An Exact Algorithm for Difficult Detailed Routing Problems Kolja Sulimma Wolfgang Kunz J. W.-Goethe Universität Frankfurt

Motivation: routing for automated leaf cell generation  1-D and 1.5-D cells  channel routing seems more adequate than maze routing  Scenario somewhat different from conventional inter cell channel routing

Motivation: channel routing for automated leaf cell generation  a typical routing channel in a standard cell layout looks like this:  long channel with many tracks

Motivation: channel routing for automated leaf cell generation  a typical routing channel in a standard cell layout looks like this:  long channel with many tracks  fast algorithms like “Greedy Router” perform very well: they need only one or two tracks more than the optimal solution

Motivation: channel routing for automated leaf cell generation  a typical routing channel in a standard cell layout looks like this:  long channel with many tracks  fast algorithms like “Greedy Router” perform very well: they need only one or two tracks more than the optimal solution  some algorithms need a few extra columns

Motivation: channel routing for automated leaf cell generation  a channel in a leaf cell is much smaller

Motivation: channel routing for automated leaf cell generation  a channel in a leaf cell is much smaller  the use of extra area is relatively expensive

Motivation: channel routing for automated leaf cell generation  a channel in a leaf cell is much smaller  the use of extra area is expensive  or might simply be not feasible

Motivation channel routing for automated leaf cell generation small problem instances:  small constant area overhead of heuristical algorithms becomes a large relative area overhead  asymptotically slow algorithms can be practical  difficult instances  obstacles / keep-out regions  dense terminal placement

Problem Formulation routing model  grid-based  all elements are aligned to a routing grid  over-the-cell-routing  terminals can be located anywhere in the channel, not only at the top or bottom boundary  restricted two layer model  one layer for horizontal connections (trunks)  one layer for vertical connections (doglegs)  arbitrary obstacles  Each signal crosses each column at most once  (no forks or detours) Grid-based over the cell channel routing in the restricted two layer routing model

Problem Formulation example channel segment  Connect all terminals that belong to the same net in a channel of length n and a given height of t tracks

Problem Formulation example channel segment  Connect all terminals that belong to the same net in a channel of length n and a given height of t tracks

Algorithmic Approach decision problem  algorithm does not search for a minimum height routing  tries to find a routing for a given channel height of t tracks  this is all that is needed for constant height leaf cells  minimization is done by iterating over t  obviously increases the runtime  in the paper we show that the overhead is only O(1)

Algorithmic Approach dynamic programming  Sweep over the channel from left to right column by column like the “Greedy Router”  Use track assignments to record for each track the signal that leaves the current column on this track  “Greedy Router” heuristically selects one track assignment for each column  We enumerate all reachable track assignments for the column  Dynamic programming prevents an exponential growth of the search space with the channel length

Algorithm pseudo code for (each column) { for (each possible track assignment m of the previous column) {  check if m valid in this column with respect to terminals and obstacles; use m to generate new track assignments by adding all combinations of doglegs; }

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

1 1 Algorithm example  Denote a track assignment by a t-tupel of integers in the range [0,..., t], where 0 means an empty track  How can can signals 1 and 2 leave column 5 and enter column 6?

Algorithm details of column processing // takes as input a set M i-1 of all track assignments of column i-1 and // produces a set M i of all track assignments for column i next_column(M i-1, i) { M i := Ø; for (each m  M i-1 ) { m := removeRightEdgeTerminals(m, i-1); m := addLeftEdgeTerminals(m, i); m := connectTerminals(m, i); m := checkForObstacles(m, i); if (m ≠ 0) { M i := M i  {m}; M i := M i  doglegs(m); }

removeRightEdgeTerminals() addLeftEdgeTerminals()  Remove all signals that had their last terminal on the previous column.  Add signals to the assignment that have their first terminal on the current column.  Delete assignment if above is impossible

removeRightEdgeTerminals() addLeftEdgeTerminals()  Remove all signals that had their last terminal on the previous column.  Add signals to the assignment that have their first terminal on the current column.  Delete assignment if above is impossible

removeRightEdgeTerminals() addLeftEdgeTerminals()  Remove all signals that had their last terminal on the previous column.  Add signals to the assignment that have their first terminal on the current column.  Delete assignment if above is impossible

connectTerminals()  Existing signals might have terminals in this column that are not connected by the incoming assignment.  Add dogleg to connect the signal to the terminal.  Delete the assignment if the track with the terminal is blocked by another signal

doglegs()  Enumerate all the possible dogleg combinations for this assignment.  Multiple layouts might generate the same track assignment.  These layouts are equivalent under routability aspects  only one is selected by dynamic programming  selection can be used to minimize number of vias, wire length,

Computational Complexity worst case  Up to t! possible assignments per column  dogleg() dominates all other subroutines  less than 2 t/2 doglegs per assignment  worst case runtime per column O(t!2 t )  O(nt!2 t ) for a channel of length n  fixed parameter tractable:  overall runtime linear for a fixed channel height  runtime exponential in the channel height

Data Structures MDDs  Represent the set of track assignments by its characteristic function f:[0,..., t] t  {0, 1}  represent f by an MDD (Multi Valued Decision Diagram)  reduces the size of the representation  typically 1M assignments require 80K MDD-nodes  allows efficient manipulation of the assignments in the set

Data Structures MDD example  MDD representation of the example set in column 6  connections to the 0-terminal are omitted for clarity Track 0 Track 1 Track 2 Track

Data Structures MDD example Track 0 Track 1 Track 2 Track  Each path from the root to the 1-terminal represents a track assignment

Data Structures MDD example Track 0 Track 1 Track 2 Track  Each path from the root to the 1-terminal represents a track assignment

Data Structures MDD example Track 0 Track 1 Track 2 Track  Each path from the root to the 1-terminal represents a track assignment

Data Structures MDD example Track 0 Track 1 Track 2 Track  Each path from the root to the 1-terminal represents a track assignment

Data Structures MDD example Track 0 Track 1 Track 2 Track  Each path from the root to the 1-terminal represents a track assignment

Data Structures MDD example Track 0 Track 1 Track 2 Track  Each path from the root to the 1-terminal represents a track assignment

Data Structures MDD example: LeftEdgePorts()  To be able to connect a new signal to track 2 the subset of assignments with an empty track 2 must be found  calculate cofactor f| track2=0 4 1 Track 0 Track 1 Track 2 Track

Data Structures MDD example: LeftEdgePorts()  To be able to connect a new signal to track 2 the subset of assignments with an empty track 2 must be found  calculate cofactor f| track2=0 4 1 Track 0 Track 1 Track 2 Track

Data Structures MDD example: LeftEdgePorts()  To be able to connect a new signal to track 2 the subset of assignments with an empty track 2 must be found  calculate cofactor f| track2=0 4 1 Track 0 Track 1 Track 2 Track

Data Structures MDD example: LeftEdgePorts()  To insert signal 4 in the empty track move track 2 edges from 0 to Track 0 Track 1 Track 2 Track

Data Structures MDD example: LeftEdgePorts()  To insert signal 4 in the empty track move track 2 edges from 0 to Track 0 Track 1 Track 2 Track

Data Structures MDD example: LeftEdgePorts()  To insert signal 4 in the empty track move track 2 edges from 0 to Track 0 Track 1 Track 2 Track

Experimental results runtime per column in milliseconds #tracksHashset MDD

Conclusion  Intra cell channel routing can be solved exactly by exhaustive enumeration of reachable track assignments  dynamic programming yields a runtime linear in the channel length  Possible application to conventional channel routing:  less than exhaustive search  more than one assignment per column as “Greedy Router” does.  => heuristically select a set of maybe a few hundred track assignments for each column