Copyright A. A. Frempong

Transcription

1 Copyright.. Frempong P vs NP: Solutions of NP Problems bstract The simplest solution is usually the best solution---lbert Einstein est news. fter over 30 years of debating, the debate is over. Yes, P is equal to NP. For the first time, NP problems have been solved in this paper. Techniques and formulas were developed and used to solve these problems as well as produce simple equations to help programmers apply the techniques. The techniques and formulas are based on an extended shanti fairness wisdom as exemplified below. If two people and are to divide items of different sizes which are arranged from the largest size to the smallest size, the procedure will be as follows. In the first round, chooses the largest size, followed by choosing the next largest size. In the second round, chooses first, followed by. In the third round, chooses first, followed by and the process continues up to the last item. To abbreviate the sequence in the above choices, one obtains the sequence ",,,, ". Let and divide the sum of the whole numbers, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 as equally as possible, by merely always choosing the largest number. Then chooses 10, chooses 9 and 8, followed by choosing 7 and 6; followed by choosing 5 and 4; followed by choosing 3 an 2; and finally, chooses 1. The sum of 's choices is = 28; and the sum of 's choices is = 27, with error, plus or minus 0.5. Observe the sequence ",,,, ". Observe also that the sequence is not ",,,, as one might think. If one were to use,,,,, the sum for would be = 30 and the sum for would be = 25, with error, plus or minus 2.5. The reason why the sequence is ",,,, ", and not ",,,, " is as follows. In the first round, when chooses first, followed by, has the advantage of choosing the larger number and has the disadvantage of choosing the smaller number. In the second round, if were to choose first, would have had two consecutive advantages, and therefore, in the second round, will choose first to produce the sequence, or. In the third round, chooses first, because chose first in the second round. fter three rounds, the sequence would be,,. When this technique was applied to 100 items of different masses, by mere combinations, the total mass of 's items was equal to the total mass of 's items. Similarly, for 1000 items of different masses, the total mass of 's items was equal to the total mass of 's items. y hand, the techniques can be used to prepare final exam schedules for 100 or 1000 courses. School secretaries, office assistants can learn and apply the techniques covered. The author also confirmed the notion that a method that solves one of these problems can also solve other NP problems. Consider the following two problems which are modifications of suggested NP problems from the Wikipedia (Simple English) website. Problem 1: businessman wants to take 100 items of different masses to the market. These items are to be packed into boxes. Each box can only hold up to 560 units. The businessman would like to know if 10 boxes would be sufficient to carry all 100 items to the market. Problem 2: school offers 100 different courses, and each course requires one hour for the final exam. For each course, all students registered for that course must take the final exam at the same time. Since some students take more than one course, the final exam schedule must be such that students registered for two or more courses will be able to take the exams for all their registered courses. teacher would like to know if it is possible to schedule all of the exams for the same day so that every student can take the exam for each course registered for. fter solving Problem 1, one was able to solve Problem 2 by mere inspection of the solutions to Problem 1. Since a method that solves a single NP problem can solve other NP problems, and six problems have been solved in this paper, all NP problems (well-posed problems) can be solved. If all NP problems can be solved, then all NP problems are P problems and therefore, P is equal to NP. The CMI Millennium Prize requirements have been satisfied. 1

2 Solutions of NP Problems The following sample problems will be solved and analyzed. They are based on the suggested sample problems from the Wikipedia (Simple English) website. Many Thanks to Wikipedia. Example 1 (Preliminaries) y always choosing the largest number, and will divide the following set of numbers equally or nearly equally. 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. Example 2a Consider the existence of dollar bills with denominations $100, $99, $98,...$2, down to $1. Suppose the bills are on a table with the $100 bill at the top, followed by the $99, $98, $97 bills, and so on with the $1 bill at the bottom of the stack. Now, by mere grabbing in turns always from the top of the stack, divide the total value of these dollar bills equally between and. Example 2b Consider the existence of dollar bills with denominations $100, $99, $98,...$2, down to $1. Suppose the bills are on a table with the $100 bill at the top, followed by the $99, $98, $97 bills, and so on with the $1 bill at the bottom of the stack. Now, by mere grabbing in turns always from the top of the stack, the total value of these dollar bills is to be divided equally between and. Question: (a) If a computer costs $2,000, can afford to buy this computer? (b) If a computer costs $3,000, can afford to buy this computer? page 7 Example 3 Let one randomly delete some of the bills in Example 2a, a previous example, and divide the remaining bills between and. fter the deletion of some of the bills, there are 78 bills remaining. Example 4 businessman wants to take 100 items of different masses to the market. These items are to be packed into boxes. Each box can only hold up to 560 units. The businessman would like to know if 10 boxes would be sufficient to carry all 100 items to the market. Example 5 school offers 100 different courses, and each course requires one hour for the final exam. For each course, all students registered for that course must take the final exam at the same time. Since some students take more than one course, the final exam schedule must be such that students registered for two or more courses will be able to take the exams for all their registered courses. teacher would like to know if it is possible to schedule all of the exams for the same day so that every student can take the exam for each course registered for. Example 6 builder has 1000 concrete blocks of different masses arranged from 1000 units to one unit. The builder would like to divide the blocks into two piles and of equal masses. Prepare a list by masses of all the blocks in pile, and all the blocks in pile. 2

3 Case 1: Only two devisors and Example 1 (Preliminaries) y always choosing the largest number, and will divide the following set of numbers equally or nearly equally. 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. Solution For communication purposes, one will call the numbers to be divided the "dividends"; and one will call and the "divisors". Let the sum of 's choices be Q, and let the sum of 's choices be Q. Step 1: Check to ensure that the numbers are arranged in decreasing order. One will apply the wisdom method of the introduction. That is, one applies ",,,,,,, Method 1 Using braces Step 2: chooses the first element, 14 Step 3: chooses the next two elements, 13 and 12., Step 4: chooses the next two elements 11, and 10, and the alternating consecutive choices continue to the end. 14 {, 13, 12, 11, 10, 9{, 8, 7{, 6, 5{, 4, 3{, 2, 1{ (1) Step 5: dd the choices for and add the choices for. Q = = 53 Q = = 52 The sum for = 53; and the sum for = 52. Method 2 (Tabular form) Step 1: List the dividends as shown below Step 2: Write the divisors,,,,, etc, above the numbers, This is the choosing step Note: 14 means chooses means chooses 13. Step 3: Collect and add the corresponding (dividends) choices Q Q Total:

4 Mathematical formulas for choosing the elements a1 = 14, a2 = 13, a3 = 12, a4 = 11, a5 = 10, Let a = 9, a = 8, a = 7, a = 6, a = 5, a = 4, a = 3, a = 2, a = y experimentation, one obtains the following formulas for and. 6 Q = a1+ a2n + a2n+ 1 ( Q = a1 + a4 + a5 + a8 + a9 + a12 + a13) n= 246,,, 5 2n 2n+ 1 n= 1357,,, Q = a + a + a 14 ( Q = a2 + a3 + a6 + a7 + a10 + a11 + a14) pply the formulas to above (The above formulas are valid for only two divisors.) n= 246,, Q = a + a n + a n+ = = n 2n+ 1 n= 135,,. Q = a + a + a 14 = = 52 Note that the above formulas using the sigma notation are valid for only two divisors, and. For three divisors,, and C, different formulas would have to be derived, based on the solutions of the problem. 4

5 Example 2a Consider the existence of dollar bills with denominations $100, $99, $98,...$2, down to $1. Suppose the bills are on a table with the $100 bill at the top, followed by the $99, $98, $97 bills, and so on with the $1 bill at the bottom of the stack. Now, by mere grabbing in turns always from the top of the stack, the total value of these dollar bills is to be divided equally between and. Method 2a: Using the numerical values and braces pply,,,,,. (as in Method 1 of Example 1) Step 1: chooses the first $100 bill. (Only a single item is removed). Step 2: chooses the next two bills, the $99 and $98 bills, (two items removed consecutively) Step 3: chooses the next two bills, the $97 and $96 bills, and the alternating removal continues to the end. 100 {, 99 98, 97, 96, 95, 94, 93, 92 91, 90, 89, 88, 87 86, 85, 84, 83, 82, 81, 80 79, 78, 77, 76,, 75 74, 73, 72, 71, 70, 69, 68 67, 66, 65, 64, 63 62, 61, 60, 59, 58, 57, 56 55, 54, 53, 52,, 51, 50, 49 48, 47, 46, 45, 44, 43, 42 41, 40, 39, 38, 37 36, 35, 34, 33, 32, 31, 30 29, 28,, 27, 26, 25 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9{, 8, 7{, 6, 5{, 4 3{, 2, 1{, Step 4: Collect and add the choices (dividends) for and Q = = Q = = Conclusion: receives $2525 and receives $2525, Note the zero error for and. Method 2b: Using tabular form Step 1: Write the divisors and above the numbers (as done in Method 2 of Example 1) Step 2: Collect and add the Choices (dividends) Q = =

6 Q = = The above results are pleasantly astonishing. Of the possible ways to divide the above bills, the above technique and consequently the derived formulas divided the above mixture of bills into exactly two equal parts in value. Why has this technique been hiding for nearly 30 years? Note that the ratio Q : Q is 1: 1. Equations for above: Q = a + a + a + a n 2n+ 1 n= 246,,, n= 135,,,... and Q = a n + a n+ Method 1b: Using term numbers and braces pply,,,,,,... a1, a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a a13 a14 a15 a16 a17 a18 a {,,,,,, 12 4, 34, 12 4, 34,, 12 4, 34, 12 4, 34 19, a, a a, a , a24, a25,, a26, a27 a28, a29 a30, a31, a32, a33 a34, a35, a36, a37 a38 a39 a40, a41, a42, a , a, a , 47, 48, 49,, 50, 51, 52 53, 54, 55 56, 57 58, 59 60, 61, 62, a a a a a a a a a a a a a a a a a a a a, a, a a68, a69 a70, a71a72, a73, a74, a75, a76, a77 a78, a79 a80, a81 a82, a83, a84, a85, a86, a , a , a89, 34 a90, a91, a92, a93, a94, a95, a96, a97 a98, a99, a100, { Using the term numbers and tabular form a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11 a 12 a 13 a 14 a 15 a 16 a 17 a 18 a 19 a 20 a 21 a 22 a 23 a 24 a 25 a 26 a 27 a 28 a 29 a 30 a 31 a 32 a 33 a 34 a 35 a 36 a 37 a 38 a 39 a 40 a 41 a 42 a 43 a 44 a 45 a 46 a 47 a 48 a 49 a 50 a 51 a 52 a 53 a 54 a 55 a 56 a 57 a 58 a 59 a 60 a 61 a 62 a 63 a 64 a 65 a 66 a 67 a 68 a 69 a 70 a 71 a 72 a 73 a 74 a 75 a 76 a 77 a 78 a 79 a 80 a 81 a 82 a 83 a 84 a 85 a 86 a 87 a 88 a 89 a 90 a 91 a 92 a 93 a 94 a 95 a 96 a 97 a 98 a 99 a 100 Collect the terms for and add them ; and similarly collect the terms for and add them. Q = a + a + a + a + a + a + a + a + a + a + a + a + a + a + a + a + a + a36 + a37 + a40 + a41 + a44 + a45 + a48+ a49 + a52 + a53 + a56 + a57 + a60 + a61 + a64 + a65 + a68 + a69 + a72+ a73 + a + a + a + a + a + a + a + a + a + a + a + a + a Q = a + a + a + a + a + a + a + a + a + a + a + a + a + a + a + a + a + a35 + a38 + a39 + a42 + a43 + a46 + a47 + a50 + a51 + a54 + a55 + a58 + a59 + a62 + a63 + a66 + a + a + a + a + a + a + a + a + a + a + a + a + a ++ a + a + a + a

7 Example 2b Consider the existence of dollar bills with denominations $100, $99, $98,...$2, down to $1. Suppose the bills are on a table with the $100 bill at the top, followed by the $99, $98, $97 bills, and so on with the $1 bill at the bottom of the stack. Now, by mere grabbing in turns always from the top of the stack, the total value of these dollar bills is to be divided equally between and. Question: (a) If a computer costs $2,000, can afford to buy this computer? (b) If a computer costs $3,000, can afford to buy this computer? nswers: (a) From the solution of Example 2a, received $2,525, and therefore can afford to buy this computer. Yes. can afford to buy this $2,000 computer. (b) Since from the solution of Example 2a, received $2,525, and the computer costs $3,000, cannot afford to buy $3,000 computer No. cannot afford to buy this $3,000 computer. Example 3 Let one randomly delete some of the bills in Example 2a, a previous example, and divide as equally as possible the remaining bills between and. fter the deletion of some of the bills, there are 78 bills remaining. Solution: Using the numerical values and braces 98.97,96,95,94,93,91,90,89,88,87,86.85,.81,80,79,78,77,76, { { 75,74,73,72,69.68,67,66,65,64,62,61,60,58, 56,55,54, , 49, 47, 46, 45, 43., 41,40,39,37,36,35.34,33.32,31,30,29,26,25.24, 23,22,21,20,19,18,16,14, 13,12,11,10,9,8,7,5,4,2,1 { {{ { Q = = 1937 Q = = 1933 Total of Q and Q = Division by 2 yields For Q, relative error = 2 = or about 0.1% 1935 For Q, relative error = 2 = Using term numbers Q = a1 + a4 + a5 + a8 + a9 + a12 + a13 + a16 + a17 + a20 + a21 + a24 + a25 + a28 + a29 + a32 + a33 + a36 + a37 + a40 + a41 + a44 + a45 + a48+ a49 + a52 + a53 + a56 + a57 + a60 + a61 + a64 + a65 + a68 + a69 + a72+ a73 + a76 + a77 Q = a2 + a3 + a6 + a7 + a10 + a11+ a14 + a15+ a18 + a19 + a22 + a23 + a26 + a27 + a30 + a31 + a34 + a35 + a38 + a39 + a42 + a43 + a46 + a47 + a50 + a51 + a54 + a55 + a58 + a59 + a62 + a63 + a66 + a67 + a 70 + a71 + a74+ a75 + a78 Observe above that the last term for Q is a 77 and the last term foir Q is a 78 (there are 78 terms) 7

8 Case 2: Three or more divisors In the previous examples, for communication purposes, and were called the "divisors" and the numbers or terms to be divided were called "dividends". The concept of divisors and can be extended to three or more divisors such as,, C, or,, C, D, but in these cases, geometric figures will help keep track of the choices. Geometric figures to keep track of the order and directions of the divisors (For three or more divisors such as,, C; four divisors,, C,D) The arrows are for directions 4 1 1, 5 4, 8 C 2 3 D C C / C / C / C 2, 6 3, 7 CD/ DC / CD/ DC / CD For C: Step 1: Go Clockwise C (In the first round, chooses first and C chooses last)) Step 2: egin with C and reverse the direction in Step 1 and go C. (Since C was at the largest disadvantage in the first round, by choosing last, C chooses chooses first in the second round followed by ) Step 3: egin with and reverse previous direction (direction of C) and go clockwise C. Step 4: egin with again, change previous direction (direction of ) and go counterclockwise C. For CD Step 1: Go Clockwise CD. (first round) Step 2: egin with D and reverse the direction in Step 1 (direction of ) and go DC... Step 3: egin with C and reverse previous direction (direction of D) and go clockwise CD. Step 4: egin with reverse previous direction (direction of C) and go counterclockwise DC. Step 5: eginning again, reverse the direction but by coincidence go clockwise CD. (5th round) For five divisors,, C, D, E CDE, EDC, DEC, CED CDE Step 1: Go Clockwise CDE Step 2: egin with E and reverse the direction in Step 1 and go EDC Step 3: egin with D and reverse previous direction and go clockwise DEC Step 4: egin with C reverse previous direction and go counterclockwise CED Step 5: egin with reverse previous direction and go clockwise CDE. Step 6: eginning again with, reverse the direction but by coincidence, go clockwise CDE. E 2 1, 6 C 5 3 D CDE, EDC DEC CED, CDE 4 8

9 4a. businessman wants to take 100 items of different masses to the market. These items are to be packed into boxes. Each box can only hold up to 560 units. The businessman would like to know if 10 boxes would be sufficient to carry all 100 items to the market. Step 1: rrange the items in decreasing order of their masses. Let the mass of the first item (largest) be 100 units, and let the masses of the rest of the items be respectively, 99, 98, 97, and so on down to smallest item of mass 1 unit. Let the 10 boxes be labeled,, C, D, E, F, G, H, J, and K The ten boxes are to divide the 100 items. Imitate Example 2 but with 10 divisors. Guide1; CDEFGHJK Guide 2; KJHGFEDC Guide 3 JKCDEFGH Guide 4; HGFEDCKJ Guide 5; GHJKCDEF Guide 6; FEDCKJHG Guide 7: EFGHJKCD Guide 8; DCKJHGFE Guide 9; CDEFGHJK Guide10: KJHGFEDC C D E F G H J K K J H G F E D C a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11 a 12 a 13 a 14 a 15 a 16 a 17 a 18 a 19 a 20 J K C D E F G H H G F E D C K J a 21 a 22 a 23 a 24 a 25 a 26 a 27 a 28 a 29 a 30 a 31 a 32 a 33 a 34 a 35 a 36 a 37 a 38 a 39 a 40 G H J K C D E F F E D C K J H G a 41 a 42 a 43 a 44 a 45 a 46 a 47 a 48 a 49 a 50 a 51 a 52 a 53 a 54 a 55 a 56 a 57 a 58 a 59 a 60 E F G H J K C D D C K J H G F E a 61 a 62 a 63 a 64 a 65 a 66 a 67 a 68 a 69 a 70 a 71 a 72 a 73 a 74 a 75 a 76 a 77 a 78 a 79 a 80 C D E F G H J K K J H G F E D C a 81 a 82 a 83 a 84 a 85 a 86 a 87 a 88 a 89 a 90 a 91 a 92 a 93 a 94 a 95 a 96 a 97 a 98 a 99 a 100 Step 2: Collect the choices for,, C, D, E, F, G, H, J, K Q Q Q C Q D Q E Q F Q G Q H Q J Q K 100 a 1 99, a 2 98 a 3 97 a 4 96 a 5 95 a 6 94 a 7 93 a 8 92 a 9 91 a a 20 82, a a a a a a a a a a 23 77, a a a a a a a a a a 38 64, a a a a a a a a a a 45 55, a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a 88 9 a a 91 1 a a 99 3 a 98 4 a 97 5 a 96 6 a 95 7 a 94 8 a 93 Total: H J G K F E C D 9

10 Condition for sufficiency: The 10 boxes would be sufficient to carry all the 100 items to the market if the mass of the contents of each box is equal to or less than 560 units. Since mass of the contents in each box is 505 units, which is less than 560 units, each box satisfies this sufficiency condition. Therefore, the 10 boxes would be sufficient to carry the 100 items to the market. Note above that the ratio Q : Q : QC : QD : QE : QF : QG : QH : QJ: : QK = : : : : : : : : : Method 4b Using the term numbers C D E F G H J K K J H G F E D C a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11 a 12 a 13 a 14 a 15 a 16 a 17 a 18 a 19 a 20 J K C D E F G H H G F E D C K J a 21 a 22 a 23 a 24 a 25 a 26 a 27 a 28 a 29 a 30 a 31 a 32 a 33 a 34 a 35 a 36 a 37 a 38 a 39 a 40 G H J K C D E F F E D C K J H G a 41 a 42 a 43 a 44 a 45 a 46 a 47 a 48 a 49 a 50 a 51 a 52 a 53 a 54 a 55 a 56 a 57 a 58 a 59 a 60 E F G H J K C D D C K J H G F E a 61 a 62 a 63 a 64 a 65 a 66 a 67 a 68 a 69 a 70 a 71 a 72 a 73 a 74 a 75 a 76 a 77 a 78 a 79 a 80 C D E F G H J K K J H G F E D C a 81 a 82 a 83 a 84 a 85 a 86 a 87 a 88 a 89 a 90 a 91 a 92 a 93 a 94 a 95 a 96 a 97 a 98 a 99 a 100 Collect the terms for,, C, D, E, F, G, H, J, K Q = a1 + a20 + a23 + a38 + a45 + a56 + a67 + a74 + a89 + a92 Q = a2 + a19 + a24 + a37 + a46 + a55 + a68 + a73 + a90 + a91 QC = a3 + a18 + a25 + a36 + a47 + a54 + a69 + a72+ a81 + a100 QD = a4 + a17 + a26 + a35 + a48 + a53 + a70 + a71+ a82 + a99 QE = a5 + a16 + a27 + a34 + a49 + a52 + a61 + a80 + a83 + a98 QF = a6 + a15 + a28 + a33 + a50 + a51 + a62 + a79 + a84 + a97 QG = a7 + a14 + a29 + a32 + a41 + a60 + a63 + a78+ a85 + a96 QH = a8 + a13 + a30 + a31 + a42 + a59 + a64 + a77 + a86 + a95 QJ = a9 + a12 + a21 + a40 + a43 + a58 + a65 + a76 + a87 + a94 QK = a10 + a11 + a22 + a39 + a44 + a57 + a66 + a75+ a88 + a93 Sub-Conclusion The fairness wisdom method has performed perfectly. Observe above in Step 2 that the totals for Q, Q, Q C, Q D, Q E, Q F, Q G, Q H, Q J, Q K are all the same. The technique applied picked combinations to produce these equal totals. Note that in using the technique in this paper, the items involved must be arranged in decreasing order, preferably. Therefore, in programming, the first step should be to arrange the items in decreasing order, a task a computer performs very fast.. In the next eample, Example 5, one will confirm the notion that a method that solves one of the NP problems can be used to solve other similar problems. One will use the results of the above example Example 4b to do the next problem. 10

11 Example 5: school offers 100 different courses, and each course requires one hour for the final exam. For each course, all students registered for that course must take the final exam at the same time. Since some students take more than one course, the final exam schedule must be such that students registered for two or more courses will be able to take the exams for all their registered courses. teacher would like to know if it is possible to schedule all of the exams for the same day so that every student can take the exam for each course registered for. C D E Step1: Final Exams 8M 6PM = 8 9; = 9 10; = 10 11; = 11 12; = 12 1; F = 1 2; G = 2 3; H = 3 4; J = 4 5; K = 5 6 Let the couse numbers be a 1, a 2, a 3,... a 100 Using the result of Example 4b C D E F G H J K K J H G F E D C a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11 a 12 a 13 a 14 a 15 a 16 a 17 a 18 a 19 a 20 J K C D E F G H H G F E D C K J a 21 a 22 a 23 a 24 a 25 a 26 a 27 a 28 a 29 a 30 a 31 a 32 a 33 a 34 a 35 a 36 a 37 a 38 a 39 a 40 G H J K C D E F F E D C K J H G a 41 a 42 a 43 a 44 a 45 a 46 a 47 a 48 a 49 a 50 a 51 a 52 a 53 a 54 a 55 a 56 a 57 a 58 a 59 a 60 E F G H J K C D D C K J H G F E a 61 a 62 a 63 a 64 a 65 a 66 a 67 a 68 a 69 a 70 a 71 a 72 a 73 a 74 a 75 a 76 a 77 a 78 a 79 a 80 C D E F G H J K K J H G F E D C a 81 a 82 a 83 a 84 a 85 a 86 a 87 a 88 a 89 a 90 a 91 a 92 a 93 a 94 a 95 a 96 a 97 a 98 a 99 a 100 Step 2: Collect the choices for,, C, D, E, F, G, H, J, K Final Exam Schedule: 8 M-6 PM Q = 8 9; Q = 9 10; QC = 10 11; QD = 11 12; QE = 12 1; QF = 1 2; QG = 2 3; QH = 3 4; QJ = 4 5; QK = Q Q Q C Q D Q E Q F Q G Q H Q J Q K a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 20 a 19 a 18 a 17 a 16 a 15 a 14 a 13 a 12 a 11 a 23 a 24 a 25 a 26 a 27 a 28 a 29 a 30 a 21 a 22 a 38 a 37 a 36 a 35 a 34 a 33 a 32 a 31 a 40 a 39 a 45 a 46 a 47 a 48 a 49 a 50 a 41 a 42 a 43 a 44 a 56 a 55 a 54 a 53 a 52 a 51 a 60 a 59 a 58 a 57 a 67 a 68 a 69 a 70 a 61 a 62 a 63 a 64 a 65 a 66 a 74 a 73 a 72 a 71 a 80 a 79 a 78 a 77 a 76 a 75 a 89 a 90 a 81 a 82 a 83 a 84 a 85 a 86 a 87 a 88 a 92 a 91 a 100 a 99 a 98 a 97 a 96 a 95 a 94 a 93 The final exam for every course has been scheduled. However, if a student takes for example, Course a 1 and course a 20, because the duration for the final exams for these two courses is 8-9 M, the student cannot take the final exams for these two courses simultaneously. Therefore, it is not possible to prepare a schedule to allow every student to take the final exams for all registered courses on the same day. However, below is what is possible. 11

12 In order for every student to take the final exam for all courses registered for, ten days would be needed as shown below, where the course numbers are a 1, a 2, a 3,... a DY Q Q Q C Q D Q E Q F Q G Q H Q J Q K 1 a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 2 a 20 a 19 a 18 a 17 a 16 a 15 a 14 a 13 a 12 a 11 3 a 23 a 24 a 25 a 26 a 27 a 28 a 29 a 30 a 21 a 22 4 a 38 a 37 a 36 a 35 a 34 a 33 a 32 a 31 a 40 a 39 5 a 45 a 46 a 47 a 48 a 49 a 50 a 41 a 42 a 43 a 44 6 a 56 a 55 a 54 a 53 a 52 a 51 a 60 a 59 a 58 a 57 7 a 67 a 68 a 69 a 70 a 61 a 62 a 63 a 64 a 65 a 66 8 a 74 a 73 a 72 a 71 a 80 a 79 a 78 a 77 a 76 a 75 9 a 89 a 90 a 81 a 82 a 83 a 84 a 85 a 86 a 87 a a 92 a 91 a 100 a 99 a 98 a 97 a 96 a 95 a 94 a 93 Observe how one used the results of the previous example (Example 4b) to solve the above problem, Example 5.. In the next example, one will cover an example involving 1000 items, which will be similar to Example 2a. 12

13 Example 6 builder has 1000 concrete blocks of different masses arranged from 1000 units to one unit. The builder would like to divide the blocks into two piles and of equal masses. Prepare a list by masses of all the blocks in pile, and all the blocks in pile. Review Example 2a before proceeding

14

15

16 Concrete masses for Pile Step 2: Collect and add the Choices (dividends) : Q 1 = = 47, 525 Q = = 42, 525 Q = = 37, 525 Q = = 32, 525 Q = = 27, 525 Q = = 22,525 Q = = 17,525 Q = = 12,525 Q = = 7,525 Q = = 2,525 Total for Q = 250,250 units 16

17 Concrete masses for Pile Q = = 47, 525. Q = = 42, 525 Q = = 37, 525 Q = = 32, 525 Q = = 27, 525 Q = = 22, 525 Q = = 17, 525 Q = = 12, 525 Q = = 7,525 Q = = 2,525 Total for Q = 250,250 units 17

18 Overall Conclusion n extended shanti fairness wisdom technique was applied to solve six NP problems. In the first step, one did not think about mathematical models to apply, but rather thought about the process involved, and this led to the development of the fairness wisdom technique which was applied to a set of 100 items of different values or masses. Two people and were able to divide the items equally by merely choosing in turns from a set of ordered items. The total value or mass of 's items was found to be equal total value or mass of 's items, and these results are combinations of the items of different values or masses. This problem was followed by similar problems in which in addition to dividing the masses, certain conditions were to be satisfied before definite answers could be given to the problems. It is very pleasing that such a simple technique can produce desired combinations. Even though the solutions are combinations, no knowledge of combinatorial mathematics was involved or required. In fact, if one was asked to name the mathematics involved, the answer would be arithmetic. Thus, high school graduates or perhaps middle school graduates could be taught the technique involved. From the solutions, formulas or simple equations were produced to help programmers apply the techniques. Confirmed was the notion that a method that solves one of these problems can also solve other NP problems. Note that in using the technique in this paper, the items involved must be arranged in decreasing order, preferably. Therefore, in programming, the first step should be to arrange the items in decreasing order, a task a computer performs very fast. The technique was also applied to 1000 items; and the results were perfect, just like the results for the 100 items. For 100 items, out of the possible combinations, the technique produced the desired combinations, and similarly, for 1000 items, out of possible combinations, the technique produced the desired combinations. Therefore, the technique covered does not care whether there are or possibilities. The desired and correct combinations are always produced. This technique can divide a set of items of different lengths, masses, volumes, value, or sizes into equal parts by combinations only. There are social consequences of the method and principles used to divide the set of items into equal totals. The results can be applied by government agencies in the distribution of goods and services. Management personnel should be aware of the principles involved in the above technique. From the elementary school, through high school, and perhaps college, students should be taught the principles in the above wisdom technique, since throughout life, one is going to encounter situations in which two or more people are asked to choose in turns, from items of different values or sizes, and in this case, the sequence by which the choices are made matters; one may be either a participant or one may be in charge of the distribution process. y hand, the techniques can be used to prepare final exam schedules for 100 or 1000 courses. School secretaries and office assistants can learn and apply the techniques covered. Finally, if a method can solve one NP problem, that method can also solve other NP problems. Since six NP problems have been solved in this paper, the formerly NP problems are now P problems, and therefore, it is concluded that P is equal to NP. Perhaps, one may make the following statements. 1. NP plus human ability equals P. 2. NP plus human inability is not equal to P. 3. NP minus human inability equals P. 18