1. Shortest Path Problems
Bellman-Ford Algorithm for the Single-Source Shortest Path Problem with Arbitrary Arc Costs
Updated 18 February 2008

2. Bellman-Ford Algorithm
begin
d(i) :=  for each node i in N
d(s) := 0 and pred(s) := 0;
for k = 1 to n
for each (i, j) in A do
if d(j) > d(i) + cij then
d(j) := d(i) + cij and pred(j) := i
end;

3. Bellman-Ford Example 1  2 2 1 4 3 -2  

4. dk(i) denotes d(i) after iteration k
Bellman-Ford Example 1 -1   2
k =1 1 2
d1(1) = 0
d1(2) = -1
d1(3) = 4
d1(4) = 1 -2 1 1 3 4 3 4  1   
dk(i) denotes d(i) after iteration k

5. Bellman-Ford Example 1 -1 k = 2 2 1 2 d2(1)=0 d2(2)=-1 d2(3)=0 d2(4)=1
k = 2 2 1 2
d2(1)=0
d2(2)=-1
d2(3)=0
d2(4)=1 -2 1 1 3 4 3 1  4

6. Bellman-Ford Example 1 -1 k =3 2 1 2 d3(1)=0 d3(2)=-1 d3(3)=0 d3(4)=1
k =3 2 1 2
d3(1)=0
d3(2)=-1
d3(3)=0
d3(4)=1 -2 1 1 3 4 3
d3(i)= d2(i) 1

7. Shortest Path Tree d(j)  d(i) + cij and d(j) = d(i) + cij-1 2 1 2
d(j)  d(i) + cij
and
d(j) = d(i) + cij
for all predecessor arcs -2 1 1 3 4 3 1

8. Complexity of Bellman-Ford
There are n iterations
Each iteration inspects each arc once
Inspecting an arc is O(1)
Complexity is O(mn)

9. Bellman-Ford Example 2 
100 1 2 10
200
-150
-150 4 3  

10. Bellman-Ford Example 2 100   100 k = 1 1 2 d1(1) = 0 d1(2) = 100
100
k = 1 1 2
d1(1) = 0
d1(2) = 100
d1(3) = 
d1(4) = 200 10
-150
200
-150 4 3  
200 

11. Bellman-Ford Example 2 -100 100 100 100 k = 2 1 2 d2(1) = 0
100
k = 2 1 2
d2(1) = 0
d2(2) = -100
d2(3) = 50
d2(4) = -90 10
-150
200
-150 4 3 
100
200
-90 50
100

12. Bellman-Ford Example 2 -100 100 k = 3 1 2 d3(1) = 0 d3(2) = -100
100
k = 3 1 2
d3(1) = 0
d3(2) = -100
d3(3) = -240
d3(4) = -90 10
-150
200
-150 4 3
-90
-240
100 50

13. Bellman-Ford Example 2 -100 -390 100 k = 4 1 2 10 d4(1) = 0 -150 200
-390
100
k = 4 1 2 10
d4(1) = 0
-150
200
d4(2) = -390
d4(3) = -240
-150 4 3
d4(4) = -90
-90
-240
100
d3(2) = -100

14. Detecting Negative-Cost Cycles
If there is a negative-cost cycle in G, then dn(i) < dn-1(i) for some node i.

15. Detecting Negative-Cost Cycles
dn-1(1)
dn-1(2) 1 2
dn(4)  dn-1(1) + c14
dn(1)  dn-1(2) + c21
dn(2)  dn-1(3) + c32 4 3
dn(3)  dn-1(4) + c43
dn-1(4)
dn-1(3)

16. Detecting Negative-Cost Cycles
If the length of the cycle < 0 then dn(i) < dn-1(i)
for at least one node i in the cycle.