アルゴリズムイントロダクション第5.4章

Copyright© 2010 tniky1 All rights reserved.

2010年CS勉強会第3回 2010/06/12

tniky1

h:p://www.tniky1.com/


2010年CS勉強会アルゴリズムイントロダクション

前回の雇用問題

雇用代理店

・・・・・

1 2 3 　　 n 現在の秘書best

候補 i

面談

一番じゃないといけないんですか！？

全員と面談をするのは無駄でしょ！

毎回雇用するのは無駄でしょ！



5.4.4 オンライン雇用問題の目的

もう少し実践的な雇用問題にしよう

対応ページの記載

一言で言っちゃうとこれだけ。特に新しい概念とか技法はない(^^;)



オンライン雇用問題

•  方針 – 採用は一度きり – 最適に近い候補(なるべく優秀な人)を採用できればよい

– 面談直後に採用の合否を判断

•  今回の戦略 – 最初の一部 k人と面談し、その中で最高点を確認する。(採用はしない)

–  k+1人目以降で今までの最高点を最初に更新した人を採用

採用しない最初の面談回数は何回がよい？最良の人を雇用する確率は？ p109



オンライン雇用問題アルゴリズム

5.4 Probabilistic analysis and further uses of indicator random variables 115

then rejecting the first k applicants, and hiring the first applicant thereafter who hasa higher score than all preceding applicants. If it turns out that the best-qualifiedapplicant was among the first k interviewed, then we will hire the nth applicant.This strategy is formalized in the procedure ON-LINE-MAXIMUM(k, n), whichappears below. Procedure ON-LINE-MAXIMUM returns the index of the candidatewe wish to hire.

ON-LINE-MAXIMUM(k, n)

1 bestscore ! "#2 for i ! 1 to k3 do if score(i) > bestscore4 then bestscore ! score(i)5 for i ! k + 1 to n6 do if score(i) > bestscore7 then return i8 return n

We wish to determine, for each possible value of k, the probability that wehire the most qualified applicant. We will then choose the best possible k, andimplement the strategy with that value. For the moment, assume that k is fixed.Let M( j) = max1$i$ j {score(i)} denote the maximum score among applicants 1through j . Let S be the event that we succeed in choosing the best-qualifiedapplicant, and let Si be the event that we succeed when the best-qualified appli-cant is the i th one interviewed. Since the various Si are disjoint, we have thatPr {S} = !n

i=1 Pr {Si}. Noting that we never succeed when the best-qualified ap-plicant is one of the first k, we have that Pr {Si} = 0 for i = 1, 2, . . . , k. Thus, weobtain

Pr {S} =n"

i=k+1

Pr {Si} . (5.13)

We now compute Pr {Si}. In order to succeed when the best-qualified applicantis the i th one, two things must happen. First, the best-qualified applicant must bein position i , an event which we denote by Bi . Second, the algorithm must notselect any of the applicants in positions k +1 through i "1, which happens only if,for each j such that k + 1 $ j $ i " 1, we find that score( j) < bestscore in line 6.(Because scores are unique, we can ignore the possibility of score( j) = bestscore.)In other words, it must be the case that all of the values score(k + 1) throughscore(i "1) are less than M(k); if any are greater than M(k) we will instead returnthe index of the first one that is greater. We use Oi to denote the event that noneof the applicants in position k + 1 through i " 1 are chosen. Fortunately, thetwo events Bi and Oi are independent. The event Oi depends only on the relative

雇用代理店

・・・・・

1 2 k

候補 I (i≤k)

面談

k+1 k+2 n

・・・

・最高点を確認・最高点以上を採用・採用すれば終了・1~kに最高点者がいた場合nを採用

bestscore 候補 I (i>k)

・全員と面談しないですむ。・雇用は一回ですむ

面談

面談直後に採用の合否を判断という方針のため

最良の雇用者を得る確率は？ kをどのような値にすればよい？

bestscore



最良の人を雇用する確率

•  最良の人を雇用するという事象をS •  Sの確率を求め(kに依存)、kをどのような値にすべきか(最良のk)、またその時の確率をみる

•  i番目の応募者が最高得点の人であり、その人の雇用に成功する事象Si

–  i = 1,2,•••,k

–  i = k+1,•••,n







Pr {S} =n"

i=k+1

Pr {Si} . (5.13)








Pr {S} =n"

i=k+1

Pr {Si} . (5.13)


=

参考用の人数を何人にすればよいか

最初のk人の中に最高得点者がいる場合、成功しない。

(Bi:最高の応募者がi番目に出現)∩(Oi: k+1からi-‐1番目の応募者が不採用)

この二つの条件が成り立てばよい

=

116 Chapter 5 Probabilistic Analysis and Randomized Algorithms

ordering of the values in positions 1 through i ! 1, whereas Bi depends only onwhether the value in position i is greater than the values in all other positions. Theordering of the values in positions 1 through i ! 1 does not affect whether thevalue in position i is greater than all of them, and the value in position i does notaffect the ordering of the values in positions 1 through i ! 1. Thus we can applyequation (C.15) to obtain

Pr {Si} = Pr {Bi " Oi} = Pr {Bi} Pr {Oi} .

The probability Pr {Bi} is clearly 1/n, since the maximum is equally likely to bein any one of the n positions. For event Oi to occur, the maximum value in po-sitions 1 through i ! 1 must be in one of the first k positions, and it is equallylikely to be in any of these i ! 1 positions. Consequently, Pr {Oi } = k/(i ! 1) andPr {Si} = k/(n(i ! 1)). Using equation (5.13), we have

Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.

We approximate by integrals to bound this summation from above and below. Bythe inequalities (A.12), we have" n

k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .

Evaluating these definite integrals gives us the bounds

kn(ln n ! ln k) # Pr {S} # k

n(ln(n ! 1) ! ln(k ! 1)) ,

which provide a rather tight bound for Pr {S}. Because we wish to maximize ourprobability of success, let us focus on choosing the value of k that maximizes thelower bound on Pr {S}. (Besides, the lower-bound expression is easier to maximizethan the upper-bound expression.) Differentiating the expression (k/n)(ln n ! ln k)with respect to k, we obtain

1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

n全員が最高得点者になる確率を等分に持っている

p109-‐110




–  i = k+1,•••,n







Pr {S} =n"

i=k+1

Pr {Si} . (5.13)


= (Bi:最高の応募者がi番目に出現)∩(Oi: k+1からi-‐1番目の応募者が不採用)

=





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

n全員が最高得点者になる確率を等分に持っている

・・・・・

1 2 k k+1 k+2 n

・・・

i

・・・・・

不採用

i-‐1番目までの応募者で最高得点者がこの中





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

・・・・・

1番目

＋＋＋

2番目 k番目





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

p110








Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .







Pr {S} =n"

i=k+1

Pr {Si} . (5.13)






Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

i = k+1,•••,n

i = 1,2,•••,k

1/k + 1/(k+1) + 1/(k+2) + ••• + 1/(n-‐1)

実際に入れてみるとわかりやすいです





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

積分での挟み撃ちを利用。

今回は、成功確率（最良の人を雇用する確率）を最大にしたい。 Pr{S}に関するkの値をどのようにするのが最適化かをみる。計算も楽だし、確率の低い方で考えておけば問題ない。

p110-‐111



　　　の計算





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

1068 Appendix A Summations

n+1n–1n–2m+2mm –1

f (m)

f (m+1)

f (m+2)

f (n–2)

f (n–1)

f (n)

f (x)

x… … n

… …

(a)

m+1

n+1n–1n–2m+2mm –1

f (m)

f (m+1)

f (m+2)

f (n–2)

f (n–1)

f (n)

f (x)

x… … n

… …

(b)

m+1

Figure A.1 Approximation of!n

k=m f (k) by integrals. The area of each rectangle is shownwithin the rectangle, and the total rectangle area represents the value of the summation. The in-tegral is represented by the shaded area under the curve. By comparing areas in (a), we get" n

m!1 f (x) dx " !nk=m f (k), and then by shifting the rectangles one unit to the right, we get

!nk=m f (k) "

" n+1m f (x) dx in (b).





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

p325








Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .







Pr {S} =n"

i=k+1

Pr {Si} . (5.13)






Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

i = k+1,•••,n

i = 1,2,•••,k

1/k + 1/(k+1) + 1/(k+2) + ••• + 1/(n-‐1)

実際に入れてみるとわかりやすいです





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

積分での挟み撃ちを利用。

今回は、成功確率（最良の人を雇用する確率）を最大にしたい。 Pr{S}に関するkの値をどのようにするのが最適化かをみる。計算も楽だし、確率の低い方で考えておけば問題ない。

p111








Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .


Setting this derivative equal to 0, we see that the lower bound on the probabilityis maximized when ln k = ln n ! 1 = ln(n/e) or, equivalently, when k = n/e.Thus, if we implement our strategy with k = n/e, we will succeed in hiring ourbest-qualified applicant with probability at least 1/e.

Exercises

5.4-1How many people must there be in a room before the probability that someonehas the same birthday as you do is at least 1/2? How many people must there bebefore the probability that at least two people have a birthday on July 4 is greaterthan 1/2?

5.4-2Suppose that balls are tossed into b bins. Each toss is independent, and each ballis equally likely to end up in any bin. What is the expected number of ball tossesbefore at least one of the bins contains two balls?

5.4-3 !For the analysis of the birthday paradox, is it important that the birthdays be mutu-ally independent, or is pairwise independence sufficient? Justify your answer.

5.4-4 !How many people should be invited to a party in order to make it likely that thereare three people with the same birthday?

5.4-5 !What is the probability that a k-string over a set of size n is actually a k-permutation? How does this question relate to the birthday paradox?

5.4-6 !Suppose that n balls are tossed into n bins, where each toss is independent and theball is equally likely to end up in any bin. What is the expected number of emptybins? What is the expected number of bins with exactly one ball?

5.4-7 !Sharpen the lower bound on streak length by showing that in n flips of a fair coin,the probability is less than 1/n that no streak longer than lg n!2 lg lg n consecutiveheads occurs.





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

微分





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

≥

この式の値を最大にするkを知りたい。グラフがかければ一発。グラフかけなくても、微分をして、プラス

からマイナスへ入れ替わる所があればそこが最大値。（上に凸のグラフ）

0になり、微分した式がプラスからマイナスへ入れ替わる時のk


Setting this derivative equal to 0, we see that the lower bound on the probabilityis maximized when ln k = ln n ! 1 = ln(n/e) or, equivalently, when k = n/e.Thus, if we implement our strategy with k = n/e, we will succeed in hiring ourbest-qualified applicant with probability at least 1/e.

Exercises

5.4-1How many people must there be in a room before the probability that someonehas the same birthday as you do is at least 1/2? How many people must there bebefore the probability that at least two people have a birthday on July 4 is greaterthan 1/2?

5.4-2Suppose that balls are tossed into b bins. Each toss is independent, and each ballis equally likely to end up in any bin. What is the expected number of ball tossesbefore at least one of the bins contains two balls?

5.4-3 !For the analysis of the birthday paradox, is it important that the birthdays be mutu-ally independent, or is pairwise independence sufficient? Justify your answer.

5.4-4 !How many people should be invited to a party in order to make it likely that thereare three people with the same birthday?

5.4-5 !What is the probability that a k-string over a set of size n is actually a k-permutation? How does this question relate to the birthday paradox?

5.4-6 !Suppose that n balls are tossed into n bins, where each toss is independent and theball is equally likely to end up in any bin. What is the expected number of emptybins? What is the expected number of bins with exactly one ball?

5.4-7 !Sharpen the lower bound on streak length by showing that in n flips of a fair coin,the probability is less than 1/n that no streak longer than lg n!2 lg lg n consecutiveheads occurs.

これが最良のk!!





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

代入





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

≥

Pr{Smax} = 1/e

[結論] k=n/e を用いるのが最良。その時、最高の応募者を雇用できる確率は少なくとも 1/e p111



余談(応用) 昨日のTVでの話

•  初めての彼女と結婚するより最良の人と結婚できる方法は？ – 結婚までに付き合える人数を10人とする – 付き合っている時に返事をする(よりを戻すことは不可)

初めての彼女と結婚した場合、最良の人を得る確率は1/10 = 0.1

オンライン雇用問題と全く同じ！(n=10) 最初は結婚しないで参考値だけを得て、それ以上の人が現れたら即結婚!(^^;)

= 10/(2.7182・・・) ≒ 3.679





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

≥





Pr {S} =n!

i=k+1

Pr {Si}

=n!

i=k+1

kn(i ! 1)

= kn

n!

i=k+1

1i ! 1

= kn

n!1!

i=k

1i

.


k

1x

dx #n!1!

i=k

1i

#" n!1

k!1

1x

dx .



n(ln(n ! 1) ! ln(k ! 1)) ,


1n(ln n ! ln k ! 1) .

k = n/e

kは整数 k=3 Pr{S} ≒ 0.3611 k=4 Pr{S} ≒ 0.3665



参考

雇用問題計算用

n k k/n lgn lgk logn-logk Pr{S} 10 1 0.1 2.302585093 0 2.302585093 0.230258509 10 2 0.2 2.302585093 0.693147181 1.609437912 0.321887582 10 3 0.3 2.302585093 1.098612289 1.203972804 0.361191841 10 4 0.4 2.302585093 1.386294361 0.916290732 0.366516293 10 5 0.5 2.302585093 1.609437912 0.693147181 0.34657359 10 6 0.6 2.302585093 1.791759469 0.510825624 0.306495374 10 7 0.7 2.302585093 1.945910149 0.356674944 0.249672461 10 8 0.8 2.302585093 2.079441542 0.223143551 0.178514841 10 9 0.9 2.302585093 2.197224577 0.105360516 0.094824464 10 10 1 2.302585093 2.302585093 0 0

アルゴリズムイントロダクション第5.4章

Technology

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アルゴリズムイントロダクション 第5.4章

Technology

Transcript of アルゴリズムイントロダクション 第5.4章

アルゴリズムイントロダクション第5.4章

Transcript of アルゴリズムイントロダクション第5.4章