cycle.
• If the final graph has more than one component,
  then there is no edge joining two of them and G
  is not connected, because such an edge would be
  accepted.
• Since G is connected, some such edge exists and
  we considered it. Thus the final graph is
  connected and acyclic, which makes it a tree.

38
Theorem 2.3.3 Continue

• Proof continue
• Let T be the resulting tree, and let T be a
  spannig tree of minimum weight.
• If TT, we are done.
• If T?T, let e be the first edge chosen for T
  that is not in T. Adding e to T creates one
  cycle C. Since T has no cycle, C has an edge
  e?E(T). Consider the spanning tree Te-e.
• Since T contains e and all the edges of T
  chosen before e, both e and e are available when
  the algorithm chooses e, and hence w(e)?w(e).
  Thus Te-e is a spanning tree with weight at
  most T that agrees with T for a longer initial
  list of edges than T does.
• Repeating this argument eventually yields a
  minimum-weight spanning tree that agrees
  completely with T.
• Phrased extremely, we have prove that the minimum
  spanning tree agreeing with T the longest is T
  itself.

39
Shortest Paths

• How can we find the shortest route from one
  location to another?

40
Dijkstras Algorithm2.3.5

• Input A graph (or digraph) with nonnegtive edge
  weights and a starting vertex u. The weight of
  edge xy is w(xy) let w(xy)? if xy is not an
  edge.
• Idea Maintain the set S of vertices to which a
  shortest path from u is known, enlarging S to
  include all vertices. To do this, maintain a
  tentative distance t(z) from u to each z?S, being
  the length of the shortest u, v-path yet found.
• Initialization Set Su t(u)0 t(z)w(uz) for
  z?u

41
Dijkstras Algorithm2.3.5

• Iteration
• Select a vertex v outside S such that t(v)minz?S
  t(z).
• Add v to S.
• Explore edges from v to update tentative
  distance for each edge vz with z?S, update t(z)
  to mint(z), t(v)w(vz)
• The iteration continues until SV(G) or
  until t(z)? for every z?S. At the end, set d(u,
  v)t(v) for all v.

42
Example Find Shortest paths by using Dijkstras
Algorithm 2.3.6
t1
a
d
5
d
a
5
1
2
1
2
u
4
4
4
u
e
e
4
d0
3
6
6
3
c
b
5
c
5
b
t3
43
Example Find Shortest paths by using Dijkstras
Algorithm continue 2.3.6
d1
t5
a
4
c
1
5
2
u
e
d0
5
6
3
4
d
t6
d3
b
44
Example Find Shortest paths by using Dijkstras
Algorithm continue 2.3.6
d5
d5
c
c
d1
d1
4
4
6
a
a
t11
6
e
5
5
e
5
1
1
d0
d0
u
u
t8
5
2
2
3
3
4
d
4
t6
d
d3
d3
d6
b
b
45
Theorem Given a (di)graph G and a vertex u?V(G),
Dijkstras algorithm compute d(u,z) for every z
?V(G) 2.3.7

• Proof
• We prove the stronger statement that at each
  iteration,
• For z?S, t(z)d(u, z), and
• For z?S, t(z) is the least length of a u, z-path
  reaching z directly from S.
• We use induction on kS.
• Basis step k1. From the initialization, Su,
  d(u, u)t(u)0, and the least length of a u,
  z-path reaching z directly from S is t(z)w(u,z),
  which is infinite when uz is not an edge.

46
Theorem 2.3.7 Continue

• Induction step
• Suppose that when Sk, (1) and (2) are true.
• Let v be a vertex among z?S such that t(z) is
  smallest.
• The algorithm now choose v let SS?v. We
  first argue that d(u, v)t(v). A shortest u,
  v-path must exit S before reaching v. The
  induction hypothesis states that the length of
  the shortest path going directly to v from S is
  t(v). The induction hypothesis and choice of v
  also guarantee that a path visiting any vertex
  outside S and later reaching v has length at
  least t(v).
• Hence d(u, v)t(v), and (1) holds for S.

47
Theorem 2.3.7 Continue

• To prove (2) for S, let z be a vertex outside S
  other than v.
• By the hypothesis, the shortest u, v-path
  reaching z directly from S has length t(z) (? if
  there is no such path).
• When we add v to S, we must also consider paths
  reaching z from v. Since we have now computed
  d(u, v)t(v), the shortest such path has length
  t(v)w(vz), and we compare this with the previous
  value of t(z) to find the shortest path reaching
  z directly from S.
• We have verified that (1) and (2) hold for the
  new set S of size k1 this completes the
  induction step.

48
Prims Algorithm

• Input A graph (or digraph) with nonnegtive edge
  weights and a starting vertex u. The weight of
  edge xy is w(xy) let w(xy)? if xy is not an
  edge.
• Idea Maintain a tree T including u at the
  beginning, enlarging T to include all vertices.
  To do this, select the cheapest edge from the
  edges E(x,v) x? T, v?T .
• Initialization Set Tu.

49
Prims Algorithm

• Iteration Select a vertex v outside T such that
  w(x,v)minx? T, v?T w(x,v) . Add v to T.
• The iteration continues until
  V(T)V(G).

50
Dijkstras Algorithm v.s. Prims Algorithm
t6
d
d1
6
a
e
t5
3
c
u
d0
b
8
t3
The end vertex of the cheapest edge
The vertex which is the nearest to u
51
Algorithm - Breadth First Search 2.3.8

• Input An unweighted graph (or digraph) and a
  start vertex u.
• Idea Maintain a set R of vertices that have been
  reached but not searched and a set S of vertices
  that have been searched. The set R is maintained
  as a First-In First-Out list (queue), so the
  first vertices found are the first vertices
  explored.
• Initialization Ru, S?, d(u, u)0
• Iteration As long as R??, we search from the
  first vertex v of R. The neighbors of v not in
  S?R are added to the back of R and assigned
  distance d(u, v)1, and then v is removed from
  the front of Rand assigned distance d(u, v)1,
  and then v is removed from the front of R and
  placed in S

52
Example of BFS
a
b
d
c
e
g
j
i
f
h
Order of BFS a, b, c, d, e, f, g, h, i, j,
a, c, b, d, f, h, e, g, i,
j, b, a, g, f, e, c, d, h, i, j,
53
Breadth-F-S vs. Depth-F-S
a
a
b
b
d
d
c
c
e
e
g
g
j
j
i
i
f
h
f
h
a,b,c,d,e,f,g,h,i,j,
a,b,e,f, c, h,
54
Application Chinese Postman Problem 2.3.9

• A mail carrier must traverse all edges in a road
  network, starting and ending at the Post Office.
  The edges has nonnegative weights representing
  distance or time.
• We seek a closed walk of minimum total length
  that uses all the edges.
• This is the Chinese Postman Problem, named in
  honor of the Chinese mathematician Guan Meigu
  1962, who proposed it.

55
Application Chinese Postman Problem continue
2.3.9

• If every vertex is even, then the graph is
  Eulerian and the answer is the sum of the edge
  weights.
• Otherwise, we must repeat edges. Every traversal
  is an Eulerian circuit of a graph obtained by
  duplicating edges.
• Finding the shortest traversal is equivalent to
  finding the minimum total weight of edges whose
  duplication will make all vertex degrees even .

56
Application Chinese Postman Problem continue
2.3.9

• We say duplication because we need not use an
  edge more than twice. If we use an edge three or
  more times in making all vertices even, then
  deleting two of those copies will leave all
  vertices even. There may be many ways to choose
  the duplicated edges.

57
Application Chinese Postman Problem continue
2.3.9

• If there are only two odd vertices, then
• We can use Dijkstras Algorithm to find the
  shortest path between them and solve the problem
• If there are 2k odd vertices, then
• Use Dijkstras algorithm to find the shortest
  paths connecting each pair of odd vertices
• Use these lengths as weights on the edges of K2k
• Find the minimum total weight of k edges that
  pair up these 2k vertices. This is a maximum
  matching problem.

58
Application Chinese Postman Problem continue
2.3.9

• A vertex
• An odd vertex
• An edge between u and v
• The shortest path between u and v in the
  given graph

20
A matching in a general graph with minimum total
weight
40
10
16
15
20
59
Rooted Tree 2.3.11

• A rooted tree is a tree with one vertex r chosen
  as root.
• For each vertex v, let P(v) be the unique v,
  r-path.
• The parent of v is its neighbor on P(v)
• Its children are its other neighbors.
• Its ancestors are the vertices of P(v)-v.
• Its descendants are the vertices u such that P(u)
  contains v.
• The leaves are the vertices with no children.
• A rooted plane tree or planted tree is a rooted
  tree with a left-toright ordering specified for
  the children of each vertex.

60
Rooted Tree 2.3.11
root
Parent of v, Ancester of v
v
A leaf, A child of v
61
Binary Tree 2.3.12

• A binary tree is a rooted plane tree where each
  vertex has at most two children, and each child
  of a vertex is designated as its left child or
  right child.
• The subtrees rooted at the children of the root
  are the left subtree and the right subtree of the
  tree.
• A k-ary tree allows each vertex up to k children.

62
Huffmans Algorithm 2.3.13

• Input Weights (frequencies or probabilities)
  p1,,pn
• Output Prefix-free code (equivalently, a binary
  tree)
• Idea Infrequent items should have longer codes
  put infrequent items deeper by combining them
  into parent nodes.
• Initial case When n2, the optimal length is
  one, with 0 and 1 being the codes assigned to the
  two items (the tree has a root and two leaves
  n1 can also be used as the initial case)

63
Huffmans Algorithm 2.3.13

• Recursion
• Replace the two least likely items p, p with a
  single item q of weight pp when ngt2
• Treat the smaller set as a problem with n-1
  items.
• After solving it, give children with weights p,
  p to the resulting leaf with weight q.
• Equivalently, replace the code computed for the
  combined item with its extensions by 1 and 0,
  assigned to the items that were replaced.

64
Example of Huffman coding 2.3.14

• Consider eight items with frequencies 5, 1, 1, 7,
  8, 2, 3, 6.
• Algorithm 2.3.13 combines items according to the
  tree on the left below, working from the bottom
  up.
• First the two items of weight 1 combine to from
  one of weight 2.

65
Huffman coding 2.3.14

• Now this and the original item of weight 2 are
  the least likely and combine to form an item of
  weight 4.

66
Huffman coding 2.3.14
7
7
11
4
4
2
2
5
1
1
8
2
7
3
6
5
1
1
8
2
7
3
6
67
Huffman coding 2.3.14
68
Theorem 2.3.15

• Given a probability distribution pi on n items,
  Huffmans Algorithm produces the prefix-free code
  with minimum excepted length