Independence of P vs. NP in regards to oracle relativizations

Indep endence of P vs. NP in regards to oracle relativizat ions. JERRALD MEEK This is the third article i n a s eries of f our articles dealing with the P vs. NP question. The purpose of this w ork is to demonstrate that the methods used in the first tw o articles of this series are not affected by oracle relativizations. F urthermore, the solution to the P vs. NP problem is actually indep enden t of oracle relativizations. Categories and Sub ject Descriptors: F.2.0 [ Analy sis o f Algo rithms and Problem C omplex- ity ]: General General T erms: Al gori thms, Theory Additional Key W ords and Phrases: P vs NP , Or acle Machine 1. INTRODUCTION. Previous ly in “ P is a prop er subset of NP ” [Me e k Article 1 2008] and “ Analysis of the Deterministic P olynomial Time Solv ability of the 0 -1-Knapsack P roblem” [Meek Ar ticle 2 200 8] the present author ha s demo nstrated that some NP-c omplete problems are likely to hav e no deterministic po lynomial time solution. In this article the c o ncepts of these pr evious works will be applied to the relativiza tions o f the P vs. NP pr oblem. It has pre v iously b een shown b y Hartmanis and Hop croft [197 6 ], that the P vs. NP problem is independent of Zermelo-F r a enk el set theory . How e ver, this same discov ery had been shown b y B a k er [19 79], who also demonstrates that if P = NP then the solution is incompatible with ZF C . The purpos e of this article will b e to demonstrate that the o racle relativiza tions of the P vs. NP problem do not preclude any solution to the origina l pro blem. This will b e accomplished by constructing examples of five out of the s ix relativizing or- acles from Baker, Gill, and Solov ay [19 75], it will b e demonstra ted that inconsistent relativizatio ns do not pr eclude a proo f of P 6 = N P or P = NP . It will then b ecome clear that the P vs. NP question is indep enden t of oracle relativizations. 2. PRELIMINARIE S. Baker, Gill, and Solov ay [197 5] have previously shown that there are six t ype s of oracles that can be co nstructed to examine the P vs. NP problem. These include (1) Oracle A such that P A = NP A . (2) Oracle B s uch that P B 6 = NP B . (3) Oracle C such that NP C is not closed under c o mplemen ta tion. Pe ople who wish to remain anon ymous hav e offered comments and suggestions which ha ve im- prov ed this work. The author wishes to express his appreciation for their assistance. Jerrald M eek Copyrigh t c  2008 ArXiv, V ol. V, N o. N, Mon th 20YY, Pages 1–14. 2 · Jerrald Meek (4) Oracle D such that P D 6 = NP D but NP D is closed under complementation. (5) Oracle E such that P E 6 = NP E and P E = NP E ∩ c o-NP E . (6) Oracle F such that P F ⊆ [ NP F ∩ c o-NP F ] ⊆ NP . Theorem 4.4 fr o m P is a pr op er subset of NP . [Meek A r ticle 1 2 008] 2.1. P = NP Optim ization Theorem. The only deterministic optimization of a NP-c omplete pr oblem that c ould pr ove P = NP would b e one t hat c an always solve a NP-c omplete pr oblem by examining no m or e than a p olynomial numb er of input sets for that pr oblem. Definition 2.1. Or acle Machine An ora cle mac hine c a n be deterministic or non deter ministic. The difference betw een a n o r acle ma chine a nd a reg ular T uring Machine is tha t an o racle Machine has a yes state, a no state, and a query state. When the o racle machine is placed in the query state after pa ssing a string to the o r acle, if that string is an element of the ora cle set, then the o racle will place the T uring Machine in the y es state, other wise it will place the T uring Mac hine in the no s ta te. The pro cessing time required by the oracle can b e considered insta n ta ne o us. 2.1 Enco ding Meth ods. In this article, t wo enco ding metho ds will be us e d. One enco ding metho d will enco de the input sets for a problem and is identical to that used by Baker, Gill, and Solov ay [197 5]. This will b e iden tified as “input enco ding.” The second metho d of enco ding will b e used to identify a partition of the s et o f all p ossible input s ets for a problem. This metho d o f enco ding will b e identified as “partition enco ding.” 2.1.1 Partition Enc o ding. The concept of partition enco ding is not to enco de the input sets for a problem, but an entire partition of the set of all p ossible input sets. How ev er, for this metho d to work the partitio ning must be identified with a sp ecific problem. The metho d of partition enco ding used in this w ork will divide partitions b y the nu m be r of literals in an input set whic h ha ve a true v alue. If a pro blem has n literals, then there will be n + 1 partitio ns (one partition has no true literals). Notice that the problem [ a ∨ b ∨ c ] has differen t input sets that res ult in true ev aluations from the problem [ ¬ a ∨ b ∨ c ] even though the cardinality of these input sets are equiv alent. It will then be nece s sary that the pr o blem which a pa rtition a pplies to will need to be enco ded. This can be a ccomplished by the following metho d: —Let P b e a NP-Complete problem with k liter als. —Let g be the G¨ odel Num b er calcula ted from P . —Let σ be a set of str ings r epresent ing ea c h of the k + 1 partitions for P . —Let 1 ≤ i ≤ k + 1. —Let S b e the set of all po ssible input sets fo r P . — T ( x ) = the num b er of true literals in input s et x . ArXiv, V ol. V, No. N, Month 20YY. Indep endence of P vs. NP in regards to oracle relativizations. · 3 then ∀ i : ( σ i ≡ h i − 1 , g i ) and the partition enco ding o f S i is represented b y: σ T ( S i ) F or each par tition of P there will b e one string consisting of: (1) a n um ber in the ra nge of 0 to k representing the num b er of true literals in S i ; (2) and the G¨ odel Num ber o f P . 3. THE RELA TIVIZING ORACLES. Baker, Gill, and Solov ay [1975] ha ve shown that the c la sses of P and NP do not relativize in a cons isten t manner. The inconsistency of the relativizations of P and N P ha s often been used to justify ar gumen ts that the P vs. NP problem is un usually difficult at bes t, o r mayb e even imp ossible. By slig h tly altering the machine mo del, we can o btain differing answ ers to the relativized question. This suggests that resolving the original question r equires careful a na lysis of the computational p ow er o f ma- chines. It seems unlik ely that ordinary diagonalization methods ar e adequate for pro ducing a n example of a langua ge in NP but not in P ; such diago nalizations, we would exp ect, would apply equally well to the relativized cla sses, implying a negative answer to all r elativized P =? NP questions, a co n tr adiction. [Baker et a l. 1 975] Although it is genera lly accepted that diagona lization is not an effective method for so lving the P vs. NP problem, it is the purp ose of this article to demons trate that oracle relativizations in no w ay preven t a solution to the P vs. NP problem. It is also worth while to mention that the six oracles of Baker, Gill, and Solov ay [1975] do not represent an exclusive set o f all or acle relativizatio ns. There ha s b een extensive work to find other unusual re la tivizations of co mplex it y classes in regar ds to the P vs. NP question. F or ex ample, Bennett a nd Gill [1981 ] have found that random o racles can pr oduce the r elativization P A 6 = NP A 6 = c o-NP A . How ever, the scop e of this work will b e limited to the s ix or acles from Ba k er, Gill, and Solov ay [1975] with the exp ectation that what is learne d from these six oracles ma y b e applied to any other relativizing oracle. 3.1 A P = NP Or acle. The P = NP Optimization The or em requires that no more than a p olynomial nu m be r of input sets can b e exa mined by a deter ministic p olynomial time sea rc h algorithm. By the same lo gic it is obvious that a deterministic oracle machine can not request more than a p olynomial num b er of queries in p olynomial time. Baker, Gill, and Solov ay [197 5] de fine the oracle A suc h that A = {h i, x, 0 n i : NP A i accepts x in < n steps } . This de finitio n pro duces an or acle suc h that P A = NP A . How ever, this definition do es no t rea lly s ay anythin g ab out how the oracle functions. In this work, the oracle will b e cr eated b y partition enco ding strings, a string will b e included in or acle A ArXiv, V ol. V, No. N, Month 20YY. 4 · Jerrald Meek if ther e exists at least one elemen t o f the partition r e presen ted b y the string which results in the problem asso ciated to the str ing ev a luating t r u e . —Let L ( X ) = { x : there exists y ∈ X such that | y | = | x |} for any or acle X . —Let x be the set of all problems in L ( X ). —Let S be the set suc h that S i is the set of all possible inpu t sets for x i when 1 ≤ i ≤ | x | . The oracle set is created b y the following algorithm. (1) Let i = 1. (2) Let e = 1. (3) If x i ev aluates t rue with input S i e , then add the par titio n enco ding of S i e 7→ x i to set A , and let e = | S i | . (4) If e < | S i | , then incr emen t e a nd co n tin ue at step 3. (5) If e = | S i | and i < | x | , then increment i and contin ue at step 2. (6) If e = | S i | and i = | x | , then the set A has been found. The goa l was to pro duce an example of a n ora cle set such that P A = NP A . Here, the ora cle A is actually s uc h tha t P A = NP . Such an ora cle would also crea te the condition that P A = N P A . A nother metho d could be us ed to create an or a cle such that P X = PSP ACE , whic h w ould also satisfy P X = NP X , except that NP 6 = NP X . There are multip le such oracles, how ever a ccording to Mehlho rn [19 73] the set of computable oracles suc h that P A = NP A is meager. Notice that in this ex a mple the set of all p ossible input sets will b e exp onen tial in size, and there fo re so me partitions for a n y pro blem will b e ex ponential in size. It is then easy to see that the problem of cr eating the ora c le s et (even when the oracle is created to only work for one pro blem) is a member o f the FNP complexity class (when P A = NP ). The pro cess that a Deterministic O r acle Machine with oracle s e t A uses to so lv e problem x i which is in NP and has n literals is the algor ithm. (1) Let e = 0. (2) Enco de the partition for pr oblem x i for input sets having e true literals. (3) Pass the partition enco ding from step 2 to the o racle and enter the query state. (4) If the machine is in the y es state then ha lt in a n accepting state. (5) If the machine is in the no state and e < n then incr emen t e and contin ue at step 2. (6) If the machine is in the no s tate and e = n then halt in a non-accepting state. The algorithm of the ora cle machine solves any NP problem in polyno mial time on a Deterministic Orac le Machine. A No n Deterministic Or acle Machine could als o use this algorithm to solve the same set of pro blems in p olynomial time. Therefore, P A = NP A . This method is dep endant o n an oracle set creation algorithm r equiring de ter - ministic exp onen tial time for each problem that the o racle is capable of solving. If so desir ed, the time r e quired to co mpute the oracle set could b e treated as s eparate to the proc ess o f solving the problem. ArXiv, V ol. V, No. N, Month 20YY. Indep endence of P vs. NP in regards to oracle relativizations. · 5 Another option is that the Deterministic Oracle Mac hine could b e netw or k ed to bo th an o racle a nd a Non Deterministic T uring Machine. In this model th e non deterministic machine gener ates the oracle se t which it sends to the ora cle. This metho d will a llo w the entire pro cess to require p olynomial time. A third method is that we simply assume that the oracle set so meho w cre ates itself magically . Regardless of the metho d used to g enerate the ora cle set, all three methods inv olve somehow conce aling the ma jority of the work. This does not in any wa y reduce the co mplexit y o f the problem; it o nly r emo ves the difficult y of the pro ble m by pla cing the ma jority of the co mputational burden up on the pro cess of oracle s et creation. In t his example the oracle A w as created such that P A = NP . It would have bee n v alid to create oracle set A such that P A = PSP ACE . Howev er , it can not b e exp ected that suc h a n ora cle would b e any easier to create. 3.2 A P 6 = NP Oracle. The method of pro ducing oracle set B as described b y Baker, Gill, and Solov ay [1975] is similar to the algorithm: —Let L ( X ) = { x : there exists y ∈ X such that | y | = | x |} for any or acle X . —Let x be the set of all problems in L ( X ). —Let S be the set suc h that S i is the set of all possible inpu t sets for x i when 1 ≤ i ≤ | x | . —Let B ( i ) repres e n t the elemen ts o f B placed in B prio r to stage i . —Let p i ( n ) < | S i | b e a p olynomial f unction of n tha t represents the n um ber of computations pre fo rmed b y P X i and NP X i for all oracles X . If p i ( n ) computations hav e b een preformed, and no accepting e lemen t of S i has b een fo und, then the algorithm halts in the non-accepting sta te without examining all elements of S i . Create the oracle with the algorithm. (1) Let i = 1. (2) If P B i ( i ) rejects x i then the nex t unexamined element o f S i is an ele men t of B . (3) If i < | x | then increment i and contin ue at step 2. (4) If i = | x | then the oracle set B ha s b een found. The pro cess used by a n oracle machine with oracle se t B for finding a solution to a problem x i in N P with n i literals is preformed b y the fo llo wing algorithm. (1) Search for an element of S i which res ults in x i ev aluating true . T erminate the search if the time limit p i ( n ) has b een reached. (2) If a solution to x i was found then, halt in an accepting state. (3) If no so lution to x i was found a nd all elements o f S i were e x amined in step 1, then halt in a non a ccepting sta te. (4) If no solution to x i has b een found and not all elements of S i hav e b een exam- ined, then query the o racle B with an unexamined elemen t of S i . (5) If the machine is in the y es state then ha lt in a n accepting state. ArXiv, V ol. V, No. N, Month 20YY. 6 · Jerrald Meek (6) If the machine is in the no s tate then halt in a no n accepting state. This algorithm will allow for a p olynomial time so lution on a Non Deterministic Oracle Machine NP B bec ause suc h a mac hine will a lw ays find a solution in step one for an y problem in N P if one exists, and halt in step 2 or 3 . How ever, a Deterministic O racle Mach ine ma y not find the so lution in step 1. When a deterministic machine queries the oracle the o r acle only con tains one se t of the prop er c a rdinalit y which may or may not b e an accepting input se t (b ecause the r esult of the input set is not a requirement fo r membership in B ). The o racle will then place the machine in the yes state if the input set pa ssed to the oracle is equiv alent to the o ne input set of prop er cardinality tha t is an e lemen t o f B . It is then the ca se that the Deterministic Ora cle Machine may terminate in an a ccepting state w hen there is no accepting input to the problem. On the other hand if the deterministic o racle machine fails to find a so lution in step 1, and the input set passed to the ora cle is not an e lemen t of B , then the machine ma y terminate in a non-accepting state when an accepting input set may exist. It is then clear that oracle B do es not pro duce a re lia ble p olynomial time solution to the problem x i . It is then true that P B 6 = NP B . How ever this is only b ecause the ora c le B is dysfunctional. Ora cle B do es not actually help a Non Deterministic Ora cle Machine to solve a NP problem, because the non deterministic machine has the ability to solve the pr oblem without en tering the query sta te. The oracle is then never queried, and the dys functionalit y of t he oracle does not interfere with the op eration of the Non Deterministic Or a cle Machine. How ever, a Deterministic Oracle Machine requir es the assista nce of the oracle to ov erc o me the P = NP Optimization The or em . It is then the ca s e that a dys functiona l or acle will pr ev en t a Deterministic O racle Machine from solving a problem in polyno mial time while not affecting the per formance of a Non Deterministic Or acle Machine. It then becomes easy to see why P A = NP A oracle sets are meager [Mehlho rn 1973]. The ora cle set must be carefully fabrica ted in order to create such a condition. If g reat car e has not been taken to ensure that the oracle will b e of assistance to a deterministic machine, then the ora cle will b e dysfunctional. 3.3 An Oracle such that NP is no t clos ed under complemen tation. A complexity class is closed under complementation if the complements of all pr ob- lems in that class are members of the same class . Baker, Gill, and Solov ay [19 75] hav e s ho wn that ther e exists an oracle such that one or more pr o blems in NP C hav e a complement that is not in NP C . —Let L ( X ) = { x : there exists y ∈ X such that | y | = | x |} for any or acle X . —Let x be the set of all problems in L ( X ). —Let S be the set suc h that S i is the set of all possible inpu t sets for x i when 1 ≤ i ≤ | x | . —Let C ( i ) represent the element s of C placed in C prior to s ta ge i . —Let p i ( n ) < | S i | be a po lynomial function of n that repr esen ts the ma xim um nu m be r of computatio ns preformed by P X i and NP X i for all oracles X . If p i ( n ) computations hav e b een preformed, and no accepting elemen t o f S i has been ArXiv, V ol. V, No. N, Month 20YY. Indep endence of P vs. NP in regards to oracle relativizations. · 7 found, then the algorithm halts in the non-accepting state witho ut examining all elements of S i . Create the oracle with the algorithm. (1) Let i = 1. (2) If NP C ( i ) i accepts x i then any one elemen t of S i that is accepted by x i is a dded to the set C . (3) If i < | x | then increment i and contin ue at step 2. (4) If i = | x | then the oracle set C has been fo und. The se t C is a set containing exactly one accepting input set for each pr oblem describ ed by L ( X ) that has an accepting input set. The proc ess of finding a so lution to a problem x i which is in NP with or acle set C is preformed with the follo wing algorithm. (1) Pass each element o f S i to the oracle a nd en ter the quer y state. (2) If the machine is in the yes s tate for any quer y , then halt in an accepting state. (3) If the machine is in the no state for a ll queries, then ha lt in a non accepting state. F or a Non Deterministic Or a cle Machine, all elemen ts of S i can be ev aluated simult aneously . How e v e r, a Deter minis tic Oracle Machine must iter ate threw them one at a time. It is then eas y to see that P C 6 = NP C . T o create the oracle set ¯ C the compliment of C , use the algorithm. (1) Let i = 1. (2) If NP ¯ C ( i ) i rejects x i then add all elemen ts of S i to the set ¯ C . (3) If i < | x | then increment i and contin ue at step 2. (4) If i = | x | then the oracle set ¯ C has been found. With the oracle set ¯ C a Deterministic O racle Machine can so lv e the co mplimen t of any NP problem b y perfo r ming one query . This is b e c ause ¯ C contains all input sets fo r any pro ble m in c o-NP that ha s at least one input set that r esults in the problem ev aluating tru e . If the o racle places the machine in the yes state for any input set then an accepting input set exists for the c o-NP problem, although the accepting input s et may not be represented by the string queried. If the oracle places the ma c hine in the no state then no accepting input set exists fo r the c o-NP problem. It is then the case that P ¯ C ⊃ c o-NP . How ever, this situa tion requires that the oracle sets C and ¯ C are created non deterministically . It is then the case that the w o rk o f solving these problems has bee n done by the creation of the oracle sets, a nd this allows the Deterministic Oracle Ma c hine to av oid the limitations of the P = NP Optimization The or em when solving c o-NP problems. 3.4 A P 6 = NP oracle such that NP is cl osed under com plementation. —Let L ( X ) = { x : there exists y ∈ X such that | y | = | x |} for any or acle X . —Let x be the set of all problems in L ( X ). ArXiv, V ol. V, No. N, Month 20YY. 8 · Jerrald Meek —Let S be the set suc h that S i is the set of all possible inpu t sets for x i when 1 ≤ i ≤ | x | . —Let D ( i ) repr e s en t the elements of D placed in D pr ior to sta g e i . —Let ¯ D ( i ) repr e s en t the elements of ¯ D placed in ¯ D prio r to stage i . —Let p i ( n ) < | S i | be a po lynomial function of n that repr esen ts the ma xim um nu m be r of computatio ns preformed by P X i and NP X i for all oracles X . If p i ( n ) computations hav e b een preformed, and no accepting elemen t o f S i has been found, then the algorithm halts in the non-accepting state witho ut examining all elements of S i . Construct the oracle sets D and ¯ D with the algorithm. (1) Let i = 1. (2) Let n = 1. (3) If n is even, then let e = 1. (4) If n is even and S n e / ∈ ¯ D ( i ) then find the prefix u of S n e with length | S n e | / 2. (5) If n is even and S n e / ∈ ¯ D ( i ) and u is an input set for x | u | and NP D ( n ) i rejects x | u | , then S n e ∈ D . (6) If n is even and S n e / ∈ ¯ D ( i ) and e < | S n | then increment e and contin ue at s tep 4. (7) If n is even and S n e / ∈ ¯ D ( i ) a nd e = | S n | then br eak to step 10 . (8) If n is o dd and all elements o f ¯ D ( n ) have c a rdinalit y less than n , and p i ( n ) < 2 ( n − 1) / 2 , then add to ¯ D all strings queried by P D ( n ) i for problem x n . If P D ( n ) i rejects x n , then a dd to D the next element of S n not queried. (9) If n is odd, then incr emen t i . (10) If n < | x | , then increment n and contin ue at step 3. (11) If n = | x | , then the sets D and ¯ D have b een found. In steps 4 and 5, elements a re added to D if their prefix is rejected. Notice that if f ( x ) has no input set re sulting in a tru e ev aluation, then f ( x ) ∧ g ( x ) has no input set res ulting in a tru e ev aluation. It is then the case that in step 4 a nd 5, elements are only added to D if they repres e n t a problem that has no accepting input set. In step 8, an input s e t is added to D if P D ( n ) i rejects the problem which that input set b elongs to. Ho wever, o n the firs t iter ation, D ( n ) will b e an empty set. An empty o racle set is eq uiv alent to having no oracle, and so the p olynomial time algorithm will not chec k all p ossible input sets for the firs t problem ex amined. F or the third pro blem exa mined, D ( n ) may con tain a n input s e t if it belo ngs to a problem that alwa ys ev a luates false . It is then easy to see that step 8 adds elements to D that may or ma y no t cause a problem to ev aluate true . Because the elements o f D do not ha ve any consisten t represe ntation, it then follows that the set D will result in a dysfunctiona l o racle. A deter ministic ma c hine requires a functional oracle set to be able to solve all NP problems, but a non deterministic machine do es not rely upo n the o racle set whe n solving NP pro blems. It then follows that P D 6 = NP D . In s tep 8 , the req uiremen t that all elements of ¯ D ( n ) must have cardinality less than n will alwa ys ev alua te t r u e . The statement p i ( n ) < 2 ( n − 1) / 2 will ev a luate tru e ArXiv, V ol. V, No. N, Month 20YY. Indep endence of P vs. NP in regards to oracle relativizations. · 9 for the first v alues o f n up to so me limit. F o r pr oblems less than this limit, a p olyno- mial num b er of e le ments will b e added to ¯ D . The elements will b e added reg ardless of weather or not they result in a true ev a luation of the c o-NP pr oblem asso ciated to them. It is then clear to see that the set ¯ D will also result in a dys functiona l oracle, and P ¯ D will not pro duce reliable results for a n y c o-NP pro blem. 3.5 A P 6 = NP oracle such that P = [ NP ∩ co-N P ]. Baker, Gill, and Solov ay [19 75] construct the o racle set E by adding elemen ts to the oracle set A which w as prev iously constructed suc h that P A = NP A . The additional elemen ts in E will result in P E 6 = NP E . The tric k is to arrange the additional elemen ts suc h that if a problem is in NP , and the complimen t of the problem is also in N P , then both o f these pro blems are solv able in po lynomial time by the oracle P E . In the description of crea ting oracle E , it is allo w ed to start with “any or a cle A such that P A = NP A .” Here, the oracle A that was previo usly constructed in this article will be used. Ho w ever, remember that oracle A did not contain strings representing input sets, but instead partition enco ding was used. I t will then b e the case that the same enco ding sp ecification will need to b e us e d for the additiona l elements of E . —Let L ( X ) = { x : there exists y ∈ X such that | y | = | x |} for any or acle X . —Let x be the set of all problems in L ( X ). —Let S be the set suc h that S i is the set of all possible inpu t sets for x i when 1 ≤ i ≤ | x | . —Let e ( n ) be a function such that e (0) = 0 and e ( x ) = 2 2 e ( x − 1) ← x > 1. —Let E (0) = A and let E ( n ) be the set o f elements placed in E prior to stage n . —Let r b e the set of all po ssible v a lues for h j, k i ← j ∈ N , k ∈ N , j 6 = k . —Let r j ( i ) denote the j element of set r i , and r k ( i ) denote the k e lemen t o f r i . —Let p i ( n ) < | S i | be a po lynomial function of n that repr esen ts the ma xim um nu m be r of computatio ns preformed by P X i and NP X i for all oracles X . If p i ( n ) computations hav e b een preformed, and no accepting elemen t o f S i has been found, then the algorithm halts in the non-accepting state witho ut examining all elements of S i . Create the oracle set E with the algo rithm. (1) Let n = 1. (2) Let i = 1. (3) Let l ∈ S n . (4) If e ( n − 1) < ( l og 2 | l | ) ≤ e ( n ) ≤ max  p r j ( i ) ( l ) , p r k ( i ) ( l )  < e ( n + 1) and neither NP E ( n ) r j ( i ) or NP E ( n ) r k ( i ) . accepts x n , then h j, k i is unsatisfied. (5) If p i ( e ( n )) ≥ 2 e ( n ) , then h i, i i is unsatisfied. (6) If h j, k i was not determined unsatisfied in step 4, and h i, i i w as not determined unsatisfied in step 5, then incremen t i . ArXiv, V ol. V, No. N, Month 20YY. 10 · Jerrald Meek (7) If h j, k i w a s no t deter mined unsatisfied in s tep 4 , and h i, i i was determined unsatisfied in step 5, and P E ( n ) i rejects x n , then add to E the enco ding o f x n with the next unev a luated elemen t of S n . (8) If h j, k i w a s no t deter mined unsatisfied in s tep 4 , and h i, i i was determined unsatisfied in step 5, then incremen t i . (9) If i < | N | then incre ment n and contin ue a t step 3. (10) If i = | N | then the set E has b een found. Notice that when h i, i i is unsatisfied, then P E ( n ) i will b e r un on x n . The s e t E ( n ) will only contain elements relev ant to x n if it was inherited from set A . It is then the c ase that P E ( n ) i will only reject x n when x n has no accepting input set. In step 7, if h i, i i is unsatisfied, and P E ( n ) i rejects x n , then the input set encoding for x n will a lw ays b e an enco ding tha t repr esen ts a par tition containing no accepting input set. It will then b e the case that P E will accept x n when a Non Deterministic Oracle Machine would halt in a non accepting state without querying the oracle. Therefore, P E 6 = NP E bec ause the oracle E is dysfunctional when handling some problems. Let κ be the set of all problems in NP which hav e complements in NP . When x n ∈ κ , then h j, k i will b e satisfied. When h j, k i is satisfied then no strings applicable to x n will b e a dded to E . It is then the cas e that the s ubs e t of E that applies to all elements of κ will b e ident ical to the subs e t of A that applies to all e lemen ts of κ . So P E = NP E when the problem being ev aluated is an elemen t of κ . 3.6 An o racle set su ch that P F ⊆ [ NP F ∩ co-NP F ] ⊆ NP . Baker, Gill a nd Solov ay [197 5] do not lay out the sp ecific metho d for crea ting oracle F . Ho wev er they do pro vide an outline which includes creating tw o lang ua ges L 1 ( F ) and L 2 ( F ) such that L ( F ) = [ L 1 ( F ) ∪ L 2 ( F )]. Tw o oracle sets must also be c reated by t wo differen t methods , a nd F is the union of these tw o or acle sets. How ever, notice that if P F ⊆ [ NP F ∩ c o-NP F ], and [ NP F ∩ c o-NP F ] ⊆ NP , then P F ⊆ of NP . It has already bee n demonstrated that an oracle set can b e cre ated suc h that P A = NP . In other words the set of pr oblems s o lv able by a Deterministic Oracle Machine with ora cle set A in p olynomial time is equiv alent to the set of problems solv able by a Non Deterministic T uring Machine in p olynomial time. It is then the cas e that the only new condition o f oracle set F is that [ NP F ∩ c o-NP F ] contains all problems in NP . This is the same thing as saying that NP is strictly contained within c o-NP . Earlier , it was s ho wn that an o racle set C co uld b e created s uch that NP C is not closed under complementation. The reaso n that ora cle set C allow ed this conditio n was b ecause it r esulted in all problems in c o-NP b eing solv able in po lynomial time by P ¯ C , while problems in NP where not so lv a ble in p olynomial time b y P C . It is then easy to see that an oracle F could b e crea ted such that all pro blems in NP are so lv a ble in p olynomial time b y P F , and all problems in c o-NP are solv a ble in p olynomial time by P ¯ F . U nder such an oracle the complex it y classes NP , and c o-NP are both contained b y P . If c o-NP ⊂ P ¯ F , and N P ⊆ P F , then P F ⊆ [ NP ∩ c o-NP ]. All problems in ArXiv, V ol. V, No. N, Month 20YY. Indep endence of P vs. NP in rega rds to oracle relativizations. · 11 bo th NP and c o-NP hav e p olynomia l time solutions o n a Deterministic Or acle Machine with o r acle set F . It then follows tha t all pro blems in NP a nd c o-NP hav e po lynomial time solutions on a No n Deterministic O racle Machine with o racle set F . 4. ORACLE RELA T IVIZA TIONS AND P VS. NP SOLUTION METH ODS. Here, the relation of oracle relativizations to the P v s. NP question will b e exam- ined by crea ting an a nalog of the P vs. NP question. Two complexity classes will be c reated, P Λ and NP Λ ; the question will be a sk ed, is P Λ = NP Λ ? T o create the analog complexity class es, one problem that is ob viously in P will be excluded from P Λ . An y problem in P will due, here the problem used will b e the set-sum problem. The set-sum problem will b e defined as: —Let S b e a set of real n um ber s . —Let r b e the num b er of elemen ts in S . —Let M be a real num b er. r X i =1 S i = M Given the v alue of S and M , determine if the equalit y ev aluates true . Notice that the set-sum problem is similar to the 0-1-Knapsack problem, and could b e solved by finding the sum o f all subsets of S , but only ev aluate t rue if the one subset of S that contains all elements of S sums to M . In the ana lo g complexity mo del, it will b e assumed that this is the only known metho d o f s olving set-su m . An a nalog complexity mo del will be created whic h ignor es the ex is tence of so me NP problems (such as NP-c omplete problems). The ana log complexity class e s will be de fined a s: —Let NP Λ contain all problems known to b e in P assuming P 6 = N P . —Let P Λ contain all pro blems in NP Λ except the set-sum problem. In the analog co mputational mo del NP Λ contains a ll pr oblems known to b e solv - able in non deterministic polyno mial time, while P Λ contains all pr oblems k nown to b e solv able in deterministic poly nomial time. In the analo g ques tion P Λ = NP Λ assume tha t no metho d has yet b een found to prov e that set -sum is a member o f P Λ , yet set-sum has not bee n conclusively excluded fro m P Λ . Now the following questio ns may b e asked. —Do e s there ex is t an or acle such that P A Λ = NP A Λ ? —Do e s there ex is t an or acle such that P B Λ 6 = NP B Λ ? —Do e s there exist an ora cle such that NP C Λ is not closed under complementation? —Do e s there exist an orac le such that P D Λ 6 = NP D Λ but NP D Λ is closed under co m- plement ation? —Do e s there ex is t an or acle such that P F Λ ⊆ [ NP F Λ ∩ c o-NP F Λ ] ⊆ NP Λ ? ArXiv, V ol. V, No. N, Month 20YY. 12 · Jerrald Meek It should b e easy to see that the answer to all o f these questions is yes. If inconsistent relativizations are an indication of the difficulty of the P Λ vs. NP Λ problem, then the problem o f proving the set-sum problem solv a ble in deter minis tic po lynomial time must b e at least as difficult as proving that NP-c omplete problems are solv able in deterministic p o lynomial time. Is the set of computable oracles such that P X Λ = NP X Λ meager? If so, then do es that indicate a greater likelihoo d tha t P Λ 6 = NP Λ ? With this a nalog it happ ens to be the cas e that P Λ = NP Λ . If the P vs. NP ques- tion is dep endan t on or a cle relativiza tions, then the analog P Λ vs. NP Λ question should indicate that P = N P . If P vs. NP is dep endant on ora cle relativizations, then it sho uld b e e xpected that P = NP co uld b e proven by finding the rela tion b e- t ween the polyno mial time solv ability of set-sum and the rela tiviz ations of P Λ and NP Λ . If such a relatio n exis ts, then it should provide a n indication o f how to prov e P = NP . How e ver, if no such re lation exis ts, then there must exis t no dep endancy betw een oracle relativizatio ns and the solution to the P vs. NP question. Obviously , the pro of that P Λ = NP Λ can b e a c complished by eliminating the ev aluation of all subsets of S for the set -sum problem and only ev aluating the one relev ant subset. It is known that the only relev ant subset is the subset equiv a len t to S . This is known beca use the original definition of the problem says it is so. Is there some way that the orac le relativizatio ns als o identify the one relev ant subset of S ? If so then the relativiza tions sho uld ident ify a limited par tition of relev ant subsets for the Knapsack problem. It is then easy to see that or acle relativiza tions only tell us that the co mplexit y of any given problem can b e hidden by o racle set cr eation. Essentially , using an or acle to solve a NP-c omplete pro blem in deterministic p olynomial time is the reverse op eration o f taking a problem fro m the P complex ity clas s and pr etending that it has no deterministic p olynomial time so lutio n. When a pplied to a NP-c omplete problem, which ma y a c tually have no deterministic p olynomial time solution, an oracle s uc h tha t P A = NP A allows us to pretend that N P-c omplete pro blems have deterministic p olynomial time solutions. 5. CONCLUSION. In this article a ll six or acles form Ba k er, Gill and Solov ay [1975] hav e been examined. As a result, the follo wing things ca n b e said ab out the o racles. —Oracle set A such that P A = NP A is a functional ora cle which allows a Deter- ministic Oracle Machine to solve a n y NP problem with a p olynomial num b er of queries. —Oracle set B such that P B 6 = NP B is a dysfunctional oracle whic h fails to allow a Deterministic Oracle Machine to solve a ll NP B problems with a po lynomial nu m be r of quer ies. —Oracle se t C suc h that NP C is not close d under co mplemen ta tion is a dysfunc- tional o racle such that a Determinis tic Oracle Mac hine can no t solve all NP problems with a p olynomial num b er of que r ies, although a Deterministic Oracle Machine could solve any c o-NP pr oblem with one query . —Oracle set D such that P D 6 = N P D but NP D is clo sed under complementation is a dysfunctiona l oracle with a dysfunctiona l complement. It is then the case that ArXiv, V ol. V, No. N, Month 20YY. Indep endence of P vs. NP in rega rds to oracle relativizations. · 13 a Deterministic Oracle Machine ca n not solve any NP pro blem or an y c o-NP problem in po lynomial time. —Oracle set E such that P E 6 = NP E and P E = [ NP E ∩ c o-NP E ] is a dysfunctional oracle tha t allows a Deterministic O racle Mac hine to solv e a NP problem in po lynomial time when the co mplimen t of the problem is also in NP . —Oracle set F such that P F ⊆ [ NP F ∩ c o-NP F ] ⊆ NP is a functional ora cle which allows a Deterministic Or acle Machine to so lv e all problems in NP and all problems in c o-NP in p olynomia l time. If P = NP then all pr oblems in NP will be solv able in deter ministic polyno mia l time with o r without an oracle s e t. It would then be the case that a Deterministic Oracle Machine would only b e dep endan t on a n o racle s et to solve a problem in NP or c o-NP if the machine is not using an o ptima l algo rithm. It is p ossible to create an algo rithm for an y N P problem suc h that a solution would requir e exp onential time on a Non Deterministic T uring Machine. This howev er do es not disq ualify the problem’s member ship in NP . Likewise, just b ecause a NP problem has a determin- istic exp onen tial time so lution, this do es not disqualify that problem’s membership in P . Therefore, the oracle relativizations are no indica tor of P vs. N P . If P 6 = NP then a n y problem no t ordinar ily in P will b e solv a ble in p olynomial time by a Deterministic Or acle Ma c hine only when the o r acle set is functional for that problem. In this case if P 6 = NP then the Baker, Gill, and Solov ay mo del works exactly as would b e exp e c ted. The mea g erness of co mputable P X = N P X oracle sets a lso see ms not to indicate anything abo ut the P vs. N P pr o blem. Tha t is, we should exp ect a functional o racle for any purp ose to b e meag e r. All of these observ ations lead to the co nclusion that the P vs. NP question is indep endent of or acle rela tiv iz ations. 6. VERSION HISTORY. The author wishes to encourage further feedback which may improv e, strengthen, or pe rhaps dispr o v e the conten t o f this ar ticle. F or that rea son the author do es not publish the names o f any sp ecific p eople who may hav e sug gested, comment ed, o r criticized the ar ticle in such a wa y that r esulted in a revisio n, unles s premissio n has bee n g rant ed to do so. arXiv Current V ersion 3Sep08 Submitted to arXiv. —Revision of partition enco ding metho d. —Refrance to [Nagel and Newman 2001] was added. arXiv V ersi on 5 22Aug08 Submitted to arXiv. —Minor addition. arXiv V ersi on 4 21May08 Submitted to arXiv. —Minor error correc tio n. ArXiv, V ol. V, No. N, Month 20YY. 14 · Jerrald Meek arXiv V ersi on 3 20May08 Submitted to arXiv. —Clarifications were made for some statements. —The Or acle r elativizations and P vs. NP solution metho ds section was a dded. —Refrance to [Bennett and Gill 1981] and [Mehlho rn 1973] were added. arXiv V ersi on 2 16May08 The article was withdrawn due to misleading statemen ts and incomplete resear ch. arXiv V ersi on 1 14May08 Submitted to arXiv. REFERENCES Baker, T. 1979. On “Prov able” Analogs of P and NP . Mathematic al Systems The ory 12, 213-218. Baker, T. , Gill, J . , and Solov a y, R. 1975. Relativizations of the P =? NP question. SIAM J. Comput. 4, (Dec.), 431-442. MR395311. Zbl 0323.68033. Bennett, C. , an d Gill, J. 1981. Relativ e to a r andom oracle A, P A 6 = NP A 6 = c o-NP A with probability 1. SIAM J. Comput. 10, (F eb.), 96-113. MR 0605606. Zbl 0454.68030. Nagel, E. , and Newman, J. 2001. G¨ odel’s Pr oof New Y ork Univ ersity Pr ess . Har tmanis, J. , and Hopcroft, J. 1976. Indep ende nce results in computer science. ACM SIGA CT News 8, 4. Meek, J. 2008. P is a prop er s ubset of NP . arXiv:0804.1079 Article 1 in series of 4 . Meek, J. 2008. Analysis of the determini s tic p olynomial time solv abilit y of the 0-1-Knapsack problem. arXiv:0805.0517 A rticle 2 in series of 4 . Mehlhorn, K. 1973. On the si ze of sets of c omputable functions. 14 th Annu al Symp. on Au tomata and Switching The ory (Univ. Io w a, Io w a City , Io w a), 190-196. MR429513. Receiv ed xx/2008; revised xx/2008; accepted xx/2008 ArXiv, V ol. V, No. N, Month 20YY.