Book Stacking
Mathematics for Computer Science
MIT 6.042J/18.062J
Harmonic Series,
Integral Method,
Stirling’s Formula
table
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. .
L8-3.1
October 28,2005
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
Book Stacking
L8-3.2
October 28,2005
Book Stacking
How far out?
One book
center of mass
of book
?
overhang
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
L8-3.3
October 28,2005
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
Book Stacking
One book
L8-3.4
October 28,2005
Book Stacking
One book
center of mass
of book
center of mass
of book
1
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
October 28,2005
L8-3.5
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
October 28,2005
L8-3.6
1
n books
n books
center
of
mass
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
October 28,2005
L8-3.7
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
n books
October 28,2005
L8-3.8
n books
Need
center of mass
over table
center of mass
of the whole stack
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
October 28,2005
L8-3.9
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
October 28,2005
L8-3.10
N
n+1 books
center of mass
of all n+1 books
at table edge
N
center of mass
of top n books
at edge of book
n+1
Δ overhang ::=
Horizontal distance from
n-book to n+1-book
centers-of-mass
∆overhang
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
October 28,2005
L8-3.11
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.12
2
Δ overhang
Choose origin so center of n-stack at x = 0.
n
Now center of n+1st book is at x = 1/2, so
1
}
center of n+1-stack is at
Δ
Δ=
1/2
1
1
2 =
n + 1 2(n + 1)
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
L8-3.13
October 28,2005
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
center of mass
of all n+1 books
at table edge
}
center of mass
of top n books
at edge of book
n+1
1
2(n + 1)
H n ::= 1 +
October 28,2005
L8-3.15
1 1
1
+ +" +
2 3
n
Bn ::= overhang of n books
B1 = 1/2
1
Bn+1 = Bn +
2(n + 1)
1⎛ 1 1
1⎞
⎜1 + + + " + ⎟
2⎝ 2 3
n⎠
Bn =
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
Estimate Hn :
Integral Method
1
x+1
1
2
1
3
Bn = Hn/2
0
L8-3.17
1
3
1
2
1
October 28,2005
L8-3.16
October 28,2005
1
nth Harmonic number
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
L8-3.14
October 28,2005
Book stacking summary
n+1 books
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005. All rights reserved.
n ⋅ 0 + 1 ⋅1 / 2
1
=
n +1
2(n + 1)
x=
1
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
2
3
4
5
October 28,2005
6
7
8
L8-3.18
3
Book stacking
n
1
∫ x + 1 dx ≤
1+
0
n+1
∫
1
1 1
1
+ + ... +
2 3
n
So Hn → ∞ as n→ ∞, and so
overhang can be any desired size.
1
dx ≤ H n
x
ln(n + 1) ≤ H n
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.19
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
Book stacking
October 28,2005
L8-3.20
Team Problem
Bn ≥ 3
Hn ≥ 6
Integral bound: ln (n+1) ≥ 6
So can do with n ≥ ⎡e6−1⎤ = 403 books
Actually calculate Hn :
227 books are enough.
Overhang 3: need
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
Problem 1
L8-3.21
Crossing a Desert
April 5, 2004
Copyright © Albert R. Meyer, 2004.
L10-1.22
1 Tank of Gas
1 tank
Gas
depot
truck
How big a desert can the truck cross?
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
D1::= max distance on 1 tank = 1
L8-3.23
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.24
4
n+1 Tanks of Gas
Let Dn ::=
max distance into the
desert using n tanks of gas from the depot
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
x
1 −2x
1−2x
1−2x
1−x
L8-3.25
October 28,2005
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n+1 Tanks of Gas
Set (1−2x)n + (1−x) = n.
(1−2x)n
+ (1−x)
Set depot at x
to be n tanks;
continue from
x with n tank
method.
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
Then using n tank strategy
from position x, gives
Dn+1 = Dn + x
October 28,2005
L8-3.27
October 28,2005
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
L8-3.28
October 28,2005
1 1
1
Dn = 1 + + + " +
3 5
2n −1
(1−2x)n + (1−x) = n
1
x=
2n+1
1
Dn+1 = Dn +
2n+1
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
L8-3.26
October 28,2005
depot at x
x
So have:
n
n
1
∫ 2(x + 1) −1 dx
≤ Dn
0
ln(2n +1)
≤ Dn
2
Can cross any desert!
L8-3.29
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.30
5
Closed form for n!
Team Problem
Factorial defines a product:
n
n! :: = 1 ⋅ 2 ⋅ 3 ⋅⋅⋅ (n − 1) ⋅ n = ∏ i
i=1
Problem 2
Turn product into a sum taking logs:
ln(n!) = ln(1 · 2 · 3 · · · (n – 1) · n)
= ln 1 + ln 2 + · · · + ln(n – 1) + ln(n)
n
= ∑ ln(i)
i=1
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
L8-3.31
October 28,2005
Integral Method
Integral Method to bound
Copyright © Albert Meyer and RonittRubinfeld, , 2003.
Integral Method
n
∑ ln i
Bounds on ln(n!)
i=1
… ln (x+1)
ln 5
ln 4
ln 3
ln 2
ln 3 ln 4
ln 2
1
2
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
3
ln 5
4
5
ln ln n
n-1
n–2 n–1
October 28,2005
n
n
1
i=1
1
n
L8-3.33
Integral Method
Reminder:
October 28,2005
L8-3.34
⎛ x⎞
=
ln
x
dx
x
ln
⎜ ⎟
∫
⎝e⎠
= x ln x −
x
x(ln x − ln e) = x (ln x − 1)
= x ln x − x
October 28,2005
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
Integral Method
⎛ x⎞
ln
x
dx
x
ln
=
⎜ ⎟=
∫
⎝e⎠
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n
∫ ln( x) dx ≤ ∑ ln(i) ≤ ∫ ln( x + 1) dx
ln (x)
ln n
October 16, 2003
Quickie:
verify by differentiating.
L8-3.35
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.36
6
Integral Method
Integral Method
Bounds on ln(n!)
n
1
n
i =1
1
i =1
n
⎛n⎞
⎛ n +1⎞
n ln ⎜ ⎟ + 1 ≤ ∑ ln(i) ≤ (n + 1) ⋅ ln⎜
⎟ + 0. 6
⎝
e ⎠
⎝ e ⎠
i =1
exponentiating:
⎛n⎞
n! ≈ n / e ⎜ ⎟
⎝e⎠
⎛n⎞
1
n
So guess:
∑ ln(i) ≈ (n + 2 )ln ⎜⎝ e ⎟⎠
i =1
Copyright © Albert Meyer, 2003.
Copyright © Albert Meyer , 2003.
October 16, 2003
October 16, 2003
A precise approximation:
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
n
~ 2π n ⎛⎜ ⎞⎟
n
Asymptotic Equivalence
Stirling’s Formula
n!
⎛n⎞
∑ ln(i) ≈ (n + 2 )ln ⎜⎝ e ⎟⎠
∫
ln( x) dx ≤ ∑
ln(i) ≤ ∫
ln( x + 1) dx
n
1
n
f(n) ~ g(n)
n
⎛ f ( n) ⎞
lim ⎜
⎟ =1
n →∞
g
(
n
)
⎝
⎠
⎝e⎠
October 28,2005
L8-3.39
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.40
Team Problem
Problem 3
Copyright © Albert R. Meyer and Ronitt Rubinfeld, 2005.
October 28,2005
L8-3.41
7