Lecture 7: Shortest Paths in Graphs with Negative Arc Lengths

Lecture 7:
Shortest Paths in
Graphs with
Negative Arc
Lengths
Reading: AM&O Chapter 5
Label Correcting Methods
Assume the network G is allowed to have negative arc lengths but no directed negativelyweighted cycles. Assume that there exist a
directed (s, j)-path for every node j, and that
all arc lengths are integer. Let C be the largest
absolute value of an arc length.
Recall Bellman’s conditions for a set of node labels dj to comprise correct shortest (s, j)-path
distances:
(a) for each node j, there exists at least one
(s, j)-path of length dj
(b) Bellman’s Equations are satisﬁed:
ds = 0
dj = min{di + cij : i ∈ B(j)} j ̸= s.
Idea of label correcting: Look for arcs (i, j) for
which dj > di + cij and correct dj to satisfy
dj = di + cij
Generic Label Correcting Procedure
{
set labels dj =
0 j=s
∞ otherwise
while dj > di + cij for some arc (i, j)
{ set dj = di + cij , Pred(j)=(i, j) }
Correctness and Complexity:
1. The Pred() graph will eventually include every node, and can never produce a cycle
(Why?)
2. Thus every node i will eventually have a
path from s of length exactly di
3. Labels are nonincreasing, and each of the n
labels can become no smaller than −nC
4. Thus the algorithm must terminate in at
most n2C iterations, with labels satistying
Bellman’s conditions.
Problem: Not polynomial
Modified Label Correcting Algorithm
Again we can modify FINDPATH to improve the
complexity of the generic procedure.
{
set labels dv =
0 v=s
∞ otherwise
set LIST = {s},
while LIST ̸= ∅
begin
select k ∈ LIST
for i ∈ A(k) do
if di > dk + cki then
begin
set di = dk + cki
set Pred(i)=(k, i)
if i ̸∈ LIST than add i to LIST
end
remove k from LIST
end
Example Problem
4
2
5
4
6
0
s 1
−5
2
6
5
4
3
node
-
1
0
6
3
2
∞
4
3
∞
4
∞
5
∞
6
∞
Example Problem
LIST
1
6
2
4
5
4
6
0
s 1
−5
2
6
5
4
3
6
node
–
1
1
0
0
6
3
2
∞
6
4
3
∞
6
4
∞
∞
5
∞
∞
6
∞
∞
LIST
1
2,3
Example Problem
6
2
10
5
4
4
6
0
s 1
−5
2
6
5
4
3
6
node
–
1
2
1
0
0
0
6
3
2
∞
6
6
4
3
∞
6
6
4
∞
∞
∞
5
∞
∞
10
6
∞
∞
∞
LIST
1
2,3
3,5
Example Problem
6
2
10
5
4
4
6
0
s 1
−5
2
6
5
3
6
node
–
1
2
3
1
0
0
0
0
6
3
2
∞
6
6
6
4
3
∞
6
6
6
4
∞
∞
∞
10
4
10
5
∞
∞
10
10
6
∞
∞
∞
∞
LIST
1
2,3
3,5
5,4
Example Problem
6
2
10
5
4
4
6
0
s 1
−5
2
3
6
5
3
6
node
–
1
2
3
5
1
0
0
0
0
0
14
6
2
∞
6
6
6
6
4
3
∞
6
6
6
6
4
∞
∞
∞
10
10
4
10
5
∞
∞
10
10
10
6
∞
∞
∞
∞
14
LIST
1
2,3
3,5
5,4
4,6
Example Problem
5
2
10
5
4
4
6
0
s 1
−5
2
3
6
5
3
6
node
–
1
2
3
5
4
1
0
0
0
0
0
0
14
6
2
∞
6
6
6
6
5
4
3
∞
6
6
6
6
6
4
∞
∞
∞
10
10
10
4
10
5
∞
∞
10
10
10
10
6
∞
∞
∞
∞
14
14
LIST
1
2,3
3,5
5,4
4,6
6,2
Example Problem
5
2
9
4
5
4
6
0
s 1
−5
2
3
6
5
3
6
node
–
1
2
3
5
4
6
2
1
0
0
0
0
0
0
0
0
14
6
2
∞
6
6
6
6
5
5
5
4
3
∞
6
6
6
6
6
6
6
4
∞
∞
∞
10
10
10
10
9
4
10
5
∞
∞
10
10
10
10
10
10
6
∞
∞
∞
∞
14
14
14
14
LIST
1
2,3
3,5
5,4
4,6
6,2
2
5
Example Problem
5
2
9
4
5
4
6
0
s 1
−5
2
3
6
5
3
6
node
–
1
2
3
5
4
6
2
5
6
1
0
0
0
0
0
0
0
0
0
0
13
6
2
∞
6
6
6
6
5
5
5
5
5
4
3
∞
6
6
6
6
6
6
6
6
6
4
∞
∞
∞
10
10
10
10
9
9
9
4
10
5
∞
∞
10
10
10
10
10
10
10
10
6
∞
∞
∞
∞
14
14
14
14
13
13
LIST
1
2,3
3,5
5,4
4,6
6,2
2
5
6
–
Theorem: If elements from LIST are chosen in
FIFO order, then the modiﬁed label correcting
algorithm ﬁnds all minimum length (s, i)-paths
in O(nm) steps.
Proof: Because of the FIFO search order, a node
i is reexamined only after every other node
whose label was changed before i has been examined. Since a node that has not had its
label changed will not aﬀect any other labels,
we can assume WLOG that in a particular pass
of the algorithm all nodes are examined.
Claim: After the kth stage, the algorithm
has determined the correct shortest path
distance for all nodes i for which the shortest (s, i)-paths has at most k arcs.
Proof of claim: Clearly true for k = 0. Now
assume that the claim is true after k ≥ 0
passes. Since WLOG the algorithm examines all nodes in the (k + 1)st pass,
then any node for whom the shortest path
is of length k + 1 must be given the correct shortest path distance in the (k +1)st
pass.
Therefore the algorithm will have found the correct
shortest distance labels after at most n − 1
passes, each pass of which requires looking at
possibly all of the edges. The total complexity
is thus O(nm).
Converting Negative Arc Weights
to Nonnegative Arc Weights
Properties of shortest path distances: It follows
directly from Label-Correcting Algorithm that
the ﬁnal set of distances d∗i satisfy
(i) d∗i ≤ d∗j + cij for every arc (i, j), and
(ii) d∗i = d∗j + cij for every arc in the shortest
path tree.
Useful Fact: Let πi, i ∈ N be any set of numbers
assigned to the nodes, and set
cπ
ij = cij − πi + πj .
Then every path P between two nodes s and t
will have cπ -length
cπ (P ) = c(P ) − πs + πt.
Valuable Application
Suppose you solved the Shortest s-to-All Path
Problem for a graph G with general arc weights,
and obtained the correct shortest (s, i)-path
distances d∗i , i ∈ N . If we set π = −d∗ above,
then it follows directly from the the above two
observations that
∗ − d∗ ≥ 0 for all arcs (i, j),
(i) cπ
=
c
+
d
ij
ij
i
j
(ii) cπ
ij = 0 for all arcs in the shortest path
tree,
(iii) for any pair of nodes s and t, the shortest cπ -length (s, t)-paths and the shortest
c-length (s, t)-paths will be identical.
In particular, it follows from (i) and (iii) that
once you ﬁnd all shortest (s, i)-paths for one s
using, say label correcting methods, than any
other shortest (i, j)-paths can be found using
Dijkstra’s Algorithm (which is much faster).
Example
5
2
9
4
5
4
6
0
s 1
−5
2
13
6
3
6
5
3
6
4
10
4
Final Label−Correcting Distances
5
2
9
5
0
0
1
0
s 1
0
1
13
6
2
0
2
3
6
0
4
10
Adjusted Arc Weights
Example: Finding shortest pairs between all pairs
of nodes involves n applications of a shortest
(s, j)-path algorithm (one for each s). Using
the Modiﬁed Label Correcting Algorithm will
take O(n2m) time, but with the above procedure the time is now
O(nS(m, n))
Where S(m, n) is the fastest implementation of
Dijkstra’s Algorithm. This could be as small as
O(nm + n2 log n) (Fibonacci heaps) or O(nm +
√
2
n log C) (radix+Fibonacci heaps).
Matrix Methods for Finding
All-Pairs Shortest Paths
Given: n×n matrix C of arc costs for n-node graph
G — where cij = 0 if i = j and cij = ∞ if arc
(i, j) is not present — for which there are no
negative weight directed cycles.
Find: n × n matrix D all-pair shortest path lengths
(that is, dij = shortest path from i to j, and
∞ if no (i, j)-path exists).
General Procedure: Produce the sequence
D0, D 1, . . . , D r of n × n matrices, with
D0 = C and Dr = D.
Matrix Multiplication Method
Dk has entries
[
shortest length of an (i, j)-path
k
d [i, j] =
containing at most k + 1 arcs
]
The update from Dk−1 to Dk is the “matrix
multiplication” Dk = Dk−1 × C, with the arithmetic operations redeﬁned as follows:
• where numbers were multiplied, they are
now added;
• where numbers were added, the minimum
is now found.
and Dn−2 now represents the all-pairs shortest
path lengths.
Complexity: O(n3 log n)
(by “successive squaring”)
Example
2
1
2
-2
1
1
4
3
3

D0 =

0 ∞ ∞ 1
0 1 ∞ 
 2
 ∞ ∞ 0 ∞ 
∞ −2 3 0
D1

0 −1
4 1
0
1 3 
 2
= 
∞ ∞
0 ∞ 
0 −2 −1 0
D2

0 −1
0 1
0
1 3 
 2
= 
∞ ∞
0 ∞ 
0 −2 −1 0


Floyd-Warshall Method
Dk has entries


shortest length of an (i, j)-path


dk [i, j] = all of whose interior nodes are in
the set {1, 2, . . . , k}
The update from Dk−1 to Dk is produced by
the pivot operation
dk [i, j] = min{dk−1[i, j], dk−1[i, k] + dk−1[k, j]}
and Dn now represents the all-pairs shortest
path lengths.
Floyd-Warshall Procedure
Pred[i, j] = node preceding j on the shortest (i, j)path


 0
Initialize: set d[i, j] = cij

 ∞
i=j
(i, j) ∈ A
i ̸= j, (i, j) ̸∈ A
Pred[i, j] = i all i, j
for k = 1, . . . , n do
for i = 1, . . . , n do
for j = 1, . . . , n do
if d[i, j] > d[i, k] + d[k, j] then
begin
set d[i, j] = d[i, k] + d[k, j]
set Pred[i, j]=Pred[k, j]
end
Complexity: O(n3)
Example
1
2
2
-2
1
1
4
3
3
Predk
Dk
(
(
(
(
(
0
2
∞
∞
∞
0
∞
−2
∞
1
0
3
1
∞
∞
0
0
2
∞
∞
∞
0
∞
−2
∞
1
0
3
1
3
∞
0
0
2
∞
0
∞
0
∞
−2
∞
1
0
−1
1
3
∞
0
0
2
∞
0
∞
0
∞
−2
∞
1
0
−1
1
3
∞
0
0
2
∞
0
−1
0
∞
−2
0
1
0
−1
1
3
∞
0
)
(
k=0
)
(
k=1
)
(
k=2
(
)
k=3
(no change)
)
(
k=4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
1
3
4
1
2
3
2
1
2
3
4
1
2
3
2
1
1
3
4
1
2
3
2
1
2
3
4
1
2
3
2
1
1
3
4
1
2
3
2
4
2
3
4
2
2
3
2
1
1
3
4
)
)
)
)
)
Shortest (1,3)-path: 1=Pred(1,4)→4=Pred(1,2)→2=Pred(1,3)→3
Finding Negative Cycles
The Floyd-Warshall Algorithm will ﬁnd negative cycles naturally. The ﬁrst time a negative
cycle is encountered, it will do so by creating
a negative number in one of the diagonal elements of the array. The corresponding change
in the Pred array will allow you to reconstruct
the cycle.
1
2
2
−4
1
1
4
3
3
Predk
Dk
(
(


0
2
∞
∞
∞
0
∞
−4
∞
1
0
3
1
∞
∞
0
0
2
∞
∞
∞
0
∞
−4
∞
1
0
3
1
3
∞
0
0
2
∞
−2
∞
0
∞
−4
∞
1
0
−3
)
(
k=0
)
1
3
∞
−1
(
k=1



k=2

1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
1
1
3
4
1
2
3
2
Cycle: 4=Pred(4,2)→2=Pred(4,1)→1=Pred(4,4)→4
1
2
3
4
1
2
3
2
1
1
3
1
)
)


Application:
The Tramp Steamer Problem
A tramp steamer travels from port to port,
doing business at each of the ports it visits.
The net proﬁt — counting travel costs — of
making the trip from Port i to Port j is pij ,
and the time (in days) to get from Port i to
Port j is τij (all data are positive integers).
The steamer captain wants to choose a circuit W : i1, i2, . . . , ik = i1 which has maximum
mean daily profit
∑k−1
j=1 pij ij+1
µ(W ) = ∑k−1
.
j=1 τij ij+1
Choose a target value µ and attempt to determine if there is a route W which exceeds the
daily proﬁt target, µ(W ) > µ, or equivalently,
∑
(µτuv − puv ) < 0
(u,v)∈W
Letting cuv = µτuv − puv , we are interested in
ﬁnding a negative length cycle in G.
Cases:
G has a negative cycle: Then µ underestimates
the value of the maximum mean daily proﬁt.
Increase µ and try again.
G has only positive length cycles: Then µ overestimates the value of the maximum mean
daily proﬁt. Decrease µ and try again.
G has a minimum cycle W of zero length: Then
µ = µ∗ is the correct value for the cycle with
maximum mean daily proﬁt, and W is the
optimal cycle.
Binary Search to Find the Correct µ∗
Facts: Let P = max{puv : (u, v) ∈ A} and
τ0 = max{τuv : (u, v) ∈ A}
1. The minimum value for µ∗ is −P , and the
maximum value for µ∗ is P .
2. If W1 and W2 are two cycles in G with different mean daily proﬁts µ(W1) and µ(W2),
then
|µ(W1) − µ(W2)| ≥ 1/(nτ0)2
The binary search procedure searches for µ∗
within an interval [µ, µ
¯] = {µ : µ ≤ µ ≤ µ
¯}
known to contain µ∗, progressively halving the
interval based on the outcome of the negative
cycle procedure.
The Procedure
1. Begin with the interval [−P, P ].
Now for interval [µ, µ
¯] obtained in a general
step of the procedure:
2. Test for a negative cycle, using the value
µ0 = (µ + µ
¯)/2
3. If a negative cycle is found in G, then reset
the interval to [µ, µ
¯] = [µ0, µ
¯]
4. If only positive cycles are found, then reset the interval to [µ, µ
¯] = [µ, µ0]
5. If the minimum length cycle has zero length,
then halt the algorithm, returning the value
µ = µ∗ and the corresponding cycle of
maximum mean daily proﬁt.
Termination of the algorithm:
If the algorithm fails to ﬁnd a zero length cycle
¯] is small enough
by the time the interval [µ, µ
so that µ
¯ − µ < 1/(nτ0)2, then there must be
exactly one number µ∗ ∈ [µ, µ
¯] having denominator 1/(nτ0)2. By Fact 2 this must be the
maximum mean daily proﬁt value, and so we
run Step 5 one time to ﬁnd the optimal cycle
W.
Complexity of procedure: The number of steps
in the binary search procedure is O(log((nτ0)2P )) =
O(log((nτ0)2P )), each of which requires one
search for negative cycles. Using Floyd-Warshall,
the complexity becomes O(n3 log((nτ0)2P ))).
Example
Consider the following tramp steamer problem,
where the arcs are marked pij , τij :
1
4,2
3
2,2
2,1
4,2
4
2,2
2
3,1
5
Note that 1 − 3 − 4 − 5 − 2 − 1 is the maxp
imum length cycle using either pij or τij for
ij
arc weights. Using binary search with P = 4
and τ0 = 2, we have the following sequence of
stages:
1
−4
3
−2
−2
−4
4
−2
2
−3
5
[µ, µ
¯] = [−4, 4],
1
0
µ0 = 0
3
2
0
0
4
2
2
−1
[µ, µ
¯] = [0, 4],
5
µ0 = 2
1
2
3
4
1
2
4
4
2
5
0
[µ, µ
¯] = [2, 4],
1
1
µ0 = 3
3
3
1/2
1
4
3
2
−1/2
[µ, µ
¯] = [2, 3],
5
µ0 =
5
2
1
1/2
3
5/2
1/4
1/2
4
5/2
2
−3/4
[µ, µ
¯] = [2, 5/2],
5
µ0 = 9/4
At this point we obtain zero-length cycle
1 − 5 − 2, which is the minimum-mean-length
cycle, of mean length
2+3+4
9
= .
1+1+2
4
Note that if no zero-length cycle was found,
we would have continued until the interval size
reached
µ
¯ − µ < 1/(nτ0)2 = .01
which would have taken another 5 stages.

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