coNP and Function Problems

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coNP and Function Problems

• coNP
• definition and its relationship with NP
• the class NP ? coNP, Pratts theorem
• Function Problems
• FSAT, FTSP
• total functions

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coNP

• NP - problems with succint certificates
• coNP problems with succint disqualifications
• Examples
• VALIDITY - is the given boolean expression
  valid, i.e. is it true for all truth assignments?
• succint disqualification invalidating truth
  assignment
• HAMILTONIAN PATH COMPEMENT given graph, is it
  true that it does not have Hamiltonian path
• succint disqualification a Hamiltonian path
• what about SAT COMPLEMENT?
• VALIDITY of the negated formula

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coNP Completeness
VALIDITY is coNP-complete If L ? coNP, then coL
? NP. Therefore, there is reduction R from coL to
SAT. For any input x we have x ? coL iff R(x) is
satisfiable. Reduction from L to VALIDITY is
R(x) ?R(x). More generally Theorem If L is
NP-complete, then its complement coLS-L is
coNP-complete.
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coNP vs NP

• Is it true that coNP NP?
• open problem ( NTIME classes closed under
  complement)
• we know coNSPACE NSPACE (Immerman
  Szelepszenyi)
• Which problems in coNP are least likely to be in
  NP?
• coNP-complete problems
• Theorem If a coNP-complete problem is in NP then
  coNP NP.
• If L ?NP is coNP-complete, then coNP ? NP For
  any L ?coNP there is a reduction R to L. The
  nondeterministic polynomial machine recognizing
  L constructs R(x) from input x (reduction is
  polynomial time) and then decides in polynomial
  time (remember, L ? NP) whether R(x) is in L.

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The PRIMES Problem

• PRIMES is the input number prime?
• Is PRIMES in coNP?
• succint disqualification
• a divisor
• Is PRIMES in NP?
• we can check in time O(?n log2n) whether n is
  prime
• just by trying to divide by 2, 3, , ?n
• wait! Looks like PRIMES is in P!
• this is pseudopolynomial algorithm, it does not
  even prove that PRIMES is in NP!
• we will try to find succint certificate that n
  is prime

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The classs coNP ? NP

• NP has succing certificate, coNP has succint
  disqualification
• What does coNP ? NP have?
• both succint certificate and succint
  disqualification
• nonsense
• each instance(input) has either succint
  certificate (yes instance) or succint
  disqualification (no instance)
• From coP P we get P ? coNP ? NP.
• Is it true that P coNP ? NP?
• we know that R RE ? coRE (decidability
  analogue)
• but there are problems in coNP ? NP for which we
  do not have polynomial algorithm

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PRIMES Revisited
Pratts theorem PRIMES is in NP ? coNP. It is
sufficient to show that PRIMES is in NP. We show
that there is a succint certificate for
PRIMES. We will need the following number-theory
lemma Lemma A number pgt1 is prime if and only
if there is a number r, 1 lt r lt p such that rp-1
1 mod p, and furthermore r(p-1)/q ? 1 for all
prime divisors q of p-1.
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PRIMES is in NP

• Lemma A number pgt1 is prime if and only if there
  is a number r, 1 lt r lt p such that rp-1 1 mod
  p, and furthermore r(p-1)/q ? 1 for all prime
  divisors q of p-1.
• Basic idea r from the above lemma is the
  certificate
• We need to verify rp-1 1 mod p in time
  polynomial in log p.
• compute r2, r4, r8, , and then multiply some of
  them to get rp-1
• apply modulo after each multiplication, to keep
  all intermediate results modulo p
• multiplying/dividing two numbers ? p can be done
  in time O(log 2 p)
• O(log p) multiplication/division operations, so
  O(log3p) overall complexity

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PRIMES is in NP

• Lemma A number pgt1 is prime if and only if there
  is a number r, 1 lt r lt p such that rp-1 1 mod
  p, and furthermore r(p-1)/q ? 1 for all prime
  divisors q of p-1.
• Problem 2021-1 1 mod 21, but 21 is not prime!
• Hah, we must also use the second part of the
  lemma!
• The certificate should contain not only r, but
  also all prime divisors of p-1.
• Second try
• the certificate is (r q1, q2, .., qk)
• we must check both properties required in the
  lemma

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PRIMES is in NP

• Disaster
• C(91)(102,45), 1090 1 mod 91
• 1090/2 90 ? 1 mod 91
• 1090/45 9 ? 1 mod 91
• But 91 7 x 13!
• Where is the problem?
• misleading certificate, 45 is not prime!
• Solution?
• add primality certificates for all prime
  divisors
• needs to be done recursively
• C(p) (r q1, C(q1), q2, C(q2), , qk, C(qk))

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PRIMES is in NP

• Example
• C(67)(2 2, (1), 3, (2 2, (1)), 11, (82, (1),
  5, (3 2, (1))))
• But the size has grown a lot!
• it is still polynomial in log(p)
• Lemma C(p) is of size at most 4 log2p.
• By induction on p trivially for C(2) and C(3).
• The number k of prime divisors of p-1 is less
  then log p.
• C(p) ? 4log p 5 Si2k C(qi)) ? 4log
  p54Si2k log2qi
• However, Si2k log2qi ? (log p -1)2 (because
  Si2k log qi ? log p -1)
• Substituting we get C(p) ? 4 log2p9-4log p ? 4
  log2p for p?5

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FUNCTION PROBLEMS

• Decision problems are nice, but we often need to
  compute a function, i.e. instead of checking
  satisfiability we would like to know the
  satisfying truth assignment.
• Clearly, function problems are not easier then
  decision problems.
• However, is it possible that the function
  problems to be much harder?
• in other words, the decision problem to be much
  easier then the corresponding function problem?

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FUNCTION PROBLEMS

• Example FSAT given formula f, find a satisfying
  truth assignment, otherwise say no.
• Observation If we can solve SAT polynomially, we
  can also solve FSAT polynomially
• use SAT to decide whether fx1true is
  satisfiable
• if yes, fix x1true and proceed recursively to
  the next variable, othervise check whether
  fx1false is satisfiable. If yes, fix x1false
  and proceed recursively, othervise say no.
• with at most 2n calls to SAT we can solve FSAT
• We have seen Self-reducibility in action.

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FUNCTION PROBLEMS

• Example FTSP Given set of intercity distances,
  find the shortest TSP path.
• Observation If we can solve TSP polynomially, we
  can also solve FTSP polynomially
• use binary search using TSP as decision
  subroutine to find the smallest D such that
  TSP(B) exists.
• works if we can solve TSP polynomially, as P
  coP
• sequentially for each edge we increase its
  weight by one and ask whether there is TSP(D).
  The edge is used in the optimal TSP iff the
  answer is no

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FUNCTION CLASSES

• Let L?NP. We know that there exists
  polynomiallydecidable, polynomially bounded
  relation RL such that for all strings x There is
  a string y with RL(x,y) if and only if x ?L. The
  function problem FL associated with L is the
  following computational problem
• Given x, find a string y such that RL(x,y) if
  such string exists, otherwise return no
• FNP FL L? NP
• FP those that can be solved in polynomial time
• FSAT ?FNP, but not expected to be in FP
• given the assignment, it is easy to verify in NP
  time
• FTSP is not expected to be in FNP
• given the shortest path, there is no known way
  to verify it in NP time

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FUNCTION CLASSES REDUCTIONS
How would reductions work between function
classes? Definition A function problem A reduces
to function problem B if there are string
functions R and S, both computable in logarithmic
space, such that for any string x and z the
following holds If x is instance of A then R(x)
is instance of B. Furthermore, if z is a correct
ouput of R(x) then S(z) is correct output of x. A
function problem A is comnplete in a class FC of
function problems if A is in FC and all problems
in that class reduce to A. FP and FNP are closed
under reductions, and compositions of reductions
are reductions as well. FSAT is FNP
complete. Theorem FP FNP if and only if P NP.
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TOTAL FUNCTIONS

• Studying problems in FNP is not very interesting
  it is the same as studying NP.
• Exception Total functions for given x there is
  always an answer y
• The corresponding decision problem is trivial
  always yes.
• Still, the subject might be interesting
• FACTORING given number x, find its prime
  decomposition p1k1p2k2pmkm, together with the
  primality certificates of p1, p2, , pm
• Note Without the primality certificates it is
  not obvious that FACTORING is in FNP.
• There are no known polynomial algorithms for
  FACTORING.
• despite trying really hard
• used in many cryptography schemes

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TOTAL FUNCTIONS
Example Given undirected graph G(V,E) with
integer edge weights. State is mapping from V to
-1, 1. We say that a node v is happy in state S
if S(i)xSi,j?E S(j)wi,j?0 HAPPYNET Given a
graph with weights, find a state in which every
node is happy. Proposition All instances of
HAPPYNET have solution. Proof Consider the sum
f(S) Si,j?ES(i)S(j)wi,j. Suppose node i is
unhappy in S. That means S(i)xSi,j?E
S(j)wi,j-dlt0. Consider now S S with i
flipped (i.e. S(i)-S(i)). Then f(S)
f(S)2d. Since f(S) is bounded from below and
above by the sum of all weights, it can be
increased only a finite number of steps, i.e.
there will not be an unhappy node.
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HAPPYNET DISCUSSION
The above algorithm is pseudopolynomial, not
polynomial. HAPPYNET is equivalent to finding
stable states in neural networks in Hopfield
model. HAPPYNET is in the class TFNP the class
of total function problems. TFNP is a subclass of
FNP, and superclass of FP