CS5234
Combinatorial and Graph Algorithms
Welcome!
Administrative
 Problem Sets
 Programming Assignment
 Exams
CS5234 Overview
 Problem set 1 due
PS1:
Hand it in now!
 Problem set 2 released tonight
PS2:
Submit next week
 IVLE (optional, electronic submission)
Soon: Once registration completes…
CS5234 Overview
 Problem set grading
Distributed grading scheme:
Once registration completes, there will be an
IVLE survey where you choose a week.
Grading supervised (and verified) by the TA.
CS5234 Overview
 How to draw pictures?
By hand:
Either submit hardcopy, or scan, or take a
picture with your phone!
Or use a tablet / iPad…
Digitally:
1.
2.
3.
4.
xfig (ugh)
OmniGraffle (mac)
Powerpoint (hmmm)
???
CS5234 Overview
 Programming assignment(s):
Berth assignment problem:
Last 3-4 weeks of the semester.
Very hard algorithmic problem (related to
graph partitioning).
Design the best solution you can!
Must know or learn: C++
More details later…
CS5234 Overview
 Mid-term exam
October 8
In class
 Final exam
November 19
Reading Week
Exams will be graded and returned.
Quick Review
 Vertex Cover
Vertex Cover
 Definition:
Given: graph G = (V,E)
Find: smallest cover C ⊆ V such that every
edge e ∈ E is covered by some node v ∈ V.
 Challenge:
NP-complete: No polynomial time algorithm
unless P = NP.
Vertex Cover
Example: (Suboptimal) cover (size 9)
Vertex Cover
Example: Optimal cover? (size 6)
Vertex Cover
 Greedy approximation algorithm:
repeat: until every edge is covered
1. Let e = (u,v) be an uncovered edge.
2. Add u and v to the cover.
 Details:
Graph representation: Sorted adjacency list…
Vertex Cover
Analysis:
1. For every matching M: OPT ≥ |M|.
2. The cover C contains a matching of size |C|/2.
3. Therefore: OPT ≥ |C|/2.
Conclusion: |C| ≤ 2OPT
 2-approximation algorithm
Today’s Plan
 Vertex Cover (Review)
 Set Cover (Review + Analysis)
 Steiner Trees
Set Cover
 Definition:
Given: set X
subsets S1, S2, …, Sm where Sj ⊆ X
Find: smallest collection I ⊆ {1, …, m}
such that:
 Challenge:
NP-complete: No polynomial time algorithm
unless P = NP.
Set Cover
 Example:
X = {C, C++, Java, Ruby, Python}
Alice: C, C++
Bob:
C++, Java
Collin: C++, Ruby, Python
Dave: C, Java
 Choose a good team:
Collin and Dave: optimal solution.
Set Cover
set that contains the most
uncovered elements
Set Cover
Example:
Set
# uncovered
S1
4
S2
6
S3
3
S4
5
S5
4
1. Choose S2.
Set Cover
Example:
Set
# uncovered
S1
2
S2
0
S3
3
S4
3
S5
2
1. Choose S2.
2. Choose S3.
Set Cover
Example:
Set
# uncovered
S1
1
S2
0
S3
0
S4
2
S5
1
1. Choose S2.
2. Choose S3.
3. Choose S4.
Set Cover
Example:
Set
# uncovered
S1
1
S2
0
S3
0
S4
0
S5
0
1.
2.
3.
4.
Choose S2.
Choose S3.
Choose S4.
Choose S1
Set Cover
Example:
Set
# uncovered
S1
0
S2
0
S3
0
S4
0
S5
0
Output:
S1, S2, S3, S4
OPT:
S1, S4, S5
Set Cover
To get a good approximation, show a lower
bound on OPT.
– “OPT has to be AT LEAST this large.”
– E.g., “OPT ≥ |M|”
Set Cover
Example:
Set
# uncovered
S1
4
S2
6
S3
3
S4
5
S5
4
Note:
12 elements
≤ 6 elements per set

OPT ≥ 2
Set Cover
Example:
Set
# uncovered
S1
2
S2
0
S3
3
S4
3
S5
2
Note:
6 uncovered elements
≤ 3 elements per set

OPT ≥ 2
Set Cover
Example:
Set
# uncovered
S1
2
S2
0
S3
3
S4
3
S5
2
General rule:
k uncovered elements
≤ t elements per set

OPT ≥ (k / t)
Notation
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
xj = the jth item covered
Notation
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
S4 covers 1 new item
S3 covers 3 items
S1 covers 2 new items
S6 covers 2 new items
Notation
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
3
2
2
1
Notation
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
3
2
2
1
c1, c2, c3, c4, c5, c6, c7, c8, …, cn
3, 3, 3, 2, 2, 2, 2, 1, …,
cj = number of items covered in same step as xj
Notation
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
2
3
–
–
–
–
–
–
cost(x1) =
cost(x2) =
cost(x3) =
cost(x4) =
cost(x5) =
…
2
1
1/3
1/3
1/3
1/2
1/2
cost(xj) = 1/cj = amount payed to cover xj
Notation
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
3
2
2
1
cost(xj) = 1/cj = amount payed to cover xj
Set Cover
Example:
Set
# uncovered
S1
2
S2
0
S3
3
S4
3
S5
2
Note:
6 uncovered elements
≤ 3 elements per set

OPT ≥ 2
Analysis
Assume (x1, x2, …, xj-1) are covered:
– How does OPT cover (xj, xj+1, …, xn)?
Analysis
Assume (x1, x2, …, xj-1) are covered:
– How does OPT cover (xj, xj+1, …, xn)?
Fact:
No set covers more than c(xj) elements in
the set (xj, xj+1, …, xn).
Why? If it did, GREEDY would have chosen it!
(Tip: use the fact that algorithm is GREEDY.)
Analysis
Assume (x1, x2, …, xj-1) are covered:
– How does OPT cover (xj, xj+1, …, xn)?
As before:
n – j + 1 uncovered elements
≤ c(xj) elements per set

OPT ≥ (n – j + 1) / c(xj) = (n – j + 1)cost(xj)
Analysis
Assume (x1, x2, …, xj-1) are covered:
– How does OPT cover (xj, xj+1, …, xn)?
As before:
n – j + 1 uncovered elements
≤ c(xj) elements per set

OPT ≥ (n – j + 1) / c(xj) = (n – j + 1)cost(xj)

cost(xj) ≤ OPT / (n – j + 1)
Analysis
Assume GREEDY covers elements in order:
x1, x2, x3, x4, x5, x6, x7, x8, …, xn
3
2
2
1
Algebra
Set Cover
Conclusion:
Theorem:
Greedy-Set-Cover is an O(log n) approximation
algorithm for set cover.
Today’s Plan
 Vertex Cover (Review)
 Set Cover (Review + Analysis)
 Steiner Trees
A Warm-Up Problem
 Problem Statement:
Given: 2d (Euclidean) map &
set of points of interest
Find: shortest set of roads connecting all points
A Warm-Up Problem
 Problem Statement:
Given: 2d (Euclidean) map &
set of points of interest
Find: shortest set of roads connecting all points
A Warm-Up Problem
 Algorithm:
Compute: for every pair (u,v): distance(u,v)
Build: complete graph G = (V, E)
where w(u,v) = distance(u,v)
Find: minimum spanning tree of G.
Spanning Tree
Definition: a spanning tree is an acyclic subset
of the edges that connects all nodes
13
Weight: 32
4
5
9
1
8
9
2
Minimum Spanning Tree
Definition: a spanning tree with minimum weight
13
4
5
9
1
8
9
2
Minimum Spanning Tree
Definition: a spanning tree with minimum weight
13
Weight: 20
4
5
9
1
8
9
2
Properties of MST
Property 1: No cycles
Properties of MST
Property 2: If you cut an MST, the two pieces
are both MSTs.
Properties of MST
Property 3: Cycle property
For every cycle, the maximum weight
edge is not in the MST.
max-weight edge on cycle
Properties of MST
Property 4: Cut property
For every partition of the nodes, the
minimum weight edge across the cut
is in the MST. min-weight edge on cut
Prim’s Algorithm
Prim’s Algorithm.
(Jarnik 1930, Dijkstra 1957, Prim 1959)
Basic idea:
–
S : set of nodes connected by blue edges.
–
Initially: S = {A}
–
Repeat:
• Identify cut: {S, V–S}
• Find minimum weight edge on cut.
• Add new node to S.
5
H
15
A
3
C
12
10
8
7
G
9
6
1
4
E
16
D
20
2
B
11
13
F
Kruskal’s Algorithm
Kruskal’s Algorithm.
(Kruskal 1956)
Basic idea:
–
Sort edges by weight.
–
Consider edges in ascending order:
• If both endpoints are in the same blue
tree, then color the edge red.
• Otherwise, color the edge blue.
5
Data structure:
–
Union-Find
–
Connect two nodes if they
are in the same blue tree.
H
15
A
3
C
12
10
8
7
G
9
6
1
4
16
D
E
20
2
B
11
13
F
A Warm-Up Problem
 Algorithm: O(n2 log n)
Compute: for every pair (u,v): distance(u,v)
Build: complete graph G = (V, E)
where w(u,v) = distance(u,v)
Find: minimum spanning tree of G.
Try again…
 Problem Statement:
Given: 2d (Euclidean) map &
set of points of interest
Find: shortest set of roads connecting all points
Thinking outside the box…
 Problem Statement:
Given: 2d (Euclidean) map &
set of points of interest
Find: shortest set of roads connecting all points
Thinking outside the box…
 Euclidean Steiner Tree:
Given: 2d (Euclidean) map &
set of points of interest (POI)
Find: set of extra points
shortest set of links connecting all POI
Steiner Tree Problems
 Metric Steiner Tree:
Given: set of required points R
set of optional points S
distance metric d(., .)
v
w
u
Steiner Tree Problems
 Metric Steiner Tree:
Given: set of required points R
set of optional points S
distance metric d(., .)
Find: tree T = (V,E) : R ⊆ V ⊆ (S∪R)
• Tree includes all required points.
• Tree may include some optional points.
cost of tree is minimized:
Steiner Tree Problems
 General Steiner Tree:
Given: set of required points R
set of optional points S
set of edges E
edge weights w(.)
Find: tree T = (V,E) : R ⊆ V ⊆ (S∪R)
• Tree includes all required points.
• Tree may include some optional points.
cost of tree is minimized
Steiner Tree Problems
Three variants:
Euclidean Steiner Tree: points in a 2d-plane
Metric Steiner Tree: distances form (any) metric
General Steiner Tree: arbitrary distances
All three variants are NP-hard.
– No polynomial time solutions (unless P = NP).
– Find a good approximation algorithm.
Steiner Tree Problems
Proposed Algorithm:
Step 1:
Construct a graph G from points R.
Ignore points S.
Step 2:
Find an MST of G.
Question:
Is an MST a good approximation of a Steiner Tree?
Steiner Tree Problems
 Challenge:
Find the worst example you can for the proposed
algorithm.
– Euclidean
– Metric
– General
Worst  maximize(cost(MST) / Steiner_OPT).
Euclidean Steiner Tree
Given: A set R of points in the plane
Find: Minimum-cost tree spanning R
Euclidean metric
1
1
1
1
Cost = 2
Cost = 3
Steiner Point
1
Conjecture: the worst-case ratio of (MST-to-Steiner) is
.
Metric Steiner Tree
Given: Set R and S of points, and a distance metric
Find: Minimum-cost tree spanning R
10
10
5
5
10
10
10
5
5
10
10
5
5
5
5
10
10
10
5
5
10
MST: 20
Steiner Tree: 15
Ratio: 1.33
5
MST: 30
Steiner Tree: 20
Ratio: 1.5
Generalize: n-gon
MST: 10(n-1)
Steiner Tree: 5n
Ratio: 2(1 – 1/n)
5
10
MST: 40
Steiner Tree: 25
Ratio: 1.8
General Steiner Tree
Given: Set R and S of points, and a set of weighted edges
Find: Minimum-cost tree spanning R
1
3000
3000
1
1
MST: 6000
Steiner Tree: 3
Ratio: 2000
3000
Conclusion:
MST is NOT a good approximation for a
general
Steiner Tree.
Today’s Plan
 Euclidean Steiner Tree (skip)
(left as an exercise)
 Metric Steiner Tree
Show that MST is a 2-approximation.
 General Steiner Trees
Reduce to Metric Steiner Tree.
Metric Steiner Tree
To get a good approximation, show a lower
bound on OPT.
– “OPT has to be AT LEAST this large.”
– E.g., “OPT ≥ |M|”
Notation
Given:
– set of required points R
– set of optional points S
– distance metric d(., .)
Define:
– T = (V, E) be the optimal (minimum)
Steiner Tree.
Build a cycle…
Consider a DFS traversal of T:
(x0, x1, x2, x3, …, xm) where x0 = xm
Example:
1
1
1
1
1
1
1
1
All other pairs are distance 2.
Build a cycle…
Consider a DFS traversal of T:
(x0, x1, x2, x3, …, xm) where x0 = xm
a
Example:
1
1
g
1
d
1
f
2
e
h
1
c
1
b
Build a cycle…
Consider a DFS traversal of T:
(x0, x1, x2, x3, …, xm) where x0 = xm
a
Example:
1
1
g
1
d
1
f
2
e
h
1
c
1
b
DFS: a  g  d  g  f  g  a  h  c  h  b  h  a  e  a
Build a cycle…
Consider a DFS traversal of T:
Each edge is included in the DFS traversal twice.
cost(DFS) = 2cost(T) = 2OPT
a
1
1
g
Example:
cost(T)
=8
cost(DFS) = 16
1
d
1
f
2
e
h
1
c
1
b
DFS: a  g  d  g  f  g  a  h  c  h  b  h  a  e  a
Shortcut Steiner nodes…
Example:
Replace (a  g  d) with (a  d)
DFS: a  g  d  g  f  g  a  h  c  h  b  h  a  e  a
NEW: a  d  g  f  g  a  h  c  h  b  h  a  e  a
a
Triangle Inequality:
d(a,d) ≤ d(a,g) + d(g,d)
1
g
1
Hint: use the fact that
distance is a metric.
d
1
f
2
1
e
h
1
c
1
b
Shortcut Steiner nodes…
Example:
Replace (a  g  d) with (a  d)
DFS: a  g  d  g  f  g  a  h  c  h  b  h  a  e  a
NEW: a  d  g  f  g  a  h  c  h  b  h  a  e  a
a
Conclusion:
cost(NEW) ≤ cost(DFS)
1
g
1
Hint: use the fact that
distance is a metric.
d
1
f
2
1
e
h
1
c
1
b
Shortcut Steiner nodes…
Example:
Replace (d  g  f) with (d  f)
DFS: a  g  d  g  f  g  a  h  c  h  b  h  a  e  a
NEW: a  d  f  g  a  h  c  h  b  h  a  e  a
a
Conclusion:
cost(NEW) ≤ cost(DFS)
1
g
1
Hint: use the fact that
distance is a metric.
d
1
f
2
1
e
h
1
c
1
b
Shortcut Steiner nodes…
Example:
Continue until done…
DFS: a  g  d  g  f  g  a  h  c  h  b  h  a  e  a
NEW: a  d  f  a  c  b  a  e  a
a
Conclusion:
cost(NEW) ≤ cost(DFS)
1
g
1
d
1
f
2
1
e
h
1
c
1
b
Remove repeats…
Example:
Replace (f  a  c) with (f  c)
NEW: a  d  f  a  c  b  a  e  a
NEW2: a  d  f  c  b  a  e  a
a
Conclusion:
cost(NEW2) ≤ cost(DFS)
1
g
1
d
1
f
2
1
e
h
1
c
1
b
Remove repeats…
Example:
Replace (b  a  e) with (b  e)
NEW: a  d  f  a  c  b  a  e  a
NEW2: a  d  f  c  b  e  a
a
Conclusion:
cost(NEW2) ≤ cost(DFS)
1
g
1
d
1
f
2
1
e
h
1
c
1
b
Remove repeats…
Example:
Final: a  d  f  c  b  e  a
11 = cost(Final) ≤ cost(DFS) = 16 = 2OPT
2
a
1
2
g
1
d
e
h
1
1
1
f
2
2
1
b
c
2
1
2
Break the cycle…
Example:
Path: a  d  f  c  b  e
9= cost(Path) ≤ cost(DFS) = 16 = 2OPT
a
1
2
g
1
d
e
h
1
1
1
f
2
2
1
b
c
2
1
2
Path is a spanning tree…
Example:
Spanning tree: a  d  f  c  b  e
9= cost(Spanning tree) ≤ cost(DFS) = 16 = 2OPT
a
1
2
g
1
d
e
h
1
1
1
f
2
2
1
b
c
2
1
2
Path is a spanning tree…
Example:
Spanning tree: a  d  f  c  b  e
cost(MST) ≤ cost(Spanning tree) ≤ 2OPT
a
1
2
g
1
d
e
h
1
1
1
f
2
2
1
b
c
2
1
2
Approximation Proof
Analysis:
1. Let T be an optimal Steiner tree.
2. Let DFS be a DFS-traversal of T.
3. Let NoSteiner be DFS where we short-cut
past Steiner nodes.
4. Let Rcycle be NoSteiner where we short-cut
past repeated nodes.
5. Let Path be Rcycle where we remove the last
edge.
Approximation Proof
Analysis:
By definition of MST.
1. cost(MST) ≤ cost (Path)
2. cost(Path) ≤ cost(Rcycle)
Trivial.
3. cost(Rcycle) ≤ cost(NoSteiner)
4. cost(NoSteiner) ≤ cost(DFS)
By triangle inequality.
5. cost(DFS) ≤ 2cost(T) = 2OPT
By construction.
Metric Steiner Tree
Theorem:
A minimum spanning tree is a 2-approximation of the
optimal metric Steiner Tree.
Question: Is this analysis tight?
2(1 – 1/n) approximation?
Today’s Plan
 Euclidean Steiner Tree (skip)
(left as an exercise)
 Metric Steiner Tree
Show that MST is a 2-approximation.
 General Steiner Trees
Reduce to Metric Steiner Tree.
Steiner Tree Problems
 General Steiner Tree:
Given: set of required points R
set of optional points S
set of edges E
edge weights w(.)
Find: tree T = (V,E) : R ⊆ V ⊆ (S∪R)
• Tree includes all required points.
• Tree may include some optional points.
cost of tree is minimized
General Steiner Tree
Problem:
Minimum spanning tree of R is not a good
approximation.
General Steiner Tree
Idea: Reduction
1. Construct an instance of Metric Steiner Tree from
the input.
2. Solve the Metric Steiner Tree problem (by finding
an MST).
3. Translate the solution back.
General Steiner Tree
Beware: Reductions are tricky for approximation
algorithms
Typical example:
–
Assume two problems ABC and XYZ
–
Function f : ABC  XYZ
–
Function g : “solutions to XYZ”  “solutions to ABC”
–
Show:
If S is an optimal solution for f(A), then g(S) is an
optimal solution for A.
General Steiner Tree
Beware: Reductions are tricky for approximation
algorithms
Typical example:
XYZ
ABC
A
f
f(A)
ALG
g
g(S)
S
General Steiner Tree
Beware: Reductions are tricky for approximation
algorithms
Problem: ALG does not find optimal solution
– Function g may not preserve approximation ratio.
XYZ
ABC
f
A
f(A)
ALG
g
g(S)
S
General Steiner Tree
Idea: Reduction
1. Construct an instance of Metric Steiner Tree from
the input.
2. Solve the Metric Steiner Tree problem (by finding
an MST).
3. Translate the solution back.
Steiner Tree Problems
 General Steiner Tree:
Given: set of required points R
set of optional points S
set of edges E
edge weights w(.)
Construct a Metric
Construction:
1. Required and optional points stay the same.
2. For every pair of points (u,v) define:
d(u,v) = distance of shortest path from u to v.
Construct a Metric
Example:
– d(A,B) = 10
– d(H,E) = 11
2
H
How do we find all the shortest paths?
3
12
G
7
C
10
8
9
– d(B,H) = 12
–…
15
A
6
1
4
16
D
E
20
5
B
11
13
F
Construct a Metric
Example:
– d(A,B) = 10
– d(H,E) = 11
2
H
How do we find all the shortest paths?
• Dijkstra’s Algorithm : O(VE log V)
• Floyd-Warshall : O(V3)
3
12
G
7
C
10
8
9
– d(B,H) = 12
–…
15
A
6
1
4
16
D
E
20
5
B
11
13
F
Construct a Metric
Claim:
The function d(.,.) is a metric.
Construct a Metric
Claim:
The function d(.,.) is a metric.
Usual properties:
•
•
•
(don’t matter)
d(u, u) = 0
d(u, v) = d(v, u)
d(u, v) ≥ 0
Construct a Metric
Claim:
The function d(.,.) is a metric.
2
H
Triangle Inequality:
Fix some (u, v, w).
d(u,w) ≤ d(u,v) + d(u,w)
–
•
3
12
G
7
C
10
8
9
•
•
15
A
6
1
4
16
D
E
11
13
20
If not, find a shorter path from u to w by going u  v  w.
Shortest paths always satisfy
triangle inequality, by definition!
5
B
F
General Steiner Tree
Idea: Reduction
1. Construct an instance of Metric Steiner Tree from
the input (via shortest paths).
2. Solve the Metric Steiner Tree problem (by finding
an MST).
3. Translate the solution back.
Translate back…
Given a Steiner Tree T’ for the metric problem:
1. For every edge (u,v) in T’, add the shortest path
from (u  v) to the graph G.
(Note, G may not be a tree.)
2. Find an MST of G.
Analysis
Overview:
1. Input: IN = (R, S, G, w)
2. Construct: MET = (R, S, d) where d = shortest paths.
Show: OPT(MET) ≤ OPT(IN)
3. Solve: Let T’ be the approximately-optimal Steiner tree for
MET. Fact: cost (T’) ≤ 2OPT(MET)
4. Translate: Convert T’ into a Steiner tree T for IN.
Show: cost(T) ≤ cost(T’)
Conclude: cost(T) ≤ cost(T’) ≤ 2OPT(MET) ≤ 2OPT(IN)
Analysis
Construct: MET = (R, S, d) where d = shortest paths.
Show: OPT(MET) ≤ OPT(IN)
1. Let T be an optimal Steiner tree for IN.
2. Let T’ be the same tree in MET.
3. cost(T’) ≤ cost(T)
•
•
For every edge (u,v): d(u,v) ≤ w(u,v)
Hence the tree only costs less under the distance metric.
4. OPT(MET) ≤ cost(T’) ≤ cost(T) = OPT(IN)
Analysis
Overview:
1. Input: IN = (R, S, G, w)
2. Construct: MET = (R, S, d) where d = shortest paths.
Show: OPT(MET) ≤ OPT(IN)
3. Solve: Let T’ be the approximately-optimal Steiner tree for
MET. Fact: cost (T’) ≤ 2OPT(MET)
4. Translate: Convert T’ into a Steiner tree T for IN.
Show: cost(T) ≤ cost(T’)
Conclude: cost(T) ≤ cost(T’) ≤ 2OPT(MET) ≤ 2OPT(IN)
Analysis
Translate: Convert T’ into a Steiner tree T for IN.
Show: cost(T) ≤ cost(T’)
1. Let G be the graph constructed from T’.
2. cost(G) ≤ cost(T’)
•
•
•
Every edge in T’ corresponds to a “shortest path.”
G is constructed by adding these paths.
Not always equal due to overlapping paths.
3. cost(T) ≤ cost(G)
•
Remove edges from G to find an MST.
Analysis
Overview:
1. Input: IN = (R, S, G, w)
2. Construct: MET = (R, S, d) where d = shortest paths.
Show: OPT(MET) ≤ OPT(IN)
3. Solve: Let T’ be the approximately-optimal Steiner tree for
MET. Fact: cost (T’) ≤ 2OPT(MET)
4. Translate: Convert T’ into a Steiner tree T for IN.
Show: cost(T) ≤ cost(T’)
Conclude: cost(T) ≤ cost(T’) ≤ 2OPT(MET) ≤ 2OPT(IN)
General Steiner Tree
Theorem:
If A is a c-approximation algorithm for Metric Steiner
Tree, then we can construct a c-approximation
algorithm for General Steiner Tree.
General Steiner Tree
Theorem:
If A is a c-approximation algorithm for Metric Steiner
Tree, then we can construct a c-approximation
algorithm for General Steiner Tree.
Theorem:
There exists a 2-approximation algorithm for General
Steiner Tree.
Steiner Tree Example
2
H
15
A
3
C
12
10
8
G
7
9
6
1
4
16
D
E
20
5
B
11
13
F
Steiner Tree Example
2
H
Shortest Paths:
(A,H) = 2
(A,D) = 7
(A,E) = 9
(H,D) = 9
(H,E) = 11
(D,E) = 10
7
11
9
2
H
9
10
D
E
15
A
3
12
G
7
C
10
8
9
6
16
D
E
20
5
B
1
4
(Look ahead: ignore Steiner nodes.)
A
11
13
F
Steiner Tree Example
2
H
A
7
11
Run MST Algorithm
9
10
D
2
H
9
E
15
A
3
12
G
7
C
10
8
9
6
1
4
16
D
E
20
5
B
11
13
F
Steiner Tree Problem
2
H
A
7
11
Convert solution back
to original graph.
9
10
D
2
H
9
E
15
A
3
12
G
7
C
10
8
9
6
1
4
16
D
E
20
5
B
11
13
F
General Steiner Tree
Theorem:
If A is a c-approximation algorithm for Metric Steiner
Tree, then we can construct a c-approximation
algorithm for General Steiner Tree.
Theorem:
There exists a 2-approximation algorithm for General
Steiner Tree.
Best known approximation: 1.55
Today’s Plan
 Euclidean Steiner Tree (skip)
(left as an exercise)
 Metric Steiner Tree
Show that MST is a 2-approximation.
 General Steiner Trees
Reduce to Metric Steiner Tree.