Minimum Spanning Tree
1.
2.
3.
Properties of MST
Second-best MST
Bottleneck Spanning Tree
Property 1 of MST
Let (u, v) be a minimum-weight edge in a graph G = (V, E).
Then (u, v) belongs to some MST of G.
Proof: Let A be the empty set of edges. Then
(1) A is a subset of some MST, and A does not include (u, v).
(2) The cut (u, V\u) respects A.
(3) Edge (u, v) is a light edge crossing cut (u, V\u).
By theorem 23.1, (u, v) is a safe edge for A. Hence, (u, v)
belongs to some MST.
Another proof: Suppose T is an arbitrary MST, and (u, v) is
not in T. Then T  {(u, v)} contains a cycle. Let f be any
edge other than (u, v) on the cycle, then T  {(u, v)} – {f} is
a MST containing (u, v).
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2
Property 2 of MST
(1) Let T be a MST of a graph G, and
(2) let T’ be an arbitrary spanning tree of G,
(3) suppose the edges of each tree are sorted in non-decreasing
weight order, that is,
w(e1)  w(e2)  …  w(en - 1) and
w(e1’)  w(e2’)  …  w(en - 1’).
Then for 1  i  n -1, w(ei)  w(ei’).
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3
Proof of Property 2
Proof. Prove by contradiction.
Suppose w.l.o.g. that there exists an index i1 such that w(e1) 
w(e1’), w(e2)  w(e2’), …, w(ei-1)  w(ei-1’), but w(ei) > w(ei’).
This indicates that
w(en-1)  w(en-2)  …  w(ei) > w(ei’)  w(ei-1’)  … 
w(e1’).
So, if we add any ex’{ e1’, e2’, …, ei’} onto T, then either ex’
is an edge of {e1, e2,…, ei-1}, or a cycle is formed which
includes only edges in {ex’, e1, e2, …, ei-1}.
Thus each ex’{ e1’, e2’, …, ei’} is included in a connected
component of {e1, e2, …, ei-1}.
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4
Proof of Property 2
Thus the graph (V, {e1’, e2’, …, ei’, e1, e2, ..., ei-1}) has the same
number of connected components as graph (V, {e1, e2, ..., ei-1})
have, i.e., n-i+1.
But the graph (V, {e1’, e2’, …, ei’}) has n-i connected
components, since e1’, e2’, …, ei’ are tree edges from T’. A
contradiction.
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5
Gi-1 = (V, {e1, …, ei-1}) has n-i+1 (=3) connected components.
By assumption, {e1’, …, ei’} are all included in the connected components of Gi-1.
So, (V, {e1’, …, ei’, e1, …, ei-1}) has also n-i+1 (=3) connected components.
Impossible, since Gi-1’ = (V, {e1’, …, ei’}) has n-i (=2) connected components.
edges in {e1, …, ei-1}
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edges in {e1’, …, ei’}
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6
Property 3 of MST
Let T be a MST of a graph G, and
let L be the sorted list of the edge weights of T.
Then for any other MST T’ of G, the list L is also the sorted list of
edge weights of T’.
Proof. By Property 2.
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7
MST Property (Prob. 23-1(a))
Property 4. Let G = (V, E) be an undirected, connected graph
with weight function w: E  R+, and suppose that
(1) | E |  | V | and
(2) all edge weights are distinct.
Then the MST of G is unique.
Proof. By Property 3.
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8
Second-best MST
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9
Definition of Second-best MST
Let G = (V, E) be an undirected, connected graph with weight
function w: E  R+, and suppose that
(1) | E |  | V | and
(2) all edge weights are distinct.
A second-best MST is a spanning tree T such that
w(T) = minT’’-{T’}{w(T’’)},
where  is the set of all spanning trees of G, and T’ is a MST of
G.
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10
Examples of Second-best MST
The second-best MST need not be unique.
2
1
3
2
4
1
4
MST
5
2
1
3
4
1
2BMST
2BMST
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5
11
Property of Second-best MST
Property 5. A second-best MST can be obtained from the MST
by replacing a single edge from the tree with another edge not in
the tree.
Proof. By contradiction. Suppose T1 is the MST and T2 is a
second-best MST, and they differ by at least two edges, we will
get another spanning tree T3 such that (1) T3  T1 and (2) w(T3)
< w(T2), thus contradicts that T2 is second-best.
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12
Proof (Cont.)
Suppose e1 is the minimum edge in T1\ T2.
Then T2  {e1} contains a cycle C on which there must be some
edge e  T2 such that e  T1, since otherwise C would be
entirely in T1 (and this is impossible since C is a cycle and T1 is
a tree).
Let T3 = T2  {e1} \ {e}. Then T3 is a spanning tree.
Since T3 differs from T2 by only one edge, while T1 differs from T2 by at least
two edges, T3 is not equal to T1. (1) is satisfied.
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13
Proof (Cont.)
We claim that w(e) > w(e1). Then we have (2) w(T3) < w(T2)
since T3 = T2 \ {e}  {e1}.
Prove the claim by contradiction. Assume w(e) < w(e1).
If we add e to T1, we get a cycle C’, which contains some edge
e' in T1 but e' is not in T2. (otherwise, T2 would contain the
entire cycle C’).
Therefore, T4 = T1 \ {e'}  {e} is a spanning tree, and we must
have w(e') < w(e), since otherwise w(T4) would be less than
w(T1).
Thus, w(e') < w(e) < w(e1), which contradicts our choice of e1
as the edge in T1 \ T2 of minimum weight. 
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14
Compute the Second-best MST
Idea: use Property 5.
 (u, v)  T, try the spanning tree T’ = T \ {e}  (u, v), where e is
the edge with maximum weight in cycle C, which is formed by
adding (u, v) onto T.
 Pick the tree T’ such that w(T’) = w(T) – w(e) + w(u, v) is
minimized among all possible trees.
Let T be a spanning tree of G and for any two vertices u, v V,
let max[u, v] be the maximum weight among all edges on the
unique path between u and v in T.
Thus we need to compute max[u, v]. How?
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15
Compute the Second-best MST
Second-Best-MST(G)
1 Compute a MST T for G.
2 Compute max[u, v] for each pair (u, v) in T.
3 For each edge (u, v) not in T, compute the difference w(u, v) –
max[u, v].
4 Find the smallest positive difference and replace the
max_edge[u, v] with corresponding edge (u, v).
Note: max_edge[u, v] means the edge with maximum weight in
the path connecting u and v in T. It can be easily computed.
Compute-max(T): O(V2)
Second-Best-MST(G): O(ElogV + V2)
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16
Compute the Second-best MST
Compute-max(T)
1 for each pair of vertices
u, v V do
2
max[u, v]  0
3 endfor
4 for each vertex s V do
5
BFS (T, s)
//E(T)=O(V)
6 endfor
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17
Compute the Second-best MST
BFS(T, s)
1 for each vertex u  V do
2
color[u]  WHITE
3 endfor
4 color[s] GRAY
5 Q 
6 Enqueue(Q, s)
©2005, 2009, 2011, SDU
7 while Q   do
8 u  Dequeue(Q)
9 for each v  Adj[u] do
10
if color[v] = WHITE then
11
color[v] GRAY
12
max[s, v] 
max{w(u, v), max[s, u]}
13
Enqueue(Q, v)
14
endif
15 endfor
16 color[u]  BLACK
17 endwhile
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18
Bottleneck Spanning Tree
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19
Bottleneck Spanning Tree
A bottleneck spanning tree T of a connected, weighted and
undirected graph G is a spanning tree of G whose largest edge
weight is minimum over all spanning trees of G.
 Let T1, T2, …, Tm are all the spanning trees G, and the largest edge
of each tree is et1, et2, …, etm. If w(eti)  w(etj) for 1  j  m and j  i,
then Ti is a bottleneck spanning tree.
 The value of a BST T is the weight of the maximum-weight edge in
T.
 The bottleneck spanning tree may not be unique.
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20
BST vs MST
Property 6. Every minimum spanning tree is a bottleneck spanning tree.
Proof. Property 2 implies it.
Another proof. Let T be a MST and T’ be a BST, let the maximum-weight edge in T
and T’ be e and e’, respectively.
Suppose for the contrary that the MST T is not a BST, then we have w(e) > w(e’),
which also implies that f  T’, w(e) > w(f).
Removing e from T disconnects T into two subtrees T1 and T2. There must exist an
edge f in T’ connecting T1 and T2, otherwise, T’ is not connected.
T1T2 {f} forms a new tree T’’ with w(T’’) = w(T) - w(e) + w(f) < w(T), a
contradiction to the fact that T is MST, thus, A MST is also a BST.
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21
The BST Problem
The BST Problem:
 I: A weighted, connected and undirected graph G.
 Q: A BST T of G.
A solution:
 Kruskal’s and Prim’s Algorithm works for the BST problem.
More efficient algorithm exist for the problem?
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22
A Decision Problem
Decision version of the BST problem:
 I: A graph G = (V, E) with weight function w on E, and an integer b.
 Q: TRUE if the value of the BST(G) is at most b and FALSE
otherwise.
How to solve this problem in linear time?
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23
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
CHECK-BOTTLENECK(G, b)
for each vertex u  V[G] do
color[u] WHITE
endfor
u randomly chosen vertex in G
DFS(u, b)
//O(V+E)
for each vertex u V[G] do
if color[u] = WHITE then
return FALSE
endif
endfor
return TRUE
©2005, 2009, 2011, SDU
1.
2.
3.
4.
5.
6.
7.
DFS(u, b)
color[u] GRAY
for each v Adj[u] do
if color[v] = WHITE and
w(u, v)  b then
DFS(v, b)
endif
endfor
color[u] BLACK
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24
Solve the BST Problem
BOTTLENECK(G)
1 Sort the weights of edges of G in non-decreasing order:
e[1]  e[2]  …  e[ | E | ]
2 start  1, end  | E |
4 while start < end do
5 middle  (start + end)/2
6 if CHECK-BOTTLENECK(G, e[middle]) = FALSE then
7
start middle
8 else
9
end middle //in the end there is a BST with value e[end]
10 endif
11 endwhile
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25
Solve the BST Problem
12 u  randomly chosen a vertex of G
13 call DFS(u, e[start]) to build a spanning tree
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26
Solve the Problem in Linear Time
Time complexity:
Sorting:O(E log E)
Finding the value of BST: O((V+E) log E)=O(E log E)
Total:O(E log E)
Can the problem be solved in linear time?
Please think it carefully, anyone who finds the solution will be given an extra
bonus.
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27
A Linear Time Algorithm for BST
BST(G)
1 T  (V, )
2 while there exist 2 types of edge costs in G do
3 Find the median of all types of edge costs, say b
4 if G|{e: w(e)  b} is connected then
5
Remove all edges with cost > b from G
6 else
7
Contract-Merge(G, b, T)
8 endif
9 endwhile
10 Contract-Merge(G, any edge cost in G, T)
11 return T
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28
A Linear Time Algorithm for BST
Contract-Merge(G, b, T)
1 while there exists an edge e with w(e)  b in G do
2 T  T  {e}
3 Contract e. Denote by e1, e2, ..., ek the edges merged
because of the contraction.
4 Keep only the edge with minimum cost in e1, e2, ..., ek,
and remove all the other edges.
5 endwhile
6 return
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29
An Example
b
4
a
c
8
4
i
7
8
h
1 2 4 6 7 8 9 10 11 14
9
2
11
d
7
14
6
1
10
g
2
f
•There is no BST with value 
e 7.
•Thus all edges with cost  7
are contracted.
•We begin from (a, b).
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
30
An Example
a
c
8
4
i
7
h
1 2 4 6 7 8 9 10 11 14
9
2
8
d
7
14
6
1
•Next contract (i, c).
e
10
g
2
f
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
31
An Example
a
c
8
d
7
1 2 4 6 7 8 9 10 11 14
9
4
7
8
•Next contract (h, c).
14
6
e
10
h
1
g
2
f
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
32
An Example
a
c
8
d
7
1 2 4 6 7 8 9 10 11 14
9
4
•Next contract (c, g).
14
1
e
10
g
2
f
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
33
An Example
a
8
c
d
7
1 2 4 6 7 8 9 10 11 14
9
2
•Next contract (c, f).
14
e
10
f
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
34
An Example
a
8
c
d
7
1 2 4 6 7 8 9 10 11 14
9
10
•Next contract (c, d).
e
f
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
35
An Example
a
8
c
8 9
•There is no BST with value 
e 8.
•We contract (a, c).
9
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
36
An Example
a
9
9
•There is only one type of edge
e costs.
•We contract (a, e).
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
37
An Example
a
•This is our BST.
b
4
a
7
4
i
11
7
h
d
9
2
8
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c
8
14
6
1
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e
10
g
2
f
38
Exercises
Page 43-44




23.1-2
23.1-3
23.1-4
23.1-5
Page 50-51
 23.2-1
 23.2-4
 23.2-8
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39
Steiner Tree and Steiner Forest
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40
The Steiner Tree Problem
Instance: An undirected graph G = (V, E) with nonnegative costs
{c(e)} defined on edges, a required vertex set D  V.
Query: a minimum cost subgraph T of G such that T spans D.
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41
Approximating Steiner Tree
The Steiner Tree problem is NP-hard.
A MST-based algorithm achieves approximation ratio 2.
A 1.55-approximation algorithm is known for a long time.
The current best known approximation ratio for Steiner is 1.39,
due to
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42
The Steiner Forest Problem
Instance: An undirected graph G = (V, E) with nonnegative costs
defined on edges, a source-sink terminal pair set D = {(s1, t1), (s2,
t2), ..., (sk, tk)}.
Query: A minimum cost subgraph F of G such that every pair (s,
t)  D is connected in F.
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43
Approximating Steiner Forest
A factor 2 approximation algorithm was due to
The algorithm in [AKR95] was simplified and generalized by
The factor 2 is currently the best for Steiner Forest.
R. Ravi
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Michel Goemans
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44
k-MST
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45
The k-MST Problem
Instance: An undirected graph G = (V, E), a positive integer k.
Query: find a MST spanning at least k vertices of G.
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46
Approximating k-MST
The k-MST problem is NP-hard.
Factor of 3k1/2:
O(log2 k):
O(log k):
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47
Approximating k-MST
35:
3:
2 + :
2:
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48
Two Men
Naveen Garg
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Sanjeev Arora
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49
Thanks for attention!
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50