Unit 5 session 2 MinimumSpanningTrees.ppt

Minimum Spanning Tree
Minimum Spanning Trees (MST)

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Minimum Spanning Trees
• Spanning Tree
– A tree (i.e., connected, acyclic graph) which contains
all the vertices of the graph
• Minimum Spanning Tree
– Spanning tree with the minimum sum of weights
• Spanning forest
– If a graph is not connected, then there is a spanning
tree for each connected component of the graph
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Applications of MST
– Find the least expensive way to connect a set of
cities, terminals, computers, etc.

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Example
Problem
• A town has a set of houses
and a set of roads
• A road connects 2 and only
2 houses
• A road connecting houses u and v has a repair
cost w(u, v)
Goal: Repair enough (and no more) roads such
that:
1. Everyone stays connected
i.e., can reach every house from all other houses
2. Total repair cost is minimum
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Minimum Spanning Trees
• A connected, undirected graph:
– Vertices = houses, Edges = roads
• A weight w(u, v) on each edge (u, v) in E
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Find T  E such that:
1. T connects all vertices
2. w(T) = Σ(u,v)T w(u, v) is
minimized

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Properties of Minimum Spanning Trees
• Minimum spanning tree is not unique
• MST has no cycles – see why:
– We can take out an edge of a cycle, and still have
the vertices connected while reducing the cost
• # of edges in a MST:
– |V| - 1

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Growing a MST – Generic Approach
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• Grow a set A of edges (initially
empty)
• Incrementally add edges to A
such that they would belong
to a MST
– An edge (u, v) is safe for A if and
only if A  {(u, v)} is also a subset
of some MST
Idea: add only “safe” edges

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Generic MST algorithm
1. A ← 
2. while A is not a spanning tree
3. do find an edge (u, v) that is safe for A
4. A ← A  {(u, v)}
5. return A
• How do we find safe edges?
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S
V - S
Finding Safe Edges
• Let’s look at edge (h, g)
– Is it safe for A initially?
• Later on:
– Let S  V be any set of vertices that includes h but not
g (so that g is in V - S)
– In any MST, there has to be one edge (at least) that
connects S with V - S
– Why not choose the edge with minimum weight (h,g)?
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Definitions
• A cut (S, V - S)
is a partition of vertices
into disjoint sets S and V - S
• An edge crosses the cut
(S, V - S) if one endpoint is in S
and the other in V – S
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S
V- S 
S
 V- S

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Definitions (cont’d)
• A cut respects a set A
of edges  no edge
in A crosses the cut
• An edge is a light edge
crossing a cut  its weight is minimum over all
edges crossing the cut
– Note that for a given cut, there can be > 1 light
edges crossing it
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S
V- S 
S
 V- S

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Theorem
• Let A be a subset of some MST (i.e., T), (S, V - S) be a
cut that respects A, and (u, v) be a light edge crossing
(S, V-S). Then (u, v) is safe for A .
Proof:
• Let T be an MST that includes A
– edges in A are shaded
• Case1: If T includes (u,v), then
it would be safe for A
• Case2: Suppose T does not include
the edge (u, v)
• Idea: construct another MST T’
that includes A  {(u, v)}
u
v
S
V - S

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u
v
S
V - S
Theorem - Proof
• T contains a unique path p between u and v
• Path p must cross the
cut (S, V - S) at least
once: let (x, y) be that edge
• Let’s remove (x,y)  breaks
T into two components.
• Adding (u, v) reconnects the components
T’ = T - {(x, y)}  {(u, v)}
x
y
p

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Theorem – Proof (cont.)
T’ = T - {(x, y)}  {(u, v)}
Have to show that T’ is an MST:
• (u, v) is a light edge
 w(u, v) ≤ w(x, y)
• w(T ’) = w(T) - w(x, y) + w(u, v)
≤ w(T)
• Since T is a spanning tree
w(T) ≤ w(T ’)  T’ must be an MST as well
u
v
S
V - S
x
y
p

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Theorem – Proof (cont.)
Need to show that (u, v) is safe for A:
i.e., (u, v) can be a part of an MST
• A  T and (x, y)  T 
(x, y)  A  A T’
• A  {(u, v)}  T’
• Since T’ is an MST
 (u, v) is safe for A
u
v
S
V - S
x
y
p

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Prim’s Algorithm
• The edges in set A always form a single tree
• Starts from an arbitrary “root”: VA = {a}
• At each step:
– Find a light edge crossing (VA, V - VA)
– Add this edge to A
– Repeat until the tree spans all vertices
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How to Find Light Edges Quickly?
Use a priority queue Q:
• Contains vertices not yet
included in the tree, i.e., (V – VA)
– VA = {a}, Q = {b, c, d, e, f, g, h, i}
• We associate a key with each vertex v:
key[v] = minimum weight of any edge (u, v)
connecting v to VA
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w1
w2
Key[a]=min(w1,w2)
a

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How to Find Light Edges Quickly?
(cont.)
• After adding a new node to VA we update the weights of all
the nodes adjacent to it
e.g., after adding a to the tree, k[b]=4 and k[h]=8
• Key of v is  if v is not adjacent to any vertices in VA
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Example
0        
Q = {a, b, c, d, e, f, g, h, i}
VA = 
Extract-MIN(Q)  a
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key [b] = 4  [b] = a
key [h] = 8  [h] = a
4      8 
Q = {b, c, d, e, f, g, h, i} VA = {a}
Extract-MIN(Q)  b
  
 
  
 
 
 

4

8

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4 

8  
8

Example
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key [c] = 8  [c] = b
key [h] = 8  [h] = a - unchanged
8     8 
Q = {c, d, e, f, g, h, i} VA = {a, b}
Extract-MIN(Q)  c
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key [d] = 7  [d] = c
key [f] = 4  [f] = c
key [i] = 2  [i] = c
7  4  8 2
Q = {d, e, f, g, h, i} VA = {a, b, c}
Extract-MIN(Q)  i


4 

8  
8
7
4
2

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Example
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key [h] = 7  [h] = i
key [g] = 6  [g] = i
7  4 6 8
Q = {d, e, f, g, h} VA = {a, b, c, i}
Extract-MIN(Q)  f
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key [g] = 2  [g] = f
key [d] = 7  [d] = c unchanged
key [e] = 10  [e] = f
7 10 2 8
Q = {d, e, g, h} VA = {a, b, c, i, f}
Extract-MIN(Q)  g
4 7

8  4
8
2
7 6
4 7

7 6 4
8
2
2
10

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Example
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key [h] = 1  [h] = g
7 10 1
Q = {d, e, h} VA = {a, b, c, i, f, g}
Extract-MIN(Q)  h
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7 10
Q = {d, e} VA = {a, b, c, i, f, g, h}
Extract-MIN(Q)  d
4 7
10
7 2 4
8
2
1
4 7
10
1 2 4
8
2

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Example
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key [e] = 9  [e] = f
9
Q = {e} VA = {a, b, c, i, f, g, h, d}
Extract-MIN(Q)  e
Q =  VA = {a, b, c, i, f, g, h, d, e}
4 7
10
1 2 4
8
2 9

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PRIM(V, E, w, r)
1. Q ← 
2. for each u  V
3. do key[u] ← ∞
4. π[u] ← NIL
5. INSERT(Q, u)
6. DECREASE-KEY(Q, r, 0) ► key[r] ← 0
7. while Q  
8. do u ← EXTRACT-MIN(Q)
9. for each v  Adj[u]
10. do if v  Q and w(u, v) < key[v]
11. then π[v] ← u
12. DECREASE-KEY(Q, v, w(u, v))
O(V) if Q is implemented
as a min-heap
Executed |V| times
Takes O(lgV)
Min-heap
operations:
O(VlgV)
Executed O(E) times total
Constant
Takes O(lgV)
O(ElgV)
Total time: O(VlgV + ElgV) = O(ElgV)
O(lgV)

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Using Fibonacci Heaps
• Depending on the heap implementation, running time
could be improved!

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Prim’s Algorithm
• Prim’s algorithm is a “greedy” algorithm
– Greedy algorithms find solutions based on a sequence
of choices which are “locally” optimal at each step.
• Nevertheless, Prim’s greedy strategy produces a
globally optimum solution!
– See proof for generic approach (i.e., slides 12-15)

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Problem 4
• (Exercise 23.2-2, page 573) Analyze Prim’s
algorithm assuming:
(a) an adjacency-list representation of G
O(ElgV)
(b) an adjacency-matrix representation of G
O(ElgV+V2)

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PRIM(V, E, w, r)
1. Q ← 
2. for each u  V
3. do key[u] ← ∞
4. π[u] ← NIL
5. INSERT(Q, u)
7. while Q  
9. for each v  Adj[u]
10. do if v  Q and w(u, v) < key[v]
as a min-heap
Executed |V| times
Takes O(lgV)
Min-heap
operations:
O(VlgV)
Executed O(E) times
Constant
Takes O(lgV)
O(ElgV)
Total time: O(VlgV + ElgV) = O(ElgV)
O(lgV)

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PRIM(V, E, w, r)
1. Q ← 
2. for each u  V
3. do key[u] ← ∞
4. π[u] ← NIL
5. INSERT(Q, u)
7. while Q  
9. for (j=0; j<|V|; j++)
10. if (A[u][j]=1)
11. if v  Q and w(u, v) < key[v]
as a min-heap
Executed |V| times
Takes O(lgV)
Min-heap
operations:
O(VlgV)
Executed O(V2) times total
Constant
Takes O(lgV) O(ElgV)
Total time: O(VlgV + ElgV+V2) = O(ElgV+V2)
O(lgV)

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Problem 5
• Find an algorithm for the “maximum” spanning
tree. That is, given an undirected weighted
graph G, find a spanning tree of G of maximum
cost. Prove the correctness of your algorithm.
– Consider choosing the “heaviest” edge (i.e., the edge
associated with the largest weight) in a cut. The
generic proof can be modified easily to show that this
approach will work.
– Alternatively, multiply the weights by -1 and apply
either Prim’s or Kruskal’s algorithms without any
modification at all!

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Problem 6
• (Exercise 23.1-8, page 567) Let T be a MST of
a graph G, and let L be the sorted list of the
edge weights of T. Show that for any other MST
T’ of G, the list L is also the sorted list of the
edge weights of T’
T, L={1,2} T’, L={1,2}

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