The document discusses algorithms for solving satisfiability problems on conjunctive normal form (CNF) formulas. It first describes a trivial algorithm that tries all possible assignments, with a runtime of O(2^n*n^3). It then discusses using a recursive approach, but shows this results in an even worse runtime of O(3^n*n^3). Finally, it proposes a recursive algorithm that runs in O(1.8^n) by recursively trying assignments on subsets of variables.

1. CS 321. Algorithm Analysis & Design
Lecture 9
Recursion
Satisfiability

4. The Birthday Problem

5. The Birthday Problem
CENTRAL EUROPEAN OLYMPIAD IN INFORMATICS

6. name is a request. This request is satisfied if
and only if John invites name.
—name is a request. This request is satisfied if
and only if John doesn’t invite name.

7. If R1, . . . , Rk are requests, then
(R1 & ... & Rk) is a request.
This request is satisfied if and only if
all requests R1, . . . , Rk are satisfied.

8. If R1, . . . , Rk are requests, then
(R1 | ... | Rk) is a request.
This request is satisfied if and only if
at least one of the requests R1, . . . , Rk is satisfied.

9. If R1, R2 are requests, then (R1 R2) is a request.
This request is not satisfied if and only if
R1 is satisfied and R2 is not satisfied.

10. Google Code Jam: Round 1A, 2008

11. n flavours of milkshake, each of two kinds: malted and unmalted.
Google Code Jam: Round 1A, 2008

12. n flavours of milkshake, each of two kinds: malted and unmalted.
Every customer likes a certain set of milkshakes.
Google Code Jam: Round 1A, 2008

13. n flavours of milkshake, each of two kinds: malted and unmalted.
Every customer likes a certain set of milkshakes.
Goal: Make n milkshakes so that every customer can get at least
one of the flavours that they like.
Google Code Jam: Round 1A, 2008

14. Satisfiability

15. Input: A set of boolean variables X, and a formula F over X using
conjunctions (AND), disjunctions (OR) and negations (NOT).
Satisfiability

16. Input: A set of boolean variables X, and a formula F over X using
conjunctions (AND), disjunctions (OR) and negations (NOT).
Task: Determine if there is an assignment (t: X ⟼{0,1})
such that F evaluates to one.
Satisfiability

17. Input: A set of boolean variables X, and a formula F over X using
conjunctions (AND), disjunctions (OR) and negations (NOT).
Task: Determine if there is an assignment (t: X ⟼{0,1})
such that F evaluates to one.
Literal: A variable or it’s negation.
Satisfiability

18. Input: A set of boolean variables X, and a formula F over X using
conjunctions (AND), disjunctions (OR) and negations (NOT).
Task: Determine if there is an assignment (t: X ⟼{0,1})
such that F evaluates to one.
Literal: A variable or it’s negation.
CNF: Conjuctions of disjunctions of literals.
(x + y + z) . (w + x + (1-z)) . (x + (1-w))
Satisfiability

19. Input: A set of boolean variables X, and a formula F over X using
conjunctions (AND), disjunctions (OR) and negations (NOT).
Task: Determine if there is an assignment (t: X ⟼{0,1})
such that F evaluates to one.
Literal: A variable or it’s negation.
CNF: Conjuctions of disjunctions of literals.
(x + y + z) . (w + x + (1-z)) . (x + (1-w))
DNF: Disjunctions of conjunctions of literals.
(xz(1-z)) + (wx(1-z)) + (xwyz) + (y(1-y))
Satisfiability

20. It is widely believed that there is no polynomial time algorithm for
determining if a 3-CNF formula over n variables is satisfiable.

21. It is widely believed that there is no polynomial time algorithm for
determining if a 3-CNF formula over n variables is satisfiable.
That doesn’t stop us from thinking about algorithms anyway!

22. 3CNF - WHAT’S THE BEST WE CAN DO?

23. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.

24. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
O(2nn3)

25. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
O(2nn3)
Recursion?

26. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
O(2nn3)
Recursion?
Pick a clause: (x + y + (1-z))

27. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
O(2nn3)
Recursion?
Pick a clause: (x + y + (1-z))
SAT(F(x=1)), SAT(F(y=1)), SAT(F(z=0))

28. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
O(2nn3)
Recursion?
Pick a clause: (x + y + (1-z))
SAT(F(x=1)), SAT(F(y=1)), SAT(F(z=0))
T(n) < 3T(n-1) + poly(n)

29. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
O(2nn3)
Recursion?
Pick a clause: (x + y + (1-z))
SAT(F(x=1)), SAT(F(y=1)), SAT(F(z=0))
T(n) < 3T(n-1) + poly(n)
O(3nn3) - this is even worse!

30. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
Recursion?
O(2nn3)

31. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
Recursion?
Pick a clause: (x + y + (1-z))
O(2nn3)

32. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
Recursion?
Pick a clause: (x + y + (1-z))
SAT(F(x=1)), SAT(F(x = 0, y=1)), SAT(F(x = 0, y = 0, z=0))
O(2nn3)

33. 3CNF - WHAT’S THE BEST WE CAN DO?
Trivial Algorithm: Try all possible assignments.
Recursion?
Pick a clause: (x + y + (1-z))
SAT(F(x=1)), SAT(F(x = 0, y=1)), SAT(F(x = 0, y = 0, z=0))
T(n) < T(n-1) + T(n-2) + T(n-3) + poly(n)
O(2nn3)

35. If T(n) was bounded by O(cn),
for what c will an inductive proof work?
Let’s reverse-engineer it!

36. T(n) < T(n-1) + T(n-2) + T(n-3) + poly(n)
If T(n) was bounded by O(cn),
for what c will an inductive proof work?
Let’s reverse-engineer it!

37. T(n) < T(n-1) + T(n-2) + T(n-3) + poly(n)
c(n) < c(n-1) + c(n-2) + c(n-3) + poly(n)
If T(n) was bounded by O(cn),
for what c will an inductive proof work?
Let’s reverse-engineer it!

38. T(n) < T(n-1) + T(n-2) + T(n-3) + poly(n)
c(n) < c(n-1) + c(n-2) + c(n-3) + poly(n)
c3 < c2 + c1 + 1 + poly(n)
If T(n) was bounded by O(cn),
for what c will an inductive proof work?
Let’s reverse-engineer it!

39. T(n) < T(n-1) + T(n-2) + T(n-3) + poly(n)
c(n) < c(n-1) + c(n-2) + c(n-3) + poly(n)
c3 < c2 + c1 + 1 + poly(n)
Solving for c gives us c = 1.8-ish.
If T(n) was bounded by O(cn),
for what c will an inductive proof work?
Let’s reverse-engineer it!

40. T(n) < T(n-1) + T(n-2) + T(n-3) + poly(n)
c(n) < c(n-1) + c(n-2) + c(n-3) + poly(n)
c3 < c2 + c1 + 1 + poly(n)
Solving for c gives us c = 1.8-ish.
More details for the proof of correctness of the algorithm
are in the next set of slides.
If T(n) was bounded by O(cn),
for what c will an inductive proof work?
Let’s reverse-engineer it!

41. Note that the algorithm we discussed
is meant for 3-CNF formulas.
Can be easily generalised to d-CNF formulas.
For a general CNF formula
(where the size of the clauses are not bounded)
the number of recursive sub-instances is not bounded!