アルゴリズムイントロダクション 5章の勉強会資料 下記でPPT版，KEY版も配布してます． https://tanaike.jp

1. 2010 CS
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3. 2010 CS
• 5.1
•
• 5.2
•
• 5.3
•
• 5.4
•
p87-­‐113
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4. 2010 CS
(^^)
•
–
•
– n
– 1
– ,
–
–
– (
)
p87
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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
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6. 2010 CS
•
•
2
1 n
•
• n!
p88-­‐89
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7. 2010 CS
m( )
•
–
• ( ) ( :
A: )
• ( )
– Xi
Xi
• A i
A i
• Xi I
{
i
}
Xi I
{
i
}
p90
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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
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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
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simplify the calculation. We shall use indicator random variables through-
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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.
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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
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≤ 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
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which yields the bound Page
12

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( )
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15. 2010 CS
• Hire-­‐Assistant
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