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Figure 9.5 Some data points and the optimal k = 4 clusters. Figure 9.6 (a) Four centers chosen by farthest-first traversal. (b) The resulting clusters. (a) 2 (b) 4 1 3 It’s clear that this algorithm returns a valid partition. What’s more, the resulting diameter is guaranteed to be at most twice optimal. Here’s the argument. Let x ∈ X be the point farthest from µ 1 , . . . , µ k (in other words the next center we would have chosen, if we wanted k + 1 of them), and let r be its distance to its closest center. Then every point in X must be within distance r of its cluster center. By the triangle inequality, this means that every cluster has diameter at most 2r. But how does r relate to the diameter of the optimal clustering? Well, we have identified k + 1 points {µ 1 , µ 2 , . . . , µ k , x} that are all at a distance at least r from each other (why?). Any partition into k clusters must put two of these points in the same cluster and must therefore have diameter at least r. This algorithm has a certain high-level similarity to our scheme for VERTEX COVER. Instead of a maximal matching, we use a different easily computable structure—a set of k points that cover all of X within some radius r, while at the same time being mutually separated by a distance of at least r. This structure is used both to generate a clustering and to give a lower bound on the optimal clustering. We know of no better approximation algorithm for this problem. 278
9.2.3 TSP The triangle inequality played a crucial role in making the k-CLUSTER problem approximable. It also helps with the TRAVELING SALESMAN PROBLEM: if the distances between cities satisfy the metric properties, then there is an algorithm that outputs a tour of length at most 1.5 times optimal. We’ll now look at a slightly weaker result that achieves a factor of 2. Continuing with the thought processes of our previous two approximation <strong>algorithms</strong>, we can ask whether there is some structure that is easy to compute and that is plausibly related to the best traveling salesman tour (as well as providing a good lower bound on OPT). A little thought and experimentation reveals the answer to be the minimum spanning tree. Let’s understand this relation. Removing any edge from a traveling salesman tour leaves a path through all the vertices, which is a spanning tree. Therefore, TSP cost ≥ cost of this path ≥ MST cost. Now, we somehow need to use the MST to build a traveling salesman tour. If we can use each edge twice, then by following the shape of the MST we end up with a tour that visits all the cities, some of them more than once. Here’s an example, with the MST on the left and the resulting tour on the right (the numbers show the order in which the edges are taken). Wichita Wichita Albuquerque Amarillo Dallas Tulsa Little Rock Albuquerque 8 11 9 10 Amarillo 7 12 6 Dallas 5 Tulsa 4 13 14 3 Little Rock El Paso Houston El Paso 16 15 2 Houston San Antonio 1 San Antonio Therefore, this tour has a length at most twice the MST cost, which as we’ve already seen is at most twice the TSP cost. This is the result we wanted, but we aren’t quite done because our tour visits some cities multiple times and is therefore not legal. To fix the problem, the tour should simply skip any city it is about to revisit, and instead move directly to the next new city in its list: Wichita Albuquerque Tulsa Amarillo Little Rock Dallas El Paso Houston San Antonio 279
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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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from the algorithm, which in such c
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Figure 7.10 A generic primal LP in
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Now suppose the two of them play th
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In LP form, this is min w −3y 1 +
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7.6.2 The algorithm On each iterati
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Figure 7.13 Simplex in action. Init
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close to one another? Unboundedness
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Gaussian elimination Under our alge
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output AND NOT OR AND OR NOT true f
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Exercises 7.1. Consider the followi
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
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
Page 257 and 258: est of the graph as shown in Figure
Page 259 and 260: RUDRATA CYCLE−→TSP Given a grap
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Page 263 and 264: Exercises 8.1. Optimization versus
Page 265 and 266: Input: An undirected graph G = (V,
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Page 269 and 270: Chapter 9 Coping with NP-completene
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Page 275 and 276: 9.2 Approximation algorithms In an
Page 277: 2. It can be used to build a vertex
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Page 283 and 284: iterations will be needed: whether
Page 285 and 286: Figure 9.8 Local search. Figure 9.7
Page 287 and 288: What can be done about such subopti
Page 289 and 290: Figure 9.10 The search space for a
Page 291 and 292: Exercises 9.1. In the backtracking
Page 293 and 294: start with any S ⊆ V while there
Page 295 and 296: Chapter 10 Quantum algorithms This
Page 297 and 298: Figure 10.2 Measurement of a superp
Page 299 and 300: Figure 10.3 A quantum algorithm tak
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Page 303 and 304: The Fourier transform of a periodic
Page 305 and 306: Yet another basic gate, the control
Page 307 and 308: 10.6 Factoring as periodicity We ha
Page 309 and 310: Figure 10.6 Quantum factoring. 0 QF
Page 311 and 312: Exercises 10.1. ∣ ψ 〉 = 1 √2
Page 313 and 314: Historical notes and further readin
Page 315 and 316: Index O(·), 13 Ω(·), 14 Θ(·),
Page 317 and 318: interior-point method, 217 interpol
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