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Randy J. Chen Jan. 14, 2013

Randy J. Chen Jan. 14, 2013. Cyclic Node Merging Outline. Motivation Problem Formulation Cyclic Substitute Node (CSN) Added Cyclic Substitute Node (ACSN) Multiple Loops Handling Proposed Flow Future Work. Cyclic Node Merging Motivation (1/4). Yung- Chih Chen (2009):

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Randy J. Chen Jan. 14, 2013

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  1. Randy J. Chen Jan. 14, 2013

  2. Cyclic Node Merging Outline • Motivation • Problem Formulation • Cyclic Substitute Node (CSN) • Added Cyclic Substitute Node (ACSN) • Multiple Loops Handling • Proposed Flow • Future Work

  3. Cyclic Node Merging Motivation (1/4) • Yung-Chih Chen (2009): Mergers can be efficiently found by logic implications. a (0,X) v4 (1,0) v1 1. Compute MAs (T = sa0) 2. Compute MAs (T = sa1) 3. Sustitude_Node nodes having different values in MAs (T = sa0) and MAs (T = sa1), and not in the transitive fanout cone of T (1,1) b (1,1) v2 (0,0) v5 (1,0) c (1,0) v3 (1,0) d MAs(T=sa0): {v3=1, c=0, d=1, v2=1, v5=1,b=1,v1=1, v4=0} MAs(T=sa1): {v3=0, v5=0,v2=1, c=0, b=1, d=0, v1=0}

  4. Cyclic Node Merging Motivation (2/4) • Is it possible that we merge T with S which is in the transitive fanout cone (TFC) of T? a (0,X) v4 (1,0) v1 (1,1) b (1,1) v2 (0,0) v5 (1,0) c (1,0) v3 (1,0) d Replacing v3 with v5?

  5. Cyclic Node Merging Motivation (3/4) • Marc D. Riedel (2003): Introducing feedback in substitution phase can further reduce the area. • a • d • c • g • a = x1c + c’d • b = x0’(x1d + c) • c = x0x2x3’ + x2’(x0’x1’ + d) • d = x1’x2’x3 + x3’(x1x2’ + x2(x0’ + x1’)) • e = b + x3’cd • f = x1’x2’ + a’c + de’ • g = a + b’f • a = x1’c + x0’x3’c’ • b = x0’e • c = x0x2x3’ + x2’(x1’x3 + e) • d = x1e+ (x2 + x3)a • e = x3’f’ +x2’(x0’ + x1)f • f = x3’a’ + (x2’ + x0’x1’)g • g = a + x3’b’ • b • f • d • a • e • e • f • b • c • g • Acyclicform of 7-segment display • with literal count = 37 • Cyclicform of 7-segment display • with literal count = 34

  6. Cyclic Node Merging Motivation (4/4) • But their approach spends large amount of time on trial and error. • f1 = x2’x3+ f2’f3 • f2= x1f3’ + x3’f1’ • f3= f1f2’ + x2’x3’ f1 f1 f1 … • f1 = x2’x3+ f2’f3 • f2= x1f3’ + x3’f1’ • f3 = x1’f1 + x2’x3’ cost 12 cost 13 cost 12 f3 f3 f3 f2 f2 f2

  7. Cyclic Node Merging Problem Formulation • Given: a combinational circuit • Derive: an area-optimized circuit • By means of: merging a target node and a substitute node in its transitive fanout cone • With: efficient logic implications

  8. Cyclic Node Merging Cyclic Substitute Node (1/7) • The most important factor of the cyclic combinational circuit is that all loops in the circuit are false. c b a a(b+c) b+ac c(a+b) a+bc b(a+c) c+ab

  9. Cyclic Node Merging Cyclic Substitute Node (2/7) • If both D and D’ can NOT be propagated from T to S, then the circuit after merging T and S will still be combinational. • Proof: 1. As D and D’ cannot be propagated from T to S, for any input assignment, S evaluates to a definite Boolean value other than D and D’, which means S does not depend onT.

  10. Cyclic Node Merging Cyclic Substitute Node (3/7) • If both D and D’ can NOT be propagated from T to S, then the circuit after merging T and S will still be combinational. • Proof: 2. Thus even if we replace T with S, S will still evaluate to a definite Boolean value (1 or 0), so do all the nodes originally depend on T. 3. Then the circuit after merging T and S will still be combinational since all the nodes have definite outputs for any input assignment.

  11. Cyclic Node Merging Cyclic Substitute Node (4/7) D and D’ can NOT be propagated from T to S Still combinational Nodes having different values in MAs (T = sa0) and MAs (T = sa1) Identical behavior 1. (T = D & S = 1) and (T = D’ & S = 0) 2. (T = D & S = 0) and (T = D’ & S = 1) Functionally equivalent

  12. Cyclic Node Merging Cyclic Substitute Node (5/7) • Finding MAs when T is stuck-at 0. a b 1 1 0 1 8 0 0 0 b a c 1 3 1 D S f 1 1 T d D D 1 a 1 D’ 2 D 0 5 9 D’ c 0 0 1 0 1 c b 1 7 0 e 0 b 1 6 a 1 1 0 4 0 0 c MAs(T=sa0): {a=1, b=0, c=1, n1=0, n2=D’, n3=D, n4=0, n5=0, n6=1, n7=0, n8=0, n9=0, S=1}

  13. Cyclic Node Merging Cyclic Substitute Node (6/7) • Finding MAs when T is stuck-at 1. a b 1 0 8 0 b a c 1 3 D S f 0 T d D D’ a D 2 D 5 9 D c 1 c b 7 e b 1 6 a 4 c MAs(T=sa1): {b=0, n1=0, n2=D, n3=D, S=0}

  14. Cyclic Node Merging Cyclic Substitute Node (7/7) • Replacing T with S dose not change the functionality. a b 1 8 b a c 3 S f T d a 2 5 9 c c b 7 e b 6 a 4 c MAs(T=sa0): {a=1, b=0, c=1, n1=0, n2=D’, n3=D, n4=0, n5=0, n6=1, n7=0, n8=0, n9=0, S=1} MAs(T=sa1): {b=0, n1=0, n2=D, n3=D,S=0}

  15. Cyclic Node Merging Added Cyclic Substitute Node (1/4) • Though few cyclic substitute nodes exist in practice, there are still potential cyclic substitute nodes. 1 8 3 10 T 2 5 9 (D,D’) (D,0) 7 6 (D’,1) 4

  16. Cyclic Node Merging Added Cyclic Substitute Node (2/4) • These potential cyclic substitute nodes can be further divided into 2 types. 1 8 3 10 T 2 5 9 (D,D’) (D,0) 7 6 (D’,1) 4 Zero-masked One-masked

  17. Cyclic Node Merging Added Cyclic Substitute Node (3/4) • For one-masked nodes, we can let it drive an added node with no inverter to produce a masked zero. (D’,1) (1,D’) (D,1) (1,D’) (1,D) (D,1) (D’,1) (1,D) S S S S S S S S (1,0) (1,0) (0,1) (1,0) (1,0) (0,1) (0,1) (0,1) A A A A A A A A M M M M M M M M (1,0) (1,0) (0,1) (1,0) (0,1) (0,1) (1,0) (0,1)

  18. Cyclic Node Merging Added Cyclic Substitute Node (4/4) • For zero-masked nodes, we can let it drive an added node with an inverter to produce a masked one. (D’,0) (0,D’) (D,0) (0,D’) (0,D) (D,0) (D’,0) (0,D) S S S S S S S S (1,0) (1,0) (0,1) (1,0) (1,0) (0,1) (0,1) (0,1) A A A A A A A A M M M M M M M M (1,0) (1,0) (0,1) (1,0) (0,1) (0,1) (1,0) (0,1)

  19. Cyclic Node Merging Multiple Loops Handling (1/7) • Once the first set of loops is constructed, we will start finding another one. x0 x1 x2 x3 S1 T1

  20. Cyclic Node Merging Multiple Loops Handling (2/7) • But how can we guarantee the second set of loops will not affect the previous one? x0 S2 x1 x2 x3 ? T2

  21. Cyclic Node Merging Multiple Loops Handling (3/7) • Method 1: keeping the conflict conditions leading to T1 not observable at S1. x0 S2 x1 4 x2 x3 3 ? 4 2 5 6 1 T2 7 7

  22. Cyclic Node Merging Multiple Loops Handling (4/7) • Method 2: seeing if the functional change caused by the merging destroys previouscondition. x0 S2 x1 x2 x3 ? T2

  23. Cyclic Node Merging Multiple Loops Handling (5/7) • Method 3: performing an obsevrability test once the implication touched a loopnode. x0 S2 x1 x2 x3 ? T2

  24. Cyclic Node Merging Multiple Loops Handling (6/7) • If we can guarantee the newly-added loops do not affect previous loops, theseloops can beaddedsafely.

  25. Cyclic Node Merging Multiple Loops Handling (7/7) • If we can guarantee the newly-added loops do not affect previous loops, theseloops can beaddedsafely.

  26. Cyclic Node Merging Proposed Flow Target node Ti Derive MAs of Ti Find cyclic substitute node Ci YES Found? NO Find added cyclic substitute node Ci YES Merge Ti and Ci Found? NO YES More Ti? NO Termination

  27. Cyclic Node Merging Future Work • Coming up with a efficient and effective conditionthat are able to deal successfully with multiple combinational loops.

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