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Path: ...!eternal-september.org!feeder3.eternal-september.org!news.eternal-september.org!.POSTED!not-for-mail From: wij <wyniijj5@gmail.com> Newsgroups: comp.theory Subject: Re: Improved =?UTF-8?Q?=E2=84=99=E2=89=A0=E2=84=95=E2=84=99?= proof Date: Fri, 31 May 2024 23:07:13 +0800 Organization: A noiseless patient Spider Lines: 98 Message-ID: <eed8bcaeee2337a7d842401b4bd6e2e409ddc213.camel@gmail.com> References: <108d3c553ccae9c7e6eeb1b8b1a85a52b8b0b78d.camel@gmail.com> MIME-Version: 1.0 Content-Type: text/plain; charset="UTF-8" Content-Transfer-Encoding: quoted-printable Injection-Date: Fri, 31 May 2024 17:07:15 +0200 (CEST) Injection-Info: dont-email.me; posting-host="58e01b5a2bbe399e2a0ce9c1509683fb"; logging-data="2399439"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1/qN0Ujo3Je5qZgCEJGF2Ga" User-Agent: Evolution 3.50.2 (3.50.2-1.fc39) Cancel-Lock: sha1:EgJ7lhgYF5Vwqjl8AZFTTxVPhuA= In-Reply-To: <108d3c553ccae9c7e6eeb1b8b1a85a52b8b0b78d.camel@gmail.com> Bytes: 4970 Woo, the updated proof is even lots shorter (and intuitive?). ---------------------------------------------------------------------------= -- This file is intended a proof that =E2=84=99=E2=89=A0=E2=84=95=E2=84=99. Th= e contents may be updated anytime. https://sourceforge.net/projects/cscall/files/MisFiles/PNP-proof-en.txt/dow= nload ---------------------------------------------------------------------------= -- Algorithmic problem::=3D Problems that can be processed by asymptotic analy= sis. ANP::=3D Another NP is a set defined as {q| q is a description of the algor= ithmic decision problem that provides 1. A set of certificate data C 2. A Ptime (Polynomial time) verification method v 3. The answer of q is 'yes' iff there exists a certificate c, c=E2=88=88C, such that v(c) is true 4. q c= an be=20 computed in at most O(2^|q|) steps. }. More precisely, ANP is the set of problems that can be solved by the=20 following pseudo-C/C++ program temp_anp(n): bool temp_anp(Problem q) { // Problem: Description of the prob= lem Certificate c,begin,end; // Certificate data can be accessed= by begin=3D get_begin_certificate(q); // iteration, at least. end =3D get_end_certificate(q); for(c=3Dbegin; c!=3Dend; c=3Dnext(c)) { // O(2^|n|) loop (see Note2) if(v(c)) return true; // v:Certificate->{true,false}, Pti= me // verification function. } return false; } Note1: Relative to the Turing Machine language for =E2=84=95=E2=84=99, t= he reason using pseudo-C/C++ is that 1.C-code (almost all high level programming language not involving special hardware features) and TM language = are computationally interchangeable. 2.It describes the problem more clearly (but not always) 3.The result 'false' is visible 4. =E2=84= =95=E2=84=99 definition does not say the certificate C and the verication v are given, fixed arguments. In ANP, v(c) is explicitly spedified a Pti= me function 5.Easier for machine aided verification. Note2: The semantics of the for loop in temp_anp(n) includes nested loop= s for sets like C=3DC1=C3=97C2=C3=97C3=C3=97... Polynomial time procedure::=3D (or polynomial time function) A procedure co= mposed of sequential execution of O(P) number of fixed-time procedures. (Therefore, O(P) number of sequential Ptime procedure is equivalent to a single Ptime procedure) Size-1-subproblem::=3D The problem whose size of input is less by one than = that of the orginal problem. Theorem: ANP problem can be divided into two size-1-subproblems. Proof: By spliting the certificate as follow: bool temp_anp(Problem q) { if(q.certificate().size()<Thresh) { // Thresh is a small consta= nt return solve_thresh_case(q); } Problem q1,q2; split_certificate(q,q1,q2); // split the certificate in = q return temp_anp(q1) || temp_anp(q2); // to form q1,q2 } The proof above also illustrates that ANP is merely a kind of class of prob= lems that temp_anp can handle independent certificates. I.e. the result of v(c1)= and v(c2) may be completely unrelated. As there are n number of such certificat= es, at most such n number of steps are required. This conclusion applies to any number of n (this view can also be formed by the definition of temp_anp, th= e proof just provides another view in subproblems). Because there are O(2^N) certificates, ANP=E2=89=A0=E2=84=99. If ANP=E2=8A= =86=E2=84=95=E2=84=99, =E2=84=99=E2=89=A0=E2=84=95=E2=84=99 can be conclude= d. ---------------------------------------------------------------------------= --