アルゴリズムイントロダクション 第5.4章 勉強会資料 下記でPPT版，KEY版も配布してます． https://tanaike.jp

1. 2010 CS 3
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5.4.4
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– ( )
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– k+1
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5. 2010 CS
5.4 Probabilistic analysis and further uses of indicator r
the first k applicants, and hiring th
then rejecting
a higher score than all preceding applicants. If
applicant was among the first k interviewed, th
This strategy is formalized in the procedure O
appears below. Procedure O N -L INE -M AXIMUM
1
2
k
k+1
k+2
n
we wish to hire.
O N -L INE -M AXIMUM (k, n)
1 bestscore ← −∞
2 for i ← 1 to k
3 do if score(i) > bestscore
4 then bestscore ← score(i)
5 for i ← k + 1 to n
I
I
bestscore
6 do if score(i) > bestscore
(i≤k)
bestscore
(i>k)
7 then return i
8 return n
We wish to determine, for each possible va
1 k
hire the most qualified applicant. We will the
n
k
implement the strategy with that value. For th
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Let M( j ) = max1≤i≤ j {score(i)} denote the m
Page
5
through j . Let S be the event that we succe

6. 7 then return i
n8
i return n
2010 CS
We wish to determine, for each possible value
Chapter 5 Probabilistic Analysis and Randomized Algorithms of k, the probability that
Chapter Probabilistic Analysis
e, hire the most qualified applicant. We will thenthat we the best possible k, a
for each5 possible value of k, and Randomized Algorithms
the probability choose
applicant.• We will 5 Probabilisticthe bestand Randomized Algorithms
implementChapter then positions Analysis possiblewhereas Biassume that k on fixe
116 thevalues in choose 1value. For the moment, depends only is
strategy with that through i S
1, k, and
ordering of the −
with that value.max1≤i≤ j moment, assume that maximum score among applicants
Let M( the = For the {score(i)} ) k denote the k is fixed.
whether • ) Svalue in position i is greater than the values in all other positions. The
j (k (
ordering of. of values inthe event 1 throughsucceed − not affect whether the
score(i)} denotetheSvalues in positionswe through iin choosing the Bi depends
ordering k)
through j the Let be positions that 1
i − 1 does 1,
the maximum score among applicants 1 whereas best-qualifi
event positionletis in position is greater than value in − in all does positi
evalue in thatand valueiof the valuesi in positions 1 through ithe 1, whereas not app
applicant, iorderingSgreater than all of them, best-qualified position i other Bi dep
whether the isucceed the choosing the and the the values best-qualified
we be in event that we succeed when
event • thatith theofthe value in the best-qualified ithanare does not we other p
the ordering whether values when position 1 1throughappli- 1Thus we can all have th
cant the orderingsucceed in positions i is greater Si the values in apply
we various −
affect is the of one Sinterviewed. positionsthethrough i − 1. disjoint, affect wh
the values in Since
i
viewed. =(C.15) theobtain the Si are disjoint, wesucceed when− 1 best-qualified a
Pr {S} Since to {Si }. Noting that we k of them, and the valuedoes not affec
ordering of
n
Pr various values in positions have that i the in position i
equation in position i is greater than all never 1 through
value value in,2,•••,k
i is greater than all of them, and the value in posit
i=1 =
1
–
never position when the best-qualified ap-
Noting that one of the succeedwe have that Pr {S } = 0 for i = 1, 2, . . . , k. Thus,
plicant is we
first the values in positions 1 through i − 1. Thus we c
k,
affect the{B ∩ Oi } = Pr {Bi i {Oi } 2, . . . k. Thus, we
} = Praffect {S of 0 for Pr the .
ordering =
Pr {Sihave thati Pr thei }ordering} of= 1,values,ini positions 1 through i − 1. Thus
k, we
obtain
equation equationto obtain obtain
(C.15)+1,•••,n
to
The probability Prk i(C.15)
–
=
{B } is clearly 1/n, since the maximum is equally likely to be
n
inPr {Sone ofPr {B{Si } O{B= Pr {Bi } PrOi to}Pr ){O i:
the maximum value in po- (5.1
Pr any i= = thei }ni= Pr. i
ii: ∩For }event i {Bi } occur,i }k+1 i-­‐1
{S} } Pr {S positions. Oi = Pr {Oi . (O .
Pr ∩ }(B )
sitions 1 through i − 1 must be in one of the first k (5.13)
i=k+1 positions, and it is equally
n
likely to be The probability− 1 clearly
Consequently, Pr {Oi } maximum isand lik
The probability Pr {Bii} is {Bi } is clearly 1/n, the maximum is − 1) equall
in any of these Pr positions. 1/n, since since the = k/(i equally
compute
Pr {Si }. n order to succeed
Pr {Sany k/(n(i − onenUsing equation (5.13),For Oi when ithe best-qualified applica
We nowin any 1)). of the In positions. we event O to occur, the maximum
i } = one of the positions. For event have
Si }.in p109-­‐110
to succeed when the best-qualifiedthe to occur, the maximum valu
is the ith one, two1things must− 1 mustFirst, applicant first k positions, and
In order
i happen. be of must the
best-qualified applicant must
sitions the best-qualifiedin one in one ofbe k positions, and it i
n First, throughmust be applicant the first
s must happen. an eventll
− 1eserved.
we denote by B . Second, the algorithm must n
in {S} =1likely010
tniky1
i
A in any of these i − 1 positions. Consequently, Pr {O } =
sitions Copyright©
2
position throughbe} rights
r
Pr we denote by {Si . which the algorithm imust not
which
i, Pr to B Second, Page
6
likely to be in any iof these i − 1 positions. Consequently, Pr {O } = k/(i i

7. 7 then returnPr {Si } = Pr {Bi }∩ Oi } {BiPr {Bii{Si } {O{Bi.{Bi ∩ O}} .= Pr {Bi } Pr {
i {S = Pr = ∩ Pr } Pr Pr iPr } Pr {Oi i
O = = }
8 return n
2010 CS The probability Pr {Bi } is Pr The} probability1/n, maximummaxim
The probability clearly 1/n, since Pr {Bi } is clearly equ
{Bi is clearly the since the is 1/
116 any one of theonepositions. positions. ofOthe noccur,i the occur, e
in inChapter 5 the n For event For to and Randomize
any n of Probabilistic Analysis positions.maxim
in any one i event O to For
Chapter 5 Probabilistic Analysissitions each possible value ofmust the in one of mustfirst in p
We wish to determine, and Randomized Algorithms isitions k,through i first1k the that
for 1 through 1i through − 1in 1 be the − positions,
sitions − 1 must be one of probability be k
be We to be in any − 1 positions. Consequently,− 1{Oi a
likely will then of these ibe 1 any of possible k, }
likely toandin any of these Algorithms − in positions. Consequentl
hire the most qualified applicant. Randomized i likely to the best these i Pr posi
Chapter 5 Probabilistic Analysis choose
implementChapter 5 Probabilistic= k/(n(i} andi RandomizediUsing(5.13),− 1)).only wethro
116
ordering of the strategy with that1value. = 1)). of the {S } = Biassume (5.13), on fixe
Pr {Si } Pr {Si
thevalues in positions Analysis −For
1,−Prequationk/(n(ipositionsk1ishave
k/(n(i 1)).
Using
moment, equation that equati
ordering−the whereas in
through values depends
Algorithms
we have
Using
n n n
whether the = max1≤i≤ j {score(i)} Prwhether ithe value all=position {Sis applicants
Let M( j ) value in position i {S}greater than the }maximum score among} greater th
denote= {S
Pr is = {S} Prthe values{Si } in other positions. The Pr in
Pr {S} Pr i i
ordering Let be positions that 1 i i=k+1 the − not whereas best-qualifi
ordering of. of theSvalues in positionswe throughdoes choosing positions the thro
through j the values inthe event 1 throughsucceed iinvalues i=k+1 the Bi depends
ordering − 1
i=k+1 of 1, affect whether 1 in
−
applicant, orderingSgreater
values
i
ofk+1
weink
position
i
values
n
best-qualifiedi app
whether the valueiof
the than
all invalue+2
succeed
when
i
the 1,n
whereas not dep
and letis
in
position
k is positions
n1
through
greater
k allBof th
1
the
event that
k and
the
the
is
in all other positi
be 2
greater
than value
in position than
n
value in position i them,
k
i does
cant the ordering values in in position 1 1through thanofThus we can1) positio
ordering whether the value positions in(i − 1) n(i
Si1) = does n(i inaffect whp
ith one the values in =Since =the ordering 1.1 i=k+1 not all other
is greater
various − thedisjoint,− in
affect is the of theof interviewed. positionsthethrough ii−−arethe valueswe have th
affect i=k+1 through i − 1 does not affec
values
apply
{S} =(C.15) to obtain the values inwe nnever(C.15) 1 whenvalue in1position ia
ordering of Noting that positionssucceed 1
n i=k+1
equation in position ii }. greater than all of them, and obtain n best-qualified
Prvalue i=1 Pr {S is equation1 n
k k to the kthe
plicant is one of in i-­‐1first thei values
n positions 10throughn1,
2, 1. −,1k.in posit
value theposition we have thatthan all =
them,i and the
. .
k, is greater Pri {Si1 of i − 1 =
i − value Thus, c
affect the{Bi ∩ the} ordering} of =
i }in = in i=k+1 for 1 = i=k+1.ii − 1. Thus
ordering= Pr {Bi Pr {O values − }
Pr {Si } = Praffect Oi of thePri=k+1} = Pr {Bi ∩ Oi } = Pr {BiThus we .
.{S n positions through
} Pr {Oi }
obtain
equation equationto obtain obtain k n−1 1 k n−1 1
(C.15) (C.15) to i
1
2
k
The probability Prk+1,•••,n
k n−1 likely to be
–
=
{Bi } is clearly 1/n, since= . maximum is equally .
the 1
n = The probability Pr {B } is clearly 1/n, sin
. =
any i } = the } i= Pr. i }(Bi: ∩For }event O } occur,i }k+1 i-­‐1 i n i=k i
inPr {Sone ofPr {B{Si } O{B= Pr {Bi }n ii=ki ito}Pr )i=k(Oi:
the maximum value in po- (5.1
Pr {B n. of .
Pr {S} = Pr {Si n positions. Oi =in anyi ione{Oi the n positions. For event
Pr ∩
i Pr {O )
sitions 1 through i − 1 must be in one of the first k positions, and it is equally
i=k+1
n sitions since since −i 1 summation from lik
The probability Pr {Bii}We clearly approximate bytheapproximate k/(i besummationo
in any of these is {Bi }We clearly 1/n,We Pr {Oto maximum in equall
is 1/n, 1 through i the} = by this is one
must equally
likely to be The probability− 1approximate Consequently,bound thisboundis − 1) and abov
Pr positions. by integrals to maximum integrals to boun
integrals
compute
Pr {Sithe In positions. For event O of best-qualified applica
}. n order(5.13), we have we to occur, the maximum
We nowin any 1)). of the inequalities (A.12), we in any thethese i − positions
to succeed when have
Pr {Sany k/(n(i − onenUsing equationlikely to be(A.12),inequalities (A.12),1we have valu
in i } = one of the
the inequalities have
the
positions. For event Oi to occur, the maximum
i
is the p110
one, two1things must− 1 mustFirst,n−111 best-qualified1k positions, and
ith sitions throughn i1 happen. 1 ibe= k/(n(i 1− 1)).n−1 applicant must
n−1
n inthe first k positions, and it(5
Pr1{S } n−1 one nofn−1 1 first
the Using equation i
n−1
sitions 1 through i
All
− 1eserved.
we ≤ indx≤ of .the . dx ≤dx . algorithm 1must n
in {S} = Copyright©
2010
tniky1
}in any x dx be x − 1 positions.x Consequently, Pr {Odx .
positionlikely Pr {S rights
r must denote by Bi i x≤n
i, an to be whichof these i i one
n
event ≤ Second, the i ≤
dx
x }=
Page
7
Pr x
k
positions. Consequently, Pr {O } = k/(i
likely to be in any iof these i − 1k i=k k−1
i=k k k−1 i=k k−1 i

8. n
Let Pr {S}j )== max{Si } {Bji }{score(i)} denotePr {Smaximum score among appli
M( probability1≤i≤ is clearly 1/n, since thethe } =isPr {B likely to be Pr {B }
The Pr Pr maximum equally
2010 CS any one of the n positions. For event Oi to occur,i the maximumi value in}po-
in
∩ Oi = i
through j . 1Let i=k+1 S be the event that we succeed in choosing the best-q
sitions through i − 1 must be in one of the first k positions, and it is equally
n
applicant,=and let k iofbe the−event thatConsequently, Pr {Oi } = k/(i − 1) iandis clearly
likely to be in any 1) these i 1 positions. we succeed when Pr {B }
S The probability the best-qualified
cant is the
n(i −
= k/(n(i interviewed. Since in any
i=k+1
Pr {Si } ith one− 1)). Using equation (5.13), thehave one of the n positions. ha
we various Si are disjoint, we F
k n n 1
n
Pr {S} = = i=1 Pri {S1 }. Noting
i sitions 1 through − best-qualifi
that we never succeed wheni the 1 must be
Pr {S} n i=k+1 −Pr {Si }
=
plicant is one n−1i=k+1 first k, we have that likely}to be forany of these. i, − 1
of the =
1,2,•••,k
Pr {Si = 0 in i = 1, 2, . . k. Th
k 1
obtain =
n i=k i
n.
k =
k+1,•••,n
Pr {Si } = k/(n(i − 1)). Using eq
=
n n(i − 1)
i=k+1 n
Pr {S} approximatePr {Si } .1to bound this summation from{S} and below. By Pr
{S }
We = by integrals
k n
Pr
above
=
=
the inequalities (A.12), we have i
i=k+1 n i=k+1 i − 1
n n−1 i=k+1
1/k
+
1/(k+1)
+
1/(k+2)
+
•••
+
1/(n-­‐1)
1 1 n−1 n−1 1
We now≤compute 1 .x{Si .}. In order to succeed when the best-qualified
k x
dx
=
i=k
i
k≤ Pr dx
k−1
n
k ap
is the ith one, two things mustus the bounds First, the=
n i=k i
Evaluating these definite integrals gives happen. best-qualified applicant m
n(i − 1)
i=k+1the algorithm m
in position i, an event which bounddenote by BfromSecond, below. By
k We approximate by integrals to we this summation i . above and
k
(ln n − ln k) ≤ Pr {S} ≤ (ln(n − 1) − ln(k − 1)) , n
select anyinequalities (A.12), we have positions k + 1 through ik− 1, which happens
n the of the applicants in n 1
which provide a that k + 1 ≤ j ≤ i Because we wish to =
for each nj1such rather1tight bound1 for Pr {S}.− 1, we find that score( j ) < bestscore in maximize our
probability of success, let us focus on choosing the value of k that maximizesi=k+1 i − 1n the
n−1 n−1
dx ≤ are unique, we. can ignore the possibility of score( j ) = bes
≤ dx
(Because scores {S}.i (Besides, the lower-bound expression is easier to
maximize
lowerkbound on Pr k
Pr{S}
x k−1 x
than the upper-bound must be the case that all of the n k n−1 1
i=k
In other words, it expression.) Differentiating the expression (k/n)(ln values score(k + 1) t
−ln k)
score(i − 1) are less than M(k); if any are greater = n
p110-­‐111
Evaluatingk, we obtain
with respect to these definite integrals gives us the bounds .
than M(k) we will instead
i
k k
1 Copyright©
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rights
reserved.
i=k Page
8
the index of−the .first one (ln(n 1) − ln(k We use Oi to denote the event th
(ln n(ln ln k − 1) ≤ Pr
− n ln k) {S} ≤ that is−greater. − 1)) ,

9. Pr {S} = Pr {Si }
i=k+1
We approximate by integrals to bound this summation from above and below. By
2010 CS n
the inequalities (A.12), we have k
= 1068 Appendix A Summations
i=k+1
n(i − 1)
n n−1 n−1
1 1 1
k x
dx ≤
i=k
i
=
≤ k n dx .
k−1 x
1
n i=k+1 i − 1 f (x)
Evaluating these definite integrals gives us the bounds
k n−1 1
k = k .
(ln n − ln k) ≤ Pr {S} i=k i (ln(n − 1) − ln(k − 1)) ,
n ≤
n n
which provide a rather tight bound for Pr {S}. Because we wish to maximize our
We approximate by integrals to bound this summation from above … below. By and
f (m+1)
f (m+2)
f (n–2)
f (n–1)
f (m)
f (n)
…
probability of success, let us focus on choosing the value of k that maximizes the
the inequalities (A.12), we have
lower bound on Pr {S}. (Besides, the lower-bound expression is easier to maximize x
… …
than the upper-bound n−1 1
n
1 expression.) Differentiating themexpression (k/n)(ln n −ln k) n–1 n
n−1
1
–1 m m+1 m+2 n–2 n+1
with respect to dx we obtain≤
k, ≤ dx . (a)
k x i=k
i k−1 x
1
(ln n −Evaluating. these definite integrals gives us the bounds
ln k − 1) f (x)
n
k k
(ln n − ln k) ≤ Pr {S} ≤ (ln(n − 1) − ln(k − 1)) ,
n n
which provide a rather tight bound for Pr {S}. Because we wish to maximize our
probability of success, let us focus on choosing the value of k that maximizes the
f (m+1)
f (m+2)
f (n–2)
f (n–1)
lower bound on Pr {S}. (Besides, the lower-bound expression is … easier to maximize
f (m)
f (n)
…
than the upper-bound expression.) Differentiating the expression (k/n)(ln n −ln k) x
with respect to k, we obtain m –1 m m+1 m+2 … … n–2 n–1 n n+1
(b)
p325
1
(ln n − ln k − 1) .
n Copyright©
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reserved.
Figure A.1 Approximation of n Page
9
k=m f (k) by integrals. The area of each rectangle is shown
within 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

10. n
Let Pr {S}j )== max{Si } {Bji }{score(i)} denotePr {Smaximum score among appli
M( probability1≤i≤ is clearly 1/n, since thethe } =isPr {B likely to be Pr {B }
The Pr Pr maximum equally
2010 CS any one of the n positions. For event Oi to occur,i the maximumi value in}po-
in
∩ Oi = i
through j . 1Let i=k+1 S be the event that we succeed in choosing the best-q
sitions through i − 1 must be in one of the first k positions, and it is equally
n
applicant,=and let k iofbe the−event thatConsequently, Pr {Oi } = k/(i − 1) iandis clearly
likely to be in any 1) these i 1 positions. we succeed when Pr {B }
S The probability the best-qualified
cant is the
n(i −
= k/(n(i interviewed. Since in any
i=k+1
Pr {Si } ith one− 1)). Using equation (5.13), thehave one of the n positions. ha
we various Si are disjoint, we F
k n n 1
n
Pr {S} = = i=1 Pri {S1 }. Noting
i sitions 1 through − best-qualifi
that we never succeed wheni the 1 must be
Pr {S} n i=k+1 −Pr {Si }
=
plicant is one n−1i=k+1 first k, we have that likely}to be forany of these. i, − 1
of the =
1,2,•••,k
Pr {Si = 0 in i = 1, 2, . . k. Th
k 1
obtain =
n i=k i
n.
k =
k+1,•••,n
Pr {Si } = k/(n(i − 1)). Using eq
=
n n(i − 1)
i=k+1 n
Pr {S} approximatePr {Si } .1to bound this summation from{S} and below. By Pr
{S }
We = by integrals
k n
Pr
above
=
=
the inequalities (A.12), we have i
i=k+1 n i=k+1 i − 1
n n−1 i=k+1
1/k
+
1/(k+1)
+
1/(k+2)
+
•••
+
1/(n-­‐1)
1 1 n−1 n−1 1
We now≤compute 1 .x{Si .}. In order to succeed when the best-qualified
k x
dx
=
i=k
i
k≤ Pr dx
k−1
n
k ap
is the ith one, two things mustus the bounds First, the=
n i=k i
Evaluating these definite integrals gives happen. best-qualified applicant m
n(i − 1)
i=k+1the algorithm m
in position i, an event which bounddenote by BfromSecond, below. By
k We approximate by integrals to we this summation i . above and
k
(ln n − ln k) ≤ Pr {S} ≤ (ln(n − 1) − ln(k − 1)) , n
select anyinequalities (A.12), we have positions k + 1 through ik− 1, which happens
n the of the applicants in n 1
which provide a that k + 1 ≤ j ≤ i Because we wish to =
for each nj1such rather1tight bound1 for Pr {S}.− 1, we find that score( j ) < bestscore in
maximize our
probability of success, let us focus on choosing the value of k that maximizesi=k+1 i − 1
n the
n−1 n−1
dx ≤ are unique, we. can ignore the possibility of score( j ) = bes
≤ dx
(Because scores {S}.i (Besides, the lower-bound expression is easier to
maximize
lowerkbound on Pr k
Pr{S}
x k−1 x
than the upper-bound must be the case that all of the n k n−1 1
i=k
In other words, it expression.) Differentiating the expression (k/n)(ln values score(k + 1) t
−ln k)
score(i − 1) are less than M(k); if any are greater = n
p111
Evaluatingk, we obtain
with respect to these definite integrals gives us the bounds .
than M(k) we will 1instead
i
k k
1 Copyright©
2010
tniky1
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rights
reserved.
i=k Page
0
the index of−the .first one (ln(n 1) − ln(k We use Oi to denote the event th
(ln n(ln ln k − 1) ≤ Pr
− n ln k) {S} ≤ that is−greater. − 1)) ,

11. k We i=k i k−1
approximate by integrals to bound this summation from above and below. B
in any onethe inequalities (A.12), we For event Oi the bounds the maximum valu
of the these definite integrals gives us to occur,
Evaluating
nn positions. have
2010 CS Pr {S} = Pr {Si }
i −in
ordering 1kthrough i=k+1 11positions in through the firstwhereas Bi depends is
sitions of the valuesn−1 mustn−1 1 one of i − 1, k positions, and it o
n
be
1 k1
likely the value ink) ≤ Pr ≤ − (ln(n. − 1) − ln(k values in Pr {Oi } position
whetherto n (ln nx− ln≤positioni ik−1 1greater thanConsequently,all other = k/(i
be in any of these ≤ positions. the − 1)) ,
dx n {S} is dx
i k nx
k = − 1)). Using equation (5.13), we have
Pr {Si } of the values n(i − 1)
ordering = k/(n(i i=k+1in positions 1 through{S}.− 1 does not affect whet
i=k
Pr i Because we wish to maximiz
which provide a rather tight bound for us the bounds
Evaluating these definite integrals gives
value in positionn i is greater than all of them, and the value of k position i do
probability of success, let us focus on choosing the value in that maximize
k n 1 k
Pr {S} = k bound k) ≤ {S}. ≤ in(ln(n − lower-bound expression easier to can
lower = ln the values
k
affect the≥
(ln n −ofon Pr}Pr− 1(Besides, the 1) − ln(k − 1)) , i − 1. isThus wemax
{Si
ordering Pri=k+1 i {S}
n positions 1 through
n i=k+1 n
upper-bound expression.) Differentiating the expression (k/n)(ln n −
than theto obtain
equation (C.15) provide a rather tight bound for Pr {S}. Because we wish to maximize ou
n−1
which n
with respect 1
k to k,
we obtain
k . let random variables
= of indicator us focus on choosing the value of k that maximizes th
istic analysis and further usesn of =i Pr {B } Pr {O } . 117
Pr {Si } = Prprobabilityi } success, i
= {Bi ∩ O i=k i
1 lower bound on Pr1) (Besides, the lower-bound expression is easier to maximiz
117− ln k − 1) {S}.
(ln i=k+1 n(i − .
n
The probability Prn{Bi }by integrals to1/n, since the maximum is and below.−ln k
n than the upper-bound expression.) Differentiating the expression equallynlikel
We approximate is clearly bound this summation from above
(k/n)(ln
By
of k n 0 1
with respect to k, we obtain k
in any one to the we (A.12), we have event Oi bound on the probability
derivative equal the inequalities see thatFor lower to occur, the maximum value
= 1 0, positions. the
sitions 1 through − ln ki1− 1. n−1 1 in one of the first k positions, and it is e
n i=k+1
ed on the ln k = lnn ndx ≤ − 1 ≤ln(n/e) or, equivalently, when k = n/e.
n (ln n i n−1 − must be
1 − 1 = 1)
d whenlikely to be in any of these i − 1 positions. Consequently, Pr {O } = k/(i −
probability dx .
k xk
n−1 i
1 i k−1 x
tly, when{Siour k/(n(i − 1)).. Using= n/e, we will succeed in hiring our
implementk } = =
= strategy with k equation (5.13), we have
n/e. i=k
Pr k!!
d applicant
Evaluating these definite integrals gives us the bounds
ucceed in hiring probability at least 1/e.
with our i=k i
n
n
k [ k]
Pr {S} ≥
n = (ln n − ln k){Si Pr {S} ≤ (ln(n − 1) − ln(k − 1)) ,
Pr ≤}
k=n/e
n
We approximate by integrals to bound this summation from above and be
i=k+1
which provide a rather tight bound for Pr {S}. Because we wish to maximize our
the inequalities (A.12), welet us focus on choosing the value of k that maximizes the
p111
probability of success, have
n
k
1/e
=lowern−1 on Pr
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2010
tniky1
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rights
reserved.
{S}. (Besides, the lower-bound expression is easier to maximize
bound
n
1 than the upper-bound 1
i=k+1 1
n(i − 1) expression.) Differentiating the expression (k/n)(ln n1−ln k)
n−1 Page
1

12. 116 Chapter 5 Probabilistic Analysis and Randomized Algorithms
Pr {Si } = k/(n(i − 1)). Using equation (5.13), we have
2010 CS n
Pr {S} = ordering of the values in positions
Pr {Si }
ordering of the values in positions 1 through i − in position i Bi gre
i=k+1 whether the value 1, whereas is de
(whether the value in position i is greater than the valuesin positions
)
TVn k
ordering of the values in all other
=
ordering of the values in positions 1 in position −is greater than al
i=k+1
n(i − 1) value through i i 1 does not affe
value in position i is greater than all of them, and thethe values in value in posi
• k n 1 affect the ordering of
affect the ordering of the values in positions 1 through i − 1. Thus
=
equation
(C.15) to obtain i − 1 equation (C.15) to obtain
n i=k+1
– k n−1 1 10 Pr {Si } = Pr {Bi ∩ Oi } = Pr {Bi } Pr
= .
Pr {Si } = Pr {Bi ∩ On}i=k iPr {Bi } Pr {Oi } .
i =
– (The probability Pr {B )
is clearly 1/
i}
The probability Pr {Bi } by integrals to bound this summation from aboveequal
We approximate
is clearly 1/n, one ofthe maximum is and b
in any since the n positions. For
in any onethe inequalities (A.12), we For event through
0.1
1the maximum
of the n positions. sitions 1 Oi1/10
occur, must be in
have to = i −
sitions 1 through i − 1 mustn−1 likely to be in any of k positions,pos
n n−1 be in one of the first these i − 1 an
1 1 1
dx of ≤ positions. Consequently, Pr {Oi } =
(n=10)
dx .
likely to be in any ≤ these i − 1Pr {S } = k/(n(i − 1)). Using equat
k
k x i k−1 x
i
Pr {Si } = k/(n(i − 1)). Using equation (5.13),k=3
have
i=k
!(^^;)
we
Evaluating these definite integrals gives us the
nbounds
0.3611
Pr{S}
n
= k k {S} =
Pr
Pr {S} ≥
(ln n − ln k) ≤ }Pr {S} ≤ (ln(n − 1) − ln(k
Pr{S}
,
0.3665
Pr {Si
k=4
Pr {Si }
−
i=k+11))
n i=k+1 n
n
which provide a rather/e
10/(2.7182 Pr {S}. .679
kwe wish to maxi
n k
=
n tight bound for )
3 Because
=
probability of success, let us focus on=
k choosing the value of Page
12
maxim
n(i − 1)k that
=
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2010
tniky1
All
rights
reserved.
i=k+1
lower bound on Pr {S}. (Besides, the lower-bound expression is easier to m

13. 2010 CS
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
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2010
tniky1
All
rights
reserved.
Page
13