Multi-Layer Channel Routing
Complexity and Algorithms
Theorem 5.6, 5.7, 5.8, 5.9 (page 166-171)
Presented by –
Md. Salahuddin
# 040805012P
How do we solve problem?
Polynomial Algorithm, O(nk)
Problem
P
Polynomial Time Verifiable
3 SAT
NP
NP-Completeness
“I can’t find an efficient algorithm, I guess I’m just too dumb.”
“I can’t find an efficient algorithm, because no such algorithm is possible!”
NP-Completeness
“I can’t find an efficient algorithm, but neither can all these famous people.”

To prove NP-Complete
◦ Polynomial time verifiable
◦ Polynomial Reduction to a known hard problem
Our Problem
Multi-layer Channel Routing
Is there a
independent
set of size 3?
Independent Set in a Graph
3 SAT
Multi-Layer Channel Routing
Channel
Multi-Layer Channel Routing
Terminal
Hi
Vi
Channel
H2
V2
H1
V1
Terminal
Example
0
1
0
2
3
3
4
1
2
2
3
0
4
0
VH Solution
0 1 0
1 2 2 3 0 0 4 0
2 3 3 0
4
0
1
0
2
3
3
4
1
2
2
3
0
4
0
Example
0
0
1
2
1
2
1
2
2
1
0
0
VHV Solution
0
1 2
2
1
0
0
1
0
2
1
2
0
0
1
2
1
2
1
2
2
1
0
0
Previous Idea
Problem : ISi , i ≥ 4
 Instance :
An undirected graph G = (V, E). Here
the number of vertices in G is n.
 Question :
n
Is there an independent set of size  i  ?

ISi , i ≥ 4 is NP-Complete

ISi NP
 Given a 3SAT instance with a collection F
= {c1, c2, …, cq} of q clauses
 We construct a graph G = (V, E)

◦ Total vertices, n = iq + i

We prove that F has a satisfiable truth
assignment iff G has an independent set of
n
size q + 1 =  i 
Graph Construction
(i – 3)q + i
A set
3q
B set
MNViHi, 4 ≤ i < dmax is NP-Complete
Terminal
Hi
Vi
H2
V2
H1
V1
Terminal
Proof Overview

MNViHi NP
 3SAT instance to construct Graph
 An instance I of CRP where dmax = n
Three conditions are equivalent:
(i) F is satisfiable
n
(ii) G has an independent set of size  i 
(iii) I has a ViHi routing solution using  d
 i
tracks

max



MNViHi  NP

We can verify in polynomial time
◦ Each track in each horizontal layer is assigned
a set of mutually non over-lapping intervals
◦ RVCGCC is cycle free
Hi
Hi
Vi
H2
V2
H1
V1
3SAT instance to Graph
(i – 3)q + i
3q
A set
B set
So, (i) implies (ii)
CRP instance I , dmax = n
A channel with 2(n + e) columns
(ii) Implies (iii)
n = dmax
n
d 
  i  =>  i 


 CCH be the set of
d 
 i  nets


 RVC has no edge
 RVC is cycle free
d 
  i  tracks in Hi

max
Hi
Vi
i
H2
max
max

Rest in (i -1) H. layers
V2
H1
V1
dmax –  d max 
 i
 d max
= (i -1) 
 i




(iii) Implies (ii)





Consider CCH of nets
assigned to Hi
Each track has one net
RVCGCC cycle free
By construction it is
edge free
So, G has an independent
set of size  n 
i
Hi
 d max 
 i 


 d max 
 i 


i 
So, MNViHi, 4 ≤ i < dmax is NP-Complete. (proved)
Hi
Vi
H2
V2
H1
V1
2ISi , i ≥ 3
Problem : 2ISi , i ≥ 3
 Instance :
An undirected graph G = (V, E). Here
the number of vertices in G is n.
 Question :
Is there two disjoint independent set of
size  i n1 ?



2ISi , i ≥ 3 is NP-Complete

2ISi
NP
 Given a 3SAT instance with a collection F
= {c1, c2, …, cq} of q clauses
 We construct a graph G = (V, E)

◦ Total vertices, n = (i + 1)(q + 1)

We prove that F has a satisfiable truth
assignment iff G has two disjoint
independent sets of size q + 1 each
Graph Construction
(i – 3)q + i
A set
3q
q+1
B set
C set
Proof
GB can’t have an independent set of size
larger than q
 G A B have an independent set of size q+1
 GB has an independent set of size q
 F must be satisfiable

MNViHi+1, 3 ≤ i < dmax – 1 is NP-Complete
Terminal
Hi+1
Vi
Channel
H2
V1
H1
Terminal
Proof Overview

MNViHi+1 NP
 3SAT instance to construct Graph
 An instance I of CRP where dmax = n
Three conditions are equivalent:
(i) F is satisfiable
(ii) G has two disjoint independent set of
 n 
size  i 1
(iii) I has a ViHi+1 routing solution using
d 
 i 1  tracks



max
Graph Construction
(i – 3)q + i
A set
3q
q+1
B set
C set
CRP instance I , dmax = n
A channel with 2(n + e) columns
(ii) Implies (iii)
n = dmax
 d max 
 n 
   => 

 i 1
 i 1 
 CCH and CCH be
the set of  d max  nets
 i 1 
 RVC has no edge
 RVC is cycle free
 d max 
  i 1  tracks in H1 and

1
Vi
i 1
CCH i1

Hi+1
Rest in other H. layers
H2
V1
H1
(iii) Implies (ii)

Consider CCH of nets
assigned to Hi+1and CCH
of nets assigned to H1
Each track has one net
RVCGCC and RVCGCC
are cycle free
By construction they are
edge free
So, G has two disjoint  n 
independent sets of size  i 1
i 1
1




H1
Hi 1

 d max 
 i 1 


 d max 
 i 1 


 d max 
 i 1 



So, MNViHi+1, 3 ≤ i < dmax – 1 is NP-Complete. (proved)
Hi+1
Vi
H2
V1
H1
2IS2??
We see 2ISi , i ≥ 3
 If i = 2 then what?
 2IS2 is also NP-Complete
 Different Graph construction than 2ISi ,
i≥3

2IS2 is NP-Complete

2IS2 NP
 Given a 3SAT instance with a collection F
= {c1, c2, …, cq} of q clauses
 We construct a graph G = (V, E)

◦ Total vertices, n = 6q

We prove that F has a satisfiable truth
assignment iff G has two disjoint
independent sets of size 2q each
Graph Construction
q
3q
2q
A set
B set
C set
2IS2 and MNV2H3 are NP-Complete

Similarly we can show that MNV2H3 is
NP-Complete because:
◦ 2IS2 is NP-Complete
Any Question??
Thank You