# Bellman-Ford's Shortest Pathby Isai Damier, Android Engineer @ Google

```#=======================================================================
# Author: Isai Damier
# Title:  Bellman-Ford's Single Source Shortest Path
# Project: geekviewpoint
# Package: algorithm.graph
#
# Statement:
#   Given a source vertex and a directed, weighted graph; find the
#   shortest path between the source vertex and each of the other
#   vertices. Some of the edges may have negative weights. If the graph
#   contains any negative weight cycle, throw an exception.
#
# Time-complexity: O(V*E)
#   where V is the number of vertices and E is the number of edges.
#
# The Bellman-Ford Strategy:
#
#   The Bellman-Ford argument is that the longest path in any graph
#   can have at most V-1 edges, where V is the number of vertices.
#   Furthermore, if we perform relaxation on the set of edges once, then
#   we will at least have determined all the one-edged shortest paths;
#   if we traverse the set of edges twice, we will have solved at least
#   all the two-edged shortest paths; ergo, after the V-1 iteration thru
#   the set of edges, we will have determined all the shortest paths
#   in the graph.
#
#   But then Bellman-Ford continues: If after V-1 iterations we are still
#   able to relax any path, then the graph has a negative edge cycle.
#
#   From that argument, even before looking at any code, we can see
#   that the time complexity is (V-1)*E +1*E = V*E. The 1*E is the
#   cycle-detection pass.
#
#   The argument also implicitly tells us that this algorithm does not
#   mind handling negatives edges -- only negative cycles are scary.
#
#   Hence, while Bellman-Ford takes longer than all versions of
#   Dijkstra's algorithm, it has the advantage of accepting all
#   directed graphs.
#
#   I use the term relaxation a few times. Here is the explanation.
#   The computation starts by estimating that all vertices are infinitely
#   far away from the source vertex. Then as we compute, we relax that
#   outrageous assumption by checking if there is actually a shorter
#   distance. For instance, say we have already computed the shortest
#   distance from s to v as sv and now we are seeking the solution from
#   s to x, where x is adjacent to v. Then we know sx <= sv + vx; which
#   simply means, either the path from s to v to x is the shortest path
#   or there is yet a shorter path than that. If however we find that
#   our "outrageous" estimate so far is that sx > sv+vx, then we can
#   set sx = sv+vx as our new better estimate. That's it. That's
#   relaxation.
#=======================================================================

from sys import maxint
class BellmanFord( object ):

def __init__( self ):
'''
Constructor
'''

def singleSourceShortestPath( self, weight, source ) :
# auxiliary constants
SIZE = len( weight )
EVE = -1; # to indicate no predecessor
INFINITY = maxint

# declare and initialize pred to EVE and minDist to INFINITY
pred = [EVE] * SIZE
minDist = [INFINITY] * SIZE

# set minDist[source] = 0 because source is 0 distance from itself.
minDist[source] = 0

# relax the edge set V-1 times to find all shortest paths
for i in range( 1, SIZE - 1 ):
for v in range( SIZE ):
for x in self.adjacency( weight, v ):
if minDist[x] > minDist[v] + weight[v][x]:
minDist[x] = minDist[v] + weight[v][x]
pred[x] = v

# detect cycles if any
for v in range( SIZE ):
for x in self.adjacency( weight, v ):
if minDist[x] > minDist[v] + weight[v][x]:
raise Exception( "Negative cycle found" )

return [pred, minDist]

#=====================================================================
# Retrieve all the neighbors of vertex v.
#=====================================================================
def adjacency( self, G, v ) :
result = []
for x in range( len( G ) ):
if G[v][x] is not None:
result.append( x )

return result;
```
```import unittest
from graph.BellmanFord import BellmanFord

class Test( unittest.TestCase ):

def testBellmanFordWithPositiveEdges( self ):
fordAlgorithm = BellmanFord()
weight = [
[None, 10, None, None, 3],
[None, None, 2, None, 1],
[None, None, None, 7, None],
[None, None, 9, None, None],
[None, 4, 8, 2, None]
]
source = 0
expResult = [[-1, 4, 1, 4, 0], [0, 7, 9, 5, 3]]
result = fordAlgorithm.singleSourceShortestPath( weight, source )
self.assertEquals( expResult, result )

def testBellmanFordWithNegativeEdges( self ):
fordAlgorithm = BellmanFord()
weight = [
[None, -1, 4, None, None],
[None, None, 3, 2, 2],
[None, None, None, None, None],
[None, 1, 5, None, None],
[None, None, None, -3, None]
]
source = 0
expResult = [[-1, 0, 1, 4, 1], [0, -1, 2, -2, 1]]
result = fordAlgorithm.singleSourceShortestPath( weight, source )
self.assertEquals( expResult, result )

def testBellmanFordWithNegativeCycle( self ):
fordAlgorithm = BellmanFord()
weight = [
[None, -1, 4, None, None],
[None, None, 3, 2, 2],
[None, -6, None, None, None],
[None, 1, 5, None, None],
[None, None, None, -3, None]
]
source = 0
try :
fordAlgorithm.singleSourceShortestPath( weight, source )
except:
pass
else:
self.fail( "Should have thrown an exception: Negative cycle" )
```