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Np cooks theorem

The document discusses the theory of NP-completeness. It begins by defining the complexity classes P, NP, NP-hard, and NP-complete. It then explains the concepts of reduction and how none of the NP-complete problems can be solved in polynomial time deterministically. The document provides examples of NP-complete problems like satisfiability (SAT), vertex cover, and the traveling salesman problem. It shows how nondeterministic algorithms can solve these problems and how they can be transformed into SAT instances. Finally, it proves that SAT is the first NP-complete problem by showing it is in NP and NP-hard.

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8- 1 Chapter 8 The Theory of NP- Completeness
8- 2  P: the class of problems which can be solved by a deterministic polynomial algorithm.  NP : the class of decision problem which can be solved by a non-deterministic polynomial algorithm.  NP-hard: the class of problems to which every NP problem reduces.  NP-complete (NPC): the class of problems which are NP-hard and belong to NP.
8- 3 Some concepts of NPC  Definition of reduction: Problem A reduces to problem B (A ∝ B) iff A can be solved by a deterministic polynomial time algorithm using a deterministic algorithm that solves B in polynomial time.  Up to now, none of the NPC problems can be solved by a deterministic polynomial time algorithm in the worst case.  It does not seem to have any polynomial time algorithm to solve the NPC problems.
8- 4  The theory of NP-completeness always considers the worst case.  The lower bound of any NPC problem seems to be in the order of an exponential function.  Not all NP problems are difficult. (e.g. the MST problem is an NP problem.)  If A, B ∈ NPC, then A ∝ B and B ∝ A.  Theory of NP-completeness: If any NPC problem can be solved in polynomial time, then all NP problems can be solved in polynomial time. (NP = P)
8- 5 Decision problems  The solution is simply “Yes” or “No”.  Optimization problems are more difficult.  e.g. the traveling salesperson problem  Optimization version: Find the shortest tour  Decision version: Is there a tour whose total length is less than or equal to a constant c ?
8- 6 Solving an optimization problem by a decision algorithm :  Solving TSP optimization problem by a decision algorithm :  Give c1 and test (decision algorithm) Give c2 and test (decision algorithm)  Give cn and test (decision algorithm)  We can easily find the smallest ci
8- 7 The satisfiability problem  The satisfiability problem  The logical formula : x1 v x2 v x3 & - x1 & - x2 the assignment : x1 ← F , x2 ← F , x3 ← T will make the above formula true . (-x1 , -x2 , x3 ) represents x1 ← F , x2 ← F , x3 ← T
8- 8  If there is at least one assignment which satisfies a formula, then we say that this formula is satisfiable; otherwise, it is unsatisfiable.  An unsatisfiable formula : x1 v x2 & x1 v -x2 & -x1 v x2 & -x1 v -x2
8- 9  Definition of the satisfiability problem: Given a Boolean formula, determine whether this formula is satisfiable or not.    A literal : xi or -xi  A clause : x1 v x2 v -x3 ≡ Ci  A formula : conjunctive normal form (CNF) C1 & C2 & … & Cm
8- 10  Resolution principle C1 : x1 v x2 C2 : -x1 v x3 ⇒ C3 : x2 v x3  From C1 &C2 , we can obtain C3 , and C3 can be added into the formula.  The formula becomes: C1 &C2 &C3 The resolution principle x1 x2 x3 C1 &C2 C3 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 0 1 1 1 1 1 0 0 0 0 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1
8- 11  Another example of resolution principle C1 : -x1 v -x2 v x3 C2 : x1 v x4 ⇒ C3 : -x2 v x3 v x4  If no new clauses can be deduced, then it is satisfiable. -x1 v -x2 v x3 (1) x1 (2) x2 (3) (1) & (2) -x2 v x3 (4) (4) & (3) x3 (5) (1) & (3) -x1 v x3 (6)
8- 12  If an empty clause is deduced, then it is unsatisfiable. - x1 v -x2 v x3 (1) x1 v -x2 (2) x2 (3) - x3 (4)   ⇓ deduce   (1) & (2) -x2 v x3 (5) (4) & (5) -x2 (6) (6) & (3) □ (7)
8- 13 Semantic tree  In a semantic tree, each path from the root to a leaf node represents a class of assignments.  If each leaf node is attached with a clause, then it is unsatisfiable.
8- 14 Nondeterministic algorithms  A nondeterminstic algorithm consists of phase 1: guessing phase 2: checking  If the checking stage of a nondeterministic algorithm is of polynomial time-complexity, then this algorithm is called an NP (nondeterministic polynomial) algorithm.  NP problems : (must be decision problems)  e.g. searching, MST sorting satisfiability problem (SAT) traveling salesperson problem (TSP)
8- 15 Decision problems  Decision version of sorting: Given a1 , a2 ,…, an and c, is there a permutation of ai ′ s ( a1 ′ , a2 ′ , … ,an ′ ) such that∣a2 ′ –a1 ′ ∣+∣a3 ′ –a2 ′ ∣+ … +∣an ′ –an-1 ′ ∣ ＜ c ?  Not all decision problems are NP problems  E.g. halting problem :  Given a program with a certain input data, will the program terminate or not?  NP-hard  Undecidable
8- 16 Nondeterministic operations and functions [Horowitz 1998]  Choice(S) : arbitrarily chooses one of the elements in set S  Failure : an unsuccessful completion  Success : a successful completion  Nonderministic searching algorithm: j ← choice(1 : n) /* guessing */ if A(j) = x then success /* checking */ else failure
8- 17  A nondeterministic algorithm terminates unsuccessfully iff there does not exist a set of choices leading to a success signal.  The time required for choice(1 : n) is O(1).  A deterministic interpretation of a non- deterministic algorithm can be made by allowing unbounded parallelism in computation.
8- 18 Nondeterministic sorting B ← 0 /* guessing */ for i = 1 to n do j ← choice(1 : n) if B[j] ≠ 0 then failure B[j] = A[i] /* checking */ for i = 1 to n-1 do if B[i] > B[i+1] then failure success
8- 19 Nondeterministic SAT /* guessing */ for i = 1 to n do xi ← choice( true, false ) /* checking */ if E(x1, x2, … ,xn) is true then success else failure
8- 20 Cook’s theorem NP = P iff the satisfiability problem is a P problem.  SAT is NP-complete.  It is the first NP-complete problem.  Every NP problem reduces to SAT. Stephen Arthur Cook (1939~)
8- 21 Transforming searching to SAT  Does there exist a number in { x(1), x(2), …, x(n) }, which is equal to 7?  Assume n = 2. nondeterministic algorithm: i = choice(1,2) if x(i)=7 then SUCCESS else FAILURE
8- 22 i=1 v i=2 & i=1 → i≠2 & i=2 → i≠1 & x(1)=7 & i=1 → SUCCESS & x(2)=7 & i=2 → SUCCESS & x(1)≠7 & i=1 → FAILURE & x(2)≠7 & i=2 → FAILURE & FAILURE → -SUCCESS & SUCCESS (Guarantees a successful termination) & x(1)=7 (Input Data) & x(2)≠7
8- 23  CNF (conjunctive normal form) : i=1 v i=2 (1) i≠1 v i≠2 (2) x(1)≠7 v i≠1 v SUCCESS (3) x(2)≠7 v i≠2 v SUCCESS (4) x(1)=7 v i≠1 v FAILURE (5) x(2)=7 v i≠2 v FAILURE (6) -FAILURE v -SUCCESS (7) SUCCESS (8) x(1)=7 (9) x(2)≠7 (10)
8- 24  Satisfiable at the following assignment : i=1 satisfying (1) i≠2 satisfying (2), (4) and (6) SUCCESS satisfying (3), (4) and (8) −FAILURE satisfying (7) x(1)=7 satisfying (5) and (9) x(2)≠7 satisfying (4) and (10)
8- 25 The semantic tree i=1 v i=2 (1) i≠1 v i≠2 (2) x(1)≠7 v i≠1 v SUCCESS (3) x(2)≠7 v i≠2 v SUCCESS (4) x(1)=7 v i≠1 v FAILURE (5) x(2)=7 v i≠2 v FAILURE (6) -FAILURE v -SUCCESS (7) SUCCESS (8) x(1)=7 (9) x(2)≠7 (10)
8- 26 Searching for 7, but x(1)≠7, x(2)≠7  CNF (conjunctive normal form) : i=1 v i=2 (1) i≠1 v i≠2 (2) x(1)≠7 v i≠1 v SUCCESS (3) x(2)≠7 v i≠2 v SUCCESS (4) x(1)=7 v i≠1 v FAILURE (5) x(2)=7 v i≠2 v FAILURE (6) SUCCESS (7) -SUCCESS v -FAILURE (8) x(1) ≠ 7 (9) x(2) ≠ 7 (10)
8- 27  Apply resolution principle : (9) & (5) i≠1 v FAILURE (11) (10) & (6) i≠2 v FAILURE (12) (7) & (8) -FAILURE (13) (13) & (11) i≠1 (14) (13) & (12) i≠2 (15) (14) & (1) i=2 (11) (15) & (16) □ (17) We get an empty clause ⇒ unsatisfiable ⇒ 7 does not exit in x(1) or x(2).
8- 28  CNF: i=1 v i=2 (1) i≠1 v i≠2 (2) x(1)≠7 v i≠1 v S U C C E S S (3 ) x(2)≠7 v i≠2 v S U C C E S S (4 ) x(1)=7 v i≠1 v FA I L U R E ( 5 ) x(2)=7 v i≠2 v FA I L U R E ( 6 ) SUCCESS (7) -SUCCESS v -FAILURE (8) x(1)=7 (9) x(2)=7 (10) Searching for 7, where x(1)=7, x(2)=7
8- 29 The semantic tree It implies that both assignments (i=1, i=2) satisfy the clauses.
8- 30 The node cover problem  Def: Given a graph G=(V, E), S is the node cover if S ⊆ V and for every edge (u, v) ∈ E, either u ∈ S or v ∈ S. node cover : {1, 3} {5, 2, 4}  Decision problem : ∃ S ∋ | S | ≤ K ?
8- 31 Transforming the node cover problem to SAT BEGIN i1 ← choice({1, 2, …, n}) i2 ← choice({1, 2, …, n} – {i1})  ik ← choice({1, 2, …, n} – {i1, i2, …, ik-1}). For j=1 to m do BEGIN if ej is not incident to one of νit (1≤t≤k) then FAILURE END SUCCESS
8- 32 i1 = 1 v i1 = 2… v i1 = n (i1≠ 1→i1＝2 v i1＝3…v i1 = n) i2 = 1 v i2 = 2… v i2 = n  ik = 1 v ik = 2… v ik = n i1 ≠1 v i2 ≠1 (i1=1 →i2≠1 & ... & ik≠1) i1 ≠1 v i3 ≠1  ik-1 ≠n v ik ≠n νi1 ∈e1 v νi2 ∈e1 v … v νik ∈e1 v FAILURE (νi1 ∉e1&νi2 ∉e1&…&νik ∉e1→ Failure) νi1 ∈e2 v νi2 ∈e2 v … v νik ∈e2 v FAILURE  νi1 ∈em v νi2 ∈em v … v νik ∈em v FAILURE SUCCESS CN F: (To be continued)
8- 33 -SUCCESS v -FAILURE νr1 ∈e1 νs1 ∈e1 νr2 ∈e2 νs2 ∈e2  νrm ∈em νsm ∈em
8- 34 SAT is NP-complete (1) SAT has an NP algorithm. (2) SAT is NP-hard:  Every NP algorithm for problem A can be transformed in polynomial time to SAT [Horowitz 1998] such that SAT is satisfiable if and only if the answer for A is “YES”.  That is, every NP problem ∝ SAT .  By (1) and (2), SAT is NP-complete.
8- 35 Proof of NP-Completeness  To show that A is NP-complete (I) Prove that A is an NP problem. (II) Prove that ∃ B ∈ NPC, B ∝ A. ⇒ A ∈ NPC  Why ?
8- 36 3-satisfiability problem (3- SAT)  Def: Each clause contains exactly three literals.  (I) 3-SAT is an NP problem (obviously)  (II) SAT ∝ 3-SAT Proof: (1) One literal L1 in a clause in SAT : in 3-SAT : L1 v y1 v y2 L1 v -y1 v y2 L1 v y1 v -y2 L1 v -y1 v -y2
8- 37 (2) Two literals L1 , L2 in a clause in SAT : in 3-SAT : L1 v L2 v y1 L1 v L2 v -y1 (3) Three literals in a clause : remain unchanged. (4) More than 3 literals L1 , L2 , …, Lk in a clause : in 3-SAT : L1 v L2 v y1 L3 v -y1 v y2  Lk-2 v -yk-4 v yk-3 Lk-1 v Lk v -yk-3
8- 38 SAT transform 3-SAT S S′  The instance S′ in 3- SAT : x1 v x2 v y1 x1 v x2 v -y1 -x3 v y2 v y3 -x3 v -y2 v y3 -x3 v y2 v -y3 -x3 v -y2 v -y3 x1 v -x2 v y4 x3 v -y4 v y5 -x4 v -y5 v y6 x5 v x6 v -y6  An instance S in SAT : x1 v x2 -x3 x1 v -x2 v x3 v -x4 v x5 v x6 Example of transforming SAT to 3- SAT
8- 39  Proof : S is satisfiable ⇔ S′ is satisfiable “⇒” ≤ 3 literals in S (trivial) consider ≥ 4 literals S : L1 v L2 v … v Lk S′: L1 v L2 v y1 L3 v -y1 v y2 L4 v -y2 v y3  Lk-2 v -yk-4 v yk-3 Lk-1 v Lk v -yk-3
8- 40  S is satisfiable ⇒ at least Li = T Assume : Lj = F ∀ j ≠ i assign : yi-1 = F yj = T ∀ j < i-1 yj = F ∀ j > i-1 ( Li v -yi-2 v yi-1 ) ⇒ S′ is satisfiable.  “⇐” If S′ is satisfiable, then assignment satisfying S′ can not contain yi ’s only. ⇒ at least one Li must be true. (We can also apply the resolution principle). Thus, 3-SAT is NP-complete.
8- 41 Comment for 3-SAT  If a problem is NP-complete, its special cases may or may not be NP-complete.
8- 42 Chromatic number decision problem (CN)  Def: A coloring of a graph G=(V, E) is a function f : V → { 1, 2, 3,…, k } such that if (u, v) ∈ E, then f(u)≠f(v). The CN problem is to determine if G has a coloring for k. <Theorem> Satisfiability with at most 3 literals per clause (SATY) ∝ CN. 3-colorable f(a)=1, f(b)=2, f(c)=1 f(d)=2, f(e)=3
8- 43 Proof : instance of SATY : variable : x1, x2, …, xn , n ≥ 4 clause : c1, c2, …, cr instance of CN : G=(V, E) V={ x1, x2, …, xn }∪{ -x1, -x2, …, -xn } ∪{ y1, y2, …, yn }∪{ c1, c2, …, cr }  newly added E={ (xi, -xi) | 1≤ i ≤ n }∪{ (yi, yj) | i ≠ j } ∪{ (yi, xj) | i ≠ j }∪{ (yi, -xj) | i ≠ j } ∪{ (xi, cj) | xi ∉ cj }∪{ (-xi, cj) | -xi ∉ cj } SATY ∝ CN
8- 44 x1 v x2 v x3 (1) -x3 v -x4 v x2 (2) ⇓ Example of SATY ∝ CN True assignment: x1 =T x2 =F x3 =F x4 =T E={ (xi , -xi ) | 1≤ i ≤ n }∪{ (yi , yj ) | i ≠ j } ∪{ (yi , xj ) | i ≠ j }∪{ (yi , -xj ) | i ≠ j } ∪{ (xi , cj ) | xi ∉ cj }∪{ (-xi , cj ) | -xi ∉ cj }
8- 45  Satisfiable ⇔ n+1 colorable  “⇒” (1) f(yi ) = i (2) if xi = T, then f(xi ) = i, f(-xi ) = n+1 else f(xi ) = n+1, f(-xi ) = i (3)if xi in cj and xi = T, then f(cj ) = f(xi ) if -xi in cj and -xi = T, then f(cj ) = f(-xi ) ( at least one such xi ) Proof of SATY ∝ CN
8- 46  “⇐” (1) yi must be assigned with color i. (2) f(xi ) ≠ f(-xi ) either f(xi ) = i and f(-xi ) = n+1 or f(xi ) = n+1 and f(-xi ) = i (3) at most 3 literals in cj and n ≥ 4 ⇒ at least one xi , ∋ xi and -xi are not in cj ⇒ f(cj ) ≠ n+1 (4) if f(cj ) = i = f(xi ), assign xi to T if f(cj ) = i = f(-xi ), assign -xi to T (5) if f(cj ) = i = f(xi ) ⇒ (cj , xi ) ∉ E ⇒ xi in cj ⇒ cj is true if f(cj ) = i = f(-xi ) ⇒ similarly
8- 47 Set cover decision problem  Def: F = {Si } = { S1 , S2 , …, Sk } = { u1 , u2 , …, un } T is a set cover of F if T ⊆ F and The set cover decision problem is to determine if F has a cover T containing no more than c sets.  Example F = {(u1 , u3 ), (u2 , u4 ), (u2 , u3 ), (u4 ), (u1 , u3 , u4 )} s1 s2 s3 s4 s5 T = { s1 , s3 , s4 } set cover T = { s1 , s2 } set cover, exact cover FS i i S ∈  FS i TS i ii SS ∈∈ =
8- 48 Exact cover problem (Notations same as those in set cover.) Def: To determine if F has an exact cover T, which is a cover of F and the sets in T are pairwise disjoint. <Theorem> CN ∝ exact cover (No proof here.)
8- 49 Sum of subsets problem  Def: A set of positive numbers A = { a1 , a2 , …, an } a constant C Determine if ∃ A′ ⊆ A ∋  e.g. A = { 7, 5, 19, 1, 12, 8, 14 }  C = 21, A′ = { 7, 14 }  C = 11, no solution <Theorem> Exact cover ∝ sum of subsets. Ca Aa i i =∑′∈
8- 50  Proof : instance of exact cover : F = { S1 , S2 , …, Sk } instance of sum of subsets : A = { a1 , a2 , …, ak } where where eij = 1 if uj ∈ Si eij = 0 if otherwise.  Why k+1? (See the example on the next page.) Exact cover ∝ sum of subsets { } FS ni i uuuS ∈ = ...,,2,1 j nj iji kea )1( 1 += ∑≤≤ kkkkC n nj j /)1)1)((1()1( 1 −++=+= ∑≤≤
8- 51 Example of Exact cover ∝ sum of subsets  Valid transformation: u1=6, u2=8, u3=9, n=3 EC: S1={6,8}, S2={9}, S3={6,9}, S4={8,9} k=4 SS: a1=51 +52 =30 a2=53 =125 a3=51 +53 =130 a4=52 +53 =150  Invalid transformation: EC: S1={6,8}, S2={8}, S3={8}, S4={8,9}. K=4 Suppose k-2=2 is used. SS: a1=21 +22 =6 a2=22 =4 a3=22 =4 a4=22 +23 =12 C=21 +22 +23 =14 { } { }9,8,6...,,2,1 == ∈  FS ni i uuuS
8- 52 Partition problem  Def: Given a set of positive numbers A = { a1 ,a2 ,…,an }, determine if ∃ a partition P, ∋  e. g. A = {3, 6, 1, 9, 4, 11} partition : {3, 1, 9, 4} and {6, 11} <Theorem> sum of subsets ∝ partition ∑∑ ∉∈ = Pa i Pa i ii aa
8- 53 Sum of subsets ∝ partition proof : instance of sum of subsets : A = { a1, a2, …, an }, C instance of partition : B = { b1, b2, …, bn+2 }, where bi = ai, 1≤ i ≤ n bn+1 = C+1 bn+2 = ( ∑ ai )+1−C 1≤i≤n C = ∑ai ⇔ ( ∑ai )+bn+2 = ( ∑ai )+bn+1 ai∈S ai∈S ai∉S ⇔ partition : { bi | ai∈S}∪{bn+2} and { bi | ai∉S }∪{bn+1} S S’ A C
8- 54  Why bn+1 = C+1 ? why not bn+1 = C ?  To avoid bn+1 and bn+2 to be partitioned into the same subset.
8- 55 Bin packing problem  Def: n items, each of size ci , ci > 0 Each bin capacity : C  Determine if we can assign the items into k bins, ∋ ∑ci ≤ C , 1≤j≤k. i∈binj <Theorem> partition ∝ bin packing.
8- 56 VLSI discrete layout problem  Given: n rectangles, each with height hi (integer) width wi and an area A Determine if there is a placement of the n rectangles within the area A according to the rules : 1. Boundaries of rectangles parallel to x axis or y axis. 2. Corners of rectangles lie on integer points. 3. No two rectangles overlap. 4. Two rectangles are separated by at least a unit distance. (See the figure on the next page.)
8- 57 A Successful Placement <Theorem> bin packing ∝ VLSI discrete layout.
8- 58 Max clique problem  Def: A maximal complete subgraph of a graph G=(V,E) is a clique. The max (maximum) clique problem is to determine the size of a largest clique in G.  e. g. <Theorem> SAT ∝ clique decision problem. maximal cliques : {a, b}, {a, c, d} {c, d, e, f} maximum clique : (largest) {c, d, e, f}
8- 59 Node cover decision problem  Def: A set S ⊆ V is a node cover for a graph G = (V, E) iff all edges in E are incident to at least one vertex in S. ∃ S, ∋ |S| ≤ K ? <Theorem> clique decision problem ∝ node cover decision problem. (See proof on the next page.)
8- 60 Clique decision ∝ node cover decision  G=(V,E) : clique Q of size k (Q⊆V) G’=(V,E’) : node cover S of size n-k, S=V-Q where E’={(u,v)|u∈V, v ∈V and (u,v)∉E}
8- 61 Hamiltonian cycle problem  Def: A Hamiltonian cycle is a round trip path along n edges of G which visits every vertex once and returns to its starting vertex.  e.g. Hamiltonian cycle : 1, 2, 8, 7, 6, 5, 4, 3, 1. <Theorem> SAT ∝ directed Hamiltonian cycle ( in a directed graph )
8- 62 Traveling salesperson problem  Def: A tour of a directed graph G=(V, E) is a directed cycle that includes every vertex in V. The problem is to find a tour of minimum cost. <Theorem> Directed Hamiltonian cycle ∝ traveling salesperson decision problem. (See proof on the next page.)
8- 63 Proof of Hamiltonian ∝ TSP
8- 64 0/1 knapsack problem  Def: n objects, each with a weight wi > 0 a profit pi > 0 capacity of knapsack : M Maximize ∑pi xi 1≤i≤n Subject to ∑wi xi ≤ M 1≤i≤n xi = 0 or 1, 1≤ i ≤n  Decision version : Given K, ∃ ∑pi xi ≥ K ? 1≤i≤n  Knapsack problem : 0 ≤ xi ≤ 1, 1≤ i ≤n. <Theorem> partition ∝ 0/1 knapsack decision problem.
8- 65  Refer to Sec. 11.3, Sec. 11.4 and its exercises of [Horowitz 1998] for the proofs of more NP-complete problems.  [[Horowitz 1998] E. Howowitz, S. Sahni and S. Rajasekaran, Computer Algorithms, Computer Science Press, New York, 1998, 「台北圖書」代 理 , 02-23625376