Lecture 7: BDD Construction and Manipulation

1. Lecture 7: BDD Construction and Manipulation 1. BDD construction2. Boolean operations on BDDs3. BDD-Based configuration

2. Today’s Program • [12:00-13:15] • Unique table • Build(t) • Apply(op,u1,u2) • BDD-Based configuration (SJHAHM04) • [13:20-14:00] Henrik Hulgaard, Configit A/S • A configuration technology company 2011 Rune Møller Jensen

3. BDD Construction 2011 Rune Møller Jensen

4. BDD construction Last week: • Make a Decision Tree of formula • Keep reducing it until no further reductions are possible This week: • Reduce the decision tree to a BDD while building it 2011 Rune Møller Jensen

5. Reduce decision tree to BDD during construction • Represent BDD by a table of unique nodes (UT) • Build BDDs recursively, i.e. to add a new node u: • Compute high(u) andlow(u) and store them in UT • Maintain BDD reductions when adding u to UT: • Only extend UT with u if high(u)  low(u) (non-redundancy test) • Only extend UT with u if u UT(uniqueness) 2011 Rune Møller Jensen

6. Unique Table Representation Node Attributes Represent Unique Tableby two tables T and H T : u (i,l,h) H: (i,l,h)  u uunique node identifier {0,1,2,3,…} i = var(u)variable index {1,2,…,n,n+1}l = low(u) node identifier of low h = high(u) node identifier of high u var(u) low(u) high(u) H is the inverse of T: T(u) = (i,l,h)  H(i,l,h) = u 2011 Rune Møller Jensen

7. Primitive Operations on T and H 2011 Rune Møller Jensen

8. Unique Table Interface: MakeNode (Mk) 2011 Rune Møller Jensen

9. Build Idea: Construct the BDD recursively using theShannon Expansion t = x t[1/x], t[0/x] • Terminal casesBuild(0) = 0Build(1) = 1 • Recursive case Build(t(xi,xi+1,…,xn)) = Mk( ) xi Build(t(0,xi+1,…,xn)) Build(t(1,xi+1,…,xn)) 2011 Rune Møller Jensen

10. Build 2011 Rune Møller Jensen

11. BDD Manipulation 2011 Rune Møller Jensen

12. Multi-Rooted BDD Unique Table contains many BDDs x1 x3 x1 x2 x1 5 6 x2 x2 3 4 x3 2 1 0 2011 Rune Møller Jensen

13. Apply • Apply(op,u1,u2):computes the BDD of u1 op u2where op : any of the 16 Boolean operatorsu1, u2: root nodes of BDDs • Relies on the Shannon expansion properties:(xt1,t0) op (xt’1,t’0)  x(t1 opt’1),(t0opt’0)(xt1,t0) optx(t1 opt),(t0opt) 2011 Rune Møller Jensen

14. Apply with op =  • Terminal case: u {0,1} u’ {0,1}App(uu’) = u u’ • Recursive case: u = xv u1, u0 u’ = xw u’1, u’0App(u u’) =Mk(xv, App(u0  u’0), App(u1  u’1) ) if v = w Mk(xv, App(u0  u’), App(u1  u’) ) if v < w Mk(xw, App(u  u’0), App(u  u’1) ) if w < v 2011 Rune Møller Jensen

15. 2011 Rune Møller Jensen

16. Construct BDDs from expression tree u6  u5 u4 u5  u2 u3 u4 u2 u3 x1 x2 x3 x1 x2 x3 2011 Rune Møller Jensen

17. Properties of Apply • Improvements? • Early termination. E.g., no reason to keep recursing if the left side in a conjunction is 0 • Complexity : O(|u1||u2|) , due to dynamic programming • So a BDD of any formula can be computed in poly time? 2011 Rune Møller Jensen

18. BDDs • Compact • Equality check easy • Easy to evaluate the truth-value of an assignment • Boolean operations efficient • SAT check efficient • Tautology check efficient • Easy to implement 2011 Rune Møller Jensen

19. BDD-Based Configuration 2011 Rune Møller Jensen

20. Configuration Problems A configuration problem C is a triple (Y,D,F) • Y is a set of variables y1, y2, … ,yn • D is the Cartesian product of their finite domainsD = D1 D2 …  Dn • F = {f1,f2,…,fm} is a set of propositional formulas over atomic propositions yi = v, where v  Di, specifying the conditions that the variable assignments must satisfy. Each formula is inductively defined byf  yi = v | f  g | f  g | f 2011 Rune Møller Jensen

21. T-Shirt Example1 • y1 {black, white, red, blue} : Color y2  {small, medium, large} : Sizey3  {“Men in black” - MIB, “Save the whales” -STW} : Print • f1  (y3 = MIB)  (y1 = black)f2  (y3 = STW)  (y2  small) • y1y2y3black small MIBwhite medium STW red largeblue 2011 Rune Møller Jensen 1: Due to Erik van der Meer, now at Microsoft

22. T-Shirt Example1 • y1 {black, white, red, blue} : Color y2  {small, medium, large} : Sizey3  {“Men in black” - MIB, “Save the whales” -STW} : Print • f1  (y3 = MIB)  (y1 = black)f2  (y3 = STW)  (y2  small) • y1y2y3black small MIBwhite medium STW red largeblue 2011 Rune Møller Jensen 1: Due to Erik van der Meer, now at Microsoft

23. T-Shirt Example1 • y1 {black, white, red, blue} : Color y2  {small, medium, large} : Sizey3  {“Men in black” - MIB, “Save the whales” -STW} : Print • f1  (y3 = MIB)  (y1 = black)f2  (y3 = STW)  (y2  small) • y1y2y3blacksmallMIBwhitemediumSTWredlargeblue 2011 Rune Møller Jensen 1: Due to Erik van der Meer, now at Microsoft

24. T-Shirt Example1 • y1 {black, white, red, blue} : Color y2  {small, medium, large} : Sizey3  {“Men in black” - MIB, “Save the whales” -STW} : Print • f1  (y3 = MIB)  (y1 = black)f2  (y3 = STW)  (y2  small) • y1y2y3black small MIBwhite medium STW red largeblue 2011 Rune Møller Jensen 1: Due to Erik van der Meer, now at Microsoft

25. T-Shirt Example1 • y1 {black, white, red, blue} : Color y2  {small, medium, large} : Sizey3  {“Men in black” - MIB, “Save the whales” -STW} : Print • f1  (y3 = MIB)  (y1 = black)f2  (y3 = STW)  (y2  small) • y1y2y3blacksmallMIBwhite medium STWred largeblue 2011 Rune Møller Jensen 1: Due to Erik van der Meer, now at Microsoft

26. Interactive Product Configurator IPC(C) • R Compile(C) • while|R| > 1 do • choose(yi = v)  ValidAssignments(R) • R  R  (yi = v) 2011 Rune Møller Jensen

27. BDD-based configuration Idea • Use a BDD to represent R • Use a polynomial-time BDD algorithm to compute ValidAssignments(R) 2011 Rune Møller Jensen

28. Represent R by a BDD • Define domains in binary00 : black, 01 : white, 10 : red, 11 : blue00 : small, 01 : medium, 10 : large0 : MIB, 1 : STW • Build a BDD of the rules

29. Compute ValidAssignments(R) • Trace paths for each variable layer in the BDD

30. Henrik Hulgaardconfigit 2011 Rune Møller Jensen