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Sequential Optimization without State Space Exploration

Sequential Optimization without State Space Exploration. A. Mehrota, S. Qadeer, V. Singhal, R. Brayton, A. Sangiovanni-Vincentelli, A. Aziz Presented by: Andrew Mihal. Outline. Goal Definitions Combinational Redundancies Sequential Redundancies Experimental Results Future Work.

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Sequential Optimization without State Space Exploration

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  1. Sequential Optimizationwithout State Space Exploration A. Mehrota, S. Qadeer, V. Singhal, R. Brayton, A. Sangiovanni-Vincentelli, A. Aziz Presented by: Andrew Mihal

  2. Outline • Goal • Definitions • Combinational Redundancies • Sequential Redundancies • Experimental Results • Future Work

  3. Main Idea • Optimize a sequential circuit for area • Locate and remove redundancies in the circuit • Avoid exploring state space (exponential) • Search for redundancies using a method based on implications and recursive learning • Result is a safe delayed replacement of original circuit • Scales well and works on large circuits

  4. Definitions • Redundancy: a net that does not affect circuit operation • Similar to an untestable stuck-at fault • Compatible redundancies: a set of redundant nets that are independent • Remove one redundancy and the other redundancies don't go away • Can simultaneously remove all redundancies

  5. Example • n1 = 0  n2 unobservable • n1 = 1  n2 = 1 • Therefore n2 is stuck-at-1 redundant • Can replace n2 with constant 1. • Similarly, n1 is stuck-at-1 redundant • But these redundancies are not compatible • Cannot be simultaneously replaced n1 n o n2

  6. Finding Combinational Redundancies • Choose a net • Assign it a value v and do implications. • Switch to v’ and do implications. • Find a commonality between the sets of implications. • Redundancy if: • A net n is constant b in both sets of implications • n is constant in one set and unobservable in the other

  7. Implication Rules b b’ 1 1 0 1 0 b’ b C1 C2 C3 0/1 1 1 0 1 0 1 0/1 C4 C5 C6

  8. Implication Rules •  means that a net is unobservable  0    O1 O2

  9. Recursive Learning • When no implication rules apply, recursively make another assumption • Helps to find more redundancies

  10. d f g c a e b Recursive Example • a = 0 0

  11. d f g c a e b Recursive Example 0 • a = 0  d = 0  0

  12. d f g c a e b Recursive Example 0 • a = 0  d = 0  f = 0 0 0

  13. d f g c a e b Recursive Example • a = 0  d = 0  f = 0 • a = 1  ? 1

  14. d f g c a e b Recursive Example 0 • a = 0  d = 0  f = 0 • a = 1  ? • d = 0  1

  15. d f g c a e b Recursive Example 0 • a = 0  d = 0  f = 0 • a = 1  ? • d = 0 f = 0 0 1

  16. d f g c a e b Recursive Example 1 • a = 0  d = 0  f = 0 • a = 1  ? • d = 0  f = 0 • d = 1  1

  17. d f g c a e b Recursive Example 1 • a = 0  d = 0  f = 0 • a = 1  ? • d = 0  f = 0 • d = 1  b = 1 1 1

  18. d f g c a e b Recursive Example 1 • a = 0  d = 0  f = 0 • a = 1  ? • d = 0  f = 0 • d = 1  b = 1  e = 1 1 1 1

  19. d f g c a e b Recursive Example 1  • a = 0  d = 0  f = 0 • a = 1  ? • d = 0  f = 0 • d = 1  b = 1  e = 1  f =  • f is stuck-at-0 redundant 1 1 1

  20. d f g c a e b Recursive Example • a = 0  f = 0 • a = 1  f is stuck-at-0 redundant •  f is stuck-at-0 redundant and can be replaced with constant 0

  21. Problem • This algorithm can find incompatible redundancies 1  0 1 1 1 0 0 1 

  22. Solution • Don’t let the algorithm overwrite an existing label b d a2 e a a1 c • a=0  c=1 • a=1  c=0, d=1  e=0 • No redundancies found

  23. Solution • We missed a redundancy! b d a2 e a a1 c • a=1  d=1 c=0  d=  a2=  • A2 is stuck-at-0 redundant

  24. Solution • It is safe to overwrite a constant label with a  • The authors prove it • But first, we need more definitions

  25. Definitions • Circuit • A graph G = (V, E) • V = PI’s, PO’s, gates, latches • E = wires • Assumption • A labeling of the nets P  E • Each net n  P is labeled with value v  {0,1}

  26. Definitions • AP • The set of all possible assumptions on P • Consistent • An assumption A  AP is consistent if there exists an assignment to the PI’s that satisfies it PI’s PO’s 1 0 Inconsistent assumption

  27. Definitions • Parent and Child nets • Share a common node v • Parent is an input to v • Child is an output of v • Sibling nets Parent v Child v Siblings

  28. Definitions • Implication graph • A DAG that details the implications leading from an assumption to a label a=1 Assumption b d a2 e c=0 a a1 c d= Label

  29. Definitions • Compatible labels • A set of labels C derived from an assumption A is compatible if: • Each label c  C has a valid implication graph Gc • Each label in Gc C • This means that no two implication graphs may have contradicting nodes

  30. Label Compatibility a1  a 0 0  a2 a=0 a1=0 a2= Incompatibility! a=0 a2=0 a1=

  31. Label Compatibility • One way to ensure compatible labels is to never switch a label • But this misses some redundancies, as we saw before • Now we will prove that it is safe to replace a constant label with a  label

  32. Proof by Contradiction • Assume that replacing a constant label by a  label creates an incompatibility Assumption New Implication Graph m = Incompatibility! ni Existing ImplicationGraph m = a

  33. Proof by Contradiction • Show that the existing implication graph can always be rewritten to not use m=a • 3 cases to consider: • ni-1 is a child of ni • ni-1 and ni are siblings • ni-1 is a parent of ni • These are the only implication relationships allowed given our implication rules ni-1 ni m = a

  34. Case 1: Child • We have two pieces of information • 1. ni-1 implies ni=a using some rule • 2. something implies ni= using some rule • No assignment satisfies both 1 and 2 • Thus, case 1 cannot happen ni=1 ni-1=1 ni= ni= ni-1=0 ni-1= 0

  35. Case 2: Sibling • 1. ni+1 is a sibling • Rewrite without using ni=a ni-1=a ni=a ni+1=a ni-1=a ni=a ni+1=a

  36. Case 2: Sibling • 2. ni+1 is a parent • Rewrite without using ni=a ni-1=a ni+1=a ni=a ni-1=a ni+1=a ni=a

  37. Case 2: Sibling • 3. ni+1 is a child • Rewrite without using ni=a ni= ni= ni+1=0 ni+1= 0 n’ ni= ni+1=0 0 n’

  38. Case 2: Sibling • What about this case? • If ni+1=, then ni+2, ni+3, … must also be  because a  can only imply a  ni= ni+1= ni-2=ci-2 ni-1=ci-1 ni=ci ni+1= ni+2=  Contains no  labels Can be modified not to use m=a

  39. Case 3: Parent • Same as case 2: sibling ni-1=a ni=a ni+1=a ni-1=a ni=a ni+1=a

  40. Case 3: Parent • Same as case 2: parent ni+1=a ni=a ni-1=a ni+1=a ni=a ni-1=a

  41. Case 3: Parent • Same as case 3: child ni= ni= ni+1=0 ni+1= 0 n’ ni= ni+1=0 0 n’

  42. Proof Summary • In every case, existing implication graph can be modified to be compatible with m= • Therefore no incompatibility arises when replacing m=a with m=, given a consistent initial assumption

  43. Original Algorithm for Optimization • Now we have formalized the way implications are found • Next, apply to sequential circuits • Choose a net • Assign it a value v and do implications. • Switch to v’ and do implications. • Find a commonality between the sets of implications. • Redundancy if: • A net n is constant b in both sets of implications • n is constant in one set and unobservable in the other

  44. Sequential Redundancies • Extend labeling algorithm to include time stamp with each label • New implication rule for latches: • Gate implication rules are the same • Implicator time stamps must be the same nt=a nt+1=a

  45. Sequential Redundancies • Time t is the time of the assumption • Implications can go forward across latches • t+1, t+2, ... • And backwards • t-1, t-2, ... • We can still overwrite a constant label with a  label • But we cannot overwrite a constant label with a different constant, even at a different time step

  46. C-delayed Replacement • A net may become stuck-at-n redundant at some time t+c • Removing the redundancy may change circuit behavior in the time between t and t+c • This is because latch outputs will initialize non-deterministically. • Assume no designated reset state • After time t+c the optimized circuit will behave the same as the unoptimized circuit

  47. C-delayed Replacement • This is allowed • The initial c period corresponds to a reset period, before the optimized circuit can be used • Most designers use lengthy resets anyway • In experimental results, c < 10000 • 10000 clocks = 100us at 100MHz • Not a severe restriction

  48. When to Remove a Sequential Redundancy • Replace net n with constant v if • nt’ = v • or nt’ is unobservable for all assumptions • For each assumption, there is an implication graph that shows nt’=v or nt’=

  49. When to Remove a Sequential Redundancy • Let t’’ be the least time offset on any label in those graphs • If t’’ > t’ then c = 0 • Else c = t’ - t’’ • n is c-cycle stuck-at-v redundant • Replace n with constant v to get ac-delayed safe replacement of the circuit

  50. Experimental Results • First run script.rugged on test circuit • Map to two-input gates and inverters • Run recursive learning redundancy removal algorithm • Compare optimized circuit area • CPU times from << 1 minute to 100 minutes on a dual Alpha 300 with 2GB RAM

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