Logik 2

1
Logik 2

• More propositional logic

2
Example Magic Square

• Every element is 1,2, , or 9
• Each row, each column, and each diagonal sums to
  15
• Can I present this requirement as a propositional
  logic formula?

? 15
? 15
? 15
?15
?15
?15
?15
?15
3
First Steps
15 159 168 249 258 267
348 357 456

• Introduce variables x11x19, x21x29,,x19x99
• xij is supposed to mean xi j
• Let r1 (x11?x25?x39) ? (x11?x26?x38) ?
  (x11?x28?x36) ? row 1 sums to 15
• Similar for r2, r3, c1, c2, c3, d1, d2 rows,
  columns, and diagonals
• A position holds exactly one number
• Let vij be an abbreviation for xij ? ?k ?j ? xik
• Let vi ?j vij ? ?j xij exactly one number for
  square i
• Let v ?i vi exactly one number in every
  position
• Each number occurs once
• Let ui ?j ?k ?j ? (xji ? xki) number i occurs
  once in square
• Let u ?i ui Each number occurs once in square
• The formula is
• v ? u ? r1 ? r2 ? r3 ? c1 ? c2 ? c3 ? d1 ? d2

4
Next

• Write a script to generate the formula
• Download limboole and a SAT solver like Precosat
  from Armin Bieres home page
• Run the formula through precosat to get an answer
• Then, try the same for Sudoku

5
Some Rules

• ?(a ? b) is equivalent to ?a ? ?b
• ?(a ? b) is equivalent to ?a ? ?b
• ? ? ? is equivalent to ?? ? ?
• We can see ? as an abbreviation, leaving us only
  with ?, ?, and ?.
• (Prove by drawing truth tables)
• Proof by induction. Let P be a claim about
  formulas
• If P(p) and P(?p) and
• P(?) and P(?) implies P(? ? ?)
• P(?) and P(?) implies P(? ? ?)
• P(?) implies P(??)

6
Equivalence of Circuits

• Circuits can safely be modeled at Boolean level
• There is a big market for tools that optimize
  circuits.
• But they make mistakes!
• How do we check whether two circuits are
  equivalent?

7
Example

• Let ? ? ? abbreviate (? ? ?) ? (? ? ?)
• Let ? ? ? abbreviate ?(? ? ?)
• Compare
• c ? (?a ? b) ? ( a ? ? b)
• c ? (a ? b) ? (?a ? ?b)
• Rename
• ? c ? (?a ? b) ? ( a ? ? b)
• ? c ? (a ? b) ? (?a ? ?b)
• Introduce a comparison
• c ? c
• Is the following satisfiable?
• ? (c ? (?a ? b) ? ( a ? ? b)) ? (c ? (a ? b)
  ? (?a ? ?b)) ? (c ? c)
• Claim ? is satisfiable iff ? and ? are
  equivalent

8
CNF

• SAT solvers decide satisfiability very quickly
• For simplicity, sat solvers only take conjunctive
  normal form (CNF)
• Definitions
• A literal is a propositional symbol or its
  negation (p, ?p)
• A clause is a disjunction of literals (p ? q ?
  ?r)
• A formula in CNF is a conjunction of clauses
  ((?p?q) ? (?q ? r) ? (?r ? ?p))
• Claim I can rewrite any formula ? to a formula
  CNF(?) in CNF such that ? is satisfiable iff
  SAT(?) is satisfiable

9
Rewriting to CNF, Conjunction

• c ? a ? b
• is equivalent to
• c ? a and
• c ? b and
• a ? b ? c
• is equivalent to
• (?c ? a) ? (? c ? b) ? (?a ? ?b ? c)

10
Rewriting to CNF

• Idea
• For every subformula introduce a new variable
• ? (p ? q) becomes CNF(?) (xp?q ? p ? q) ?
  xp?q
• ? (p ? q) ? ?p becomes CNF(?) (xp?q ? p ? q)
  ? (x?p ? ?p) ? (x? ? xp?q ? x?p) ? x?
• Note
• ? and CNF(?) are not equivalent!
• ? and CNF(?) are equisatisfiable
• CNF(?) can easily be written as CNF
• Only a linear blowup
• Rewriting to CNF
• r ? (p ? q) is equivalent to (p ? r) ? (q ? r) ?
  (r ? p ? q) is equivalent to (?p ? r) ? (?q ? r)
  ? (?r ? p ? q)
• Similar for conjunction and negation

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Rewriting a formula to CNF

• For any subformula ?,
• If ? is a literal, let x? ?
• Otherwise, let x? be a new variable
• Definition CNF(?) EXT(?) ? x?
• Base EXT(p) p EXT(?p) ?p
• Recursion
• EXT(? ? ?) EXT(?) ? EXT(?) ? (?x? ? x?) ? (?x?
  ? x?) ? (?x? ? x? ? x?)
• Note x? ? ? ?
• EXT(? ? ? ) EXT(?) ? EXT(?) ? (x? ? x?) ? (x? ?
  x?) ? (?x? ? ?x? ? x?)
• Note x? ? ? ?
• EXT(??) EXT(?) ? (x? ? x?) ? (?x? ? ?x?)
• Note x? ??

The formula is satisfied
The assignment follows the rules of the formula
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Proof of Equisatisfiability

• Theorem ? is satisfiable iff CNF(?) is
  satisfiable
• Lemma 1 If CNF(?) is satisfiable, then ?
  satisfiable
• Lemma 2 If ? is satisfiable, then CNF(?)
  satisfiable

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Lemma 1

• Lemma 1 If CNF(?) is satisfiable, then ?
  satisfiable
• Proof
• We will prove if e CNF(?) then e ?
• Proof by induction. Suppose e CNF(?). Then, e
  EXT(?) ? x?
• base case ? p, ? ?p. Clear because EXT(?)
  ?
• Induction step
• Suppose ? ? ? ?
• EXT(?) EXT(?) ? EXT(?) ? (?x? ? x?) ? (?x? ?
  x?) ? (?x? ? x? ? x?)
• e x? so
• e EXT(?) ? x? or e EXT(?) ? x? so
• e CNF(?) or e CNF(?), so
• e ? or e ?, so
• e ? ? ? ?
• Similar for conjunction and negation

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Rewriting to CNF

• You can do the same for circuits. Try with
  example

15
SAT Solving Simple Observations

• Definition a partial assignment assign values to
  some variables.
• An assignment satisfies a formula in CNF if it
  makes all clauses true
• A partial assignment does not satisfy a formula
  if there is a clause that it makes false
• Boolean Constraint propagation
• For a clause (a ? b), if a partial assignment e
  sets a FALSE, then any full assignment e that
  extends e and satisfies the formula sets b
  TRUE.
• Example a ? b ? (a ? b)

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SAT Solving

• DPLL branch and bound (Davis, Putnam, Logeman,
  Loveland, 1960, 1962)
• Call SAT(?)
• SAT(e)
• e BCP(e)
• If e makes ? true return true
• If e makes ? false, return false
• Pick a variable v
• r1 SAT(e e ? (var ? 1)
• r0 SAT(e e ? (var ? 0)
• return r0 r1.
• Example (? a ? b) ? (a ? ? b) ? (? a ? ? b)

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• more details see Biere, http//fmv.jku.at/amc/,
  from whence the examples.