Lemma 24.17 (Predecessor-subgraph property)
Let
G
D
.V; E/
be a weighted, directed graph with weight function
w
W
E
!
R
,
let
s
2
V
be a source vertex, and assume that
G
contains no negative-weight cycles
that are reachable from
s
. Let us call I
NITIALIZE
-S
INGLE
-S
OURCE
.G; s/
and then
execute any sequence of relaxation steps on edges of
G
that produces
:
d
D
ı.s; /
for all
2
V
. Then, the predecessor subgraph
G
is a shortest-paths tree rooted
at
s
.
Proof
We must prove that the three properties of shortest-paths trees given on
page 647 hold for
G
. To show the first property, we must show that
V
is the set
of vertices reachable from
s
. By definition, a shortest-path weight
ı.s; /
is finite
if and only if
is reachable from
s
, and thus the vertices that are reachable from
s
are exactly those with finite
d
values. But a vertex
2
V
f
s
g
has been assigned
a finite value for
:
d
if and only if
:
¤
NIL
. Thus, the vertices in
V
are exactly
those reachable from
s
.
The second property follows directly from Lemma 24.16.
It remains, therefore, to prove the last property of shortest-paths trees: for each
vertex
2
V
, the unique simple path
s
p
;
in
G
is a shortest path from
s
to
in
G
. Let
p
D h
0
;
1
; : : : ;
k
i
, where
0
D
s
and
k
D
. For
i
D
1; 2; : : : ; k
,
we have both
i
:
d
D
ı.s;
i
/
and
i
:
d
i
1
:
d
C
w.
i
1
;
i
/
, from which we
conclude
w.
i
1
;
i
/
ı.s;
i
/
ı.s;
i
1
/
. Summing the weights along path
p
yields
w.p/
D
k
X
i
D
1
w.
i
1
;
i
/
k
X
i
D
1
.ı.s;
i
/
ı.s;
i
1
//
D
ı.s;
k
/
ı.s;
0
/
(because the sum telescopes)
D
ı.s;
k
/
(because
ı.s;
0
/
D
ı.s; s/
D
0
) .
Thus,
w.p/
ı.s;
k
/
. Since
ı.s;
k
/
is a lower bound on the weight of any path
from
s
to
k
, we conclude that
w.p/
D
ı.s;
k
/
, and thus
p
is a shortest path
from
s
to
D
k
.
Exercises
24.5-1
Give two shortest-paths trees for the directed graph of Figure 24.2 (on page 648)
other than the two shown.
24.5
Proofs of shortest-paths properties
677
24.5-2
Give an example of a weighted, directed graph
G
D
.V; E/
with weight function
w
W
E
!
R
and source vertex
s
such that
G
satisfies the following property: For
every edge
.u; /
2
E
, there is a shortest-paths tree rooted at
s
that contains
.u; /
and another shortest-paths tree rooted at
s
that does not contain
.u; /
.
24.5-3
Embellish the proof of Lemma 24.10 to handle cases in which shortest-path
weights are
1
or
1
.
24.5-4
Let
G
D
.V; E/
be a weighted, directed graph with source vertex
s
, and let
G
be initialized by I
NITIALIZE
-S
INGLE
-S
OURCE
.G; s/
. Prove that if a sequence of
relaxation steps sets
s:
to a non-
NIL
value, then
G
contains a negative-weight
cycle.
24.5-5
Let
G
D
.V; E/
be a weighted, directed graph with no negative-weight edges. Let
s
2
V
be the source vertex, and suppose that we allow
:
to be the predecessor
of
on
any
shortest path to
from source
s
if
2
V
f
s
g
is reachable from
s
,
and
NIL
otherwise. Give an example of such a graph
G
and an assignment of
values that produces a cycle in
G
. (By Lemma 24.16, such an assignment cannot
be produced by a sequence of relaxation steps.)
24.5-6
Let
G
D
.V; E/
be a weighted, directed graph with weight function
w
W
E
!
R
and no negative-weight cycles. Let
s
2
V
be the source vertex, and let
G
be initial-
ized by I
NITIALIZE
-S
INGLE
-S
OURCE
.G; s/
. Prove that for every vertex
2
V
,
there exists a path from
s
to
in
G
and that this property is maintained as an
invariant over any sequence of relaxations.
24.5-7
Let
G
D
.V; E/
be a weighted, directed graph that contains no negative-weight
cycles. Let
s
2
V
be the source vertex, and let
G
be initialized by I
NITIALIZE
-
S
INGLE
-S
OURCE
.G; s/
. Prove that there exists a sequence of
j
V
j
1
relaxation
steps that produces
:
d
D
ı.s; /
for all
2
V
.
24.5-8
Let
G
be an arbitrary weighted, directed graph with a negative-weight cycle reach-
able from the source vertex
s
. Show how to construct an infinite sequence of relax-
ations of the edges of
G
such that every relaxation causes a shortest-path estimate
to change.
678
Chapter 24
Single-Source Shortest Paths
Problems
24-1
Yen’s improvement to Bellman-Ford
Suppose that we order the edge relaxations in each pass of the Bellman-Ford al-
gorithm as follows. Before the first pass, we assign an arbitrary linear order
1
;
2
; : : : ;
j
V
j
to the vertices of the input graph
G
D
.V; E/
. Then, we parti-
tion the edge set
E
into
E
f
[
E
b
, where
E
f
D f
.
i
;
j
/
2
E
W
i < j
g
and
E
b
D f
.
i
;
j
/
2
E
W
i > j
g
. (Assume that
G
contains no self-loops, so that every
edge is in either
E
f
or
E
b
.) Define
G
f
D
.V; E
f
/
and
G
b
D
.V; E
b
/
.
a.
Prove that
G
f
is acyclic with topological sort
h
1
;
2
; : : : ;
j
V
j
i
and that
G
b
is
acyclic with topological sort
h
j
V
j
;
j
V
j
1
; : : : ;
1
i
.
Suppose that we implement each pass of the Bellman-Ford algorithm in the fol-
lowing way. We visit each vertex in the order
1
;
2
; : : : ;
j
V
j
, relaxing edges of
E
f
that leave the vertex. We then visit each vertex in the order
j
V
j
;
j
V
j
1
; : : : ;
1
,
relaxing edges of
E
b
that leave the vertex.
b.
Prove that with this scheme, if
G
contains no negative-weight cycles that are
reachable from the source vertex
s
, then after only
dj
V
j
=2
e
passes over the
edges,
:
d
D
ı.s; /
for all vertices
2
V
.
c.
Does this scheme improve the asymptotic running time of the Bellman-Ford
algorithm?
24-2
Nesting boxes
A
d
-dimensional box with dimensions
.x
1
; x
2
; : : : ; x
d
/
nests
within another box
with dimensions
.y
1
; y
2
; : : : ; y
d
/
if there exists a permutation
on
f
1; 2; : : : ; d
g
such that
x
.1/
< y
1
,
x
.2/
< y
2
, . . . ,
x
.d /
< y
d
.
a.
Argue that the nesting relation is transitive.
b.
Describe an efficient method to determine whether or not one
d
-dimensional
box nests inside another.
c.
Suppose that you are given a set of
n d
-dimensional boxes
f
B
1
; B
2
; : : : ; B
n
g
.
Give an efficient algorithm to find the longest sequence
h
B
i
1
; B
i
2
; : : : ; B
i
k
i
of
boxes such that
B
i
j
nests within
B
i
j
C
1
for
j
D
1; 2; : : : ; k
1
. Express the
running time of your algorithm in terms of
n
and
d
.
Problems for Chapter 24
679
24-3
Arbitrage
Arbitrage
is the use of discrepancies in currency exchange rates to transform one
unit of a currency into more than one unit of the same currency. For example,
suppose that
1
U.S. dollar buys
49
Indian rupees,
1
Indian rupee buys
2
Japanese
yen, and
1
Japanese yen buys
0:0107
U.S. dollars. Then, by converting currencies,
a trader can start with
1
U.S. dollar and buy
49
2
0:0107
D
1:0486
U.S. dollars,
thus turning a profit of
4:86
percent.
Suppose that we are given
n
currencies
c
1
; c
2
; : : : ; c
n
and an
n
n
table
R
of
exchange rates, such that one unit of currency
c
i
buys
RŒi; j
units of currency
c
j
.
a.
Give an efficient algorithm to determine whether or not there exists a sequence
of currencies
h
c
i
1
; c
i
2
; : : : ; c
i
k
i
such that
RŒi
1
; i
2
RŒi
2
; i
3
RŒi
k
1
; i
k
RŒi
k
; i
1
> 1 :
Analyze the running time of your algorithm.
b.
Give an efficient algorithm to print out such a sequence if one exists. Analyze
the running time of your algorithm.
24-4
Gabow’s scaling algorithm for single-source shortest paths
A
scaling
algorithm solves a problem by initially considering only the highest-
order bit of each relevant input value (such as an edge weight). It then refines the
initial solution by looking at the two highest-order bits. It progressively looks at
more and more high-order bits, refining the solution each time, until it has exam-
ined all bits and computed the correct solution.
In this problem, we examine an algorithm for computing the shortest paths from
a single source by scaling edge weights. We are given a directed graph
G
D
.V; E/
with nonnegative integer edge weights
w
. Let
W
D
max
.u;/
2
E
f
w.u; /
g
. Our
goal is to develop an algorithm that runs in
O.E
lg
W /
time. We assume that all
vertices are reachable from the source.
The algorithm uncovers the bits in the binary representation of the edge weights
one at a time, from the most significant bit to the least significant bit. Specifically,
let
k
D d
lg
.W
C
1/
e
be the number of bits in the binary representation of
W
,
and for
i
D
1; 2; : : : ; k
, let
w
i
.u; /
D
w.u; /=2
k
i
˘
. That is,
w
i
.u; /
is the
“scaled-down” version of
w.u; /
given by the
i
most significant bits of
w.u; /
.
(Thus,
w
k
.u; /
D
w.u; /
for all
.u; /
2
E
.) For example, if
k
D
5
and
w.u; /
D
25
, which has the binary representation
h
11001
i
, then
w
3
.u; /
D
h
110
i D
6
. As another example with
k
D
5
, if
w.u; /
D h
00100
i D
4
, then
w
3
.u; /
D h
001
i D
1
. Let us define
ı
i
.u; /
as the shortest-path weight from
vertex
u
to vertex
using weight function
w
i
. Thus,
ı
k
.u; /
D
ı.u; /
for all
u;
2
V
. For a given source vertex
s
, the scaling algorithm first computes the
680
Chapter 24
Single-Source Shortest Paths
shortest-path weights
ı
1
.s; /
for all
2
V
, then computes
ı
2
.s; /
for all
2
V
,
and so on, until it computes
ı
k
.s; /
for all
2
V
. We assume throughout that
j
E
j j
V
j
1
, and we shall see that computing
ı
i
from
ı
i
1
takes
O.E/
time, so
that the entire algorithm takes
O.kE/
D
O.E
lg
W /
time.
a.
Suppose that for all vertices
2
V
, we have
ı.s; /
j
E
j
. Show that we can
compute
ı.s; /
for all
2
V
in
O.E/
time.
b.
Show that we can compute
ı
1
.s; /
for all
2
V
in
O.E/
time.
Let us now focus on computing
ı
i
from
ı
i
1
.
c.
Prove that for
i
D
2; 3; : : : ; k
, we have either
w
i
.u; /
D
2w
i
1
.u; /
or
w
i
.u; /
D
2w
i
1
.u; /
C
1
. Then, prove that
2ı
i
1
.s; /
ı
i
.s; /
2ı
i
1
.s; /
C j
V
j
1
for all
2
V
.
d.
Define for
i
D
2; 3; : : : ; k
and all
.u; /
2
E
,
y
w
i
.u; /
D
w
i
.u; /
C
2ı
i
1
.s; u/
2ı
i
1
.s; / :
Prove that for
i
D
2; 3; : : : ; k
and all
u;
2
V
, the “reweighted” value
y
w
i
.u; /
of edge
.u; /
is a nonnegative integer.
e.
Now, define
y
ı
i
.s; /
as the shortest-path weight from
s
to
using the weight
function
y
w
i
. Prove that for
i
D
2; 3; : : : ; k
and all
2
V
,
ı
i
.s; /
D y
ı
i
.s; /
C
2ı
i
1
.s; /
and that
y
ı
i
.s; /
j
E
j
.
f.
Show how to compute
ı
i
.s; /
from
ı
i
1
.s; /
for all
2
V
in
O.E/
time, and
conclude that we can compute
ı.s; /
for all
2
V
in
O.E
lg
W /
time.
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