Routing

Robust Channel Router

Figures taken from S. Gerez, “Algorithms for VLSI Design Automation”, Wiley, 1998

Channel Routing Algorithms   Previous algorithms we considered only work

when one of the types of constrains is present   Horizontal or   Vertical

  Dealing with a case where both constraints are present is a challenge

  In fact in such a situation, the problem becomes NP-complete

Robust Channel Router   Many heuristic algorithms were proposed to

address this issue   One of them is the algorithm proposed by

Yoeli and is called robust channel router   Performs well in practice and is easier to

understand but is quite involved   Uses the concept of dynamic programming

  Optimal solution of some problem can be efficiently defined in terms of problem instances of smaller size

Robust Channel Router   First we identify the number of nets present in the

channel and identify them by numbering them   Then a weight wi is assigned to each net i   This is done using a set of rules; to define these

rules, we need to define a term called “current side”   The algorithm iterates over rows; once on the top

and then on the bottom   The “side” that the algorithm is presently in is called

the “current side”

Rules for assigning weights   The weights wi for each net i are assigned based

on the following rules 1.  For all nets i whose intervals contain the columns

of maximal density, add a large number B to the weights wi

2.  For each net i that has a current side terminal at the column position x, add to wi the local density d(x)

3.  For each column x for which an assignment of some net i to the current side will create a vertical constraint violation, subtract Kd(x) from wi

Rules ..

  Typical “rules” of thumb like this are heuristics

  The performance of this algorithm is realtively insensitive to minor modifications of these rules

  Hence this algorithm gets the name “robust” router

Maximal-weight subsets   The algorithm then finds the maximal-weight subset

of nets that can be assigned to the same row   In doing so there must not be any horizontal constraint

violations   This problem can be formulated using the interval

graph discussed previously   This is an NP-complete problem   Fortunately, we don’t need to solve this problem for

general graphs   For intervals, there is a way of solving this problem

using dynamic programming

Dynamic Programming

  Define a subinstance identified by a single parameter   And

  Definition of subinstance with   From a list of all intervals, remove intervals that

extend beyond column c

Subinstance Example

Interval 0 empty 1 empty 2 empty 3 empty 4 i1 5 i1

6 i1, i6

Consider a set of intervals

i1=[1,4] i2=[12,15] i3=[7,13] i4=[3,8] i5=[5,10]

i6=[2,6] i7=[9,14]

So on and so forth

Cost of Optimal Solutions

  Cost of optimal solution for   Stored in total[c]   Can be derived from optimal costs of

subinstances with   and the weights of nets that have their right most

terminals at position c   Let’s call one such net as net n   Net n is part of the optimal solution if:

Selection of nets   If any net satisfies the prev condition it is stored in

an array selected_net[c]   In this manner the array total[] is filled up; while

entries in selected_net[] array are entered if nets are found to satisfy the condition

  The right-most selected_net is always included in the solution

  Moving left, all the other selected_nets that do not overlap are considered in the solution for the first row

  The nets assigned in the first row are removed from the list and this procedure is repeated

Rip-up and reroute   Using this procedure

  Horizontal constraints will be met   Some vertical constraints might not be met (most of them

will be met)   If vertical constraints are not met

  Rip up and reroute   Selectively undo the assignment of some nets   Reroute through area routing (also called maze routing)   If that also does not give a solution; add another row and

run the algorithm again

41 32 30 5

(a) (b)41 2

41 32 3

41 2(c)

41 32 3

41 2

Example   4 nets; numbered   Local densities

  d(1) = 1   d(2) = 2   d(3) = 2   d(4) = 3   d(5) = 2

  B = 1000   K = 5

Example – 1st iteration w1 = (0) + (1) + (-5*2) = -9

w2 = (1000) + (2) + (-5*3) = 987 w3 = (1000) + (2+2) + (0) = 1004

w4 = (1000) + (3) + (-5*2) = 993

total[1] = 0

total[2] = max(0,0-9) = 0

total[3] = 0

total[4] = max(0,0+987) = 987

total[5] = max(987,0+1004,0+993)=1004

selected_net[1] = 0

selected_net[2] = 0

selected_net[3] = 0

selected_net[4] = 2

selected_net[5] = 3 Net 3 only will occupy row 1

Example – 2nd iteration   We switch the current

side to the bottom row   Net 3 is left out of the

list of nets   Calculate densities

again   d(1) = 1   d(2) = 2   d(3) = 1   d(4) = 2   d(5) = 1

41 32 30 5

(a) (b)41 2

41 32 3

41 2(c)

41 32 3

41 2

Example – 2nd iteration w1 = (1000) + (2) + (0) = 1002

w2 = (1000) + (2) + (-5*2) = 992

w4 = (1000) + (1) + (-5*2) = 991

total[1] = 0

total[2] = max(0,0+1002) = 1002

total[3] = 1002

total[4] = max(1002,0+992) = 1002

total[5] = max(1002,1002+991)=1993

selected_net[1] = 0

selected_net[2] = 1

selected_net[3] = 0

selected_net[4] = 0

selected_net[5] = 4

Net 4 and net 1 will occupy bottom row

Solution   3rd iteration is trivial; Net 2 is assigned to a single

row; the second one   This is not a solution (net 4 obstructs) so rip up and

reroute   Shift net 4 to middle row and area route net 2

41 32 30 5

(a) (b)41 2

41 32 3

41 2(c)

41 32 3

41 2

Global Routing   Precedes local routing and follows placement   Detailed routing is not done at this stage   Roughly determines the wiring channels a

connection will run

(a) (b)

6 (= n)1 2 3 40 5 6 (= n1 2 3 40 50

1

2

3

4

5 (= m)

Global Routing   Application to standard cells   Characterized by rows of cells seperated by wiring channels   If terminals are in same channel facing each other; local

routing is performed; otherwise global routing is used   Compute a minimum rectilinear Steiner tree; connect centers

(a) (b)

Feedthrough wires   Steiner trees have vertical segments that cross rows

of standard cells   They are realized in the following ways

  By using a wiring layer not used by standard cells   Using feedthrough cells; non-functional cells inserted to

realize vertical connections   Such cells are used in conjunction with placement

  Using feedthrough wires available within standard cells

  We refer to all these vertical connections as feedthrough cells; they have to be minimized

Feedthrough wires   Detailed routing such as channel routing will be

used to fix exact locations of vertical connections   The vertical wires might have to be shifted to be

aligned with exact feedthrough connections   Segments at approx same positions can be

permuted to reduce densities in channels   If feedthrough resources are scares a Steiner tree

with vertical connections having a higher cost can be made

  Alternative solution (b)

Timing   The minimal length Steiner tree might not be what

we are for all nets after considering timing issues   For timing; critical path is important

  Critical path determines the system’s operation speed

  Cells connected to critical nets get priority to be placed together during placement

  Better delay models required than length of wires   Length of wire metric: equivalent to 1R and 1C   Long wires behave as transmission lines   In practice Elmore delay or better delay models are used

Timing   Steiner trees are constructed to optimize lengths of

critical connections   If Steiner tree of one wire is constructed before the

other; it prevents others from being laid out optimally   Global routers should optimize the overall cost

functions   In std-cell layout the overall area is minimized if the

sum of all channel widths is minimized   Width of the channel is estimated by the density of

the channel

Building block layout   More complex than std

cells   Use slicing floorplans   The “channels” are

picked from the slicing tree

  To assign channels   Do a depth-first search   Start numbering from the

deepest channel

Building block layout   Channel routing is used to wire up within the

channels   Steiner trees are used to wire-up connections not

restricted to channels   If additional wiring layer is not available

  All Steiner trees run along wiring channels   A graph is constructed to represent the structure of routing

area   Channels are edges; their meeting points vertices   Problem now is to find a Steiner tree within the graph as

opposed to a plane

Problem Definition

  Local vertical density dv(i,j)   Number of wires crossing vertical line j between horizontal lines

i-1 and i   Local horizontal density dh(i,j)

  Number of wires crossing horizontal line i between vertical lines j-1 and j

(a) (b)

6 (= n)1 2 3 40 5 6 (= n1 2 3 40 50

1

2

3

4

5 (= m)

Problem Definition The density of the channel between grid lines i-1 and i is

The goal of global routing is to minimize total channel density:

Subject to

is the maximum feedthroughs that can be accomodated per horizontal grid segment

Algorithms for global routing   Solve sequentially; one net after another using Lee’s

algorithm   Con: Strong dependency on net ordering

  Construct Steiner trees for all nets   Such that the trees avoid congested areas

  Other way is to construct all Steiner trees   And modify shapes of wires that cause overcongestion

  We will follow the last method   Hierarchical approaches can be followed for

simplicity

Steiner trees

  Start with spanning trees   Flip the L shape to get (b); s1 is Steiner point   Flip the L shape again to get (c); s2 is another Steiner point   s1 is a redundant Steiner point; it has only two connections   A Steiner point has three or more connections   Steiner trees are usually 2/3 of the length of spanning trees

p1

p2p3

p1

p2

p3s1 s2

(a) (b) (c)

p1

p2

p3

s1

Prim - min spanning tree

O(n2)

v3v1

v2

v4

v51

2

3

1

22

5

2

4 v3v1

v2

v4

v51

(a) (b)

21

2

vi .distance, vi .via edge, for i =iteration u 1 2 3 4 50 v1 2, (v1, v2) 4,(v1, v3) 3,(v1, v4) +∞,?1 v2 1, (v2, v3) 2,(v2, v4) 5,(v2, v5)2 v3 1, (v3, v4) 2,(v3, v5)3 v4 2, (v3, v5)4 v5

1-Steiner tree problem

  Finding minimal Steiner tree for a set of points in NP-complete

  Define 1-Steiner problem   Say you have the minimum spanning tree for set

of points P (Prim’s algorithm; in P complexity)   Add one Steiner point (called s) to this and find

the minimum spanning tree   Steiner tree heuristic is to solve the 1-Steiner

problem repeatedly

1-Steiner tree problem   Implementation

  Visit all candidate points to get the minimum spanning tree for points in P union s

  Pick the point that lead to the cheapest spanning tree

  Two issues 1.  Identify the candidate points 2.  Efficient method to incrementally construct a new

spanning tree without calling Prim’s algorithm again

Candidate points- Hanan Points   Hanan solved this in 1966   Proved that an optimal rectilinear Steiner tree can

be embedded in the grid composed of grid lines having points in P

(a) (b)

Updating the spanning tree

  Addressed by Griffith et al 1.  Select candidate point 2.  Divide grid into 4 parts diagonally 3.  Find closest points in each part 4.  Connect to candidate point; avoid (remove)

cycles; to get a spanning tree 5.  Repeat for all candidate points 6.  Pick the one with minimum length

Example of updating

Steiner tree heuristic 1.  Find minimum spanning tree 2.  Use 1-Steiner algorithm to add one point 3.  If gain in step 2 is positive accept the new tree 4.  Goto step 2; repeat as long as gain is positive

p1

p2

p3

p4p5

st

Global routing   Generate Steiner trees

for all nets   Congested areas have

to be cleaned up   Use local

transformations of Steiner trees for clean up

  Transformations may increase tree length but decrease total area