6.042/18.062J
Mathematics
for
Computer
Science
March 15, 2005
Srini
Devadas
and
Eric
Lehman
Lecture
Notes
Sums,
Approximations,
and
Asymptotics
II
1
Block
Stacking
How
far
can
a
stack
of
identical
blocks
overhang
the
end
of
a
table
without
toppling
over?
Can
a
block
be
suspended
entirely
beyond
the
table’s
edge?
Table
Physics
imposes
some
constraints
on
the
arrangement
of
the
blocks.
In
particular,
the
stack
falls
off
the
desk
if
its
center
of
mass
lies
beyond
the
desk’s
edge.
Moreover,
the
center
of
mass
of
the
top
k
blocks
must
lie
above
the
(
k
+
1)
st
block;
otherwise,
the
top
k
would
fall
over.
In
order
to
find
the
best
configuration
of
blocks
satisfying
these
constraints,
we’ll
need
a
fact
about
centers
of
mass.
Fact
1.
If
two
objects
have
masses
m
1
and
m
2
and
centersofmass
at
positions
z
1
and
z
2
,
then
the
center
of
mass
of
the
two
objects
together
is
at
position:
z
1
m
1
+
z
2
m
2
m
1
+
m
2
Define
the
offset
of
a
particular
configuration
of
blocks
to
be
the
horizonal
distance
from
its
center
of
mass
to
its
rightmost
edge.
offset
?
center
of
mass
s

?
Table
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2
Sums,
Approximations,
and
Asymptotics
II
The
offset
determines
precisely
how
far
that
configuration
can
extend
beyond
the
desk
since
at
best
the
center
of
mass
lies
right
above
the
desk’s
edge.
So
hereafter
we
can
focus
on
maximizing
the
offset
of
a
stack
of
n
blocks
rather
than
the
overhang
.
As
a
result,
we
need
only
be
concerned
with
whether
the
stack
is
internally
stable
and
not
with
whether
the
whole
configuration
is
too
far
to
the
right
and
falls
off
the
table.
We
can
establish
the
greatest
possible
offset
of
a
stack
of
n
blocks
with
an
inductive
argument.
This
is
an
instance
where
induction
not
only
serves
as
a
proof
technique,
but
also
turns
out
to
be
a
nice
tool
for
reasoning
about
the
problem.
Theorem
1.
The
greatest
possible
offset
of
a
stack
of
n
≥
1
blocks
is:
1
1
1
1
X
n
=
+
+
+
. . .
+
2
4
6
2
n
Proof.
We
use
induction
on
n
,
the
number
of
blocks.
Let
P
(
n
)
be
the
proposition
that
the
greatest
possible
offset
of
a
stack
of
n
≥
1
blocks
is
1
/
2
+
1
/
4
+
. . .
+
1
/
(2
n
)
.
Base
case:
For
a
single
block,
the
center
of
mass
is
distance
X
1
= 1
/
2
from
its
rightmost
edge.
So
the
offset
is
1
/
2
and
P
(1)
is
true.
Inductive
step:
For
n
≥
2
,
we
assume
that
P
(
n
−
1)
is
true
in
order
to
prove
P
(
n
)
.
A
stack
of
n
blocks
consists
of
the
bottom
block
together
with
a
stack
of
n
−
1
blocks
on
top.
In
order
to
acheive
the
greatest
possible
offset
with
n
blocks,
the
top
n
−
1
blocks
should
themselves
have
the
greatest
possible
offset,
which
is
X
n
−
1
;
otherwise,
we
could
do
better
by
replacing
the
top
n
−
1
blocks
with
a
different
configuration
that
has
greater
offset.
Furthermore,
the
center
of
mass
of
the
top
n
−
1
blocks
should
lie
directly
above
the
right
edge
of
the
bottom
block;
otherwise,
we
could
do
better
by
sliding
the
top
n
−
1
blocks
farther
to
the
right.
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 Spring '11
 Dr.EricLehman
 Computer Science, Big O notation, Analysis of algorithms, Asymptotic analysis, Upper and lower bounds, asymptotic notation, Asymptotics II

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