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

2. 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

3. 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

4. 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

5. 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

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

7. 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

8. 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

9. 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

10. 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

11. Problem Formulation example channel segment 1 1 33 242 1 4 5 5  Connect all terminals that belong to the same net in a channel of length n and a given height of t tracks

12. 5 5 1 1 33 242 1 4 Problem Formulation example channel segment 1 3 4  Connect all terminals that belong to the same net in a channel of length n and a given height of t tracks

13. 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)

14. 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

15. 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; }

16. 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? 5 5 1 33 242 4 3 4 1 2 1

17. 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? 5 5 1 33 242 1 4 3 4 1 10201020 1 2

18. 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? 5 5 1 33 242 1 4 3 4 1 10201020 01200120 1 2

19. 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? 5 5 1 33 242 1 4 3 4 1 10201020 01200120 00210021 1 2

20. 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? 5 5 1 33 242 1 4 3 4 1 10201020 01200120 00210021 12001200 1 2

21. 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? 5 5 1 33 242 1 4 3 4 1 10201020 01200120 00210021 12001200 10021002 1 2

22. 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? 5 5 1 33 242 1 4 3 4 1 10201020 01200120 00210021 12001200 10021002 01020102 1 2

23. 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); }

24. 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

25. 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 10201020 01200120 00210021 12001200 10021002 01020102 4 1

26. 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 10201020 01200120 00210021 12001200 10021002 01020102 12401240 10421042 01420142 4 1

27. 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. 4 1 12401240 10421042 01420142 1 4 4 1 1 4 2 12401240 10421042 01420142 2

28. 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,... 10421042 4 1 4 2 10421042 1 4 1 1 4 2 10421042 12401240

29. 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

30. 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

31. Data Structures MDD example  MDD representation of the example set in column 6  connections to the 0-terminal are omitted for clarity. 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4

32. Data Structures MDD example 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4  Each path from the root to the 1-terminal represents a track assignment

33. Data Structures MDD example 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4  Each path from the root to the 1-terminal represents a track assignment

34. Data Structures MDD example 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4  Each path from the root to the 1-terminal represents a track assignment

35. Data Structures MDD example 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4  Each path from the root to the 1-terminal represents a track assignment

36. Data Structures MDD example 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4  Each path from the root to the 1-terminal represents a track assignment

37. Data Structures MDD example 10201020 01200120 00210021 12001200 10021002 01020102 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4  Each path from the root to the 1-terminal represents a track assignment

38. 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 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4 10201020 01200120 00210021 12001200 10021002 01020102

39. 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 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4 10201020 01200120 00210021 12001200 10021002 01020102

40. 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 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4 12001200 10021002 01020102

41. Data Structures MDD example: LeftEdgePorts()  To insert signal 4 in the empty track move track 2 edges from 0 to 4 4 1 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4 12001200 10021002 01020102

42. Data Structures MDD example: LeftEdgePorts()  To insert signal 4 in the empty track move track 2 edges from 0 to 4 4 1 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4 12001200 10021002 01020102

43. Data Structures MDD example: LeftEdgePorts()  To insert signal 4 in the empty track move track 2 edges from 0 to 4 4 1 Track 0 Track 1 Track 2 Track 3 0 1 2 3 4 1 0 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 40 1 2 3 4 12401240 10421042 01420142

44. Experimental results runtime per column in milliseconds #tracksHashset MDD 3 2 5 5 4 5 7 100 5 9 6000 20 11 300 138000

45. 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