1. Theorem 9.3 A problem Π is in P if and only if there exist a polynomial p( ) and a constant c such that ICp (x|Π) ≤ c holds for all instances x of Π.
Proposition 9.2 A set X is a complexity core for problem Π if and only if, for any constant c and polynomial p( ), ICp(x|Π) > c holds for almost every instance x in X.
Proof. Let X be a complexity core; in particular, there are infinitely many instances x in X for which ICp (x|Π) ≤ c holds for some constant c. Then there must be at least one machine M of size not exceeding c that solves Π and runs in polynomial time on infinitely many instances x in X. But a complexity core can have only a finite number of instances solvable in polynomial time, so that X cannot be a core – hence the desired contradiction.

2. Assume that X is not a complexity core
Assume that X is not a complexity core. Then X must have an infinite number of instances solvable in polynomial time, so that there exists a machine M that solves Π and runs in polynomial time on infinitely many instances x in X. Let c be the size of M and p( ) its polynomial bound; then, for these infinitely many instances x, we have ICp (x|Π) ≤ c, which contradicts the hypothesis.
Theorem 9.4 Let Π be a problem not in P. Then, for any polynomial p( ), there exist a constant c such that Π has infinitely many (p,c)-hard instances.

3. Theorem 9.5 Let Π1 and Π2 be two problems such that Π1 many-one reduces to Π2 in polynomial time through mapping f; then there exist a constant c and a polynomial q( ) such that ICq+p.q(x| Π1) ≤ ICp (f(x)|Π2)+c holds for all polynomials p() and instances x.
Proof. Let Mf be the machine implementing the transformation and let q( ) be its polynomial time bound. Let p( ) be any non-decreasing polynomial. Finally, let Mx be a minimal machine that solves Π2 and runs in no more than p(|f(x)|) steps on input f(x). Now define M’x to be the machine resulting from the composition of Mf and Mx. M’x solves Π1 and, when fed instance x, runs in time bounded by q(|x|)+p(|f(x)|), that is, bounded by q(|x|)+p(|q(x)|). Now we have

4. ICq+p.q(x| Π1) ≤ size(M’x) ≤ size(Mf)+size(Mx)+c
But Mf is a fixed machine, so that we have
size(Mf)+size(Mx)+c’ = size(Mx)+c = ICp (f(x)| Π2) +c
which completes the proof.
Theorem 9.6 In any polynomial transformation f from Π1 to Π2, for each constant c and sufficiently large polynomial p( ), only finitely many (p,c)-hard instances x of Π1 can be mapped to a single instance y=f(x) of Π2.
9.3 Average-Case Complexity Average-Case Complexity is a very difficult problem when compared to worst case complexity. Yet it is worth the trouble, if only because we know of NP-hard problems that turn out to be “easy on average” under reasonable distributions, while

5. Other NP-hard problems appear to resist such an attack.
Example 9.1 Consider a graph coloring problem; a simple backtracking algorithm that attempt to color with some fixed number k of colors a graph of n vertices chosen uniformly at random among all such graphs runs in constant average time! The basic reason is that most of the graphs on n vertices are dense (there are far more choices for the selection of edges when the graph has Θ(n2) edges than when it has only O(n) edges), so the most of these graphs are in fact not k-colorable for fixed k – in other words, the backtracking algorithm runs very quickly into a clique of size k+1. The computation of the constant is very complex; for k=3, the size of the backtracking tree averages around 200 – independently of n.

6. Definition 9.4 A function f is polynomial on µ-average if there exists a constant ε > 0 such that the sum
∑xf ε(x)µ(x)/|x| converges.
Proposition 9.3 Given a function f, the following statements are equivalent
There exists a positive constant ε such that the sum of ∑xf ε(x)µ(x)/|x| converges.
There exists positive constants c and d such that, for any positive real number r, we have µ[f(x)>rd|x|d]<c/r.
There exists positive constants c and ε such that we have ∑|x|≤n f ε(x)µ≤n(x) ≤ cn for all n, where µ≤n(x) is the conditional probability of x, given that its length does not exceed n.

7. Definition 9. 5 A real-valued function f : ∑
Definition 9.5 A real-valued function f : ∑* [0,1] is polynomial-time computable if there exists a deterministic algorithm and a bivariate polynomial p such that, for any input string x and natural number k, the algorithm outputs in O(p(|x|,k)) time a finite fraction y obeying |f(x)-y| ≤ 2-k
Theorem 9.7 Let µ be a polynomial-time computable distribution. There exists a constant c ϵ N and an injective, invertible, and polynomial-time computable function g : ∑*  ∑* such that, for all x, we have µ(x) ≤ c.2-|g(x)|. If, in addition, µ(x) exceeds 2-p(|x|) for some polynomial p and for all x, then there exists a second constant b ϵ N such that, for all x, we have b.2-|g(x)| ≤ µ(x) ≤ c.2-|g(x)|.

8. Definition 9. 7 Let µ and v be two distributions
Definition 9.7 Let µ and v be two distributions. We say that µ is dominated by v if there exists a polynomial p such that, for all x, we have µ(x) ≤ p(|x|)v(x). Now let (Π1,µ) and ( Π2,v) be two distributional problems and f a transformation from Π1 to Π2. We say that µ is dominated by v with respect to f if there exists a distribution µ’ on Π1 such that µ is dominated by µ’ and we have v(y)= ∑f(x)=y µ’(x) .
Definition 9.8 We say that (Π1,µ) is polynomial-time reducible to ( Π2,v) if there is a polynomial-time transformation from Π1 to Π2 such that µ is dominated by v with respect to f.

9. Definition 9.9 An instance of the Distributional Bounded Halting Problem for AP is given by a triple, (M, x, 1n ) where M is index of a deterministic Turing machine, x is a string (the input for M), and n is a natural number. The question is “Does M, run on x, halt in at most n steps?” The distribution µ for the problem is given by µ(M, x, 1n ) = c.n-2|x|-2|M|-2.2-|M|-|x|, where c is a normalization constant (a positive real number).
Theorem 9.8 Distribution Bounded Halting is DISTNP-complete.
Proof. Let (Π,µ) be an arbitrary problem in DISTNP; then Π belongs to NP. Let M be a nondeterministic Turing machine for Π and let g be the function of Theorem 9.7, so that we have µ(x) ≤ 2-|g(x)|. We define a new machine

10. M’ as follows. On input y if g-1(y) is defined, then M’ simulates M on g-1(y); it rejects y otherwise. Thus M accepts x if and only if M’ accepts g-1(x); moreover, there exists some polynomial p such that M’, run g(x), completes in p(|x|) time for all x. Then we define our transformed instance as the triple (M’, g(x), 1p(|x|)), so that the mapping is injective and polynomial time computable. Our conclusion follows easily.

11. Thank You!!!