Routing. Robust Channel Router. Figures taken from S. Gerez, Algorithms for VLSI Design Automation, Wiley, 1998

Transcription

1 Routing Robust Channel Router Figures taken from S. Gerez, Algorithms for VLSI Design Automation, Wiley, 1998

2 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

3 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

4 Robust Channel Router First we identify the number of nets present in the channel and identify them by numbering them Then a weight w i 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

5 Rules for assigning weights The weights w i 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 w i 2. For each net i that has a current side terminal at the column position x, add to w i 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 w i

6 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

7 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

8 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

9 Subinstance Example Consider a set of intervals i 1 =[1,4] i 2 =[12,15] i 3 =[7,13] i 4 =[3,8] i 5 =[5,10] i 6 =[2,6] i 7 =[9,14] Interval 0 empty 1 empty 2 empty 3 empty 4 i 1 5 i 1 6 i 1, i 6 So on and so forth

10 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:

11 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

12 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

13 Example (a) 4 4 nets; numbered Local densities 1 2 d(1) 3 4= 13 d(2) = 2 d(3) = 2 d(4) = 3 d(5) = 2 B = 1000 K 1= (b) (c)

14 Example 1 st iteration w 1 = (0) + (1) + (-5*2) = -9 w 2 = (1000) + (2) + (-5*3) = 987 w 3 = (1000) + (2+2) + (0) = 1004 w 4 = (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 selected_net[1] = 0 selected_net[2] = 0 selected_net[3] = 0 selected_net[4] = 2 total[5] = max(987,0+1004,0+993)=1004 Net 3 only will occupy row 1 selected_net[5] = 3

15 Example 2 nd 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) = (a)

16 Example 2 nd iteration w 1 = (1000) + (2) + (0) = 1002 w 2 = (1000) + (2) + (-5*2) = 992 w 4 = (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, )=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

17 Solution 3 rd 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 (a) (b) (c)

18 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 (= n) (= n (a) 5 (= m) (b)

19 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)

20 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

21 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)

22 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

23 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

24 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

25 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

26 Problem Definition (= n) (= n (a) 5 (= m) Local vertical density d v (i,j) Number of wires crossing vertical line j between horizontal lines i-1 and i Local horizontal density d h (i,j) Number of wires crossing horizontal line i between vertical lines j-1 and j (b)

27 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

28 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

29 Steiner trees p 2 p3 p 2 s 1 p 3 p 2 s2 s 1 p 3 p 1 p 1 p 1 (a) (b) (c) Start with spanning trees Flip the L shape to get (b); s 1 is Steiner point Flip the L shape again to get (c); s 2 is another Steiner point s 1 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

30 Prim - min spanning tree v v v v 5 2 v 1 2 v v 2 v 5 v 4 (a) (b) v 4 v i.distance, v i.via edge, for i = iteration u v 1 2,(v 1,v 2 ) 4,(v 1,v 3 ) 3,(v 1,v 4 ) +,? 1 v 2 1,(v 2,v 3 ) 2,(v 2,v 4 ) 5,(v 2,v 5 ) 2 v 3 1,(v 3,v 4 ) 2,(v 3,v 5 ) 3 v 4 2,(v 3,v 5 ) 4 v 5 O(n 2 )

31 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

32 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

33 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)

34 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

35 Example of updating

36 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 p 2 p 3 p 1 t s p 4 p 5

37 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