multi-layer channel routing complexity and algorithms rajat k. pal annajiat alim rasel 0409052002...
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Multi-Layer Channel RoutingComplexity and Algorithms
Rajat K. Pal
Annajiat Alim Rasel0409052002
annajiat@gmail.comCSE 6404 VLSI Layout Algorithms
Page 67 -> 78as a part of MS at BUET, 2009
Cell
• Circuit element• Simple• Pre-designed
C
EA D
B
F
ASIC (Standard Cell) Design Example
D C C B
A C C
D C D B
BCCC
CellMetal1Metal2
GNDVDD
C D
A BCell library
Placement [©Sherwani]
Channel
C
EA D
B
F
VLSI Layout
Channel
A D
Channelnets: set of terminals to be connected
Channel routing
Transitively Oriented Graph/ Comparability Graph
u
v w wv
u x
Transitively Oriented Graph/ Comparability Graph ??
u
v w wv
u x
Transitively Oriented Graph/ Comparability Graph
- Vertex Types(intermediate)
u (source)u
V(source)
w(sink)
w(source)
v(sink)
(sink)x
Channel Specification
5 3 2 6 1 7 4
5 3 6 2 1 7 4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
VH routin
g
Channel Specification
5 3 2 6 1 7 4
5 3 6 2 1 7 4
I 5
I 3I 2
I 6
I 1
I 7
I 4
A
D
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
Channel Intervals
Channel Intervals
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
HCG
2
3
1
45
6
7
Channel Intervals
HCG and
HNCG
2
3
1
45
6
7
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
Channel Intervals
HNCG
2
3
1
45
6
7
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
clique cover
3
4
1
2
5
7
6
While G ≠ empty
choose a maximal clique
delete the vertices of the clique
algorithm
clique cover
{1, 4, 5}
3
4
1
2
5
7
6
clique cover
{1, 4, 5}
3
7
6{ 2, 7 }
2
clique cover
{1, 4, 5}
36{ 2, 7 }
{ 3 }
clique cover
{1, 4, 5}
6{ 2, 7 }
{ 3 }
{ 6 } But, the cliquecover may not beminimum one
clique cover
{1, 4, 5}
{ 2, 7 }
{ 3 }
{ 6 } But, the cliquecover may not beminimum one
3
4
1
2
5
7
6
clique cover
{1, 4, 5} { 2, 7 }
{ 3 } { 6 }
Independent set 3
4
1
2
5
7
6
{5, 3, 6} {2, 3, 6}
{2, 1, 6}{7, 1}{7, 4}
clique cover
{1, 4, 5} { 2, 7 }
{ 3 } { 6 }
Independent set 3
4
1
2
5
7
6
{5, 3, 6} {2, 3, 6}
{2, 1, 6}{7, 1}{7, 4}
clique cover
{1, 4, 5} { 2, 7 }
{ 3 } { 6 }
Independent set
{5, 3, 6} {2, 3, 6}
{2, 1, 6}{7, 1}
I2
I7
I5
I4
I1
I3
I6
{7, 4}
For perfect graph
(G)=κ(G)
(G)= size of the max independent setκ(G) = size of the min clique cover
Minimum clique cover
3
4
1
2
5
7
6
Sort the intervals w. r. t. Li
Orient the edges b. t. r. Ri < Lj While G ≠ empty
choose a maximal clique from sorted list
delete the vertices of the clique
algorithm
3
4
1
2
5
7
6
I2
I7
I5
I4
I1
I3
I6
5
6
3
1 7
2
4
1 2
3 4
5 6 7
3
4
1
2
5
7
6
I2
I7
I5
I4
I1
I3
I6
5
6
3
1 7
2
4
3
4
1
2
5
7
6
5
6
3
1 7
2
4
5 2 7
3
4
1
2
5
7
6
5
6
3
1 7
2
4
5 2 7
3
4
1
6
5
3
7
2
5 2 7
3 1 4
3
4
1
6
5
3
7
2
5 2 7
3 1 4
63
5 2 7
3 1 4
6
Obviously, this is the minimum clique
Lemma
Lemma 3.1
For any vertices vi , vj and vk of the HNCG,if vi vj and vj vk , then vi vk
Proof From definition,
Ri < Lj and Rj < Lk
From the interval,
Lj < Rj
So, we have
Ri < Lk
So, the HNCG is acomparability graph
Lemma
Lemma 3.2
Each maximal clique computed by thealgorithm MCC1 in step 3 has exactly one
vertex from each maximal independent set of G
case 1:
There is no interval ending on the left side ofthe starting column of any interval in A
We consider any maximal independent set A
Independent set
{5, 3, 6} {2, 3, 6}
{2, 1, 6}{7, 1}
I2
I7
I5
I4
I1
I3
I6
{7, 4}
Lemma 3.2 (cont.) 5 2 7
3 1 4
6
In this case, vertices in A will be selected as the source vertex of successive maximal cliques
case 2:
There is an interval ending on theleft side of the starting column of at least
one of the intervals in A
Lemma 3.2 (cont.)
Independent set
{5, 3, 6}{2, 3, 6}
{2, 1, 6}{7, 1}
I2
I7
I5
I4
I1
I3
I6
{7, 4}
Lemma 3.2 (cont.) 5 2 7
3 1 4
6
Let, I be the rightmost intervalselected by MCC1 in C1 such
that at least one interval of A starts after I ends &
J be the interval of A with leftmost end point among the intervals
of A , starting after I ends
Independent set
{5, 3, 6}{2, 3, 6}
{2, 1, 6}{7, 1}
I2
I7
I5
I4
I1
I3
I6
{7, 4}
Lemma 3.2 (cont.) 5 2 7
3 1 4
6
So, one interval of A is chosen bythe algorithm MCC1 in C1
So, exactly one interval of A is chosen in C1
Lemma
Lemma 3.3
MCC1 computes a minimum clique cover of theHNCG, G=(V, E’) consisting of dmax cliques
Proof
dmax = size of the maximum
independent set
From, lemma 3.2
A will become emptywhen Cl is computed
Let, the size of the maximalindependent set A be l,
where l < dmax
Lemma 3.3 (cont.)
{5, 3, 6}{2, 3, 6}
{2, 1, 6}{7, 1}{7, 4}
5 2 7
3 1 4
6The last clique computed by MCC1will include the last interval fromeach maximum independent set
Lemma 3.3 (cont.)
{5, 3, 6}{2, 3, 6}
{2, 1, 6}{7, 1}{7, 4}
5 2 7
3 1 4
6The last clique computed by MCC1will include the last interval fromeach maximum independent set
So, it follows that
the number of cliques computed
by MCC1 is exactly dmax
Theorem 3.1
The construction of the HNCG, G=(V, E’) andits transitively oriented graph G*=(V, F) fromthe intervals belonging to a channel can beimplemented in O(n+e) time, where n is the
number of nets and e is the size of the HNCG
Theorem 3.1 (cont.)
orienting of the edges can be done by
scanning all the intervalsexactly once from left to right
The following cases mayoccur at a column
position while scanning
an interval starts
an interval terminates
all the intervals continue
Theorem 3.1 (cont.)
case 1: while Ii starts, if Ij S1 terminatedbefore Ii ; orient an edge vj vi in G*
case 2:If Ii terminates at current column
position, we add Ii in the set S1
case 3: No action required
Theorem 3.1 (cont.)
In any instance of the CRP, at most two intervalsmay start or terminate at the same column.In this case, both the intervals are processedone after another.
The scanning of the columns requires O (n) time
The algorithm spends constant time in
introducing each edge of G and G*
The introducing time of all the edges is O (e)
Theorem 3.2
The algorithm MCC1 correctly computes arouting solution for the two layer CRPwithout vertical constraints using dmax
tracks. MCC1 can be executed in O(n+e) time.
O (n)
O (n+e)
O (n+e)
In total,
O (n+e) time
Theorem 3.3
The algorithm MCC2 correctly computes arouting solution for the two layer CRPwithout vertical constraints using dmax
tracks. MCC2 can be executed in O(nlogn) time
Self-balancing binary search tree
Operation Big-O time
Lookup O(log n)
Insertion O(log n)
Removal O(log n)
In-order iteration over all elements O(n)
The overall worst-case Build Time is O(nlog n)
Times for various other operations in terms of number of nodes in the tree n:For some implementations these times are worst-case, while for others they are amortized.
Self-balancing binary search tree
Operation Big-O time
Lookup O(log n)
Insertion O(log n)
Removal O(log n)
In-order iteration over all elements O(n)
The overall worst-case Build Time is O(nlog n)
Times for various other operations in terms of number of nodes in the tree n:For some implementations these times are worst-case, while for others they are amortized.
In computer science, especially analysis of algorithms, amortized analysis finds the average running time per operation over a worst-case sequence of operations. Amortized analysis differs from average-case performance in that probability is not involved; amortized analysis guarantees the time per operation over worst-case performance.
O (n)
O (n+e)
O (nlogn)
O (logn)
Building Self Balancing BST of
intervals sorted on their starting column numbers - O (nlogn)
Total=nlogn+(n + e) + n + n X (logn + logn)=2n + e + nlogn + 2nlogn=2n + e + 3nlogn=nlogn
O (logn)
Vertex Representation• Class Vertex {
– int key; //starting column and sequence number in transitive orientation
– int n; //vertex name: v1, v2 … etc. Prefix “v” will be used for now on for clarity
– double endingPosition; // ending column, double value such as 3.5 indicates that it ended before column 4. Thought it should be int, but due to lack of numbers between 3 and 4, double values are used.
– int outDegree;}
Input / Initialization
• Assuming for the given example following declarations and initializations have been done1) Vertex v1, v2, v3, v4, v5, v6, v7;2) v1.n=1; v2.n=2; …………. V7.n=7;3) All vertexes know their key (startingPosition) and
endingPosition
Step 1 - Building BST
• Insert v5
k=1end=3.5
outDeg=4n = v5
Step 1 - Building BST
• Insert v3
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
Step 1 - Building BST
• Insert v6
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Step 1 - Building BST
The balance factor of a node is the height of its right subtree minus the height of its left subtree. A node with balance factor 1, 0, or -1 is considered balanced. A node with any other balance factor is considered unbalanced and requires rebalancing the tree. The balance factor is either stored directly at each node or computed from the heights of the subtrees.
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Here v5 is unbalanced
Step 1 - Building BST
In right right case, tree is balanced using left rotation. So, left rotating v5, v3, v6 from v3
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Here v5 is unbalanced
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Here v5 is now balanced
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Inserted v2
k=4end=5.7
outDeg=0n = v2
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Inserted v5
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Tree now has become unbalanced and v6, v2, v1 now forms a right right case
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Lets balance the tree using a left rotation
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Tree is now balanced
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Changed for a better view
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Inserted v7
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Tree is unbalenced. v3, v2, and v1 forms a right right case which is to be solved using a left rotation
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Tree is now balenced
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Insert v7
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
V1, v7, v4 forms a right right case requiring to be balanced
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 1 - Building BST
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Tree is now balanced
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 1 - Building BST, Done!
k=1end=3.5
outDeg=0n = v5
k=2end=4.5
outDeg=0n = v3
k=3end=5.5
outDeg=0n = v6
Better view
k=4end=5.7
outDeg=0n = v2
K=5End=6.5
outDeg=0n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Channel Intervals
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
HCG
2
3
1
45
6
7
Step 2 – Construct HNCG
Channel Intervals
HCG and
HNCG
2
3
1
45
6
7
Step 2 – Construct HNCG
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
Channel Intervals
HNCG
2
3
1
45
6
7
Step 2 – Construct HNCG
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
Channel Intervals
HNCG
2
3
1
45
6
7
Step 2 – Transitive Orientation
S5 S3 S6 S2 S1 S7 S4
I 5
I 3
I 2
I 6
I 1
I 7
I 4
A
D
• Similarly, whole graph is made transitively oriented and “outDegree” field of all vertexes are now updated
v3
v4
v1
v2
v5
v7
v6
5
6
3
1 7
2
4
Step 2 – Transitive Orientation
• While (! tree.empty()){– min = tree.findMinVertex();– C = C U min– Tree.delete(min)– If min.outDegree <> 0 then // not a sink
• Next = tree.findGreaterThan(min.endingPosition)
– Else • CliqueCover = CliqueCover U C• C = Empty
}
Step 3 – Overview
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Find min
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Find min
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Min = v5Add v5 to Clique, C
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Tree.delete (v5)
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Tree.delete (v5)
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
v5.outDeg > 0, so, its not sink, continuetree.findGreaterThan(v5.end)
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
min = tree.findGreaterThan(v5.end)// 4 (v2) is greater than v5.end = 3.5
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Add min to Clique
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Add min to Clique
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Now to delete v2, we need to swap v6 (3) and v2 (4)
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=3end=5.5
outDeg=2n = v6
Now to delete v2, we need to swap v6 (3) and v2 (4)
k=4end=5.7
outDeg=2n = v2
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
swapped v6 (3) and v2 (4)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Now delete v2
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Deleted v2
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
V2.outDeg > 0, so, not a sink, continueMin = tree.FindGreaterThan(v2.end)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Min = tree.FindGreaterThan(v2.end)// 6 (v7) is greater than 5.7 (end of v2)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Min = tree.FindGreaterThan(v2.end)// 6 (v7) is greater than 5.7 (end of v2)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Add min to clique
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Add min to clique
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
To delete 6 (v7), we need to swap5 (v1) and 6 (v7)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=6End=8
outDeg=0n = v7
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
Swapped 5 (v1) and 6 (v7)
k=3end=5.5
outDeg=2n = v6
K=6End=8
outDeg=0n = v7
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
delete 6 (v7)
k=3end=5.5
outDeg=2n = v6
K=6End=8
outDeg=0n = v7
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
Step 3 - Details
k=1end=3.5
outDeg=4n = v5
k=2end=4.5
outDeg=3n = v3
k=4end=5.7
outDeg=2n = v2
V7.outDeg is 0, so, v7 is a sink. So, current clique is complete. Hiding deleted vertices andGoing to build next clique …
k=3end=5.5
outDeg=2n = v6
K=6End=8
outDeg=0n = v7
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Min = tree.FindMin()We get 2 (v3)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Added 2 (v3) to cliqueNow delete 2 (v3)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }{ v3 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
deleted 2 (v3)Tree is now unbalanced in a right right case
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }{ v3 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Rebalancing using a left rotation
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }{ v3 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Rebalance using a left rotation
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }{ v3 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
2 (v3) was not a sink, so continueTree.findGreaterThan(v3.end)// 5 (v1) is greater than 4.5 (end of v3)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }{ v3 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Tree.findGreaterThan(v3.end)
// 5 (v1) is greater than 4.5 (end of v3)
Add v1 to clique
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }{ v3, v1 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Swap 3 (v6) and 5 (v1)
k=3end=5.5
outDeg=2n = v6
K=5End=6.5
outDeg=1n = v1
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Swapped 3 (v6) and 5 (v1)
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Now we can easily delete 5 (v1)
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
deleted 5 (v1)Min=tree.FindGreaterThan(v1.end)// 7 (v4) is greater than 6.5
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
v1 was not a sink, so we can continueAdd v4 to clique
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1, v4 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Delete v4
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1, v4 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Delete v4
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1, v4 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
Deleted v4
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1, v4 }
Step 3 - Details
k=2end=4.5
outDeg=3n = v3
V4.outDeg was zero, hence currentclique is complete. Hiding deleted vertices for clarity
K=5End=6.5
outDeg=1n = v1
k=3end=5.5
outDeg=2n = v6
K=7End=9
outDeg=0n = v4
{ v5, v2, v7 }
{ v3, v1, v4 }
Step 3 - Details
Min=tree.findMin()// we get 3 (v6)Add to clique
k=3end=5.5
outDeg=2n = v6
{ v5, v2, v7 }
{ v3, v1, v4 }
{ v6 }
Step 3 - Details
delete 3 (v6)
k=3end=5.5
outDeg=2n = v6
{ v5, v2, v7 }
{ v3, v1, v4 }
{ v6 }
Step 3 - Details
• deleted 3 (v6)• Tree is now empty, so we stop and we are
done!• We have found 3 minimum cliques in our
clique cover!
{ v5, v2, v7 }
{ v3, v1, v4 }
{ v6 }
Reference• Multi-Layer Channel Routine Complexity and Algorithms by
Rajat K. Pal• Graph theory and its applications to problems of society by
Fred S. Roberts• Introduction to Graph Theory by Douglas B. West• Slide 2 -> 3 is from teacher.buet.ac.bd/saidurrahman/cse6404• Slide 4 -> 5 from Mr. Jawaherul and 14 -> 47 from Mr. Emran• From en.wikipedia.org, Binary_search_tree, Tree_rotation,
Self-balancing_binary_search_tree• webpages.ull.es/users/jriera/Docencia/AVL/AVL%20tree
%20applet.htm
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