Rajat K. Pal

Chapter 3 Emran Chowdhury # P Presented by

C E A D B F Pre-requisite VLSI Layout

Pre-requisite Net Cells have rectangular shapes with terminals located on their border. A set of terminals that must be electronically connected together, constitute a net. Terminals of the same net are given the same integer level

C E A D B F C E A D B F VLSI Layout Pre-requisite Net

Pre-requisite Channel If terminals are assigned to fixed positions only two opposite sides of a rectangular region is called a channel The problem of routing within a channel is called the channel routing problem

Pre-requisite

Three dimensional view of a channel

Pre-requisite Column Row Three dimensional view of a channel

Pre-requisite Column Row

Pre-requisite

Pre-requisite

Pre-requisite

Pre-requisite A routing solution using two layers of interconnection

Pre-requisite Channel Routing Problem The channel routing problem is the problem of computing a feasible route for the nets so that the number of tracks required is minimized

Pre-requisite Let, L i be the leftmost column position of net n i Let, R i be the rightmost column position of net n i I i =(L i, R i ) is known as the Interval or span of the net n i

Pre-requisite

L4L4 R4R4 I4I4 I8I8

Pre-requisite I4I4 I 10 I8I8

Pre-requisite I4I4 I 10 I8I8 Channel Specification I9I9 I2I2 I1I1 I3I3 I7I7 I5I5 I6I6

Pre-requisite If we use a single horizontal wire segments for routine each net This wire segment spans the entire interval I i of the net n i So, routing the nets amounts to assigning intervals to horizontal tracks of the channel The horizontal constraints determine whether two intervals I i & I j of the two different nets n i & n j respectively, are assignable to the same track

Pre-requisite The horizontal constraints are represented by horizontal constraint graph (HCG) Horizontal constraints can be represented using an HCG, HC=(V, E), where a vertex v i  V corresponds to the interval I i of the net n i in the channel An undirected edge {v i, v j }  E, if the intervals I i & I j, corresponding to the nets n i & n j, intersect at a column An edge {v i, v j } in the HCG indicates that I i & I j are not assignable to the same track

I8I8 I9I9 I2I2 I7I7 I5I5 I4I4 I 10 I1I1 I3I3 I6I

HCG

The horizontal non-constraint graph (HNCG) is denoted by HNC=(V, E’)

Chapter 3 Objectives of the Chapter

We concentrate on the problem of resolving horizontal constraints for routing channels with minimum possible area We consider routing in the two layer VH routing model where we assume that there are no vertical constraints

Comparability graph An undirected graph can be converted to a directed graph by orienting each edge of this graph An undirected graph is said to be transitively orientable if there exists an orientation of all the edges such that for any three vertices v i, v j & v k : (v i  v j & v j  v k ) => v i  v k Comparability graph

source sink transitively orientable not comparability graph

Channel Routing Problem assign the horizontal wire segments of the nets to tracks so that two nets assigned to the same track do not overlap Input Channel specification or net list which introduce no vertical constraints Objective # of such mutually non-overlapping sets is minimum

Input HNCG I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6

I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 Input HNCG

I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 Input HNCG

I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 Input HNCG

I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 Input HNCG

Intuitive idea a set of non overlapping intervals forms a clique in the HNCG So, try to find the minimum clique cover of the HNCG any set of intervals forming a clique in the HNCG can be safely assigned to the same track in any routing solution

clique cover While G ≠ empty choose a maximal clique delete the vertices of the clique algorithm

clique cover {1, 4, 5}

clique cover {1, 4, 5} { 2, 7 } 2

clique cover {1, 4, 5} 3 6 { 2, 7 } { 3 }

clique cover {1, 4, 5} 6 { 2, 7 } { 3 } { 6 } But, the clique cover may not be minimum one

clique cover {1, 4, 5} { 2, 7 } { 3 } { 6 } But, the clique cover may not be minimum one

clique cover {1, 4, 5}{ 2, 7 } { 3 }{ 6 } Independent set {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}{7, 4}

clique cover {1, 4, 5}{ 2, 7 } { 3 }{ 6 } Independent set {5, 3, 6}{2, 3, 6} {2, 1, 6}{7, 1} I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 {7, 4} For perfect graph  (G)=κ(G)  (G)= size of the max independent set κ(G) = size of the min clique cover

Minimum clique cover Sort the intervals w. r. t. L i Orient the edges b. t. r. R i < L j While G ≠ empty choose a maximal clique from sorted list delete the vertices of the clique algorithm

I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I

I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I

Obviously, this is the minimum clique

Lemma Lemma 3.1 For any vertices v i, v j and v k of the HNCG, if v i  v j and v j  v k, then v i  v k Proof From definition, R i < L j and R j < L k From the interval, L j < R j So, we have R i < L k So, the HNCG is a comparability graph

Lemma Lemma 3.2 Each maximal clique computed by the algorithm 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 of the 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} I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 {7, 4} Lemma 3.2 (cont.) 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 the left 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} I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 {7, 4} Lemma 3.2 (cont.) Let, I be the rightmost interval selected by MCC1 in C 1 such that at least one interval of A starts after I ends & J be the interval of A with left most end point among the intervals of A, starting after I ends

Independent set {5, 3, 6}{2, 3, 6} {2, 1, 6}{7, 1} I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 {7, 4} Lemma 3.2 (cont.) So, one interval of A is chosen by the algorithm MCC1 in C 1 So, exactly one interval of A is chosen in C 1

Lemma Lemma 3.3 MCC1 computes a minimum clique cover of the HNCG, G=(V, E’) consisting of d max cliques Proof d max = size of the maximum independent set From, lemma 3.2 A will become empty when C l is computed Let, the size of the maximal independent set A be l, where l < d max

Lemma 3.3 (cont.) {5, 3, 6}{2, 3, 6} {2, 1, 6}{7, 1}{7, 4} The last clique computed by MCC1 will include the last interval from each maximum independent set

Lemma 3.3 (cont.) {5, 3, 6}{2, 3, 6} {2, 1, 6}{7, 1}{7, 4} The last clique computed by MCC1 will include the last interval from each maximum independent set So, it follows that the number of cliques computed by MCC1 is exactly d max

Theorem 3.1 The construction of the HNCG, G=(V, E’) and its transitively oriented graph G*=(V, F) from the intervals belonging to a channel can be implemented 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 intervals exactly once from left to right The following cases may occur at a column position while scanning an interval starts an interval terminates all the intervals continue

Theorem 3.1 (cont.) case 1: while I i starts, if I j  S 1 terminated before I i ; orient an edge v j  v i in G* case 2: If I i terminates at current column position, we add I i in the set S 1 case 3: No action required

Theorem 3.1 (cont.) In any instance of the CRP, at most two intervals may start or terminate at the same column. In this case, both the intervals are processed one 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 a routing solution for the two layer CRP without vertical constraints using d max tracks. MCC1 can be executed in O(n+e) time.

O (n) O (n+e) In total, O (n+e) time

Theorem 3.3 The algorithm MCC2 correctly computes a routing solution for the two layer CRP without vertical constraints using d max tracks. MCC2 can be executed in O(nlogn) time

O (n) O (n+e) maintaining a balanced binary search tree of intervals sorted on their starting column numbers

O (n) O (nlogn) maintaining a balanced binary search tree of intervals sorted on their starting column numbers O (nlogn) O (nlogn) time

ThankYou