1. The Polynomial Hierarchy 7.3.2 Enumeration Problems 7.3.3
R. Anderson, S. Dugginapeddi,
and V. Gadde

2. 7.3.2
Polynomial Hierarchy

3. Polynomial Hierarchy Class coNP NPcoNP and NPcoNP Dp

4. What about? But how? Not clear on…Hmmm
Is problem A in NP?
Turing reductions
Many-one reductions
What about ~A (coA)?
Reductions
But how to distinguish the differences?
In general, difference between NP and coNP
Reductions, no clear distinction
Hmmm….let’s classify coNP

5. The Class coNP Class coNP (Definition 7.2)
Composed of the complements of problems in NP
The natural complement of NP problems
Same set of valid problem instances
“yes” and “no” NP problem answers reversed
Negation of the problem question
Conjecture
NP and coNP are distinct

6. Decision problems around NP
Distinction generalized to any nondeterministic class
NP differs from coNP, NExp from coNExp, etc.
Cases of intersection
NL = coNL
Questions:
What is the class NP  coNP?
Is (NP  coNP) = P?
NP  coNP = NP-easy?
NP-easy NP
coNP co
NPC P

7. The NP  coNP class What is NP  coNP? Is (NP  coNP) = P?
problems in NP whose compliment is also in NP
Linear programming, primality testing
Is (NP  coNP) = P?
Not proven
Conjecture P  (NP  coNP)
Duality of linear programs
Compositness and primality
Could indicate problem is in P

8. The Dp class NP  coNP = NP-easy? Class Dp (Definition 7.3 p.268)No Dp
Class Dp (Definition 7.3 p.268)
Class of all sets Z that can be written as Z=X  Y, for X  NP and Y  coNP.
i.e. for any X in NP there is a problem Y in coNP where the intersect of the two result in a “yes” instance.
Dp contains both NP and coNP

9. The Dp class Proper superset of NP  coNP
NP-easy
Proper superset of NP  coNP
Dp = NP  coNP iff NP = coNP
Separates NP  coNP and NP-easy
Completion problems
Many-one TSM
SAT-UNSAT
Two sets of clauses on two disjoint sets of variables
Question: is the first set satisfiable (SAT) and the second UNSAT
Dp-complete (Theorem 7.19)

10. SAT-UNSAT is Dp-complete Proof
Prove that:
SAT-UNSAT is in Dp
Any problem in Dp reduces to it.
Simply the intersection of SAT and UNSAT sets
SAT  UNSAT
First set/half satisfiable? Yes
Second set/half unsatisfiable? Yes

11. SAT-UNSAT is Dp-complete Proof
Any problem in Dp reduces to it
Any problem XDp can be written as X=Y1  Y2
Problem of Y1NP and problem of Y2coNP
SAT is NP-complete while UNSAT coNP-complete
Y1 ≤ SAT and Y2 ≤ UNSAT
SAT ≤ SAT-UNSAT
Adding an UNSAT set of clauses of a different set of variables
Basically same for UNSAT ≤ SAT-UNSAT

12. SAT-UNSAT is Dp-complete Proof
Given and instance of X, x
x has an instance of Y1 and Y2
Many-one reductions
Y1 to SAT yields x1
Y2 to UNSAT yields x2
The concatenation of yields a new instance of SAT-UNSAT
z = x1 #x2

13. Polynomial Hierarchy
Do NP-easy problems constitute the set of all problems
solvable in polynomial time if P equals NP?
Any class with the property that collapses into P if
P equals NP, exists as a separate entity only under
assumption.
Potentially infinite hierarchy of such classes is
Polynomial hierarchy.

14. Polynomial Hierarchy  Consider a class of NP easy decision problems.
-- is a class of all decision problems solvable in polynomial time with the help of one oracle for some suitable NP-complete problem.
-- all problems in NP are solvable by an oracle.
Denoted by PNP.

15. Polynomial Hierarchy If P=NP, an oracle for NP would be oracle for P.
We can always solve problems in polynomial time
i.e.., PNP = PP = P.
Assume P and NP differ and combine non-determinism,
co-non-determinism and oracle mechanism and define further classes.

16. Polynomial Hierarchy
Initially use a non deterministic polynomial time TM with oracle for NP.
The resulting class would be NPNP.
If P = NP then,
NPNP = NPP = NP = P

17. Polynomial Hierarchy
Problems in NPNP were solvable in polynomial time, so
would be problems in NP if and only if P = NP.
Consider class containing complements of NPNP, co-NPNP.
NPNP and co-NPNP , both are one level higher in hierarchy.
The hierarchies thus can extend to infinite levels.

18. Polynomial Hierarchy Definition:
Is formed of three types of classes, each defined recursively: the deterministic classes, the non-deterministic classes and the co-non-deterministic classes which are recursively defined as follows

19. Polynomial Hierarchy
The Polynomial Hierarchy – One level

20. Polynomial Hierarchy
If the hierarchy is infinite no complete problem can exist for PH itself.
It is not known whether the hierarchy is truly infinite or collapses into some non deterministic class.
We could have P =/= NP but NP = coNP, where the whole hierarchy collapses to NP.

21. Polynomial Hierarchy
On the whole Polynomial Hierarchy explains the complexity of issues and surrounding the fundamental question of the relationship between P and NP.

22. 7.3.3
ENUMERATION PROBLEMS

23. CONTENTS Enumerations versions #P class and significance
Completeness for #P
- Using Parsimonious transformations
#P Complete related to NP and P
Comparison of #P complete problems

24. Enumeration vs Decision
Decision computes boolean valued function
Enumeration includes all integer valued functions
Presents number of Optimal or feasible solutions for given instance of problem
Defined for either decision, search or optimization problems
Enumeration leads to uncountable infinity of different solutions

25. Enumeration versions are significantly harder
- Hamiltonian circuits for a graph
Simple decision problems leads to harder enumeration
-counting different spanning trees compared to finding single spanning tree
Some enumeration tasks can be solved in polynomial time
-counting eulerian paths of graph
-counting number of spanning trees

26. #P class
Def 7.5
An integer-valued function f belongs to # p if there exists a deterministic Turing machine T and a polynomial p() such that , for each input string x, the value of the function f(x), is exactly equal to the number of distinct concise certificates for x.
NTM accepts x in exactly f(x) different ways
#P is the enumeration version of any problem in NP

27. -How many Hamiltonian cycles in a given graph have cost less than 100?
Set of counting problems associated with decision problems in set NP
NP is related with format “are there any”
-Are there any subsets of a list of integers that add up to zero?
-Are there any Hamiltonian cycles in a given graph with cost less than 100? (traveling salesman problem)
Corresponding #P asks “how many”
-How many subsets of a list of integers add up to zero?
-How many Hamiltonian cycles in a given graph have cost less than 100?

28. #P - Complete
Completeness of #P is defined based on polynomial reductions and transformations
-What is the permanent of given matrix?
-How many perfect matchings are there for a given bipartite graph?
Parsimonious defines the equality of solutions between original and transformed instance
Parsimonious transformation acts both reduction and transformation
Weakly parsimonious serves to prove #P completeness

29. #P complete defines enumeration version of all NP-complete problems
Decision problems in P which are #P complete rises particular interest
Finding one perfect matching is solvable in polynomial time
Counting number of perfect matching in bipartite graph is #P complete

30. Comparing #P Complete #P- complete problems are NP- hard
Solvability is based on the equality of P and NP
If P= NP , #P complete may remains intractable
#P-hardness acts as strong evidence of intractability
#P is difficult to compare with other complexity classes
#P-easy is used to compare

31. Many counting problems in #P , hard don’t seem to #P complete
Counting the number of distinct isomorphism's between two graphs are NP-easy