1. Three background concepts: 1.Decision Problems: output yes/no 2. Nondeterministic algorithm: certificate additional input 3. Polynomial transformation: one problem input to another type problem input in poly-time (C) Debasis Mitra

2. 1. DECISION PROBLEMS Decision Problems: output is “True/False,” “Yes/No” Example (sorting-decision problem): –Input: a list of numbers L –Output: Answer, is L sorted? Complexity of Decision Problem? –Sorting problem P: lower bound Ω(n logn), but –Sorting-decision problem P D : O(n) (C) Debasis Mitra

3. ADVANTAGE OF DECISION PROBLEMS Every problem P has a corresponding related decision problem P D, –Has a lesser or equal complexity –Logical output to work with makes it formal Hence, complexity theory (or, any computer science formal theory) uses only decision problems But, the results are valid for all types of problems (C) Debasis Mitra

4. Example (0-1 Knapsack-decision problem): –Input: a list of objects with (weight, profit), a knapsack upper bound by weight M, and a profit lower bound t –Output: Does there exist a ks_profit ≥ t by choosing some objects with ks_wt ≤ M? Complexity? Which one is lower – original 01KS or 01KS D ? 01KS D algorithm: Backtrack just like before for 01KS but may terminate as soon as a valid ks_profit ≥ t found, need not go over all branches. EXAMPLE DECISION PROBLEM (C) Debasis Mitra

5. USE OF DECISION PROBLEMS Algorithm A(P D ) for a decision problem P D may be used to solve original problem P, and vice versa Example of sorting, using sorting-decision algorithm: –Sort (input L) : For each permutation p of input list L if AlgoSorting D (p) == True then return p; –Sorting D (input L): If any comparison within the algorithm Sorting(L) leads to an actual exchange of elements, then return “False” (L was not sorted) Similarly Alg(01KS D ) may be used to solve 0-1KS: Hint: binary search over ks_profit bound variable t Reverse is easy: ?? (C) Debasis Mitra

6. MORE CONCEPTS::: 2. NON-DETERMINISTIC ALGORITHMS (C) Debasis Mitra

7. NON-DETERMINISIM Non-deterministic version of a problem: Presume the solution exists: “certificate” Problem is simplified: Check only the certificate, Is the certificate correct answer? – return “True/False” If an algorithm can check certificate in polynomial time => The problem is Non-deterministically Polynomial, in NP-class (C) Debasis Mitra

8. NON-DETERMINISTIC PROBLEMS: EXAMPLE Sorting D is an ND problem: –Input: (1) a list of numbers L, and (2) a certificate c (another list) –Output: Is c a sort of L? –Algorithm? … –Can you find a polynomial-time algorithm? –If your answer is yes: Sorting D problem is in NP-class Is Sorting D also a P-class problem? (C) Debasis Mitra

9. NON-DETERMINISIM IS NOT A MISTERY (w=8+8, ---) pruned [ - - - ] (w=5, p=3) OptP=0 O 1 : Yes O 1 : No (w=5+3, p=33) OptP=0 O 2 : Yes O 3 : Yes B =33, B == OptP O 3 : No (w=8, p=33) OptP=33, updated Pruned because OptP cannot be improved (w=0, p=0) OptP=33 (w=3, p=30) OptP=33 X (w=3+8, ---) pruned (w=3, p=30) OptP=33, not updated OptP is still 33 B =65 > OptP (w=8, p=56) OptP=56, updated B =56, but OptP now =56 B > OptP Deterministic algorithm: Traverses all nodes of the tree in worst case Deterministic Machine versus Non-deterministic Machine (C) Debasis Mitra Non-deterministic algorithm: Given the certificate, it traverses only one path from root to a leaf: bounded by depth of the tree

10. CLASSIFICATION OF PROBLEMS BY COMPLEXITY (C) Debasis Mitra

11. COMPLEXITY OF PROBLEMS Polynomial algorithms are “faster” than Exponential algorithms, or more accurately P-algorithms has slower growth rate than that of E- algorithms, with respect to input size n O(n k ) 1 and k≥0 are constants (C) Debasis Mitra

12. COMPLEXITY THEORY Each Problem has a lower bound, for Any algorithm –What is the lower bound for Comparison Sort? Problems may be classified by such bounds An important, but “unknown” classification would be: –Polynomial-class versus “Exponential-class” Unfortunately: Life is much more complicated! –There is no E-class of problems (with exponential lower bounds) (C) Debasis Mitra

13. P-CLASS PROBLEMS AND BEYOND … P-class of problems: which have asymptotic Polynomial- time algorithms O(n k ), for k≥0 There is no “E-class” of problems: No problem S may be designated as in E-class just because no polynomial algorithm for S has been found yet! Even if you have an exponential algorithm O(a n ), for a>1, that does not mean a polynomial algorithm will also not exist! We are talking about provable lower bound Ω (C) Debasis Mitra

14. P-CLASS PROBLEMS AND BEYOND … Example: No polynomial algorithm for Multi-processor Last Finish Time (MPLFT) problem is found yet. –That does not mean in future a polynomial alg will not be found! –Although, it is not yet in P-class, it is not E-class either –No exponential lower bound has been found yet as well –That is all you can say about MPLFT However, ….continued, we use some new concepts …. (C) Debasis Mitra

15. P VERSUS NP Say, problem X is in P-class (read: has a polynomial algorithm) Is X in NP-class as well? –Read : has non-deterministic Polynomial algorithm also? –Read: given a certificate, can you check it in polynomial-time? (C) Debasis Mitra

16. P VERSUS NP Which one is faster? Polynomial Alg, or Non-det polynomial alg? ≡ What is more? Number of nodes in the tree? or the height of the tree? [ - - - ] (C) Debasis Mitra

17. P VERSUS NP P  NP True NP-class P-class (C) Debasis Mitra

18. P VERSUS NP Say, X is in NP-class –Read: has a non-det polynomial alg Is X in P-class as well? –Read: does it have a regular polynomial alg? [ - - - ] (C) Debasis Mitra

19. P VERSUS NP NP  P True? P-class NP-class? (C) Debasis Mitra

20. P VERSUS NP P  NP is True. But is NP  P also true? Meaning: NP ≡ P? A million dollar question: http://www.claymath.org/millennium-problems Jury is out there: –NP ≡ P (i.e., NP  P and P  NP), XOR, NP ≠ P? (C) Debasis Mitra

21. P VERSUS NP How would you prove: NP  P not True? Or, that the truth is P  NP, a proper subset (C) Debasis Mitra NP-class P-class X

22. P VERSUS NP LEADS TO COMPLEXITY HIERARCHY Jury is out there: –Prove, NP  P, and so, NP ≡ P? –Or, show P ≠ NP? Did not happen yet! But, can we find a “hardest” problem H in NP class? And, then maybe look for a Poly-algorithm for that H? Or, find exponential lower bound for H? Looking for “hardest” needs some concepts on Polynomial Transformation (another new concept!!) (C) Debasis Mitra

23. ANOTHER NEW CONCEPT! 3. POLYNOMIAL PROBLEM-TRANSFORMATION (C) Debasis Mitra

24. POLYNOMIAL PROBLEM-TRANSFORMATION How do you formally express a problem? Problem: (Input, Output) unambiguously expressed Problem instance: Specific values for Input, –Remember: Output is T/F in decision problems (C) Debasis Mitra

25. POLYNOMIAL PROBLEM-TRANSFORMATION Problem X can be transformed to another problem Y Problem Transformation: an algorithm –T XY (input of X) := (input of Y) –Such that, corresponding output of X and Y are same (T/F) –T XY is an algorithm, like any other ones we have seen WAIT TO SEE EXAMPLE! X Y T XY (C) Debasis Mitra

26. POLYNOMIAL PROBLEM-TRANSFORMATION Problem Transformation: an algorithm T XY –T XY (input of X) => (input of Y) –Such that, output T/F for (input of X) = output for T XY (input X) If there exists a T XY, and its time-complexity is Polynomial, then X is polynomially transformed to Y X Y T XY (C) Debasis Mitra

27. SIGNIFICANCE OF POLYNOMIAL TRANSFORMATION If there exists a T XY, and time-complexity of it is Polynomial, then X is polynomially transformable to Y Suppose, Y has a polynomial algorithm A y but, NO polynomial algorithm exists for X, YET! However, now you can solve X by –A y ( T XY (X) )  T/F –What is the time-complexity of this dual algorithm? X: no poly alg yet Y: apply poly alg A Y on Y T XY X is TrueX is False (C) Debasis Mitra

28. A CATCH IN POLYNOMIAL TRANSFORMATION A Y (T XY (X))  T/F –What is the time-complexity of this dual algorithm? Note: |(Y  output(T XY (X))| is a polynomial function of |X|, –where | | indicates problem size A y (Y) takes polynomial-time with respect to |Y| But, |Y| is a polynomial of |X|, as said above Hence: A y (Y) above takes polynomial of polynomial-time over |X|, Or, polynomial-time with respect to |X| (C) Debasis Mitra

29. SIGNIFICANCE OF POLYNOMIAL TRANSFORMATION A y (T XY (X))  T/F –Time-complexity of this dual algorithm is polynomial So, X is now in P-class –Because Y was in P-class and polynomial transformation T XY is found –So, X has an indirect polynomial algorithm now Note, this is not true in the opposite direction (C) Debasis Mitra

30. SIGNIFICANCE OF POLYNOMIAL TRANSFORMATION A y (T XY (X)) => T/F –If A y is polynomial, then the time-complexity of this dual algorithm is also polynomial If Y is in P-class, so is X Y is at least as “hard” as X Note, opposite direction is not necessarily true: –If, source problem X is already in P-class (has poly alg), –Then, polynomial transformation T XY does not say anything about the classification of target problem Y (C) Debasis Mitra

31. COOK’S THEOREM All NP-class problems may be “expressed” generically as a machine: a Non-deterministic Turing Machine There exists a polynomial-transformation from this NDTM to the SAT problem: Cook et al.’s algorithm (C) Debasis Mitra

32. COOK’S THEOREM NP-class P-class SAT problem Poly-trans (C) Debasis Mitra

33. COOK’S THEOREM Find a polynomial algorithm for SAT, and win a million dollar from Clay Institute! Then, Cook et al.’s polynomial transformation can be used to solve any NP-class problem in polynomial-time using “your polynomial-SAT” algorithm, or P=NP Alternatively, SAT is a good candidate for finding an exponential lower bound Then, P ≠NP will be proved (C) Debasis Mitra

34. BOOLEAN SATISFIABILITY PROBLEM in PROPOSITIONAL LOGIC (SAT) (C) Debasis Mitra

35. “SAT” PROBLEM Input: (U, V) –U: a set of n Boolean variables {v 1, v 2, … v n }(value: T, F) –V: a conjunctive normal formula (CNF) over U –Conjunctive normal formula is conjunction of disjunctions: { (. v. v.) ^ (. v. v. v.) ^ …} Output: Does there exist a satisfiable assignment to make C True? ≡ can you find values of U such that V becomes True? V: expressed as a set of m Clauses {C 1, C 2, … C m } C i : is a list (disjunction) of literals (a variable, or its negation) What is the complexity of finding above output? (C) Debasis Mitra

36. EXAMPLE: SAT PROBLEM Input: (U, V) –U: {a, b, c, d} –V: {{a, b, c}, {a, d}, {~b, ~d}}, ~ indicates NOT – in CNF: (a V b V c) ˄ (a V d) ˄ (~b V ~d) Output: Does there exist a satisfiable assignment to make V True? Answer: Yes –A=T, b=F, c=?, d=?, –question-mark ? means “don’t care” or T and F both correct What is the complexity of answering the above question? (C) Debasis Mitra

37. SAT is in WHICH-CLASS? Input: (U, V) –U: {a, b, c, d} –V: {{a, b, c}, {a, d}, {~b, ~d}} – a certificate: (a=F, b=T, c=T, d=T) Output: Is the certificate a satisfiable assignment to make V True? Answer: No –C 1 = T, C 2 = T, C 3 = F –V = False What is the complexity of answering the above question? (C) Debasis Mitra

38. SAT is in NP-CLASS Input: (U, V) –U: {a, b, c, d} –V: {{a, b, c}, {a, d}, {~b, ~d}} – a certificate: (a=F, b=T, c=T, d=T) Output: Is the certificate a satisfiable assignment to make V True? Answer: No –C 1 = T, C 2 = T, C 3 = F: makes V = False Complexity: O(nm) So, SAT is in NP-class But not in P-class, “yet”: no one found a polynomial alg (C) Debasis Mitra

39. COOK ET AL’S THEOREM: SAT IS NP-HARD Polynomial Transformation from NDTM to SAT SAT is NP-hard Either find a polynomial algorithm for SAT: then P=NP Or, find exponential lower bound for SAT: then P≠NP (C) Debasis Mitra

40. COOK ET AL’S THEOREM: SAT IS NP-COMPLETE Polynomial Transformation from NDTM to SAT SAT is NP-hard NP-hard: poly-trans from NDTM NP-complete: NP-hard & NP-class SAT is NP-class => SAT is NP-complete (C) Debasis Mitra

41. COOK’S THEOREM NP-class P-class * SAT Poly-trans Significance: IF there exists a poly-alg for SAT, THEN P=NP or, all NP-class problems WILL have poly-alg (C) Debasis Mitra

42. COOK ET AL’S MORE RESULT: 3-SAT IS NP-HARD Polynomial Transformation from SAT to 3-SAT Significance: IF there exists a polynomial alg for 3-SAT, Then, SAT would have an indirect poly-alg, So, P=NP would be true, or, all NP-class problems WOULD have poly-algs Polynomial Transforms are Transitive –A  B, and B  C, implies A  C Chain of poly-transformation creates NP-hard class (C) Debasis Mitra

43. NP-class COOK’S THEOREM LEADS TO NP-HARD CLASS NP-hard P-class Poly-trans SAT 3-SAT 3-DM (C) Debasis Mitra

44. NP-complete COOK’S THEOREM LEADS TO NP-COMPLETE CLASS NP-class P-class Poly-trans SAT 3-SAT 3-DM (C) Debasis Mitra NP  NP-hard = NP-complete

45. COOK’S MORE RESULTS: 3-SAT IS NP-HARD Chain of poly-transformation creates NP-hard class Any poly-alg found for a problem in NP-hard will solve P=NP question (for “yes”) Overlap of NP-hard and NP-class makes NP-complete NP-class problem has an exponential alg. O(a n ), and So, is any NP-complete problem –Branching factor of decision tree is bounded by a constant a (C) Debasis Mitra

46. NP-hard NP-class COOK’S THEOREM LEADS TO NP-COMPLETE CLASS P-class NP-complete (C) Debasis Mitra

47. WHAT IS 3-SAT? SAT with limit on clause-size ≤ 3 Input: (U, V) –U: {a, b, c, d} –V: {{a, b, c}, {a, d}, {~b, ~d}} This is 3-SAT ok We will use clause size = 3, rather than ≤ 3 (C) Debasis Mitra

48. 3-SAT IS NP-HARD Theorem: There exists a poly-trans from SAT to 3-SAT T XY : Given ANY given SAT input, you can create a size-fixed set of clauses, –by introducing new variables and clauses, … such that, T/F of the is preserved! And you can do that in polynomial time with respect to SAT’s input size However, 2-SAT is in P-class: David-Putnam poly algorithm –You cannot find a polynomial-trans from SAT to 2-SAT! (C) Debasis Mitra

49. 3-SAT IS NP-HARD Classify SAT clauses into 4 groups: 1-clauses, 2-clauses, 3-clauses, and p-clauses with p>3 literals in these clauses We will (1) show transformations (to the corresponding 3-clauses) for each of these 4 types of clauses (2) prove that the transformations are correct, i.e., truth preserving; and then (3) we will discuss the polynomial-time nature of the aggregate-transformation at the end (C) Debasis Mitra

50. 3-SAT IS NP-HARD For each clause do Input SAT clause-type Input SAT- clause Output 3SAT clause Output 3- SAT new variables For each 1- clause {u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} New variables for each such 1-clause: z1, z2, Remember: 3 clauses are created, and 2 NEW variables added in target 3-SAT for EACH 1-clause in the source SAT u is a literal WHY? (C) Debasis Mitra

51. 3-SAT IS NP-HARD For each clause do Input SAT clause-type Input SAT- clause Output 3SAT clause Output 3- SAT new variables For each 1-clause{u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} z1, z2, For each 2- clause {u1, u2}{u1, u2, z1} {u1, u2, ~z1} New variable for each such clause: z1 Target: 2 clauses created, and 1 new variable added Source: for each 2-clause u1, u2 are literals (C) Debasis Mitra

52. 3-SAT IS NP-HARD For each clause do Input SAT clause-type Input SAT- clause Output 3SAT clause Output 3- SAT new variables For each 1-clause{u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} z1, z2, For each 2-clause{u1, u2}{u1, u2, z1} {u1, u2, ~z1} z1 For each 3- clause ? clauses created, and ? new variable added for each 3-clause (C) Debasis Mitra

53. 3-SAT IS NP-HARD For each clause do Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3- SAT new variables For each 1-clause{u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} z1, z2, For each 2-clause{u1, u2}{u1, u2, z1} {u1, u2, ~z1} z1 For each 3-clause{u1, u2, u3} No new variable for any 3- clause For each p- clause (p>3) {u1, … u k-1,u k,u k+1,… u p } {u 1, u 2, z 1 }, {~ z 1, u 3, z 2 }, {~ z 2, u 4, z 3 }, …, {~z k-3,u k-1,z k-2 }, {~ z k-2, u k, z k-1 },{~z k-1,u k+1,z k }, …, {~z p-3, u p-1, u p } New variable for each such clause: z 1, z 2, …, z p-3 Target: (p-2) new clauses created, and (p-3) new variable added Source: for each 3-clause (C) Debasis Mitra

54. 3-SAT IS NP-HARD For each clause do Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3- SAT new variables For each 1-clause{u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} z1, z2, For each 2-clause{u1, u2}{u1, u2, z1} {u1, u2, ~z1} z1 For each 3-clause{u1, u2, u3} - For each p-clause (p>3){u1, … u k-1,u k,u k+1,… u p }{u 1, u 2, z 1 }, {~ z 1, u 3, z 2 }, {~ z 2, u 4, z 3 }, …, {~z k-3,u k-1,z k-2 }, {~ z k-2, u k, z k-1 },{~z k-1,u k+1,z k }, …, {~z p-3, u p-1, u p } z 1, z 2, …, z p-3 Complexity? Input SAT problem size: n variables in U, and m clauses in V k 1 of 1-clauses, k 2 of 2-clauses, k 3 of 3-clauses, k p for each p-clauses m= k 1 +k 2 +k 3 + ∑ p k p (C) Debasis Mitra

55. 3-SAT IS NP-HARD Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1-clause{u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} z1, z2, For each 2-clause{u1, u2}{u1, u2, z1} {u1, u2, ~z1} z1 For each 3-clause{u1, u2, u3} - For each p-clause (p>3) {u1, … u k-1,u k,u k+1,… u p }{u 1, u 2, z 1 }, {~ z 1, u 3, z 2 }, {~ z 2, u 4, z 3 }, …, {~z k-3,u k-1,z k-2 }, {~ z k-2, u k, z k-1 },{~z k- 1,u k+1,z k }, …, {~z p-3, u p-1, u p } z 1, z 2, …, z p-3 Complexity: Input SAT problem size: n variables in U, and m clauses in V where m= k 1 +k 2 +k 3 + ∑ p k p k 1 of 1-clauses, k 2 of 2-clauses, k 3 of 3-clauses, k p for each p-clauses Output of T XY : Nunmber of 3-variables: n + 2k 1 + k 2 + ∑ p (p-3)k p Number of 3-clauses: 4k 1 + 2k 2 + k 3 + ∑ p (p-2)k p Time-complexity of transformation = One step per variable or clause creation: Add above two, = a polynomial with respect to n & m (C) Debasis Mitra

56. Is 3-SAT also NP-complete? Trivial, SAT is in NP-class, so is 3-SAT: 3-SAT is a set of 3-clauses over a set of Boolean variables Certificate is a given value for each variable. In O(nm) time the clauses may be checked for satisfiability  3-SAT is NP We already proved 3-SAT is NP-hard => 3-SAT is NP-complete (C) Debasis Mitra

57. EXAMPLE: SAT TO 3-SAT Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1-clause{u}{u, z1, z2}, {u, z1, ~z2} {u, ~z1, z2} {u, ~z1, ~z2} z1, z2, For each 2-clause{u1, u2}{u1, u2, z1} {u1, u2, ~z1} z1 For each 3-clause{u1, u2, u3} - For each p-clause (p>3) {u1, … u k-1,u k,u k+1,… u p }{u 1, u 2, z 1 }, {~ z 1, u 3, z 2 }, {~ z 2, u 4, z 3 }, …, {~z k-3,u k-1,z k-2 }, {~ z k-2, u k, z k-1 },{~z k- 1,u k+1,z k }, …, {~z p-3, u p-1, u p } z 1, z 2, …, z p-3 Input SAT problem: U = {a, b, c, d, e, f, g}, |U| ≡ n = 7 V = {(~a, b), (a, c, d), (b, ~c, d, e, ~f), (~g), (a, c, b, e, f, ~g)}, |V| ≡ m = 5 Output 3-SAT problem: U 3 = {a, b, c, d, e, f, g, … V 3 = { (C) Debasis Mitra

58. EXAMPLE: SAT TO 3-SAT Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1- clause (~g){~g, z1, z2}, {~g, z1, ~z2} {~g, ~z1, z2} {~g, ~z1, ~z2} z1, z2, For each 2-clause For each 3-clause For each p-clause (p>3) Input SAT problem: U = {a, b, c, d, e, f, g}, |U| ≡ n = 7 V = {(~a, b), (a, c, d), (b, ~c, d, e, ~f), (~g), (a, c, b, e, f, ~g)}, |V| ≡ m = 5 1-clauses = {(~g)} Output 3-SAT problem: U3 = { a, b, c, d, e, f, g, z1, z2, … V3 = {(~g, z1, z2), (~g, z1, ~z2), (~g, ~z1, z2), (~g, ~z1, ~z2), … (C) Debasis Mitra

59. EXAMPLE: SAT TO 3-SAT Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1- clause (~g){~g, z1, z2}, {~g, z1, ~z2} {~g, ~z1, z2} {~g, ~z1, ~z2} z1, z2, For each 2- clause (~a, b)(~a, b, z3), (~a, b, ~z3)z3 For each 3-clause For each p-clause (p>3) Input SAT problem: U = {a, b, c, d, e, f, g}, |U| ≡ n = 7 V = {(~a, b), (a, c, d), (b, ~c, d, e, ~f), (~g), (a, c, b, e, f, ~g)}, |V| ≡ m = 5 2-clauses = {(~a, b)} Output 3-SAT problem: U3 = { a, b, c, d, e, f, g, z1, z2, z3, … V3 = {(~g, z1, z2), (~g, z1, ~z2), (~g, ~z1, z2), (~g, ~z1, ~z2), (~a, b, z3), (~a, b, ~z3)… (C) Debasis Mitra

60. EXAMPLE: SAT TO 3-SAT Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1- clause (~g){~g, z1, z2}, {~g, z1, ~z2} {~g, ~z1, z2} {~g, ~z1, ~z2} z1, z2, For each 2- clause (~a, b)(~a, b, z3), (~a, b, ~z3)z3 For each 3- clause (a, c, d) For each p-clause (p>3) Input SAT problem: U = {a, b, c, d, e, f, g}, |U| ≡ n = 7 V = {(~a, b), (a, c, d), (b, ~c, d, e, ~f), (~g), (a, c, b, e, f, ~g)}, |V| ≡ m = 5 3-clauses = {(a, c, d)} Output 3-SAT problem: U3 = { a, b, c, d, e, f, g, z1, z2, z3, … V3 = {(~g, z1, z2), (~g, z1, ~z2), (~g, ~z1, z2), (~g, ~z1, ~z2), (~a, b, z3), (~a, b, ~z3), (a, c, d), … (C) Debasis Mitra

61. EXAMPLE: SAT TO 3-SAT Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1- clause (~g){~g, z1, z2}, {~g, z1, ~z2} {~g, ~z1, z2} {~g, ~z1, ~z2} z1, z2, For each 2- clause (~a, b)(~a, b, z3), (~a, b, ~z3)z3 For each 3- clause (a, c, d) For each p- clause (p>3) (b, ~c, d, e, ~f)(b, ~c, z4), (~z4, d, z5), (~z5, e, ~f) z4, z5 Input SAT problem: U = {a, b, c, d, e, f, g}, |U| ≡ n = 7 V = {(~a, b), (a, c, d), (b, ~c, d, e, ~f), (~g), (a, c, b, e, f, ~g)}, |V| ≡ m = 5 5-clauses = {(b, ~c, d, e, ~f)} Output 3-SAT problem: U3 = { a, b, c, d, e, f, g, z1, z2, z3, z4, z5… V3 = {(~g, z1, z2), (~g, z1, ~z2), (~g, ~z1, z2), (~g, ~z1, ~z2), (~a, b, z3), (~a, b, ~z3), (a, c, d), (b, ~c, z4), (~z4, d, z5), (~z5, e, ~f), … (C) Debasis Mitra

62. EXAMPLE: SAT TO 3-SAT Input SAT clause-type Input SAT-clauseOutput 3SAT clauseOutput 3-SAT new variables For each 1- clause (~g){~g, z1, z2}, {~g, z1, ~z2} {~g, ~z1, z2} {~g, ~z1, ~z2} z1, z2, For each 2- clause (~a, b)(~a, b, z3), (~a, b, ~z3)z3 For each 3- clause (a, c, d) For each p- clause (p>3) (b, ~c, d, e, ~f)(b, ~c, z4), (~z4, d, z5), (~z5, e, ~f)z4, z5 (a, c, b, e, f, ~g)(a, c, z6), (~z6, b, z7), (~z7, e, z8), (~z8, f, ~g) z6, z7, z8 Input SAT problem: U = {a, b, c, d, e, f, g}, |U| ≡ n = 7 V = {(~a, b), (a, c, d), (b, ~c, d, e, ~f), (~g), (a, c, b, e, f, ~g)}, |V| ≡ m = 5 6-clauses = {(a, c, b, e, f, ~g)} Output 3-SAT problem: U 3 = { a, b, c, d, e, f, g, z1, z2, z3, z4, z5, z6, z7, z8 }, |U3| = 7 + 1*2 + 1*1 + 1*0 + (2 + 3) = 16 V 3 = {(~g, z1, z2), (~g, z1, ~z2), (~g, ~z1, z2), (~g, ~z1, ~z2), (~a, b, z3), (~a, b, ~z3), (a, c, d), (b, ~c, z4), (~z4, d, z5), (~z5, e, ~f), (a, c, z6), (~z6, b, z7), (~z7, e, z8), (~z8, f, ~g)}, |V 3 |=1*4 + 1*2+1*1 + (3+4) = 14 (C) Debasis Mitra

63. SUMMARY ON COMPLEXITY P-class problemPolynomial algorithm exists NP-class problemCertificate checking in polynomial. Must have exponential algorithm. P in NP?Yes. Poly-alg for original indicates poly-alg for certificate checking. NP in P?Not known. Polynomial transformation.X to Y poly-trans means X can be solved by solving Y. If Y is in P, then so is X. (C) Debasis Mitra

64. SUMMARY ON COMPLEXITY NP-hardAll NP-class problems can be poly-transformed to any NP-hard problem NP-completeNP-hard and NP-class When to prove a problem Y is NP-complete? Intuition tells so. How to prove Y is NP- complete? (a) Find a poly-trans from an NP-hard problem X to Y: - (1) truth-preserving, and (2) poly-time. & (b) Prove Y is NP-class. Any other way to prove Y is NP- hard? Yes, many. E.g. prove a restricted version of Y is an NP-hard problem. Not the other way round: 2-SAT is P-class. https://en.wikipedia.org/wiki/DPLL_algorithmhttps://en.wikipedia.org/wiki/DPLL_algorithm is poly-time for 2-SAT A poly-alg is found for an NP- hard problem (by a nut/ a genius!). What would that mean? (1)Not truth preserving, or (2) not poly-trans, or (3) source problem (X) was not really NP-hard, or, Wrong proof. Otherwise, (3) NP=P is just being proved! (C) Debasis Mitra

65. SUMMARY ON COMPLEXITY Examples of NP-complete problems: 1. 0-1 Knapsack decision problem 2. Multi-processor Last-time-finish scheduling problem 3. Hamiltonian Circuit decision problem: Given a graph G, does there exists a cycle involving each node once and only once? [“Does there exist a cycle in G involving each arc once and only once?” Euler circuit problem: P-class] 4. 3 dimensional matching (3DM): Given three equal size sets X, Y, Z of three types, does there exists a perfect matching? Matching set: A, a set of 3 elements (marriage), one from each of X, Y, and Z. Perfect matching (subset of A): each element of X, Y, Z appears in one and only one marriage in A. [2DM is P-class, but 3DM is NP-complete] (C) Debasis Mitra

66. HOW TO DEAL WITH NP-COMPLETE PROBLEM Y is proved NP-complete Restrict Y to something Z, then see if poly-alg may be developed for Z SAT to 2-SAT Approximate solution for Y may be acceptable e.g. Multi-processor LFT scheduling problem A pseudo-poly alg may work most of the time for Y 0-1 Knapsack Dynamic Programming Apply pruning in Branch- and-bound exponential- time alg for Y We use such smart exp algorithms for most NP-complete problems. Do not care about the worst-case complexity, that is still exponential! NP-complete problems have additional inherent structures! How about parallel computing? Does not solve an NP-comp problem in poly-time (C) Debasis Mitra

67. MORE ON COMPLEXITY ISSUES Co-NPComplementary to NP-class. Overlaps NP Co-NP completePoly trans from all Co-NP, and inside Co- NP P-spaceRemember, space complexity is less than or equal to time complexity! P-space completeHarder than NP-complete. Infinite hierarchy… Quantum computing- hierarchy Inherently Non-deterministic computing model, but has its own hierarchy. Some NP-complete problems (TSP) do get solved in poly-time (non-deterministically) Undecidable problemDefinition: No halting Turing machine exists, i.e. No algorithm exists in general. This is NOT about complexity! (C) Debasis Mitra