From Newsgroup: comp.lang.c


The script of proof is formal enough to submit to professional institute, I think.
Anything unclear or defect?
The following is form https://sourceforge.net/projects/cscall/files/MisFiles/PNP-proof-en.txt/download
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Algorithmic Problem::= A computer problem in which the computational steps are a
function of the problem statement's length (size). This problem can be
described asymptotically as the relationship between the problem size and
the computational steps.
Polynomial-time procedure (or Ptime procedure)::= O(P) number of consecutive,
fixed-sized basic operations (because the procedure is a deterministic
process, it is sometimes called a "function" or "operation"). Therefore, as
defined by O(P), O(P) number of Ptime procedures executed consecutively can
also be considered as a single Ptime procedure.
Reduction::= The algorithm for computation problem A can be transformed into
computation problem B by a Ptime procedure, denoted as A≤B (because the
Ptime transformation itself includes the computation of problem A, any two
Ptime problems can be reduced to each other).
ANP::= {q| q is a statement of a decision problem that a computer can solve in
O(2^|q|) steps using the following fnp algorithmic template. q contains a
certification dataset C, card(C)<=O(2^|q|), and a Ptime verification
function v:C->{true,false}. If ∃c,v(c)=true, then the answer to problem q is
true; otherwise, it is false.}
// begin_certificate is a Ptime function that retrieves the first
// Certificate element from the problem statement q. If this element does
// not exist, it returns a unique and virtual EndC element.
Certificate begin_certificate(Problem q);
// end_certificate is a Ptime function that retrieves the element EndC from
// the problem statement q.
Certificate end_certificate(Problem q);
// next_certificate is a Ptime function that retrieves the next element of
// c from the certification dataset C. If this element does not exist,
// return the EndC element.
Certificate next_certificate(Problem q, Certificate c);
// v is a Ptime function. v(c)==true if c is the element expected by the
// problem.
bool v(Certificate c);
bool fnp(Problem q) {
Certificate c, begin, end; // Declare the certification data variable
begin= begin_certificate(q); // begin is the first certification data
end= end_certificate(q); // end is the false data EndC used to
// indicate the end.
for(c = begin; c != end;
c = next_certificate(q, c)) { // At most O(2^|q|) steps.
// next_certificate(c) is a Ptime
// function to get the next
// certification data of c
if(v(c) == true) return true; // v: C->{true, false} is a polynomial
// time verification function.
}
return false;
}
Since a continuous O(P) number of Ptime functions (or instructions) can be
combined into a single Ptime function, at roughly this complexity analysis,
any Ptime function can be added, deleted, merged/split in the algorithm
steps without affecting the algorithm's complexity. Perhaps in the end, only
the number of decision branches needs to be considered.
An ANP problem q can be expressed as q<v, C>, where v = Ptime verification
function, and C = certification dataset (description).
[Note] Also testified by Church-Turing conjecture, no formal language can
surpass the expressive power of a Turing machine (or algorithm, i.e.,
the decisive operational process from parts to the whole). C
language can be regarded as a high-level language of Turing machines,
and as a formal language for knowledge or proof.
Prop1: ANP = ℕℙ
Proof: ℕℙ ⊆ ANP and ANP ⊆ ℕℙ, therefore ANP=ℕℙ
(The proof of the equivalence of ANP and the traditional
Turing machine definition of ℕℙ is straightforward and lengthy, and
not very important to most people. Although the definition of ANP does
not depend on NDTM theory, there are thousands of real-world ℕℙℂ
problems available for verification and reference)
Prop2: An ANP problem q<v,C> can be arbitrarily partitioned into two subproblems
q1<v,C1>, q2<v,C2), where C = C1∪C2.
Proof: The certification dataset can be recursively partitioned as follows:
bool banp(Problem q) {
if(certificate_size(q)<Thresh) { // Thresh is a small constant
return solve_thresh_case(q); // Solve q in constant time
}
Problem q1,q2;
split_certificate(q,q1,q2); // Split the certification dataset C
// to form q1,q2, in which the number of
// certificates are roughly the same.
return banp(q1) || banp(q2); // Compute the subproblems respectively
}
Prop3: Any two ANP problems q1 and q2 can be synthesized into another ANP
problem q, denoted as q = q1 ⊕ q2. The certification dataset C and the
verification function v of q are defined as follows:
C = C1 ∪ C2 // C1 and C2 are the certification datasets of q1 and q2
// respectively
bool v(x) {
return v1(x) || v2(x); // v1 and v2 are the verification functions of
// q1 and q2 respectively
}
Therefore, we have the identity: q1<v1,C1>⊕ q2<v2,C2> = q<v1||v2, C1∪C2>.
Prop4: ℙ=ℕℙ iff the algorithm fnp in the ℕℙ definition (or ℕℙℂ algorithm) can be
replaced by a Ptime algorithm.
Proof: Omitted (A very direct semantic of 'equivalence')
Prop5: For ℕℙℂ subproblems q1<v,C1> and q2<v,C2> (same v), if C1 ∩ C2 = ∅, then
there is no information in problem q1 that is sufficient in terms of
order of complexity to speed up the computation of q2.
Proof: Since ℕℙℂ is the most complex class of ℕℙ problems, by the definition
of ANP, the order of card(C) of the ℕℙℂ problem must be of order O(2^|q|)
no lower (any lower, ANP/ℕℙ problem cannot be completely reduced).
We can add an AuxInfo object called 'auxiliary information' to banp to
store the results after calculating a certain ℕℙℂ problem q, and rewrite
it as banp2. Depending on the content of the auxiliary information,
banp2 can represent possible algorithms for ANP/ℕℙℂ problems with
complexities ranging from O(N) to O(2^N).
bool banp2(Problem q, AuxInfo* ibuf) { // Calculate q and write the
// obtained auxiliary information to *ibuf
// Check and initialize *ibuf
if(certificate_size(q)<Thresh) {
return solve_thresh_case(q,ibuf);
}
Problem q1,q2;
split_certificate(q,q1,q2);// q can only be reduced to ℕℙℂ subproblems
// q1 and q2.
// card(C1) and card(C2) are roughly equal.
AuxInfo I; // I stores information to help solve problems.
if(banp2(q1,&I)==true) { // banp2(q1,I) recursively computes subproblem
// q1 and stores any auxiliary information
// that can be derived from q1 into I.
write_ibuf1(ibuf,q); // Write auxiliary information
return true; // The information obtained from subproblem q1
// is valid for any problem q.
}
bool rv=banp2_i(q2,&I); // Solve q2 with given information I of which
// any source is effective for q2.
write_ibuf2(ibuf,q);
return rv;
}
[Note] From banp2(q1,&I), we know that the auxiliary information I is only
meaningful if it contains information related to the certification
dataset C1 of problem q1. This is because if it is not related, the
computation of other subproblems can generate the auxiliary
information by itself.
The above `banp2_i` does not care which ANP problem (including trivial
problems) the given information `I` originates from. The `I` generated can
also provide auxiliary information for almost all other ANP problems.
Since the auxiliary information I obtained from computing a fixed
subproblem (including trivial problem) cannot provide sufficient
information to accelerate computation to the extent of improving the
complexity order for any other subproblem (would require infinite space/
time), the proposition is thus proved (this proof also demonstrates that
the complexity of the ℕℙℂ problem is O(2^N)).
Conclusion: According to Proposition 5, the certification data for the ℕℙℂ
problem are basically independent and must be tested one by one. Since the
ℕℙℂ problem has O(^|q|) certification data, the complexity of the ℕℙℂ
problem is O(2^N), therefore ℙ≠ℕℙ. --------------------------------------------------------------------------------
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