Does P=NP? Karlen G. Gharibyan. SPIRIT OF SOFT LLC, 4-th lane 5 Vratsakan 45, 0051, Yerevan, Armenia

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1 1 Does P=NP? Om tat sat Karlen G. Gharibyan SPIRIT OF SOFT LLC, 4-th lane 5 Vratsakan 45, 0051, Yerevan, Armenia karlen.gharibyan@spiritofsoft.om Abstrat-The P=NP? problem was emerged in This is one of the longstanding unresolved problems in omputer siene. Its solution is diretly related to the determination of time omplexity of NP-omplete problems. The purpose of this work is to provide the solution of P=NP? problem. KEY WORDS: algorithms and omplexity, timetable sheduling, omputer siene. I. INTRODUCTION The Peaeful Coexistene Algorithm (PCA) is introdued in this work. PCA is a polynomial time algorithm that the General Timetable Problem (GTP) solves for any instanes. The GTP is NP-omplete. This means, that eah problem in NP may be solved in polynomial time, inluding 3SAT problem. II. FORMULATION OF GTP As the matter of fat, the problem itself is very versatile and omplex, that is why, for the sake of simplifiation a ertain seondary shool is observed as an objet of implementation. In reality, the generality is kept, beause all ideas, denotations, onsiderations done for the shool, in essene are the same for the general problem and may be generalized without any diffiulty, if it is neessary. Assumptions In shool, there are N lasses, N r lassrooms, N t teahers and N s subjets are taught. In shool, there are ten different level lasses {1,2,3,4,5,6,7,8,9,10} and a few number of lasses of the same level are possible. The numbers from 1 to 10 and English letters are used in order to onstrut additional odes(ac) for identifiation of all lasses (e.g. if there are 3 of the first level lasses and 4 of the seond level lasses and et., then their addidional odes are 1A, 1B, 1C; 2A, 2B, 2C, 2D; et.). Eah subjet of N s subjets in shool is to be taught in a separate lassroom. Eah teaher in shool teahes only one subjet. Eah subjet is onsidered to be either Sharing or Unsharing. Any Sharing subjet is to be taught by two teahers in two subgroups of the same lass in two separate lassrooms. Sharing teahers is a group of exatly two teahers in a lass, who teah the same Sharing subjet in the lass. Any Unsharing subjet in a lass is to be taught by one teaher only. Any Sharing subjet in a lass is to be taught by a group of Sharing teahers only. The start up time of all first lessons of a lass is the same every day. The duration of any lesson is onsidered to be one hour. A. Input Data Variables The following variables are used for the Input Data of GTP:,r,t,s,d,l - lower ase indexes of entries of arrays, whih are usually used as odes of a lass, lassroom, teaher, subjet, day and lesson respetively. N - the number of lasses in shool. N s - the number of subjets in shool. N g =2- the number of subgroups into whih a lass is to be split if a Sharing subjet is taught. N t - the number of teahers in shool. S={S s /s=1,,n s } - the set of subjets taught in shool, where S s is the name of subjet s and index s is the ode of the subjet. C={C / =1,,N } - the set of lasses in shool, where C is AC of a lass and index is the ode of the lass. Q + ={q + / =1,,N } - q + is the total quantity of lessons of lass to be timetabled for a week. D o ={d / =1,,N } - d is the number of days, that an be timetabled for the lessons of lass for a week. B={b / =1,,N } - b is the first hour, that must be timetabled for the lessons of lass every day for a week. E={e =b +eil(q + /d )-1 / =1,,N } - e is the last possible hour, that an be timetabled for the lessons of lass every day for a week (see Appendix). F={f =q + - floor(q + /d )*d / =1,,N } - f is the number of last hours that an be timetabled for the lessons of lass for a week (see Appendix). N d =max{d 1,d 2,,d N } - the maximum number of days, that an be timetabled for the lessons of shool for a week. N l =max{e 1,e 2,,e N } - the last hour, that an generally be timetabled for the lessons of shool every day for a week T={T t / t=1,,n t } - the set of teahers in shool, where T t is the name of teaher t and index t is the ode of the teaher. Q={q,t,s / =1,,N, t=1,,n t, s=1,,n s } - q,t,s is the quantity of lessons of subjet s to be taught by teaher t in lass, suh that the following equality holds (see Appendix): q + = Nt Ns Nt Ns q ( t 1 s 1,t,s = = t= 1 s= 1 N q - the total quantity of lessons of shool: N + = q = q 1 N. (q,t, s IsSh(,s)))/N g.

2 B. Output Data Variables Z={z,l,d / =1,,N, l=1,,n l, d=1,,n d } - the timetable of the shool for a week where z,l,d is the entry of Z. At the beginning all entries of Z are 0. After onstruting Z, if z,l,d is not zero, then it is the ode of a subjet, that is to be taught by orresponding teahers in lass at hour l on day d. Z ={z,l,d / l=1,,n l, d=1,,n d } - the timetable of lass for a week. C. Deision problem of GTP Form the shool s timetable, by timetabling the entries of Z for the lessons of all subjets so that no overlapping exists: subjet to(see Appendix): No overlapping onstraint for any fixed values of t,l,d: N = 1 Lessons onstraint: N d e = 1 d= 1 l= b All the variables are integers. {t T(,l,d)? 1 : 0} 1; {z,l, d 0?1:0} = Nq. III. SOLUTION OF GTP The solution of GTP is the PCA. The PCA realizes in iterations. Some basi ideas and denotations are needed for PCA. A. Definitions Definition 1. An entry z,l,d of Z, for any values of, t and s, is alled Aessible Entry (AE) for subjet s, if q,t,s 0, z,l,d =0 and N o = 1 {ShT(, t, s) T( o, l, d)? 1:0} = 0, that is, this entry an be timetabled for subjet s(see Appendix). Definition 2. A subjet is alled Subjet Aessible to entry z,l,d of Z, if this entry an be timetabled for that subjet. Definition 3. An entry z,l,d of Z is alled Aessible Entry (AE), if this entry an be timetabled for at least one subjet. Definition 4. The number X,l,d of all subjets aessible to AE z,l,d is alled height of AE z,l,d. Definition 5. Z l,d ={z,l,d / =1,,N }is alled Pivot Entry (PE) of Z, where (l,d) is the oordinate of PE Z l,d and (,l,d) is the oordinate of entry z,l,d. Definition 6. Z l,d is alled Aessible PE (APE), if it has at least one aessible entry. Definition 7. h l,d is alled height of APE Z l,d. Its value may be found by the following way: h = N l, d = 1 X,l, d Definition 8. A number of oordinates of APE of Z whih are grouped by a ommon priniple is alled Aessible Pivot Domain(APD). Z must be split into APD, suh that D={D [m] / m=1,,n D }, where N D is the number of APD of Z and [m] is the oordinate of APD D [m].. 2 Definition 9. D [m] ={[m,j]=(l m,j,d m,j ) / j=1,,n m }, where n m is the number of APE in D [m] and [m,j] is the short ut of the oordinate of APE j in D [m]. Definition 10. D [m] ={ D,m / =1,,N }, where D,m is alled lass APD and ontains all AE of D [m] in lass. Definition 11. D,m ={,m,j = (,l m,j,d m,j ) / j=1,,n m }, where,m is the oordinate of D,m and,m,j is the short ut of the oordinate of AE j in D,m. Definition 12. Eah pair of z,l,d s, (,l,d) s or,m,j s is alled Tandem, in short T z,l,d s, T (,l,d) s or T,m,j s, if AE z,l,d or z,m,j an be timetabled for subjet s. Definition 13. Realization of T z,l,d s, T (,l,d) s or T,m,j s, in short RT z,l,d s, RT (,l,d) s or RT,m,j s, is an operation, suh that AE z,l,d, or z,m,j is in fat timetabled for subjet s. Definition 14. h,l,d,s or h,m,j,s is alled subjet s height (SH). It is the total number of AE for subjet s in APE Z l,d or Z [m,j] plus (X,l,d -1) and its value must be found by the following way(see Appendix): h,l,d,s = h,m,j,s =SH(T (,l,d) s )=SH(T,m,j s ). B. Renewable Data Variables The Renewable Data Variables (RDV) are ompletely involved in implementation of the PCA, and define the Time Complexity (TC) of the algorithm. For evaluation of TC of the algorithm the possible maximum length of Input Data N=max{N,N t,n g,n l,n d,n s,n q }is used. Both the desription and the TC of eah of Renewable Data Variables are represented below. A={a,s / =1,,N ; s=1,,n s } - a,s is the number of AE of subjet s in lass : TC=N N s N l N d N t N g O(N 6 ). P ={p,s =a,s - QL(,s) / =1,,N ; s=1,,n s } - p,s is the priority of subjet s in lass (see Appendix): TC=N N s N t O(N 3 ). G S,l,d={s,l,d,j / j=1,,x,l,d } - the group of all X,l,d subjets aessible to AE z,l,d, where X,l,d is the height of z,l,d : TC=N s N l N d N t N g O(N 5 ). G S l,d={g S,l,d / =1,,N } - the set of all subjets groups aessible to APE Z l,d : TC=N N s N l N d N t N g O(N 6 ). G S*,l,d={s j =s *,l,d,j / j=1,,x,l,d } - is the group of all x,l,d subjets aessible to AE z,l,d with the smallest priority p *,l,d and with the smallest height h *,l,d. All the subjets are arranged in desending order by lessons: G S*,l,d G S,l,d; p *,l,d=min{p,s : s G S,l,d}; h *,l,d=min{h,l,d,s : s G S,l,d}; TC=N N l N d N s O(N 4 ).

3 3 N H={ h l, d = = X 1,l, d / l=1,,n l, d=1,,n d } - h l,d is the height of APE Z l,d : TC=N N l N d O(N 3 ). D ={D [m] / m=1,,n D } - the set of all N D APD of Z, where m is the ode of D [m] of Z: TC=N N s N l N d N l N d O(N 6 ). D [m] ={[m,j] / j=1,,n m } - the set of all n m APE of D [m] of Z, suh that for any two oordinates [m,k] [m,j] D [m] the following equality is true: G S [m,k]= G S [m,j]; TC=N N s N l N d N l N d O(N 6 ). D,m ={,m,j / j=1,,n m } - the set of all n m AE of D,m of Z, suh that for any fixed values of and for any two oordinates,m,k,m,j D,m the following equalities are true: p *,m,k =p *,m,j ; G S*,m,k =G S*,m,j ; G S,m,k =G S,m,j ; X,m,k =X,m,j ; x,m,k =x,m,j ; TC=N s N l N d N l N d O(N 5 ). p,m = p *,m,1 - the urrent priority of any D,m D: TC=N N l N d O(N 3 ). P,m =p *,m,1 -n m +y,m - is the ritial priority of any D,m D, whih in advane defines the priority of all subjets s 1,s 2,,s n G S*,m,1 when all n m AE of D,m are timetabled only for those subjets, where n=x,m,1 and the value of y,m is found by funtion Y(,m) (see Appendix): TC=N N l N d N l N d O(N 5 ). C. Priniple of priority for any iteration of PCA As it was mentioned above, p,s =a,s -QL(,s) is the priority of subjet s in lass (see Appendix). This is the most important variable in searhing for the solution of GTP. On one hand, it s evident that, if p,s < 0 for any values of and s, then no solution of GTP exists. That is, the smallest priority p,s is the best ertifiate to be ertain, that no solution of GTP exists. On the other hand, the timetabling of an AE z,l,d for subjet s with the smallest priority must derease the height of the orresponding Z l,d as little as it s possible. In this way it will have the minimum impat on further onstrution of Z. Hene, it is quite reasonable to timetable all the subjets with the smallest priority and with the smallest height, first. Namely, these are two basi priniples of organizing eah iteration of PCA. D. Critial priority P,m of lass APD D,m Aording to the desription of D,m for any fixed value of and for any two oordinates, suh that,m,1,m,k and,m,1,,m,k D,m, the following equality p *,m,1 =p *,m,k is true. In this respet, the ritial priority P,m of D,m is determined by the following way: P,m =p *,m,1 -n m +y,m. The value of y,m is found by the funtion Y(,m) (see Appendix). In the Table 1, for any values of and m, the list of all subjets of s 1,s 2,,s n G S*,m,1 is represented. It illustrates how the funtion Y(,m) does work. Eah row has its denotation (0,1),,(y-1,y),,(l * -1,l * ), whih is used for definition of the values of entries in that row. For example, in the row (0,1), the value of the rightmost entry is 1 and the values of the remaining entries are 0, in the row (1,2), the value of the rightmost entry is 2 and the values of the remaining entries are 1 and so on. Similarly, in the row (y-1,y), the value of the rightmost entry is y and the values of the remaining entries are y-1, et. The funtion Y(,m) finds the value of y,m by ounting exatly n m entries starting with the row (0,1), from left to right. Keeping the generality, suppose that ounting is ended on the row (y-1,y). If Y(,m) reahes up the rightmost entry of the row (y-1,y) then y,m =y. If Y(,m) reahes up one of the remaining entries of the row (y-1,y) then y,m =y-1. As regards the ritial priority P,m of D,m, it in advane defines the priorities of subjets s 1,s 2,,s n G S*,m,1 when n m AE of D,m are timetabled only for those subjets. The timetabling of n m AE of D,m only for subjets s 1,s 2,,s n G S*,m,1 is proeeded by rows, starting with the row (0,1), from left to right (see the Table 1). Keeping the generality, suppose that for any fixed values of and m, the timetabling is ended on the row (y-1,y). Two situations are possible and both must be always analyzed in further onsiderations: Situation 1: If the timetabling reahes up the rightmost entry of the row (y-1,y), then the priority of any subjet s 1, s 2,,s k G S*,m,1 will be: P,m =p *,m,1 -n m +y. Situation 2: If the timetabling reahes up one of the remaining entries of the row (y-1,y), for example to the entry marked blak, then the priority P R of eah subjet to the right of the blak entry s j+1, s j+2,,s k G S*,m,1 may be alulated by the following expression: P R =p *,m,1 -n m +y-1=p,m, but the priority of eah subjet s 1, s 2,,s j G S*,m,1 may be alulated by the following expression: P L =p *,m,1 -n m +y=p,m +1=P R +1.

4 4 Table 1. The list of all subjets s 1,s 2,,s n G S*,m,1, for any fixed values of and m, whih are arranged in desending order by lessons. If s 1,s 2,,s n G S*,m,1 G H, then eah subjet has the ritial priority of P min =P,m and the subjet s height of H min =SH (T,m,1 s 1 ). Rows Subjets odes ( l -1, l ) l l * =QL(,s 1 ) n=x,m,1 Quantity of lessons ( y, y+1 ) y+1 ( y-1, y ) y-1 y-1 y-1 y-1 y-1 y-1 y-1 y-1 y-1 y-1 y-1 y ( y-2, y-1 ) y-1 ( y-3, y-2 ) y-2 y-3 ( 1, 2 ) ( 0, 1 ) j. k. n s 1 s 2 s j s k s n Subjets odes E. Critial priority, ritial height and objetive tandems of any iteration of PCA Definition 15. P min is alled the ritial priority of iteration (CPI) of PCA. Its value is found by the following way: P min =min{p,m :,m D}. Definition 16. H min is alled the ritial height of iteration (CHI) of PCA. Its value is found by the following way(see Appendix): G P ={G S*,m,1 /,m,1 D,m D; P min = P,m }; H min =min{sh(t,m,1 s ) : s G S*,m,1 G P }. Definition 17. Eah tandem of T z,l,d s, T (,l,d) s or T,m,j s, for any fixed values of,m and s, is alled objetive tandem (OT) of iteration of PCA and in short OT z,l,d s, OT (,l,d) s or OT,m,j s, if P min =P,m and H min =SH(T (,l,d) s )=SH(T,m,j s ).

5 Definition 18. Realization of OT z,l,d s, OT (,l,d) s or OT,m,j s in short ROT z,l,d s, ROT (,l,d) s or ROT,m,j s, is an operation, suh that AE z,l,d or z,m,j is in fat timetabled for subjet s. Definition 19. The set G H is alled the soure of all objetive tandems of iteration (SAOTI) of PCA. It is onsrtuted by the following way: G H ={G S*,m,1 / H min =SH (T,m,1 s ); s G S*,m,1 G P }. Definition 20. G OT is alled the set of all objetive tandems of iteration of PCA (AOTI). It is onstruted using SAOTI G H by the following way: G OT ={OT,m,j s k / j=1,2,,n m ; s k G S*,m,1 G H ; k=1,2,,x,m,1 }, where any subjet s G S*,m,1 G H has the priority of P min and subjet s height of H min. F. Peaeful oexistene algorithm Step 0. Initialize Input Data Variables with Input Data for an instane; Initialize Output Data Variables with 0; i= 0; ( Comment: i ounts the number of iterations ) Step 1. Determine Renewable Data Variables, taking into aount all ROT, if there are any. If N D =0 then {Timetable of shool(z) is onstruted. Print Z ; Goto Step 3;} Step 2. P min =min{p,m :,m D}; If P min <0 then {Deision problem of GTP does not have solution at all; Goto Step 3;} G P ={G S*,m,1 /,m,1 D,m D; P min = P,m }; H min =min{sh(t,m,1 s ) : s G S*,m,1 G P }; G H ={G S*,m,1 /H min =SH(T,m,1 s ); s G S*,m,1 G P }; ROT,m,1 s 1 ; (Comment:,m and s 1 are any values suh that OT,m,1 s 1 G OT, see the Table 1) i=i+1; Goto Step 1; Step 3. Stop. G. Theoretial Foundation of PCA Remember that PCA realizes in iterations. The main objetive of any iteration of PCA is to be ertain whether to interrupt or ontinue the searhing for the solution of GTP. In eah iteration, PCA splits Z into lass APD using the RDV and alulates their ritial priorities. Then PCA finds the value of P min by the following way: P min =min{p,m :,m D} and uses it as CPI. If P min <0, then no solution exists, hene the searhing for the solution of GTP must be interrupted. If P min 0 then the searhing for the solution must be ontinued. In this ase, the following two sets G P and G H are reated suh that G H G P. The seletion of elements of G P and G H proeeds in two suessive steps by using the variables P min 5 and H min respetively. First of all, the elements of G P are determined by the following way: G P ={G S*,m,1 /,m,1 D,m D; P min = P,m }, where P min guaranties that the set G P will ontain all the subjets with CPI. Then PCA finds the value of H min by the following way: H min =min{sh(t,m,1 s ) : s G S*,m,1 G P } and uses it as CHI. The elements of SAOTI G H are determined by the following way: G H ={ G S*,m,1 / H min =SH (T,m,1 s ); s G S*,m,1 G P }, where H min guaranties that the set G H will only ontain all the subjets with CPI and CHI. For ompletion of the iteration, only one ROT (,m,1 ) s 1 is to be done by PCA, where,m and s 1 are arbitrary values suh that s 1 G S*,m,1 G H (see the Table 1) and OT,m,1 s 1 G OT. In this way, the timetable of GTP will be onstruted in exatly N q iterations, if there are any. For theoretial foundation of PCA we need to prove the following auxiliary lemma. Lemma: P min of eah iteration is the best ertifiate to be ertain that: if P min <0 then no solution of GTP exists and the searhing for solution must be interrupted; if P min 0 then the searhing for the solution of GTP must be ontinued. Proof: It is evident that P min is an integer. So, three ases P min <0, P min =0 and P min >0 are to be distinguished. For eah of three ases of P min both Situation 1 and Situation 2 must be analyzed separately. In addition, we must examine the set G H only, beause for any arbitrary values of and m suh that G S*,m,1 G H, PCA guaranties that any subjet s 1,s 2,,s n G S*,m,1 has the priority of P min and the subjet s height of H min. Keeping the generality, suppose that G S*,m,1 is the only element in G H and the timetabling of all n m AE of D,m only for subjets s 1,s 2,,s n G S*,m,1 is ended on the row (y-1,y) (where n=x,m,1, see the Table 1). The timetabling has to be proeeded by rows, starting with the row (0,1). The timetabling of subjets in any row is proeeded from left to right. Case 1. P min <0. We have that G S*,m,1 is the only element in G H. That is, for the values of and m, the ritial priority of eah subjet s 1,s 2,,s n G S*,m,1 is P,m =P min <0. Situation 1: If the timetabling reahes up the rightmost entry of the row (y-1,y), then we have P min =P,m =p *,m,1 -n m +y<0. From this, it will follow, that after the timetabling of all n m AE of D,m only for subjets s 1,s 2,,s n G S*,m,1, the priority of any subjet s 1, s 2,,s k G S*,m,1 will be: P,m =p *,m,1 -n m +y<0.

6 This indiates that no solution of GTP exists, hene the searhing for the solution of GTP must be interrupted. Situation 2: If the timetabling reahes up one of the remaining entries of the row (y-1,y), for example to the entry marked blak, then we have P min =P,m =p *,m,1 -n m +y-1<0. D,m only for subjets s 1,s 2,,s n G S*,m,1, the priority of subjets s j+1, s j+2,,s k G S*,m,1 will be: P R =p *,m,1 -n m +y-1<0. This indiates that no solution of GTP exists, hene the searhing for the solution of GTP must be interrupted. Thus, it is proved, that if P min <0, then this is the best ertifiate to be ertain that the searhing for the solution of GTP must be interrupted. Case 2. P min =0. We have that G S*,m,1 is the only element in G H. That is, for the values of and m, the ritial priority of eah subjet s 1,s 2,,s n G S*,m,1 is P,m =P min =0. Situation 1: If the timetabling reahes up the rightmost entry of the row (y-1,y), then we have P min =P,m =p *,m,1 -n m +y=0. D,m only for subjets s 1,s 2,,s n G S*,m,1, the priority of any subjet s 1, s 2,,s k G S*,m,1 will be: P,m =p *,m,1 -n m +y=0. and this means that the searhing for the solution of GTP must be ontinued. Situation 2: If the timetabling reahes up one of the remaining entries of the row (y-1,y), for example to the entry marked blak, then we have P min =P,m =p *,m,1 -n m +y-1=0. D,m only for subjets s 1,s 2,,s n G S*,m,1, the priority of subjets s j+1, s j+2,,s k G S*,m,1 will be: P R =p *,m,1 -n m +y-1=0 and the priority of eah subjet s 1, s 2,,s j G S*,m,1 will be: P L =p *,m,1 -n m +y=p,m +1>0. and this means that the searhing for the solution of GTP must be ontinued. Thus, it is proved, that if P min =0, then this is the best ertifiate to be ertain that the searhing for the solution of GTP must be ontinued. Case 3. P min >0. We have that G S*,m,1 is the only element in G H. That is, for the values of and m, the ritial priority of eah subjet s 1,s 2,,s n G S*,m,1 is P,m =P min >0. Situation 1: If the timetabling reahes up the rightmost entry of the row (y-1,y), then we have 6 P min =P,m =p *,m,1 -n m +y>0. D,m only for subjets s 1,s 2,,s n G S*,m,1, the priority of any subjet s 1, s 2,,s k G S*,m,1 will be: P,m =p *,m,1 -n m +y>0. and this means that the searhing for the solution of GTP must be ontinued. Situation 2: If the timetabling reahes up one of the remaining entries of the row (y-1,y), for example to the entry marked blak, then we have P min =P,m =p *,m,1 -n m +y-1>0. D,m only for subjets s 1,s 2,,s n G S*,m,1, the priority of subjets s j+1, s j+2,,s k G S*,m,1 will be: P R =p *,m,1 -n m +y-1>0 and the priority of eah subjet s 1, s 2,,s j G S*,m,1 will be: P L =p *,m,1 -n m +y=p,m +1>0. and this means that the searhing for the solution of GTP must be ontinued. Thus, it is proved, that if P min 0, then this is the best ertifiate to be ertain that the searhing for the solution of GTP must be ontinued. The proof of the lemma is ompleted. Thus, the following theorem is the theoretial foundation of PCA: Theorem: The deision problem of GTP has at least one solution iff the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1, where, m and s 1 are any fixed values, suh that s 1 G S*,m,1 G H and OT,m,1 s 1 G OT (see Table 1). Proof: Suppose that the deision problem of GTP has at least one solution and we must prove that the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1. Indeed, if the deision problem of GTP has at least one solution then from this it will follow that the CPI of the first iteration P min 0. Aording to the lemma the searhing for the solution of GTP must be ontinued. For ompletion of the first iteration, only one ROT,m,1 s 1 is to be done by PCA, where, m and s 1 are arbitrary values suh that s 1 G S*,m,1 G H (see the Table 1) and OT,m,1 s 1 G OT. Thus, the PCA guaranties that ROT,m,1 s 1 will have the minimum impat on further onstrution of Z, beause subjet s 1 has the CPI of P min and the CHI of H min. Sine the deision problem of GTP has at least one solution then from this it will follow that the CPI of the seond iteration P min 0. Aording to the lemma the searhing for the solution of GTP must be ontinued. For ompletion of the seond iteration, only one ROT,m,1 s 1 is to be done by PCA, where, m and s 1 are arbitrary values suh that s 1 G S*,m,1 G H and OT,m,1 s 1 G OT (see the Table 1). Thus, the PCA again guaranties that

7 7 ROT,m,1 s 1 will have the minimum impat on further onstrution of Z, beause subjet s 1 has the CPI of P min and the CHI of H min. It s evident that these observations are true also for eah of the next N q -2 iterations of PCA. It s proved that if the deision problem of GTP has at least one solution then the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1. For the opposite diretion, suppose that the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1 and we must prove that the deision problem of GTP has at least one solution. Indeed, if the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1 it will follow that the CPI of the first iteration P min 0. Aording to the lemma the searhing for the solution of GTP must be ontinued. For ompletion of the first iteration, only one ROT,m,1 s 1 is to be done by PCA, where, m and s 1 are arbitrary values suh that s 1 G S*,m,1 G H (see the Table 1) and OT,m,1 s 1 G OT. Thus, the PCA guaranties that ROT,m,1 s 1 will have the minimum impat on further onstrution of Z, beause subjet s 1 has the CPI of P min and the CHI of H min. Sine the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1 it will follow that the CPI of the seond iteration P min 0. Aording to the lemma the searhing for the solution of GTP must be ontinued. For ompletion of the seond iteration, only one ROT,m,1 s 1 is to be done by PCA, where, m and s 1 are arbitrary values suh that s 1 G S*,m,1 G H and OT,m,1 s 1 G OT (see the Table 1). Thus, the PCA again guaranties that ROT,m,1 s 1 will have the minimum impat on further onstrution of Z, beause subjet s 1 has the CPI of P min and the CHI of H min. It s evident that these observations are true also for eah of the next N q -2 iterations of PCA and one of the solutions of GTP will be onstruted in exatly N q iterations. It s proved that if the CPI of eah of N q iterations P min 0 when ROT,m,1 s 1 then the deision problem of GTP has at least one solution. The proof of the theorem is ompleted. H. Time Complexity of PCA So, it is evident that TC of PCA is O(N 7 ). P=NP. IV. CONCLUSION V. RESULT OF PCA Using PCA a omputer program was developed in C++. As an instane of implementation of GTP the Input Data of a ertain shool is proessed. For this instane the program onsumed 73 seonds on IBM PC with a proessor P-VI 3000 for onstruting the timetable. ACKNOWLEDGMENTS The author thanks Poghosyan Liana, Khahatryan Alla, Gharibyan Marianna and Sahakyan Taguhi for omments that improved the manusript, Karen Gharibyan for the mathematial onept disussions. Auxiliary Funtions: APPENDIX eil(k)- returns the minimum integer that is greater than or equal to k. floor(k)- returns the maximum integer that is less than or equal to k. 1, {ondition? 1 : 0} = 0, if ondition holds; if not; T(,l,d)- returns the set of teahers odes whih teah subjet z,l,d in lass at hour l on day d, if z,l,d 0; returns, if z,l,d =0. ShT(,t,s)- returns the set of teahers odes whih teah subjet s in lass, if q,t,s 0; returns, if q,t,s =0. QL(,s)- returns the quantity of lessons of subjet s in lass. IsSh(,s)- returns 1, if s is a Sharing subjet in lass ; returns 0, otherwise. SH(T (,l,d) s )- returns the total number of AE for subjet s in APE Z l,d plus (X,l,d -1). Y(,m)- returns the value of y,m. The listing of the funtion is provided below: Y(int, int m) { int ne=n m, fl=0, yr, y=0, n=x,m,1 ; for (int i = n; i >= 1; i--) { s=s i ; ql=ql(,s); (Comment: see the Table 1) if (fl == ql) ontinue; yr = ql - fl; for (int j = yr; j >= 1; j--) { if (ne < i) return y; ne = ne - i; y++; } fl=ql; } return y; }