Shortest Paths Part 2: Negative Weights and All-pairs

Data Structures and Algorithms – COMS21103
Shortest Paths Revisited
Negative Weights and All-Pairs
Benjamin Sach
inspired by slides by Ashley Montanaro

In today’s lectures we’ll be revisiting the shortest paths problem
In particular we’ll be interested in algorithms which allow negative edge weights
in a weighted, directed graph. . .
The shortest path from MVB to Temple Meads
(according to Google Maps)
and algorithms which compute the shortest path between all pairs of vertices

Part one
Single Source Shortest Paths with negative weights

Single source shortest paths with negative weights
Bellman-Ford’s algorithm solves the single source shortest paths problem
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

It ﬁnds the shortest path from
a given source vertex
to every other vertex
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

The weights are allowed to be
positive or negative
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

The graph is stored as an Adjacency List
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

We will not need any
non-elementary data structures
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

We will not need any
non-elementary data structures
Previously we saw that Dijkstra’s algorithm (implemented with a binary heap) solves this
problem in O((|V | + |E|) log |V |) time when the edges have non-negative weights
|V | is the number of vertices and |E| is the number of edges
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

Negative weight cycles
If some of the edges in the graph have negative weights
the idea of a shortest path might not make sense:
If there is a path from s to t which includes a negative weight cycle,
A X
Z
Y B
What is the shortest path from A to B?
(-1)
1 1 1
A negative weight cycle is a path from a vertex v back to v such that
the sum of the edge weights is negative
there is no shortest path from s to t
A X
Z
Y B
-1 -1
1 1 1

Negative weight cycles
If some of the edges in the graph have negative weights
the idea of a shortest path might not make sense:
If there is a path from s to t which includes a negative weight cycle,
A X
Z
Y B
What is the shortest path from A to B?
(-1)
1 1 1
A negative weight cycle is a path from a vertex v back to v such that
the sum of the edge weights is negative
there is no shortest path from s to t
We will ﬁrst discuss a (slightly) simpler version of Bellman-Ford
that assumes there are no such cycles
A X
Z
Y B
-1 -1
1 1 1

Most of the Bellman-Ford algorithm
MOSTOFBELLMAN-FORD(s)
weight(u, v) is the weight of
the edge from u to v
(u, v) ∈ E iff there is an
edge from u to v
dist(v) is the length of the shortest
path between s and v, found so far
For all v, set dist(v) = ∞
set dist(s) = 0
For i = 1, 2, . . . , |V |,
For every edge (u, v) ∈ E
If dist(v) > dist(u) + weight(u, v)
dist(v) = dist(u) + weight(u, v)

edge from u to v
set dist(s) = 0
For i = 1, 2, . . . , |V |,
The algorithm repeatedly asks, for each edge (u, v)
“can I ﬁnd a shorter route to v if I go via u?”

edge from u to v
set dist(s) = 0
For i = 1, 2, . . . , |V |,
s
u
v

edge from u to v
set dist(s) = 0
For i = 1, 2, . . . , |V |,
This is called RELAXING edge (u, v)
s
u
v

for each vertex v, dist(v) is the length of the shortest path between s and v
Claim When the MOSTOFBELLMAN-FORD algorithm terminates,
edge from u to v
set dist(s) = 0
For i = 1, 2, . . . , |V |,
(assuming there are no-negative weight cycles)
This is called RELAXING edge (u, v)
s
u
v

RELAX(u,v)
if dist(v) > dist(u) + weight(u, v)
In each iteration we RELAX every edge (u, v)
MOSTOFBELLMAN-FORD runs |V | iterations,
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
We’re going to simulate
- the length of the best path
between s and v, found so far

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
We start by setting dist(s) = 0 and every other dist(v) = ∞. . .

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
We now start iteration 1. . .

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
∞ > 0 + (−1)

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
∞ > 0 + (−1)
-1

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
∞ > 0 + 1

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
∞ > 0 + 1
1

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1
−1 < 1 + 1

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1
−1 < 1 + 1
(so we don’t change dist)

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1
∞ > 1 + 3

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1
∞ > 1 + 3
4

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4

RELAX(u,v)
dist(v)
37 vertex v
s
(in the order they occur in the adjacency list)
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
−1 < ∞ + 2

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
−1 < ∞ + 2
(so we don’t change dist)

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
∞ < ∞ + 2

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
∞ < ∞ + 2 (dist is still ∞)

RELAX(u,v)
dist(v)
37 vertex v
s
0 ∞
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
∞ > −1 + (−2)

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
∞ > −1 + (−2)
-3

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
-3

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
1 4
-3
1 > −3 + 1

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
1 > −3 + 1
-2

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞ ∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞ ∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
This is the end of iteration 1

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
How are things looking?

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
this looks good

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
so does this

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
this doesn’t look so good

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
neither does this

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
neither does this
(it’s not the shortest path length)

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
neither does this
(it’s not the shortest path length)
This path has length −1

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
we aren’t done

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
but it seems like we made progresswe aren’t done

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
does every iteration make progress?

RELAX(u,v)
dist(v)
37 vertex v
s
0
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
-1
4
-3
-2
-2
2
does every iteration make progress?
are |V | iterations enough?

The proof idea
Imagine a different algorithm where in each iteration. . .
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
you only relax one edge (rather than all edges)

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
Further, imagine that (magically or otherwise), the edge that you relax in iteration i
is the i-th edge in a shortest path from s to some vertex t

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
This is a shortest path from s to t

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
“relax this”

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
“relax this”
-1

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
-1

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
“relax this”

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
“relax this”
-3

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
-3

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
Iteration 3:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
“relax this”
Iteration 3:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
“relax this”
-2
Iteration 3:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
Iteration 3:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
Iteration 4:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
“relax this”
Iteration 4:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
“relax this”
-1
Iteration 4:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
Iteration 4:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
Iteration 5:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
“relax this”
Iteration 5:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
“relax this”
0
Iteration 5:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
0
Iteration 5:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
0
Iteration 6:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
0
“relax this”
Iteration 6:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
0
“relax this”
-1
Iteration 6:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
-3
-2
-1
0
-1
Iteration 6:

The proof idea
s
0
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
∞
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
When this algorithm terminates. . .
dist(t) is the length of the shortest path from s to t
t
-1
-3
-2
-1
0
-1
Iteration 6:

The proof idea
Now consider the MOSTOFBELLMAN-FORD algorithm where in each iteration. . .
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
you relax every edge (rather than one edge)

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Consider the same
shortest path from s to t

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
At some point in iteration i
t
Consider the same
you relax the i-th edge in the shortest path from s to t. . .

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
Consider the same

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
we relax this
Consider the same
at some point

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
we relax this
Consider the same
at some point
Don’t worry about what ? was before. RELAX picks the smaller of
? and 0 + (−1) so. . .

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
we relax this
-1
Consider the same
at some point
Don’t worry about what ? was before. RELAX picks the smaller of
? and 0 + (−1) so. . .

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
Iteration 1:
-1
Consider the same

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
Consider the same

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
relax this
Consider the same

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
relax this
Consider the same
-3

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 2:
Consider the same
-3

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 3:
Consider the same
-3

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 3:
Consider the same
-3

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 3:
Consider the same
-3
-2

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 3:
Consider the same
-3
-2

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 4:
Consider the same
-3
-2

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 4:
Consider the same
-3
-2

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 4:
Consider the same
-3
-2
-1

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 4:
Consider the same
-3
-2
-1

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 5:
Consider the same
-3
-2
-1

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 5:
Consider the same
-3
-2
-1

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 5:
Consider the same
-3
-2
-1
0

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 5:
Consider the same
-3
-2
-1
0

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 6:
Consider the same
-3
-2
-1
0

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
relax this
Iteration 6:
Consider the same
-3
-2
-1
0
-1

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
So after enough iterations. . .
dist(t) is the length of a path from s to t
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1
which is at least as short as the shortest path. . .

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1
In other words dist(t) is the length of a shortest path from s to t

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1
In other words dist(t) is the length of a shortest path from s to t
how many iterations are needed?

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1
For this argument to hold we need to do one iteration
for every edge in the shortest path from s to t

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1
We will now prove that if there are no negative cycles in the graph,
the shortest path between s and t contains at most |V | edges

The proof idea
s
0
?
?
?
?
?
?
?
?
?
?
?
3
1
1
1
4
-2
2
2
-2
-1
1
1
-1
1
3 -1
-1
2
1
-2
1
t
-1
Iteration 6:
Consider the same
-3
-2
-1
0
-1
We will now prove that if there are no negative cycles in the graph,
the shortest path between s and t contains at most |V | edges
and therefore |V | iterations will be enough

Claim if there are no negative weight cycles in the graph,
there is a shortest path between s and t containing at most |V | edges
(or there is no path from s to t)

Consider a path between s and t with a cycle. . .

s
t
3 1 2 -1
-1
3
-1

s
t
3 1 2 -1
-1
3
-1
As there are no negative weight cycles
deleting this cycle from the path cannot increase it’s length

s
t
3 1 2 -1
-1
3
-1
path length 6

s
t
3 1 2 -1
-1
3
-1
path length 6
path length 5

s
t
3 1 2 -1
-1
3
-1
path length 6
path length 5
Therefore, there is a shortest path between s and t containing no cycles

s
t
3 1 2 -1
-1
3
-1
path length 6
path length 5
Therefore, there is a shortest path between s and t containing no cycles
A path with no cycles enters each vertex at most once
so contains at most |V | edges

MOSTOFBELLMAN-FORD Summary
for each vertex v, dist(v) is the distance between s and v

If there is no path between s and v then
at termination, dist(v) = ∞
by the algorithm description, MOSTOFBELLMAN-FORD
doesn’t ﬁnd paths that aren’t there

If there is a shortest path between s and v then
it contains at most |V | edges

After each iteration, the algorithm
has taken (at least) one more ‘step’ along this path

After each iteration, the algorithm
has taken (at least) one more ‘step’ along this path
After |V | iterations of the algorithm
dist(v) is the length of the shortest path between s and v

So what is the rest of the algorithm?
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
Output ‘Negative weight cycle found’

BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
We’ve added a ﬁnal check which determines whether
another iteration would ﬁnd an even shorter path from s to some v

BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
We’ve added a ﬁnal check which determines whether
another iteration would ﬁnd an even shorter path from s to some v
We will prove that this is happens if and only if
there is a negative weight cycle

BELLMAN-FORD without negative weight cycles
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
We ﬁrst argue that the algorithm still works when there is no negative weight cycle

BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
After the ﬁrst |V | iterations
for each vertex v,
(which haven’t changed),
dist(v) is the length
of the shortest path between s and v

BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
for each vertex v,
‘Negative weight cycle found’If the ﬁnal check outputs:
then there is a path from s to some v (via u)
with length dist(u) + weight(u, v) < dist(v)

BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
for each vertex v,
i.e. a path which is shorter than the shortest path

BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
for each vertex v,
this is a contradiction.
i.e. a path which is shorter than the shortest path

BELLMAN-FORD with a negative weight cycle
Let v1, v2, . . . vk ∈ V be a negative weight cycle
and assume for a contradiction, that it wasn’t reported by the algorithm
v1
v2
v3
v4
v5
v6
vk

v1
v2
v3
v4
v5
v6
vk
weight(v1, v2) dist(v2) − dist(v1)
We have that,

v1
v2
v3
v4
v5
v6
vk
We have that,
because we ran the following checks
but found no negative weight cycle:

v1
v2
v3
v4
v5
v6
vk
So,
We have that,
dist(v2) dist(v1) + weight(v1, v2)
(then rearrange)
because we ran the following checks
but found no negative weight cycle:

v1
v2
v3
v4
v5
v6
vk
We have that,
and by the same argument. . .

v1
v2
v3
v4
v5
v6
vk
We have that,

v1
v2
v3
v4
v5
v6
vk
weight(vj, vj+1) dist(vj+1) − dist(vj)
In general for every edge (vj, vj+1), we have
(where vk+1 = v1)

v1
v2
v3
v4
v5
v6
vk
(where vk+1 = v1)
But the cycle has negative weight so,
k
j=1
weight(vj, vj+1) < 0

v1
v2
v3
v4
v5
v6
vk
(where vk+1 = v1)
However, if we sum all the edge weights using the top expressions we get,
k
j=1
k
j=1
weight(vj, vj+1)

v1
v2
v3
v4
v5
v6
vk
(where vk+1 = v1)
k
j=1
k
j=1
weight(vj, vj+1)
k
j=1
dist(vj+1) − dist(vj)

v1
v2
v3
v4
v5
v6
vk
(where vk+1 = v1)
k
j=1
k
j=1
dist(vj+1) −
k
j=1
dist(vj)
k
j=1
weight(vj, vj+1)

v1
v2
v3
v4
v5
v6
vk
(where vk+1 = v1)
k
j=1
k
j=1
dist(vj+1) −
k
j=1
dist(vj)
k
j=1
weight(vj, vj+1) = 0

v1
v2
v3
v4
v5
v6
vk
(where vk+1 = v1)
k
j=1
k
j=1
dist(vj+1) −
k
j=1
dist(vj)
k
j=1
weight(vj, vj+1) = 0
Contradiction!

BELLMAN-FORD Summary
When the BELLMAN-FORD algorithm terminates,
(or it reports that there is a negative weight cycle)
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,

BELLMAN-FORD Summary
When the BELLMAN-FORD algorithm terminates,
(or it reports that there is a negative weight cycle)
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
What is the time complexity of the BELLMAN-FORD algorithm?

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time
(we store dist in an array of length |V |)

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time
O(1) time

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time
O(1) time
O(|E|) time for each iteration

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time
O(1) time
O(|E|) time for each iteration
. . . and there are |V | iterations

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time
O(1) time
O(|V ||E|) time overall

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V |) time
O(|E|) time
O(1) time

Time Complexity
BELLMAN-FORD(s)
set dist(s) = 0
For i = 1, 2, . . . , |V |,
O(|V ||E|) time
O(|V |) time
O(|E|) time
O(1) time
so overall the BELLMAN-FORD algorithm takes

Summary
If there is a negative weight cycle, the algorithm reports this and aborts
The BELLMAN-FORD algorithm runs in O(|V ||E|) time
This is not as good as DIJKSTRA’s algorithm which runs in O((|V | + |E|) log |V |) time
The BELLMAN-FORD algorithm solves the single source shortest paths problem
and uses no non-elementary data structures
in a directed, weighted graph with positive and negative edge weights
(in this case, the problem is not well-deﬁned)
(when implemented using a binary heap)
However, DIJKSTRA’s algorithm only works for graphs with
non-negative edge weights

Part two
All-Pairs Shortest Paths

In previous lectures, we have focused on the single source shortest paths problem
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

i.e. in ﬁnding the shortest paths from a single given source vertex
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

i.e. in ﬁnding the shortest paths from a single given source vertex
What should we do if we want to ﬁnd the shortest paths from every vertex
to every other vertex?
-21
2
1
1 2 4
3
1
2
-1
A
B E
D
C F
G

Use something we saw before

We already have two options:

If the graph has non-negative edge weights,
we could run DIJKSTRA’s algorithm |V | times, once with each vertex as the source
this takes O(|V ||E| log |V |) time
(if the priority queue is a binary heap)

we could run DIJKSTRA’s algorithm |V | times, once with each vertex as the source
this takes O(|V ||E| log |V |) time
(if the priority queue is a binary heap)
If the graph has positive and negative edge weights,
we could run BELLMAN-FORD’s algorithm |V | times, once for each vertex
this takes O(|V |2|E|) time

repeatedly running DIJKSTRA’s algorithm takes O(|V ||E| log |V |) time
repeatedly running BELLMAN-FORD’s algorithm takes O(|V |2|E|) time

Are these any good?

Are these any good? Can we do better?

Are these any good? Can we do better? How does |V | compare to |E|?

The output contains the length of the shortest path between every pair of vertices

There are |V | · (|V | − 1) pairs of vertices

so we can’t expect to do better than O(|V |2) time
There are |V | · (|V | − 1) pairs of vertices

Imagine that |E| ≈
|V |2
4 (the graph is very dense)
e.g. each vertex has an edge to about half of the other vertices

|V |2
this becomes O(|V |3 log |V |)

|V |2
this becomes O(|V |4)

imagine that |E| ≈ 5|V | (the graph is very sparse)
e.g. each vertex has an edge to about 5 other vertices
Instead,

Instead,

Instead,
O(|V |2 log |V |) is a lot better than O(|V |3)

In this lecture, we will discuss JOHNSON’s algorithm for all-pairs shortest paths
which takes O(|V ||E| log |V |) time
in graphs with positive and negative edge weights

In this lecture, we will discuss JOHNSON’s algorithm for all-pairs shortest paths
in graphs with positive and negative edge weights
It will use both DIJKSTRA’s algorithm (implemented with a binary heap)
and BELLMAN-FORD to achieve this

JOHNSON’s algorithm - the overall approach
We have already seen one algorithm for all-pairs shortest paths

If the graph had non-negative edge weights,
we could have achieved this complexity by repeatedly running DIJKSTRA’s algorithm

However, we are interested in graphs with both positive and negative edge weights,

A
B
C
D
1 4
-2
-1
0
1

The approach employed by JOHNSON’s algorithm is to reweight the edges
so that the resulting graph has non-negative edge weights,
and then (repeatedly) run DIJKSTRA’s algorithm
A
B
C
D
1 4
-2
-1
0
1

A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
5
1
0
0
0
0

A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
5
1
0
0
0
0
the shortest path length may change
but the route remains the same

A ﬁrst attempt
One natural attempt at reweighting is to increase every edge weight
by the same amount. . .

A ﬁrst attempt
A
B
C
D
1 4
-2
-1
0
1

A ﬁrst attempt
A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
14
10
11
8
9
11
if we add 10 to every edge. . .

A ﬁrst attempt
A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
14
10
11
8
9
11
The shortest path from C to B
goes this way in the original graph

A ﬁrst attempt
A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
14
10
11
8
9
11
goes this way in the reweighted graph

A ﬁrst attempt
A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
14
10
11
8
9
11
Unfortunately, if we reweight the graph like this,
both the shortest paths lengths and the routes might change

A ﬁrst attempt
A
B
C
D
1 4
becomes-2
-1
0
1
A
B
C
D
14
10
11
8
9
11
Unfortunately, if we reweight the graph like this,
both the shortest paths lengths and the routes might change
this doesn’t look good

Reweighting based on vertex potential
We are going to associate a value h(v) with each vertex v ∈ V
- we will call this the potential of v
To overcome this, we are going to reweight each edge differently
the new weight of each edge will depend on which vertices it connects

A Bweight(A,B)

A B
h(A) h(B)
weight(A,B)

A B
h(A) h(B)
weight(A,B)
becomes
weight’(A,B)
where weight’(A,B) = weight(A,B) + h(A) − h(B)
A B

A B
h(A) h(B)
weight(A,B)
becomes
weight’(A,B)
where weight’(A,B) = weight(A,B) + h(A) − h(B)
A B
Why is this better than the ﬁrst attempt?

Reweighted paths
Each each vertex v ∈ V has a value h(v) - called the potential of v
we will pick these values carefully later

Reweighted paths
Consider the following path as an example. . .

Reweighted paths
v1
v2 v3 v4 v5 v6-2 3 1 7 4

Reweighted paths
v1
v2 v3 v4 v5 v6-2 3 1 7 4
Pick some potential values

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
(for now you can think of the values as arbitrary)

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
Reweight the whole graph using weight’ (we don’t just reweight this path). . .
v1 v2 v3 v4 v5 v61 5 0 6 2

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
where weight’(vi, vi+1) = weight(vi, vi+1) + h(vi) − h(vi+1)
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2
weight’(v1, v2) = weight(v1, v2) + h(v1) − h(v2)

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2
weight’(v1, v2) = weight(v1, v2) + h(v1) − h(v2)
=(−2) + (5) − (2) = 1

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2
- this path has length 13

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2
- this path has length 14 = 13 + 5 − 4 = 13 + h(v1) − h(v6)

Reweighted paths
v1
v2 v3 v4 v5 v6
h(v1) = 5 h(v2) = 2 h(v3) = 0 h(v4) = 1 h(v5) = 2 h(v6) = 4
-2 3 1 7 4
v1 v2 v3 v4 v5 v61 5 0 6 2
- this path has length 14 = 13 + 5 − 4 = 13 + h(v1) − h(v6)
is this a coincidence?

Reweighted paths
Consider an arbitrary path. . .
v1 v2 v3 v4 v5 vk
h(v1) h(v2) h(v3) h(v4) h(v5) h(vk)
? ? ? ?

Reweighted paths
v1 v2 v3 v4 v5 vk
h(v1) h(v2) h(v3) h(v4) h(v5) h(vk)
? ? ? ?
In the original graph, the path length is
k−1
i=1
weight(vi, vi+1)

Reweighted paths
v1 v2 v3 v4 v5 vk
h(v1) h(v2) h(v3) h(v4) h(v5) h(vk)
? ? ? ?
k−1
i=1
weight(vi, vi+1)
In the reweighted graph, the path length is
k−1
i=1
weight’(vi, vi+1)

Reweighted paths
v1 v2 v3 v4 v5 vk
h(v1) h(v2) h(v3) h(v4) h(v5) h(vk)
? ? ? ?
k−1
i=1
weight(vi, vi+1)
k−1
i=1
weight’(vi, vi+1) =
k−1
i=1
weight(vi, vi+1) + h(vi) − h(vi+1)

Reweighted paths
v1 v2 v3 v4 v5 vk
h(v1) h(v2) h(v3) h(v4) h(v5) h(vk)
? ? ? ?
k−1
i=1
weight(vi, vi+1)
k−1
i=1
k−1
i=1
=
k−1
i=1
weight(vi, vi+1) +h(v1)−h(vk)

Reweighted paths
v1 v2 v3 v4 v5 vk
h(v1) h(v2) h(v3) h(v4) h(v5) h(vk)
? ? ? ?
k−1
i=1
weight(vi, vi+1)
k−1
i=1
k−1
i=1
=
k−1
i=1
weight(vi, vi+1) +h(v1)−h(vk)
So the weight of a path only changes by the potential values of the end points. . .

Reweighted shortest paths
Let the function h give a value h(v) for each vertex v ∈ V
weight’(u,v) = weight(u,v) + h(u) − h(v)
Change the weight of every edge (u, v) to be

Lemma any path is a shortest path in the original graph
if and only if it is a shortest path in the reweighted graph

u vu v
path p1
path p2

u vu v
path p1
path p2
Let l1 be the length of p1
in the original graph

u vu v
path p1
path p2

u vu v
path p1
path p2
l1 + h(u) − h(v)
in the reweighted graph
is the length of p1

u vu v
path p1
path p2
l1 + h(u) − h(v)
l2 + h(u) − h(v)
is the length of p1
is the length of p2

u vu v
path p1
path p2
l1 + h(u) − h(v)
l2 + h(u) − h(v)
is the length of p1
is the length of p2
l1 l2 if and only if
l1 + h(u) − h(v) l2 + h(u) − h(v)

Reweighted negative cycles
Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
be a negative weight cycle

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
The weight of this cycle
k
i=1
weight(vi, vi+1) < 0
in the original graph is,

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
k
i=1
The weight of this cycle in the reweighted graph is

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
k
i=1
k
i=1
weight’(vi, vi+1)

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
k
i=1
k
i=1
weight’(vi, vi+1)
=
k
i=1
weight(vi, vi+1) +h(v1)−h(v1)

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
k
i=1
k
i=1
weight’(vi, vi+1)
=
k
i=1
weight(vi, vi+1)

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
k
i=1
k
i=1
weight’(vi, vi+1)
=
k
i=1

Let v1, v2, . . . vk ∈ V
v1
v2
v3v4
v5
v6
vk
(where vk+1 = v1)
k
i=1
k
i=1
weight’(vi, vi+1)
=
k
i=1
So reweighting doesn’t affect negative cycles

Reweighting summary

Reweighting summary
Lemma Any path is a shortest path in the original graph

Reweighting summary
Fact If is the length of a path from u to v in the original graph
+ h(u) − h(v) is its length in the reweighted graph

Reweighting summary
Fact A cycle has negative weight in the original graph
if and only if it has negative weight in the reweighted graph

Reweighting summary
So if we solve the all-pairs shortest paths problem on the reweighted graph. . .
we can recover the shortest path lengths for the original graph

Reweighting summary
So if we solve the all-pairs shortest paths problem on the reweighted graph. . .
we can recover the shortest path lengths for the original graph
To take advantage of this, we need to make all the edge weights non-negative

How do we choose h?
We ﬁrst add one additional vertex called s to the original graph

How do we choose h?
s

How do we choose h?
We also add an edge (s, v) from s to each other vertex v ∈ V
s

How do we choose h?
s
0
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0

How do we choose h?
s
each of these edges has weight 00
0
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0

How do we choose h?
This does not introduce any new negative weight cycless
0
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0

How do we choose h?
0
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For each v, let δ(s, v) denote the length of the shortest path from s to v

How do we choose h?
0
0
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0
Warning: If the original graph contains a negative weight cycle,
δ(s, v) may be undeﬁned

How do we choose h?
0
0
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0
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0
Warning: If the original graph contains a negative weight cycle,
δ(s, v) may be undeﬁned
JOHNSON’s algorithm will detect this and abort
Let’s continue under the assumption that there is no negative weight cycle

How do we choose h?
0
0
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we then deﬁne h(v) to equal δ(s, v)

How do we choose h?
0
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Consider any edge (u, v) ∈ E

How do we choose h?
0
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The key observation is that δ(s, v) δ(s, u) + weight(u, v)

How do we choose h?
0
0
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This follows because there is a path from s to v via u

How do we choose h?
0
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with length δ(s, u) + weight(u, v)

How do we choose h?
0
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with length δ(s, u) + weight(u, v)
so the shortest path can’t be longer

How do we choose h?
0
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0
Rearranging we have, weight(u, v) + δ(s, u) − δ(s, v) 0

How do we choose h?
0
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The new weight of an edge (u, v) then becomes

How do we choose h?
0
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0
= weight(u, v) + δ(s, u) − δ(s, v) 0

How do we choose h?
0
0
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0
= weight(u, v) + δ(s, u) − δ(s, v) 0
So all the reweighted edge weights are non-negative

JOHNSON’s algorithm
We can now piece together JOHNSON’s algorithm which operates as follows:

Step 1: Add one additional vertex called s to the original graph

Step 2: For each vertex, add an edge (s, v) with weight 0

Step 3: Run the BELLMAN-FORD algorithm with source s
- this calculates the shortest path lengths δ(s, v) for all v

if there is a negative weight cycle in the original graph, it will be detected here

Step 4: Reweight each edge (u, v) ∈ E so that,
where for all v, h(v) = δ(s, v)

Step 5: For each vertex u ∈ V , run DIJKSTRA’s algorithm with source s = u.
- this calculates the shortest path lengths δ (u, v) for all u, v

these are the shortest path lengths in the reweighted graph

Step 6: For each pair of vertices u, v ∈ V , compute
δ(u, v) = δ (u, v) + h(v) − h(u)

Step 6: For each pair of vertices u, v ∈ V , compute
δ(u, v) = δ (u, v) + h(v) − h(u)
these are the shortest path lengths in the original graph

Time Complexity
Step 4: Reweight each edge (u, v) so that,
Step 5: For each vertex u, run DIJKSTRA’s algorithm with source s = u.
Step 6: For each pair of vertices u, v, compute
δ(u, v) = δ (u, v) + h(v) − h(u)
How long does all this take?

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
How long does all this take? O(1) time

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time
O(|V ||E|) time

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time
O(|V ||E|) time
O(|E|) time

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time
O(|V ||E|) time
O(|E|) time
O(|E| log |V |) time per iteration
(using a binary heap)

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time
O(|V ||E|) time
O(|E|) time
O(|V ||E| log |V |) time overall

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time
O(|V ||E|) time
O(|E|) time
O(|V |2) time
(The previous version of this slide said O(|E|) for Step 6 which was a typo)

Time Complexity
δ(u, v) = δ (u, v) + h(v) − h(u)
O(|V |) time
O(|V ||E|) time
O(|E|) time
So the overall time complexity is O(|V ||E| log |V |)
O(|V |2) time
(The previous version of this slide said O(|E|) for Step 6 which was a typo)
This is the complexity for (strongly) connected graphs (actually any graph without isolated vertices). . .
In such graphs we have that |E| > |V |/2 so O(|V |2) = O(|V ||E|)
where every pair of
vertices is connected For unconnected graphs there is a (non-examinable and ﬁddly) ﬁx. . .
see next slide

Johnson’s algorithm on graphs with |E| < |V |/2
This slide contains non-examinable material and was added post-lecture.
It covers the corner case that Johnson’s algorithm is run on a graph with |E| < |V |/2.
The simple answer: In this case, the algorithm (as discussed) takes O(|V |2 + |V ||E| log |V |) time
We can’t “hide” the O(|V |2) term because |V | could be much much larger than |E|
A more convoluted answer: With a simple preprocessing step, we can improve the time complexity to
O(|V |2 + |V ||E| log |V |) for this caseO(|V |2 + |V ||E| log |V |) for this case
Sketch Proof
We deﬁne a vertex to be isolated if it contains no incoming and no outgoing edges.
(in the diagram the are isolated and the are not)
Add a new step to the start of Johnson’s algorithm,
Step 0: Delete every isolated vertex O(|V | + |E|) time
In the new graph |E | = |E|, |V | |V | and |E | |V |/2
Therefore the modiﬁed algorithm takes
O(|V | + |E| + |V ||E | log |V |) = O(|V ||E| log |V |) time
Notice that the isolated vertices have distince ∞ to everywhere so deleting them doesn’t change anything
each remaining vertex is at
one end of at least one edge

JOHNSON’s algorithm summary
Fact The length of a shortest path in the original graph
can be calculated from its length in the reweighted graph
The overall approach taken by JOHNSON’s algorithm
is to reweight the edges in the graph
The key properties of the reweighted graph are:
Fact If there are no negative weight cycles in the original graph,
(using new weights picked using BELLMAN-FORD)
After reweighting, all-pairs shortest paths are calculated
using DIJSKTRAS algorithm, once with each vertex v as the source
Overall this takes O(|V ||E| log |V |) time
−ve
all of the edges in the reweighted graph have non-negative weights
+/0

Shortest paths algorithms: the summary
unweighted: Use Breadth First Search in O(|V | + |E|) time
non-negative edge weights: Use DIJKSTRA’s algorithm
positive and negative edge weights: Use BELLMAN-FORD which takes O(|V ||E|) time
To compute single source shortest paths in a directed graph which is/has:
which takes O((|V | + |E|) log |V |) time (when implemented using a binary heap)
unweighted: Use Breadth First Search once for each vertex in O(|V |2 + |V ||E|) time
non-negative edge weights: Use DIJKSTRA’s algorithm once for each vertex,
positive and negative edge weights: Use JOHNSON’S algorithm
To compute all-pairs shortest paths in a directed graph which is/has:
which takes O(|V ||E| log |V |) time (when implemented using a binary heap)
(when DIJKSTRA’s algorithm is implemented using a binary heap)

Shortest Paths Part 2: Negative Weights and All-pairs

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Shortest Paths Part 2: Negative Weights and All-pairs