1. Chapter 2 Trees and Distance 2.1 Basic properties
Graph Theory
Chapter 2
Trees and Distance
2.1 Basic properties
2.2 Spanning tree and enumeration
2.3 Optimization and trees
Ch Trees and Distance

2. Acyclic, tree2.1.1 A graph with no cycle is acyclic
Graph Theory
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
Ch Trees and Distance

3. Graph Theory
Spanning Subgraph
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 subgraph
Spanning tree
Ch Trees and Distance

4. Graph Theory
Lemma. Every tree with at least two vertices has at least two leaves
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
Maximal path
Impossible!
It is Maximal.
Ch Trees and Distance

5. Graph Theory
Lemma. Deleting a leaf from a n-vertex tree produces a tree with n-1 vertices
Proof: (2/2)
Let v be a leaf of a tree G, and that G’=G-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.
Ch Trees and Distance

6. For u, w  V(G’ ), every u, w-path in G is also in G’.
Graph Theory G G’ v w u
For u, w  V(G’ ), every u, w-path in G is also in G’.
Ch Trees and Distance

7. Graph Theory
Theorem 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 cycles B) G is connected and has n-1 edges C) G has n-1 edges and no cycles D) For u, vV(G), G has exactly one u, v-path
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
Ch Trees and Distance

8. A{B,C}. connected, acyclic  n-1 edges Theorem 2.1.4 Continue
Graph Theory
A{B,C}. connected, acyclic  n-1 edges Theorem Continue
We use induction on n.
For n=1, an acyclic 1-vertex graph has no edge
For n >1, we suppose that implication holds for graphs with fewer than n vertices
Given an acyclic connected graph G, Lemma provides a leaf v and states that G’=G-v also is acyclic and connected (see figure in previous proof)
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
Ch Trees and Distance

9. B{A, C} Connected and n-1 edges  acyclic Theorem 2.1.4 continue
Graph Theory
B{A, C} Connected and n-1 edges  acyclic Theorem continue
If G is not 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 ), 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 G’=G, and G is acyclic
Ch Trees and Distance

10. C{A, B} n-1 edges and no cycles  connected Theorem 2.1.4 continue
Graph Theory
C{A, B} n-1 edges and no cycles  connected Theorem continue
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)=I[n(Gi)-1]= n-k
We are given e(G)=n-1,so k =1, and G is connected
Ch Trees and Distance

11. Graph Theory
AD Connected and no cycles  For u, vV(G), one and only one u, v-path exists Theorem
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
Ch Trees and Distance

12. Graph Theory
DA For u, vV(G), one and only one u, v-path exists  connected and no cycles Theorem
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).
Ch Trees and Distance

13. Corollary: Every edge of a tree is a cut-edge 2.1.5
Graph Theory
Corollary: Every edge of a tree is a cut-edge
Proof:
A tree has no cycles
Theorem implies that every edge is a cut-edge.
Ch Trees and Distance

14. Corollary: Adding one edge to a tree forms exactly one cycle 2.1.5
Graph Theory
Corollary: Adding one edge to a tree forms exactly one cycle
Proof:
A tree has a unique path linking each pair of vertices (Theorem2.1.4D)
Joining two vertices by an edge creates exactly one cycle
Ch Trees and Distance

15. Corollary: Every connected graph contains a spanning tree 2.1.5
Graph Theory
Corollary: Every connected graph contains a spanning tree
Proof:
Similar to the proof of BA, C in Theorem
Iteratively deleting edges from cycles in a connected graph yields a connected acyclic subgraph
Ch Trees and Distance

16. Graph Theory
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-e+e’ is a spanning tree of G e’ e G T T’
T-e+e’
Ch Trees and Distance

17. Since T’ is connected, T’ has an edge e’ with endpoints in U and U’.
Graph Theory
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-e+e’ is a spanning tree of G
Proof:
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-e+e’ is connected, has n(G)-1 edges, and is a spanning tree of G. e’ e T T’
T-e+e’
Ch Trees and Distance

18. Distance in trees and Graphs
Graph Theory
Distance in trees and Graphs
Assume G has a u, v-path
Then the distance from u to v, written dG(u,v) or d(u,v),
is the least length of a u,v-path.
If G has no such path, then d(u,v)= 
Ch Trees and Distance

19. Distance in trees and Graphs
Graph Theory
Distance in trees and Graphs
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 is
(u) = maxvV(G) d(u,v)
Upper bound of the distance from u to the others
The radius of a graph G is rad G = minuV(G) (u)
Lower bound of the eccentricity
Ch Trees and Distance

20. Distance, Diameter, Eccentricity, and Radius
Graph Theory
Distance, Diameter, Eccentricity, and Radius a b f e c g d
Distance(f,c) : 2
Distance(g,c): 2
Distance(a,c): 3
Diameter: 3
Eccentricity(f):2
Eccentricity(a): 3
Radius: 2
Ch Trees and Distance

21. Theorem: If G is a simple graph, then diam G 3  diam  3 2.1.11
Graph Theory
Theorem: If G is a simple graph, then diam G 3  diam 
Proof: 1/3
Since diam G > 2, 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 G = 2
Not every pair is of this kind u v
A pair (u, v) of this kind exists
Ch Trees and Distance

22. Theorem: If G is a simple graph, then diam G  3  diam 3 2.1.11
Graph Theory
Theorem: If G is a simple graph, then diam G  3  diam 
Proof: 2/3
Hence every xV( ) - {u,v} has at least one of {u,v} as a nonneighbor
Either u or v is not connected to x
Equivalently, this makes x adjacent to at least one of {u,v} in
Everyone of these vertices has at least one of {u,v} as a nonneighbor u v
Ch Trees and Distance

23. Theorem: If G is a simple graph, then diam G  3  diam  3 2.1.11
Graph Theory
Theorem: If G is a simple graph, then diam G  3  diam 
Proof: 3/3
Since also u v  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
Ch Trees and Distance

24. Graph Theory
Center
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
Recall: The eccentricity of a vertex u is
the upper bound of the distance
from u to the others
Ch Trees and Distance

25. Theorem. 2.1.13 The center of a tree is a vertex or an edge 2.1.13
Graph Theory
Theorem The center of a tree is a vertex or an edge
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.
Ch Trees and Distance

26. Theorem. 2.1.13 Continue u Induction step: n(T)>2.
Graph Theory
Theorem Continue
Induction step: n(T)>2.
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).
The max path from u
The end must be a leaf u
T = Green  Pink
T’ = Green
Ch Trees and Distance

27. Theorem. 2.1.13 Continue T’(u)= T(u)-1 u
Graph Theory
Theorem Continue
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).
{v| v has min (v) in T} =
{v’| v’ has min (v’) in T’}
T’(u)= T(u)-1 u
Ch Trees and Distance

28. Theorem. 2.1.13 Continue T T’ T”
Graph Theory
Theorem Continue
It is shown T and T’ have the same center.
By the induction hypothesis, the center of T’ is a vertex or an edge. T T’ T”
Ch Trees and Distance

29. Spanning Trees and Enumeration 2.2
Graph Theory
Spanning Trees and Enumeration 2.2
There are 2c(n,2) simple graphs with vertex set [n]={1,…,n}.
since each pair may or may not form an edge.
How many of these are trees?
Max number of edges = c(n,2)
Can either appear or disappear
Ch Trees and Distance

30. Graph Theory
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.
Ch Trees and Distance

31. Algorithm for tree generation by using Prűfer code 2.2.1
Graph Theory
Algorithm for tree generation by using Prűfer code 2.2.1
Algorithm. (Prűfer code) Production of f(T)=(a1,…,an-2)
Input: A tree T with vertex set S  N
Iteration: At the i th 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 2 7 1 4 3 6 8 5
Ch Trees and Distance

32. Graph Theory
Theorem: For a set SN of size n, there are nn-2 trees with vertex set S
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
Ch Trees and Distance

33. Spanning Trees in Graphs 2.2.6
Graph Theory
Spanning Trees in Graphs
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.
Ch Trees and Distance

34. Graph Theory
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 G·e has one less edge than G u e G
G·e v
Ch Trees and Distance

35. Graph Theory
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)+ (G·e)
(G-e): The number of trees without e
(G·e): The number of trees with e
A spanning tree in G.e  A spanning tree having e in G G
G-e e
G·e =
Ch Trees and Distance

36. Need to show that G has (G·e) spanning trees containing e
Graph Theory
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)+ (G·e)
Proof: 1/2
The spanning trees of G that omit e are precisely the spanning trees of G-e
Need to show that G has (G·e) 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 G·e
Ch Trees and Distance

37. Graph Theory
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)+ (G·e)
Proof:2/2
When contract e in a spanning tree having e, obtain a spanning tree of G·e because
The resulting subgraph of G·e is spanning and connected
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 G·e by this operation.
Ch Trees and Distance

38. A Matrix Tree computation. 2.2.12
Graph Theory
A Matrix Tree computation
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 i=j
Let Q* is a matrix obtained by deleting row s and column t of Q, then
(G) = (-1)s+t detQ*
Ch Trees and Distance

39. A Matrix Tree computation. 2.2.11
Graph Theory
A Matrix Tree computation
Theorem instructs us
Form a matrix by putting the vertex degrees on the diagonal and subtracting the adjacency matrix
Delete a row and a column and take the eterminant
Ch Trees and Distance

40. A Matrix Tree computation. 2.2.11
Graph Theory
A Matrix Tree computation
Consider the kite of Example 2.2.9,
The vertex degrees are 3,3,2,2
We form the matrix on the left below and then
We take the determinant of the matrix in the middle
The result is the number of spanning trees
Ch Trees and Distance

41. Example 2.2.11 v3 v1 v4 v2 v1 v2 v3 v4 v1 v2 v3 v4 8 Graph Theory
Ch Trees and Distance

42. Graph Theory
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.
Kruskal’s Algorithm
Prim’s Algorithm
Ch Trees and Distance

43. Kruskal’s Algorithm for Minimum Spanning Tree 2.3.1
Graph Theory
Kruskal’s 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.
Ch Trees and Distance

44. Kruskal’s Algorithm for Minimum Spanning Tree 2.3.1
Graph Theory
Kruskal’s 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
Ch Trees and Distance

45. Example of using Kruskal’s Algorithm
Graph Theory
Example of using Kruskal’s Algorithm 9 9 9 8 2 2 2 4 1 8 4 1 8 4 1 7 6 7 6 7 6 5 5 5 10 11 10 3 11 10 11 3 3 12 12 12 9 9 9 8 2 1 4 2 1 8 2 4 1 8 4 7 7 6 7 6 6 5 10 10 5 5 11 3 11 3 11 3 10 12 12 12
Ch Trees and Distance

46. Graph Theory
Theorem: In a connected weighted graph G, Kruskal’s Algorithm constructs a minimum-weight spanning tree
Proof: 1/3
We show first that the algorithm produces a tree
We never choose 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
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.
Ch Trees and Distance

47. Theorem 2.3.3 Continue T: …………. e, ………… T*: …………. e*,…………
Graph Theory
Theorem Continue
Proof: continue
Let T be the resulting tree, and let T* be a spannig tree of minimum weight.
If T=T*, 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 T*+e-e’
The first edge chosen for T that is not in T*
T: …………. e, …………
T*: …………. e*,…………
Identical
Ch Trees and Distance

48. Theorem 2.3.3 Continue T: …………. e, ………… T*: …………. e*,…………
Graph Theory
Theorem Continue
Proof: continue
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 T*+e-e’ is a spanning tree with weight at most T* that agrees with T for a longer initial list of edges than T* does.
T: …………. e, …………
T*: …………. e*,…………
Identical
T: …………… e, …………
T*+e-e’: …………… e, …………
Identical
Ch Trees and Distance

49. Theorem 2.3.3 Continue Proof: continue
Graph Theory
Theorem Continue
Proof: continue
Repeating this argument eventually yields a minimum-weight spanning tree that agrees completely with T.
Phrased extremely, we have proved that the minimum spanning tree agreeing with T the longest is T itself.
Ch Trees and Distance

50. Graph Theory
Shortest Paths
How can we find the shortest route from one location to another?
Ch Trees and Distance

51. Graph Theory
Dijkstra’s Algorithm2.3.5
Input: A graph (or digraph) with nonnegative edge weights and a starting vertex u.
The weight of edge xy is w(xy)
Let w(xy)= if xy is not an edge a d 5 1 2 u 4 4 e 3 6 c b 5
Ch Trees and Distance

52. Dijkstra’s Algorithm2.3.5 Idea:
Graph Theory
Dijkstra’s Algorithm2.3.5
Idea:
Maintain the set S of vertices to which a shortest path from u is known
Enlarge 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.
The one nearest to u S S v u u
Ch Trees and Distance

53. Graph Theory
Dijkstra’s Algorithm2.3.5
Initialization: Set S={u}; t(u)=0; t(z)=w(uz) for zu
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 min{t(z), t(v)+w(vz)}
The iteration continues until S=V(G) or until t(z)= for every zS. At the end, set d(u, v)=t(v) for all v.
Ch Trees and Distance

54. Tentative distance v.s. Shortest distance 2.3.5
Graph Theory
Tentative distance v.s. Shortest distance 2.3.5
The tentative distance of a vertex may not be the shortest distance, for example:
t(d ) = 6 and not the minimum
Actually d(d ) = 5
t=6 d d
t=5
d=1 5
d=1 5 a a 1 1 2 2 u 2 u 2 e e
d=0
d=0 6 6 3 3
t=9 b
t=3 b
d=3
Ch Trees and Distance

55. Example: Find Shortest paths by using Dijkstra’s Algorithm 2.3.6
Graph Theory
Example: Find Shortest paths by using Dijkstra’s Algorithm 2.3.6
t=1 a d u a d e c b 1 5 2 6 4 3 5 1 2 u 4 e
d=0 4 3 6 c b 5
t=3
Ch Trees and Distance

56. Graph Theory
Example: Find Shortest paths by using Dijkstra’s Algorithm continue 2.3.6 u a d e c b 1 5 2 6 4 3
d=0
d=1
t=3
t=6
t=5
d=1 a 4
t=5 c 1 5 2 u e
d=0 5 3 6 4 d
t=6
d=3 b
Ch Trees and Distance

57. Graph Theory
Example: Find Shortest paths by using Dijkstra’s Algorithm continue 2.3.6
d=5
d=5 c c
d=1
d=1 4 6 4 a
t=11 a 6 5 e 5 e 1 1 5
d=0 u
d=0 u
t=8 5 2 2 3 3 4 4 d d
t=6
d=3 b
d=3 b
d=6
Ch Trees and Distance

58. Graph Theory
Theorem: Given a (di)graph G and a vertex uV(G), Dijkstra’s 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.
Ch Trees and Distance

59. Theorem 2.3.7 Continue We use induction on k=|S|. Basis step: k=1
Graph Theory
Theorem Continue
We use induction on k=|S|.
Basis step: k=1
From the initialization,
S={u}, d(u, u)=t(u)=0,
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
Ch Trees and Distance

60. Graph Theory
Theorem Continue
Induction step: Suppose that when |S|=k, (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 S’=S{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’.
Ch Trees and Distance

61. Graph Theory
Theorem 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 k+1; this completes the induction step.
Ch Trees and Distance

62. Prim’s Algorithm for Minimum Spanning Tree
Graph Theory
Prim’s Algorithm for Minimum Spanning Tree
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 }.
Ch Trees and Distance

63. Prim’s Algorithm Initialization: Set T={u}. Iteration:
Graph Theory
Prim’s Algorithm
Initialization: Set T={u}.
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).
Ch Trees and Distance

64. Dijkstra’s Algorithm v.s. Prim’s Algorithm
Graph Theory
Dijkstra’s Algorithm v.s. Prim’s Algorithm d
t=6
d=1 6 a
t=5 e c 3 u
d=0 b 8
t=3
The end vertex of the cheapest edge
The vertex which is the nearest to u
Ch Trees and Distance

65. Algorithm - Breadth First Search 2.3.8
Graph Theory
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
Ch Trees and Distance

66. Algorithm - Breadth First Search 2.3.8
Graph Theory
Algorithm - Breadth First Search 2.3.8
Initialization: R={u}, 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
Then v is removed from the front of R and placed in S
Ch Trees and Distance

67. Example of BFS a b d c e f g h i j
Graph Theory
Example of BFS a b d c e f g h i j
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, …
Ch Trees and Distance

68. Breadth-F-S vs. Depth-F-S
Graph Theory
Breadth-F-S vs. Depth-F-S a a b b d d c c e e f g h i j f g h i j
a,b,c,d,e,f,g,h,i,j, …
a,b,e,f, …c, h,…
Ch Trees and Distance

69. Application: Chinese Postman Problem 2.3.9
Graph Theory
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.
Ch Trees and Distance

70. Application: Chinese Postman Problem continue 2.3.9
Graph Theory
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 .
Ch Trees and Distance

71. Application: Chinese Postman Problem continue 2.3.9
Graph Theory
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
Ch Trees and Distance

72. Application: Chinese Postman Problem continue 2.3.9
Graph Theory
Application: Chinese Postman Problem continue 2.3.9
If there are only two odd vertices, then
We can use Dijkstra’s Algorithm to find the shortest path between them and solve the problem
If there are 2k odd vertices, then
Use Dijkstra’s 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.
Ch Trees and Distance

73. Application: Chinese Postman Problem continue 2.3.9
Graph Theory
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 10 40 16 15 20
Ch Trees and Distance

74. Graph Theory
Rooted Tree
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-to–right ordering specified for the children of each vertex.
Ch Trees and Distance

75. Rooted Tree 2.3.11 root v Parent of v, Ancester of v A leaf,
Graph Theory
Rooted Tree
root
Parent of v,
Ancester of v v
A leaf,
A child of v
Ch Trees and Distance

76. Graph Theory
Binary Tree
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
Ch Trees and Distance

77. Graph Theory
Huffman’s Algorithm
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
Ch Trees and Distance

78. Graph Theory
Huffman’s Algorithm
Initial case: When n=2, 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; n=1 can also be used as the initial case)
Ch Trees and Distance

79. Huffman’s Algorithm 2.3.13 Recursion:
Graph Theory
Huffman’s Algorithm
Recursion:
Replace the two least likely items p, p’ with a single item q of weight p+p’ when n>2
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.
Ch Trees and Distance

80. Example of Huffman coding 2.3.14
Graph Theory
Example of Huffman coding
Consider eight items with frequencies 5, 1, 1, 7, 8, 2, 3, 6
Algorithm 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 2 5 1 7 8 3 6
Ch Trees and Distance

81. Graph Theory
Huffman coding
Now this and the original item of weight 2 are the least likely and combine to form an item of weight 4. 4 2 5 1 7 8 3 6
Ch Trees and Distance

82. Graph Theory
Huffman coding 4 2 5 1 7 8 3 6 2 5 1 7 8 3 6 7 7 11 4 4 2 2 5 1 1 7 8 2 3 6 5 1 1 7 8 2 3 6
Ch Trees and Distance

83. Graph Theory
Huffman coding 33 1 1 1 19 14 1
8:11
7:01 7 1 1
3:001
6:101
5:100 11 4 1
2:0001 2
1:00000
1:00001 5 1 1 7 8 2 3 6
Ch Trees and Distance

84. Graph Theory
Theorem
Given a probability distribution {pi} on n items, Huffman’s Algorithm produces the prefix-free code with minimum excepted length
Ch Trees and Distance