CS 3240: Languages and Computation

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CS 3240 Languages and Computation

• NP-Completeness
• Overview of PSPACE

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NP-completeness

• A problem C is NP-complete if finding a
  polynomial-time solution for C would imply PNP
• Definition A language B is NP-complete if it
  satisfies two conditions
• B is in NP, and
• Every A in NP is polynomial time reducible to B

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An NP-complete problem

• A formula is Boolean if each of its variables can
  be assigned the values TRUE (1) or FALSE (0)
• A Boolean formula is satisfiable if there is some
  assignment of values that results in the formula
  evaluating to TRUE
• SAT Is a given Boolean formula satisfiable?
• SAT is NP-complete

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Examples

• (x ? y) ? (? x ? y)
• Satisfiable e.g., x y 1
• ((x?y) ? (?x?z)) ? ((?x?y) ? (?y?z))
• Satisfiable e.g., x 0, y z 1
• ((x?y) ? (?x??z)) ? ((x??y) ? (?y?z))
• Unsatisfiable

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Cook-Levin theorem

• SAT ltBgtB is a satisfiable Boolean expression
• Theorem SAT is NP-complete
• If SAT can be solved in polynomial time then any
  problem in NP can be solved in polynomial time

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Proof of Cook-Levin theorem

• First show that SAT is in NP
• Easy
• The show that SAT ? P implies P NP
• Proof converts a non-deterministic Turing machine
  M with input w into a Boolean expression f
• M accepts string w iff f is satisfiable
• ltM,wgt is converted into f in polynomial time

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Variables in proof of Cook-Levin Theorem

• Ti,j,s is true if tape position i contains symbol
  s at step j of computation
• O(p(n)2) variables
• Hi,k is true if Ms tape head is at position i at
  step k of computation
• O(p(n)2) variables
• Qq,k is true if M is in state q at step k of the
  computation
• O(p(n)) variables

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Boolean expression

• Using the variables Ti,j,k, Hi,k, and Qq,k, Cook
  Levin developed a Boolean expression that is
  satisfiable if and only if M accepts w
• Expression reflects proper behavior of M
• Qstart,0 ensures starts in start state
• H0,0 ensures starts with tape head at start of
  tape
• Ti,j,k ? ?Ti,j,k ensures one symbol per tape
  cell
• Etc

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Purpose of Boolean expression

• If the Boolean expression in the proof of the
  Cook-Levin theorem can evaluate to TRUE then
• Starting at the start state with w on the tape, M
  can take steps based on the transition function
  and end at the accept state
• I.e., M accepts w
• If the Boolean expression is unsatisfiable, then
  M rejects w

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Length of Boolean expression

• O((log p(n)) p(n)2)
• Since p(n) is polynomial in the length of the
  input, so is the Boolean expression
• If the satisfiability of the expression can be
  determined in polynomial time, we can determine
  in polynomial time whether M accepts or rejects w
• Every problem in NP would be in P if SAT is in P

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Another NP-complete problem

• 3SAT
• The Boolean expression must be in a specific form
  called 3-cnf
• Conjunctive normal form
• A literal is a Boolean variable or the negation
  of a Boolean variable
• A clause is the disjunction (OR) of literals
• A Boolean formula is in cnf if it is the
  conjunction (AND) of clauses
• It is 3-cnf if all the clauses have 3 literals

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Importance of Cook-Levin theorem

• Cook Levin proved that SAT is NP-complete
• If SAT can be solved in polynomial time, then any
  problem in NP can be solved in polynomial time
• The showed that any problem in NP can be
  polynomially reduced to the SAT problem
• The proof that 3SAT is NP-complete is similar
• Generate a 3-cnf formula from a TM

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Proving a problem is NP-complete

• A problem C is NP-complete if finding a
  polynomial-time solution for C would imply PNP
• If a polynomial-time solution is found for C,
  then that solution can be used to find a
  polynomial-time solution for any other problem in
  NP
• What does this remind you of?
• Reductions!

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Polynomial reductions

• Definition Language A is polynomial-time
  reducible to language B, written A P B, if a
  polynomial time computable function fS?S
  exists, where for every w
• w ? A iff f(w) ? B

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Reductions NP-completeness

• Theorem If A P B and B?P, then A?P.
• Proof Let M be the polynomial time algorithm
  that decides B and let f be the polynomial
  reduction from A to B. Consider the TM N
• N On input w
• Compute f(w)
• Run M on f(w) and output Ms result
• Then N decides A in polynomial time.

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Implications of NP-completeness

• Theorem If B is NP-complete and B?P, then P
  NP.
• Theorem If B is NP-complete and BPC for some C
  in NP, then C is NP-complete

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Showing a problem in NP-complete

• Two steps to proving a problem L is NP-complete
• Show the problem is in NP
• Demonstrate there is a polynomial time verifier
  for the problem
• Show some NP-complete problem can be polynomially
  reduced to L

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Space Complexity

• The space requirements of a computation are
  another important resource that we should be
  concerned about.
• It will turn out that space behaves much nicer
  than time.
• Space, unlike time, can be reused.

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Space Complexity Classes
Definition Let fN?N be a function. The space
complexity classes SPACE(f(n)) and NSPACE(f(n))
are the following sets of languages SPACE(f(n))
L there is a TM that decides the
language L in space O(f(n))
NSPACE(f(n)) L there is a
nondeterministic
TM that decides the
language L in space O(f(n))
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Savitchs Theorem
Theorem For any function fN?N with f(n)?nwe
have NSPACE(f(n)) ? SPACE((f(n))2). In other
words Nondeterminism does not give you much
extra for space complexity classes. Compare this
with our (lack of) understanding ofthe time
complexity classes TIME and NTIME.
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PSPACE
Definition 8.6 The class of languages that
canbe decided by a single tape TM in
polynomialspace is denoted by PSPACE
Savitchs theorem tells us that
nondeterminismdoesnt matter for this class
PSPACENPSPACE.
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A Hierarchy of Classes
EXPTIME
PSPACE
NP
P
P ? NP ? PSPACENPSPACE ? EXPTIME
We dont know how to prove P?PSPACEor
NP?EXPTIME. But we do know P?EXPTIME