CAD Group Meeting Prepared by Ray Cheung

1. Global Flow Optimization (GFO) in Automatic Logic Design“TCAD91”byC. Leonard Berman&Louise H. Trevillyan CAD Group Meeting Prepared by Ray Cheung

2. Outline • Introduction • Background and Definitions • Min-Cut • Main Results • Discussion & Conclusions

3. Introduction • Machine becomes more complex  more logic • Use local pattern matching to improve a circuit. • Moving signal connections from one gate to another or by deleting connection. • Advantage – Efficiency, Effectiveness • Disadvantage – Miss global information (only view at small logic windows) • Main spirit of the GFO paper • Use global flow analysis • Gather and utilize information from the entire circuit. • Especially faster than local transformation for large circuits. • Present how to use this global information to simplify logic. • Supply a firm theoretical foundation.

4. Background & Definitions • Given a circuit Cwhich have function F, which associate with every wire w, F(w). • Rearrange the connection points of wsuch that F(w) does not change. • Use data flow analysis, compute a quick approx. to F(w) and then construct a weighted, directed graph. • Every cut of the graph corresponds to a set of connection points for w. • The cost of cut equals to the number of connection points of w in the new circuit.

5. Definitions • All the logic is combinational and all functional nodes are single-outputNOR’s. • If a signal i is an input to node j, we write i  j • If there is a path from node i to node j, we write i  j which means j is reachable from i.

6. Definitions • The set of immediate successors of a signal s as • sink(s) = {t: s t} • The set of immediate predecessors of s as • input(s) = {t: t  s}

7. Definitions • If n is a node in C and S is a subset of nodes of C, then “nisblockedbyS” if every path beginning at a sink of n and ending at an output contains a node of S.

8. Definitions • For each wire, x  C, there are four sets of wires, forcing sets. • Fij(x) = {s: if x = i then s = j}, i,j  {0,1} • Below shows F10 = {n1, n2, n3, n4, n8, n9, n12, n13, n18} • There are F00, F01, F10 and F11

9. Lemma 1 • If i F10(s) and there is no path i  s, then s may be connected as an input to node i without changing the function. • No path from i to s, therefore still DAG after reconnect. • If s = 0, no problem, since they are all NOR gate, 0 is non-controlling value of NOR gate. • If s = 1, since i F10(s), therefore no change.

10. Lemma 2 • If s is connected to node j in C and j is blocked by sinks(s) then the connection of s at j can be replaced by the constant 0 without changing the function of C. • If s = 0, two circuit equivalent. • If s = 1, since all NOR gate which s is connected. Therefore, there is no difference.

11. Definitions • A frontier node is defined as from this node to any primary output, there exists a path which does not contain a node in F10 or F11. i.e. the far-most node close to PO. E.g. {n8, n9, n12, n13}

12. S-frontier of i • S = { S1, …, Sn } be a set indexed by the nodes of C, and let Si be a set of nodes of C. This is summary information. • S-frontier of i denoted as c(i,S) • Element of c(i,S) are nodes j such that • j  Si • There is a path j j1  j2 …  OUTPUT such that for no jl Si; • j is reachable in the circuit from i.

13. Theorem 3 • Let C be a circuit, and s a wire in C. Let F10 be the forcing set defined earlier, and let D be a set for which sink(s)  D  F10(s) and such that for no d  D is there a path d  s. If C’ is identical to C except that in C’, signal s is an input only to those nodes in c(s,D), then C  C’. • Figure on the left side is C, and C’ on the right side. • Use Lemma 1 and Lemma 2 and the definition of c

14. Corollary 4 • Let C and C’ be two circuits which are identical except in the set of gates to which one wire s, is an input. Let F10 and F’10 be the two forcing sets for C and C’, respectively. Assume sink(s)  D  F10(s) and sink(s)  D’  F’10(s). If c(s,D) = c’(s,D’) then C  C’. • In the figure below, C  C’.

15. Alter connection of signal s • Determine a set Ns of nodes of C such that nNs with three properties. • There is no path n  s. • Node n is blocked by c(s,D). • If C is changed to C’ in which s is connected only at the nodes of Ns, c(s,D) = c’(s,D) • Properties 1 and 2 are easy to determine. • But finding the minimal set of nodes which satisfies property 3 is NP-Hard – we call it Ns problem. • Efficient method (min-cut) to find an approximate solution to the Ns problem. • Other approach is Linear programming.

16. Construction of Gs • Add Nodes • Source node labeled SOURCE with inf weight. • Sink node labeled SINK with inf weight. • For each iC10(s) which is reachable from s, add node(i) with weight = 1. • For each node i, blocked by (s,C10), from which s is not reachable, add node(i) with weight = 1. • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO • For each node(k) with k  (s,C10), add an edge from k to the sink.

17. Construction of Gs • Add Nodes • Source node labeled SOURCE with inf weight. • Sink node labeled SINK with inf weight. • For each iC10(s) which is reachable from s, add node(i) with weight = 1. • For each node i, blocked by (s,C10), from which s is not reachable, add node(i) with weight = 1.

18. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

19. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

20. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

21. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

22. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

23. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

24. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

25. Construction of Gs • Add Edges • SET = {(s,C10)} • DO WHILE SET is not EMPTY • In reverse breadth first order, choose and remove i from SET • A. Choose k{inputs(i)} for which • k C10 • node(k) Gs • l {inputs(k)} node(l) Gs • B. For k found in A above • Remove node(k) from Gs • l {inputs(k)}: add edge(l,I)to Gs, add node(l) to SET. • If no such k exists, add edge (SOURCE, i) • End DO

26. Construction of Gs • For each node(k) with k  (s,C10), add an edge from k to the sink.

27. Derived Flow graph • Add a source node (src) and sink node (snk). • For each node in F10, add node in flow graph with weight 1.

28. Max-Flow Min-Cut Technique • Edmonds-Karp: O(|V||E|2) • Ford & Fulkerson: O(|E||f|) Note: For unit capacity, |f|  |V|, so O(|V||E|) time. 12/12 12 b b a a 20 19/20 11/16 16 s t s t 10 4 9 7 10 1/4 9 7/7 13 4 12/13 4/4 c d c d 14 11/14 min-cut = max-flow

29. Theorem 5 • Let K be any node cut separating the source and sink in Gs, and let C’ be a circuit which is identical to C except that the sinks of signal s are precisely those nodes which are cut by K, then c(s,C10) = c’(s,C10). • Proof Theorem 5 needs Lemma 6, 7, 8. • Theorem 5 allows us to rearrange the connections of signal s without changing the circuit functionality. • Given a circuit C, and wire s, construct Gs • Use min-cut algorithm to find a node cut of Gs. • Rearrange the connections of s based on the cut-set & Theorem 3.

30. Lemma 6 • Every gate in C corresponding to a node in Gs is blocked by (s,C10).

31. Lemma 7 • In Gs, every node corresponding to a gate in (s,C10(s)) is reachable from the SOURCE.

32. Lemma 8 • Every gate in C’ which corresponds to a node in Gs which is reachable from K is in C10(s). • If the longest path from K to a node is 0, then s will be the direct input of this node, there it is in C10(s). • Assume it is true for the nodes of distance less than i+1 • When the node n of distance is i+1, • By the construction of Gs, it implies • One of the input to n is in C11(s) • Therefore, n C10(s)

33. Theorem 9 • Let K be any cut separating SOURCE and SINK in Gs, then let C and C’ be related as in Theorem 5, then CC’. • Theorem 5 shows two Frontier are the same. • Theorem 9 shows two circuit are the same. • We have the following - Connection Optimization Procedure • PROC: OPTCON; /* OPTimize CONnections */ • FOR EACH SIGNAL • CONSTRUCT Gs; • FIND MINIMAL CUT, K; • CREATE C’ AS IN THEOREM 9; • C = C’; • END; • END OPTCON;

34. Results • The flow graph has the following property. If the source node is re-connected to the corresponding nodes in the cut-set, the circuit is unchanged. • Therefore, the problem finding minimum number of nodes (fan-outs of s), can be solved by finding a minimum cut-set in the flow graph. • Cut-set reduce the fanouts from 6 to 3.

35. Conclusions • Present a new approach to logic optimization. • Use global flow analysis to gather information about the relationships between wires. • Use network flow to optimize the circuit. • Decreasing the number of fan-out nodes. • Is there any better approx. than recurrences approx.? • We can use Linear programming technique to replace min-cut technique. • Another ICCAD2000 paper introduced “Implication Flow Graph” based on GFO. • Because it has a larger solution space than GFO, and for the previous example, IFG can reduce the fanout of s from 6 to 2.

36. The End • Please feel free to ask any question ! 