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It is usually cleaner to prove the contrapositive, that if I has a satisfying assignment then G has an independent set of size g. This is easy: for each clause, pick any literal whose value under the satisfying assignment is true (there must be at least one such literal), and add the corresponding vertex to S. Do you see why set S must be independent? SAT−→3SAT This is an interesting and common kind of reduction, from a problem to a special case of itself. We want to show that the problem remains hard even if its inputs are restricted somehow—in the present case, even if all clauses are restricted to have ≤ 3 literals. Such reductions modify the given instance so as to get rid of the forbidden feature (clauses with ≥ 4 literals) while keeping the instance essentially the same, in that we can read off a solution to the original instance from any solution of the modified one. Here’s the trick for reducing SAT to 3SAT: given an instance I of SAT, use exactly the same instance for 3SAT, except that any clause with more than three literals, (a 1 ∨ a 2 ∨ · · · ∨ a k ) (where the a i ’s are literals and k > 3), is replaced by a set of clauses, (a 1 ∨ a 2 ∨ y 1 ) (y 1 ∨ a 3 ∨ y 2 ) (y 2 ∨ a 4 ∨ y 3 ) · · · (y k−3 ∨ a k−1 ∨ a k ), where the y i ’s are new variables. Call the resulting 3SAT instance I ′ . The conversion from I to I ′ is clearly polynomial time. Why does this reduction work? I ′ is equivalent to I in terms of satisfiability, because for any assignment to the a i ’s, { (a1 ∨ a 2 ∨ · · · ∨ a k ) is satisfied ⎧ } ⎨ ⇐⇒ ⎩ there is a setting of the y i ’s for which (a 1 ∨ a 2 ∨ y 1 ) (y 1 ∨ a 3 ∨ y 2 ) · · · (y k−3 ∨ a k−1 ∨ a k ) are all satisfied To see this, first suppose that the clauses on the right are all satisfied. Then at least one of the literals a 1 , . . . , a k must be true—otherwise y 1 would have to be true, which would in turn force y 2 to be true, and so on, eventually falsifying the last clause. But this means (a 1 ∨ a 2 ∨ · · · ∨ a k ) is also satisfied. Conversely, if (a 1 ∨ a 2 ∨ · · · ∨ a k ) is satisfied, then some a i must be true. Set y 1 , . . . , y i−2 to true and the rest to false. This ensures that the clauses on the right are all satisfied. Thus, any instance of SAT can be transformed into an equivalent instance of 3SAT. In fact, 3SAT remains hard even under the further restriction that no variable appears in more than three clauses. To show this, we must somehow get rid of any variable that appears too many times. Here’s the reduction from 3SAT to its constrained version. Suppose that in the 3SAT instance, variable x appears in k > 3 clauses. Then replace its first appearance by x 1 , its second appearance by x 2 , and so on, replacing each of its k appearances by a different new variable. Finally, add the clauses (x 1 ∨ x 2 ) (x 2 ∨ x 3 ) · · · (x k ∨ x 1 ). And repeat for every variable that appears more than three times. 250 ⎫ ⎬ ⎭
Figure 8.9 S is a vertex cover if and only if V − S is an independent set. S It is easy to see that in the new formula no variable appears more than three times (and in fact, no literal appears more than twice). Furthermore, the extra clauses involving x 1 , x 2 , . . . , x k constrain these variables to have the same value; do you see why? Hence the original instance of 3SAT is satisfiable if and only if the constrained instance is satisfiable. INDEPENDENT SET−→VERTEX COVER Some reductions rely on ingenuity to relate two very different problems. Others simply record the fact that one problem is a thin disguise of another. To reduce INDEPENDENT SET to VERTEX COVER we just need to notice that a set of nodes S is a vertex cover of graph G = (V, E) (that is, S touches every edge in E) if and only if the remaining nodes, V − S, are an independent set of G (Figure 8.9). Therefore, to solve an instance (G, g) of INDEPENDENT SET, simply look for a vertex cover of G with |V | − g nodes. If such a vertex cover exists, then take all nodes not in it. If no such vertex cover exists, then G cannot possibly have an independent set of size g. INDEPENDENT SET−→CLIQUE INDEPENDENT SET and CLIQUE are also easy to reduce to one another. Define the complement of a graph G = (V, E) to be G = (V, E), where E contains precisely those unordered pairs of vertices that are not in E. Then a set of nodes S is an independent set of G if and only if S is a clique of G. To paraphrase, these nodes have no edges between them in G if and only if they have all possible edges between them in G. Therefore, we can reduce INDEPENDENT SET to CLIQUE by mapping an instance (G, g) of INDEPENDENT SET to the corresponding instance (G, g) of CLIQUE; the solution to both is identical. 3SAT−→3D MATCHING Again, two very different problems. We must reduce 3SAT to the problem of finding, among a set of boy-girl-pet triples, a subset that contains each boy, each girl, and each pet exactly once. In short, we must design sets of boy-girl-pet triples that somehow behave like Boolean variables and gates! Consider the following set of four triples, each represented by a triangular node joining a boy, girl, and pet: 251
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Algorithms Copyright c○2006 S. Da
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4 Paths in graphs 109 4.1 Distances
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List of boxes Bases and logs . . .
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Preface This book evolved over the
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Chapter 0 Prologue Look around you.
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The first question is moot here, as
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for addition, which works on one di
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4. Likewise, any polynomial dominat
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and in general ( Fn ) = F n+1 ( ) n
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Bases and logs Naturally, there is
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Figure 1.1 Multiplication à la Fra
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Figure 1.3 Addition modulo 8. 6 0 0
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Figure 1.4 Modular exponentiation.
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Figure 1.6 A simple extension of Eu
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We say x is the multiplicative inve
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Figure 1.7 An algorithm for testing
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Figure 1.8 An algorithm for testing
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Randomized algorithms: a virtual ch
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Alice’s encryption function is th
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Figure 1.9 RSA. Bob chooses his pub
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There is nothing inherently wrong w
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Exercises 1.1. Show that in any bas
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the depth of the circuit or the lon
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(c) Give an example of positive int
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x = x L x R = 2 n/2 x L + x R y = y
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Figure 2.2 Divide-and-conquer integ
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Binary search The ultimate divide-a
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An n log n lower bound for sorting
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The three sublists S L , S v , and
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to be the best running time possibl
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qp t AB CD EF GH IJ KL MN OP QR Why
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The equivalence of the two polynomi
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Figure 2.6 The complex roots of uni
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A matrix reformulation To get a cle
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The orthogonality property can be s
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FFT n (input: a 0 , . . . , a n−1
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The slow spread of a fast algorithm
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(e) T (n) = 8T (n/2) + n 3 (f) T (n
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2.21. Mean and median. One of the m
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(c) In fact, squaring matrices is n
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Chapter 3 Decompositions of graphs
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Therefore, each edge appears in exa
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Figure 3.3. 1 The previsit and post
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Figure 3.6 (a) A 12-node graph. (b)
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Tree edges are actually part of the
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the same thing. Since a dag is line
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Figure 3.10 The reverse of the grap
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Exercises 3.1. Perform a depth-firs
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(a) Model this as a graph problem:
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For instance, in the graph below (w
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3.30. On page 97, we defined the bi
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Chapter 4 Paths in graphs 4.1 Dista
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Figure 4.4 The result of breadth-fi
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Figure 4.6 Breaking edges into unit
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into account. This viewpoint gives
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§¦ ¥¤ £¢ ©¨ #"
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Which heap is best? The running tim
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Figure 4.11 (a) A binary heap with
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Figure 4.13 The Bellman-Ford algori
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Figure 4.15 A single-source shortes
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Input: Undirected graph G = (V, E)
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4.16. Section 4.5.2 describes a way
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Let r ∗ be the maximum ratio achi
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Chapter 5 Greedy algorithms A game
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Trees A tree is an undirected graph
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Figure 5.3 The cut property at work
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eturn x As can be expected, makeset
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Figure 5.7 The effect of path compr
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A randomized algorithm for minimum
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Figure 5.9 Top: Prim’s minimum sp
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Figure 5.10 A prefix-free encoding.
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149
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5.3 Horn formulas In order to displ
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Figure 5.11 (a) Eleven towns. (b) T
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Exercises 5.1. Consider the followi
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(a) Code: {0, 10, 11} (b) Code: {0,
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Input: Undirected graph G = (V, E);
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Chapter 6 Dynamic programming In th
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Figure 6.2 The dag of increasing su
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Programming? The origin of the term
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Figure 6.4 (a) The table of subprob
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Common subproblems Finding the righ
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6.4 Knapsack During a robbery, a bu
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Memoization In dynamic programming,
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Figure 6.7 (a) ((A × B) × C) × D
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ecause it leads right back to the O
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d ij . We must pick the best such i
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Figure 6.11 I(u) is the size of the
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6.8. Given two strings x = x 1 x 2
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Figure 6.12 Two binary search trees
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6.26. Sequence alignment. When a ne
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Chapter 7 Linear programming and re
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It is a general rule of linear prog
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the feasible region. The point of f
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Figure 7.3 A communications network
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Reductions Sometimes a computationa
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Page 201 and 202: Figure 7.5 An illustration of the m
Page 203 and 204: L e R s t e ′ We claim that size(
Page 205 and 206: Figure 7.6 Continued Current Flow R
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Page 209 and 210: Figure 7.10 A generic primal LP in
Page 211 and 212: Now suppose the two of them play th
Page 213 and 214: In LP form, this is min w −3y 1 +
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Page 217 and 218: Figure 7.13 Simplex in action. Init
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Page 221 and 222: Gaussian elimination Under our alge
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Page 227 and 228: 7.12. For the linear program max x
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Page 231 and 232: 7.28. A linear program for shortest
Page 233 and 234: Chapter 8 NP-complete problems 8.1
Page 235 and 236: Satisfiability SATISFIABILITY, or S
Page 237 and 238: cost as the budget. Fine, but how d
Page 239 and 240: Figure 8.3 What is the smallest cut
Page 241 and 242: Figure 8.4 A more elaborate matchma
Page 243 and 244: 8.2 NP-complete problems Hard probl
Page 245 and 246: I. These two translation procedures
Page 247 and 248: Figure 8.7 Reductions between searc
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Page 253 and 254: 2n − m pets will be left unmatche
Page 255 and 256: Figure 8.10 Rudrata cycle with pair
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Page 259 and 260: RUDRATA CYCLE−→TSP Given a grap
Page 261 and 262: Now that we know CIRCUIT SAT reduce
Page 263 and 264: Exercises 8.1. Optimization versus
Page 265 and 266: Input: An undirected graph G = (V,
Page 267 and 268: • (w i , w i ′ ) for all i = 1,
Page 269 and 270: Chapter 9 Coping with NP-completene
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Page 273 and 274: follow the same pattern. As before,
Page 275 and 276: 9.2 Approximation algorithms In an
Page 277 and 278: 2. It can be used to build a vertex
Page 279 and 280: 9.2.3 TSP The triangle inequality p
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Page 285 and 286: Figure 9.8 Local search. Figure 9.7
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Page 289 and 290: Figure 9.10 The search space for a
Page 291 and 292: Exercises 9.1. In the backtracking
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Page 295 and 296: Chapter 10 Quantum algorithms This
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Page 299 and 300: Figure 10.3 A quantum algorithm tak
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¡ £¢ ¥¤ §¦ ©¨ #" %
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The Fourier transform of a periodic
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Yet another basic gate, the control
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10.6 Factoring as periodicity We ha
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Figure 10.6 Quantum factoring. 0 QF
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Exercises 10.1. ∣ ψ 〉 = 1 √2
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Historical notes and further readin
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Index O(·), 13 Ω(·), 14 Θ(·),
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interior-point method, 217 interpol
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