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Figure 1.8 An algorithm for testing primality, with low error probability. function primality2(N) Input: Positive integer N Output: yes/no Pick positive integers a 1 , a 2 , . . . , a k < N at random if a N−1 i ≡ 1 (mod N) for all i = 1, 2, . . . , k: return yes else: return no This probability of error drops exponentially fast, and can be driven arbitrarily low by choosing k large enough. Testing k = 100 values of a makes the probability of failure at most 2 −100 , which is miniscule: far less, for instance, than the probability that a random cosmic ray will sabotage the computer during the computation! Carmichael numbers The smallest Carmichael number is 561. It is not a prime: 561 = 3 · 11 · 17; yet it fools the Fermat test, because a 560 ≡ 1 (mod 561) for all values of a relatively prime to 561. For a long time it was thought that there might be only finitely many numbers of this type; now we know they are infinite, but exceedingly rare. There is a way around Carmichael numbers, using a slightly more refined primality test due to Rabin and Miller. Write N − 1 in the form 2 t u. As before we’ll choose a random base a and check the value of a N−1 mod N. Perform this computation by first determining a u mod N and then repeatedly squaring, to get the sequence: a u mod N, a 2u mod N, . . . , a 2tu = a N−1 mod N. If a N−1 ≢ 1 mod N, then N is composite by Fermat’s little theorem, and we’re done. But if a N−1 ≡ 1 mod N, we conduct a little follow-up test: somewhere in the preceding sequence, we ran into a 1 for the first time. If this happened after the first position (that is, if a u mod N ≠ 1), and if the preceding value in the list is not −1 mod N, then we declare N composite. In the latter case, we have found a nontrivial square root of 1 modulo N: a number that is not ±1 mod N but that when squared is equal to 1 mod N. Such a number can only exist if N is composite (Exercise 1.40). It turns out that if we combine this square-root check with our earlier Fermat test, then at least three-fourths of the possible values of a between 1 and N − 1 will reveal a composite N, even if it is a Carmichael number. 1.3.1 Generating random primes We are now close to having all the tools we need for cryptographic applications. The final piece of the puzzle is a fast algorithm for choosing random primes that are a few hundred bits 36
long. What makes this task quite easy is that primes are abundant—a random n-bit number has roughly a one-in-n chance of being prime (actually about 1/(ln 2 n ) ≈ 1.44/n). For instance, about 1 in 20 social security numbers is prime! Lagrange’s prime number theorem Let π(x) be the number of primes ≤ x. Then π(x) ≈ x/(ln x), or more precisely, π(x) lim x→∞ (x/ ln x) = 1. Such abundance makes it simple to generate a random n-bit prime: • Pick a random n-bit number N. • Run a primality test on N. • If it passes the test, output N; else repeat the process. How fast is this algorithm? If the randomly chosen N is truly prime, which happens with probability at least 1/n, then it will certainly pass the test. So on each iteration, this procedure has at least a 1/n chance of halting. Therefore on average it will halt within O(n) rounds (Exercise 1.34). Next, exactly which primality test should be used? In this application, since the numbers we are testing for primality are chosen at random rather than by an adversary, it is sufficient to perform the Fermat test with base a = 2 (or to be really safe, a = 2, 3, 5), because for random numbers the Fermat test has a much smaller failure probability than the worst-case 1/2 bound that we proved earlier. Numbers that pass this test have been jokingly referred to as “industrial grade primes.” The resulting algorithm is quite fast, generating primes that are hundreds of bits long in a fraction of a second on a PC. The important question that remains is: what is the probability that the output of the algorithm is really prime? To answer this we must first understand how discerning the Fermat test is. As a concrete example, suppose we perform the test with base a = 2 for all numbers N ≤ 25 × 10 9 . In this range, there are about 10 9 primes, and about 20,000 composites that pass the test (see the following figure). Thus the chance of erroneously outputting a composite is approximately 20,000/10 9 = 2 × 10 −5 . This chance of error decreases rapidly as the length of the numbers involved is increased (to the few hundred digits we expect in our applications). Composites Fermat test (base a = 2) Fail Primes ¡¡¡¡ ¢¡¢¡¢¡¢¡¢¡¢ ¢¡¢¡¢¡¢¡¢¡¢ ¡ ¡ ¡ ¡ Before primality test: all numbers ≤ 25 × 10 9 Pass £¡£¡£¡£¡£¡£ ¤¡¤¡¤¡¤¡¤¡¤ ≈ 20,000 composites ≈ 10 9 ¤¡¤¡¤¡¤¡¤¡¤ primes £¡£¡£¡£¡£¡£ After primality test 37
Page 1: Algorithms Copyright c○2006 S. Da
Page 4 and 5: 4 Paths in graphs 109 4.1 Distances
Page 7 and 8: List of boxes Bases and logs . . .
Page 9 and 10: Preface This book evolved over the
Page 11 and 12: Chapter 0 Prologue Look around you.
Page 13 and 14: The first question is moot here, as
Page 15 and 16: for addition, which works on one di
Page 17 and 18: 4. Likewise, any polynomial dominat
Page 19: and in general ( Fn ) = F n+1 ( ) n
Page 22 and 23: Bases and logs Naturally, there is
Page 24 and 25: Figure 1.1 Multiplication à la Fra
Page 26 and 27: Figure 1.3 Addition modulo 8. 6 0 0
Page 28 and 29: Figure 1.4 Modular exponentiation.
Page 30 and 31: Figure 1.6 A simple extension of Eu
Page 32 and 33: We say x is the multiplicative inve
Page 34 and 35: Figure 1.7 An algorithm for testing
Page 38 and 39: Randomized algorithms: a virtual ch
Page 40 and 41: Alice’s encryption function is th
Page 42 and 43: Figure 1.9 RSA. Bob chooses his pub
Page 44 and 45: There is nothing inherently wrong w
Page 46 and 47: Exercises 1.1. Show that in any bas
Page 48 and 49: the depth of the circuit or the lon
Page 50 and 51: (c) Give an example of positive int
Page 52 and 53: x = x L x R = 2 n/2 x L + x R y = y
Page 54 and 55: Figure 2.2 Divide-and-conquer integ
Page 56 and 57: Binary search The ultimate divide-a
Page 59 and 60: An n log n lower bound for sorting
Page 61 and 62: The three sublists S L , S v , and
Page 63 and 64: to be the best running time possibl
Page 66 and 67: qp t AB CD EF GH IJ KL MN OP QR Why
Page 68 and 69: The equivalence of the two polynomi
Page 70 and 71: Figure 2.6 The complex roots of uni
Page 72 and 73: A matrix reformulation To get a cle
Page 74 and 75: The orthogonality property can be s
Page 76 and 77: FFT n (input: a 0 , . . . , a n−1
Page 78 and 79: The slow spread of a fast algorithm
Page 80 and 81: (e) T (n) = 8T (n/2) + n 3 (f) T (n
Page 82 and 83: 2.21. Mean and median. One of the m
Page 84 and 85: (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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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
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7.12. For the linear program max x
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• f (i) is a valid flow from s i
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7.28. A linear program for shortest
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Chapter 8 NP-complete problems 8.1
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Satisfiability SATISFIABILITY, or S
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cost as the budget. Fine, but how d
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Figure 8.3 What is the smallest cut
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Figure 8.4 A more elaborate matchma
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8.2 NP-complete problems Hard probl
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I. These two translation procedures
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Figure 8.7 Reductions between searc
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Figure 8.8 The graph corresponding
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Figure 8.9 S is a vertex cover if a
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2n − m pets will be left unmatche
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Figure 8.10 Rudrata cycle with pair
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est of the graph as shown in Figure
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RUDRATA CYCLE−→TSP Given a grap
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Now that we know CIRCUIT SAT reduce
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Exercises 8.1. Optimization versus
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Input: An undirected graph G = (V,
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• (w i , w i ′ ) for all i = 1,
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Chapter 9 Coping with NP-completene
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anching. We can expand either of th
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follow the same pattern. As before,
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9.2 Approximation algorithms In an
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2. It can be used to build a vertex
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9.2.3 TSP The triangle inequality p
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Let’s consider the O(nV ) algorit
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iterations will be needed: whether
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Figure 9.8 Local search. Figure 9.7
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What can be done about such subopti
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Figure 9.10 The search space for a
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Exercises 9.1. In the backtracking
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start with any S ⊆ V while there
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Chapter 10 Quantum algorithms This
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Figure 10.2 Measurement of a superp
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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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