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Figure 8.12 A gadget for enforcing paired behavior. (a) a f l m s b g k n r c h j p q d (b) a b c d (c) a b c d (d) a c b d C = {({a, b}, {c, d})} 258
RUDRATA CYCLE−→TSP Given a graph G = (V, E), construct the following instance of the TSP: the set of cities is the same as V , and the distance between cities u and v is 1 if {u, v} is an edge of G and 1 + α otherwise, for some α > 1 to be determined. The budget of the TSP instance is equal to the number of nodes, |V |. It is easy to see that if G has a Rudrata cycle, then the same cycle is also a tour within the budget of the TSP instance; and that conversely, if G has no Rudrata cycle, then there is no solution: the cheapest possible TSP tour has cost at least n + α (it must use at least one edge of length 1 + α, and the total length of all n − 1 others is at least n − 1). Thus RUDRATA CYCLE reduces to TSP. In this reduction, we introduced the parameter α because by varying it, we can obtain two interesting results. If α = 1, then all distances are either 1 or 2, and so this instance of the TSP satisfies the triangle inequality: if i, j, k are cities, then d ij + d jk ≥ d ik (proof: a + b ≥ c holds for any numbers 1 ≤ a, b, c ≤ 2). This is a special case of the TSP which is of practical importance and which, as we shall see in Chapter 9, is in a certain sense easier, because it can be efficiently approximated. If on the other hand α is large, then the resulting instance of the TSP may not satisfy the triangle inequality, but has another important property: either it has a solution of cost n or less, or all its solutions have cost at least n + α (which now can be arbitrarily larger than n). There can be nothing in between! As we shall see in Chapter 9, this important gap property implies that, unless P = NP, no approximation algorithm is possible. ANY PROBLEM IN NP−→SAT We have reduced SAT to the various search problems in Figure 8.7. Now we come full circle and argue that all these problems—and in fact all problems in NP—reduce to SAT. In particular, we shall show that all problems in NP can be reduced to a generalization of SAT which we call CIRCUIT SAT. In CIRCUIT SAT we are given a (Boolean) circuit (see Figure 8.13, and recall Section 7.7), a dag whose vertices are gates of five different types: • AND gates and OR gates have indegree 2. • NOT gates have indegree 1. • Known input gates have no incoming edges and are labeled false or true. • Unknown input gates have no incoming edges and are labeled “?”. One of the sinks of the dag is designated as the output gate. Given an assignment of values to the unknown inputs, we can evaluate the gates of the circuit in topological order, using the rules of Boolean logic (such as false ∨ true = true), until we obtain the value at the output gate. This is the value of the circuit for the particular assignment to the inputs. For instance, the circuit in Figure8.13 evaluates to false under the assignment true, false, true (from left to right). CIRCUIT SAT is then the following search problem: Given a circuit, find a truth assignment for the unknown inputs such that the output gate evaluates to true, or report that no such assignment exists. For example, if presented with the circuit in Figure 8.13 we could have 259
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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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Matrix-vector notation A linear fun
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Figure 7.5 An illustration of the m
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L e R s t e ′ We claim that size(
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Figure 7.6 Continued Current Flow R
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Page 211 and 212: Now suppose the two of them play th
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Page 231 and 232: 7.28. A linear program for shortest
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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
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Page 247 and 248: Figure 8.7 Reductions between searc
Page 249 and 250: Figure 8.8 The graph corresponding
Page 251 and 252: Figure 8.9 S is a vertex cover if a
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 263 and 264: Exercises 8.1. Optimization versus
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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 303 and 304: The Fourier transform of a periodic
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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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