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アルゴリズムイントロダクション 5.1-5.2章
1.
2010 CS
tniky1 Copyright© 2010 tniky1 All rights reserved. Page 1
2.
2010 CS
( ) Copyright© 2010 tniky1 All rights reserved. Page 2
3.
2010 CS
• 5.1 • • 5.2 • • 5.3 • • 5.4 • p87-‐113 Copyright© 2010 tniky1 All rights reserved. Page 3
4.
2010 CS
(^^) • – • – n – 1 – , – – – ( ) p87 Copyright© 2010 tniky1 All rights reserved. Page 4
5.
2010 CS
1 2 3 n i best HIRE-‐ASSISTANT(n) 1. best ← 0 => 0 2. for i ← 1 to n Ci n 3. do i 4. if i best Ch m ( ) 5. then best ← i 6. hire candidate i p88 O(nCi + mCh) m Copyright© 2010 tniky1 All rights reserved. Page 5
6.
2010 CS
• • 2 1 n • • n! p88-‐89 Copyright© 2010 tniky1 All rights reserved. Page 6
7.
2010 CS
m( ) • – • ( ) ( : A: ) • ( ) – Xi Xi • A i A i • Xi I { i } Xi I { i } p90 Copyright© 2010 tniky1 All rights reserved. Page 7
8.
2010 CS
• n/2 – ( ) (1/2 n ) • n • 1/2 • ? • E [XA] = Pr{A} – A E XA Pr – A A – {i } {i } p91 E[Xi] = Pr{i } = ½ Copyright© 2010 tniky1 All rights reserved. Page 8
9.
both sides of
the above equation to obtain We wish to compute the expected number of heads, so we take the expectation of 2010 CS n to obtain both sides of the above equation= E E [X ] X . i n i=1 E [X ] = E i=1 Xi .The left side of the above equation is the expectation of the s ables. By Lemma 5.1, we can easily compute the expectation The left side of the above variables. is the expectation of the sum of nexpectation—it i equation By equation (C.20)—linearity of random vari- • ables. By Lemma 5.1, we can easily compute sum: it equals the each of the random expectation of the the expectation of sum of the expectati variables. By equation (C.20)—linearity of expectation—it is easy thecompute the variables. Linearity of expectation makes to use of indicato – Xi = expectation of the sum: it equals analytical of the expectations even whenrandom d { i } powerful the sum technique; it applies of the n there is variables. Linearity of expectation makes the use ofcan easily compute the expected – E[Xi] = Pr { i random variables. We now indicator random variables a } = ½ powerful analytical technique; it applies even when there is dependence among the n Xi(0 or E [X = X compute i the expected number 1) heads: – random variables. We now can] easily E = X of 1 n i=1 n n E [X ] = E Xi = E [X i ] i=1 i=1 n n X = E [X i ] = 1/2 ( ) i=1 i=1 [ ] n = n/2 . = 1/2 2 i=1 Thus, compared to the method used in equation (C.36), indic = n/2 . greatly simplify the E[Y] E[X+Y] = E[X] + calculation. We shall use indicator rando out this book. p91-‐92 Thus, compared to the method used in equation (C.36), indicator random variables Copyright© 2010 tniky1 All simplify the calculation. We shall use indicator random variables through- greatly rights reserved. Page 9 out this book.
10.
= 1 ·
(1/2) + 0 · (1/2) 2010 CS = 1/2 . Thus the expected number of heads obtained by one flip of a fair E [XA] = Pr{A} ( ) the following lemma shows, the expected value of an indicator associated with an event A is equal to the probability that A occur • Lemma 5.1 Given a sample space S and an event A in the sample space S, Then E [X A ] = Pr { A}. • – Proof By the definition of an indicator random variable from eq • the definition of expected value, we have E [X A ] = E [I { A}] = 1 · Pr { A} + 0 · Pr { A} = Pr { A} , p91 where A denotes S − A, the complement of A. Copyright© 2010 tniky1 All rights reserved. Page 10
11.
n+1
n We wish to compute the expected number of heads, so we take the 2010 CS both sides the above equation to obtain f (x) dx ≤ of Approximation by integrals n m When a summation can be expressed as n k=m f (k), w k=m increasing function, we can approximate it by integra E [X ] = E Xi . i=1 f (x) dx ≤ The≤integraldxappro n f (k) f (x) . n n+1 The left side of the above equationnumber. For of the sum of is the expectation a lower b m−1 k=m m ables. By Lemma 5.1, wejustification for this approximation is shown in F The can easily compute the expectation of eac • represented as the n of expectation—it figure, variables. By equation (C.20)—linearity 1 the rectangles in theis easy area of n+1 dx region under the curve. When fthe is a monotonically expectation of the sum: it equals the sum of≥ (k) expectations o – Xi = { i variables.} use a similar method to provide the bounds Linearity of expectation makes the use of indicator ran n+1 k k=1 even when there is depend n n 1 x – E[Xi] = Pr { i powerful analytical technique; it applies f (k) ≤ } = 1 / i f (x) dx ≤ f (x) dx . random variables. We m now can easily compute= expected + 1) the ln(n numbe m−1 (i 98 ) k=m Xi(0 Chapter 5 Probabilistic Analysis and Randomized Algorithms or 1) 98 Chapter 5 Probabilistic Analysis approximationAlgorithmsgives a tight e n The integral and Randomized (A.12) 1 n 98 Chapter 5 Probabilistic Analysis and Randomized Algorithms – E [X ] = X = E [X ] = iE X (by For the upper bound, E number. For equation (5.3)) a lower bound, we obtain n X i(5.5) n n+1 i=1 i=1 n 1 n dx n E [X ] = n = E E [X i X≥ (by linearity of expectation) ] i (by equation (5.3)) i=1 k x n n E [X ] = E Xi (by equation (5.3)) i n = k=1i=1 E [X ] 1 1 dx (5.5) i=1 i=1 [ = n 1/i equation 1) ] ≤ = (byln(n +(5.4)) . For the upper bound, we derive the inequality x k n = E [X ] (by linearity of expectation) i=1 i = ln n + O(1) (by equation (A.7)) . (5.6) n X = E [X i ] (by linearity of expectation) i=1k=2 1 = 1/2 Even though we interview n people, we only actually hire approximately ln n of nn n ( ) them, on average. We summarize n result in the following lemma. 1 dx this m i=1 n i=1 = Lemma 5.2 1/i ≤ (by equation[ln] k x (5.4)) 1 = ln n , = 1/i = (5.4)) that candidates are presented in a . Assuming k=2 1 (by equation n/2 SSISTANT hasthetotal = ln nof,equationrandom order, algorithm HIRE- i=1 A = ln n + O(1) cost O(c ln n). (A.7)) . a hiring (by method usedbound weyields theand approx which onlyofactuallycost boun h i=1 Thus, compared to thewe interviewthe people,our definition(C.36), indicator ra in equation the hiring hire Proof The bound follows immediately from Even equation which yields n though = ln n + O(1) (by equation (A.7)) . (5.6). (5.6) p92-‐93 them, thep324 greatly simplifyon average. We summarize this use indicator random va calculation. We shall result in the following lemma n n this book. hiring actually ≤ ln + 1 .n 1 1 Even though we interviewoutpeople, we only cost of O(nc ).hirenapproximately ln n of The expected interview cost is a significant improvement over the worst-case Page 11 result 5.2 k=1 k Copyright© 2010 tniky1 All rights reserved. ≤ ln n + 1 . h them, on average. We summarize thisLemmain the following lemma. Exercises
12.
m
n+1 n m−1 n 2010 CS k=m f (x) dx ≤ f (k) ≤ f (x) d The integral approximation (A.12) gives a m k=m m−1 ( number. For a lower bound, we obtain ) The integral approximation (A.12) giv n n+1 HIRE-‐ASSISTANT(n) 1 number. For a dxlower bound, we obtain 1. best ← 0 => 0 ≥ k=1 kn 1 1 x dx n+1 2. for i ← 1 to n Ci n 3. do i ≥ = ln(n + 1) . x k=1 k 1 4. if i best Ch m 5. then For← i upper bound, + 1) . best the = ln(n we derive the ) ( inequality 6. hire candidate i n n 1 dx For the upper bound, we derive the inequa ≤ O(nCi + mChk n O(mCh1 (Cn<<Ch ) = ) x i ) k=2 1 dx m( ) ≤ = ln n , 1 x k=2 k (^^) which yields=Ch ln n ,) O( the bound p93 n Copyright© 2010 tniky1 All rights reserved. 1 which yields the bound Page 12
13.
2010 CS
( ) Copyright© 2010 tniky1 All rights reserved. Page 13
14.
2010 CS
Copyright© 2010 tniky1 All rights reserved. Page 14
15.
2010 CS
• Hire-‐Assistant Copyright© 2010 tniky1 All rights reserved. Page 15
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