#=======================================================================
# Author: Isai Damier
# Title: Shortest Path with Constraint
# Project: geekviewpoint
# Package: algorithm.graph
#
# Statement:
# In http://www.geekviewpoint.com/python/graph/dijkstra_shortest_path
# we talked about finding the shortest path from my local airport in
# Silicon Valley to my parents' airport in Boston, MA. In the
# excitement over going to see my parents, I forgot a very important
# detail: the landing fee each airport charges. You see, not all
# airports charge the same toll. So I can't just blindly pick the
# shortest path from Silicon Valley to Boston. I must budget for tolls.
# So here is the real problem I actually needed to solve:
#
# In addition to knowing the distance between all pairs of airports
# that my small airplane could reach in one nonstop fight, I also
# know how much each airport would charge for toll. So given that
# I only have c dollars for toll, what is the shortest path from
# my airport to my parents' airport?
#
# Time-complexity: O((V+E) log V)
# where V is the number of airports (i.e. vertices/nodes) and E is
# the number of paths connecting all of the airports.
#
# Dijkstra Constrained:
#
# It is highly recommended that you read
# http://www.geekviewpoint.com/python/graph/dijkstra_shortest_path before
# proceeding.
#
# ... waiting ... waiting ... waiting ... waiting ... waiting ...
#
# Good. So you got a taste of how life can be when money is not an issue.
# Now to my problem: How do I factor in my financial limitations?
#
# - Instead of using a vector (i.e. 1D array) to track minDist, we must
# now use a matrix (i.e. 2D array), where the rows will represent the
# distance from s to the present airport and the columns will
# represent money left for the journey.
#
# Actual Algorithm:
#
# Note: we are still using s as my local airport and e as my parents'
# airport in Boston v can be any airport.
#
# 0] If I am at airport v, then I need to know how I got there and how
# far v is from s. Therefore, declare array pred to track the
# predecessor of v and array minDist to track the distance of v
# from s so that pred[v]=u would mean I got to v from airport u,
# and minDist[v][90]=100 would mean the distance between s and v is
# 100 miles and I have $90 left in my budget after paying v.
#
# 1] Since we have not done any computation yet, we will fill pred
# with -1s and we will fill minDist with infinities. Both meaning
# undefined.
#
# 2] Since we are starting at s, we can say that the distance from
# where we are to s is zero and that we have not yet spent any money.
# Therefore set minDist[s][c]=0.
#
# 3] Create a priorityQueue containing all vertices v, sorted by
# minDist so that vertex s will head the queue since it is the
# only vertex with a minDist of zero while all others are infinity.
#
# 4] While queue is not empty, remove the head vertex as v:
#
# 5] For each vertex x that is adjacent to v and that is still on the
# queue, if x would not take us over-budget, recalculate the
# distance from s to x. To know whether x would take us over-budget,
# get the money left after v, say m, and subtract the cost of x
# from m: m - cost[x] >= 0. You may want to manually run the
# algorithm for a small graph to see how it works.
#
# 6] Re-prioritize the queue based on the recalculated values of
# minDist and repeat thru the while-loop.
#
# After step 6 the computation is done. We are left with two arrays:
# pred and minDist. pred tells us the path from s to e, and minDist
# tells us the distance along the path. I only care about the path so
# I extract the path from pred. But you can do so much more as you wish.
#=======================================================================
from sys import maxint
from Queue import PriorityQueue
class ConstrainedShortestPath( object ):
def __init__( self ):
'''
Constructor
'''
def constrainedShortestPath( self, weight, cost, money, source, destination ):
# 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 for m in range( money + 1 )] for v in range( SIZE )]
# set minDist[source][money]=0, as source is 0 distance from itself.
minDist[source][money] = 0
pq = self.createPriorityQueue( money, minDist )
while not pq.empty():
self.updatePriorityQueue( pq )
# print "LEFT OVER: %s" % pq.queue
v = pq.get()[1]
for nodeData in self.adjacency( weight, pq, v ):
x = nodeData[1]
if x is None:
continue
# find how much landing at v actually cost.
# If you can't see this, run a small graph manually.
# Recall that the first v is actuall s with all the money intact.
m = 0
while m < money and minDist[v][m] == INFINITY:
m += 1
# triangle inequality
c = m - cost[x]
if c >= 0 and minDist[x][c] > minDist[v][m] + weight[v][x]:
minDist[x][c] = minDist[v][m] + weight[v][x]
nodeData[0] = minDist[x][c]
pred[x] = v
result = []
p = destination
while -1 != pred[p]:
result.append( p )
p = pred[p]
result.append( p )
result.reverse()
return result
def createPriorityQueue( self, money, minDist ):
pq = PriorityQueue( len( minDist ) )
for v in range( len( minDist ) ):
pq.put( [minDist[v][money], v] )
return pq
#===================================================================
# Retrieve all the neighbors of vertex v that are
# in the priority queue pq.
#===================================================================
def adjacency( self, G, pq, v ) :
result = []
for ent in pq.queue: # u,key[u] list( pq.queue )
u = ent[1]
if G[v][u] is not None:
result.append( ent )
return result
#=====================================================================
# Re-prioritize the queue based on changes in the
# minDist array.
#
# Technical Details: Dijktra's algorithm requires a priority queue
# that changes continuously to reflect changes in minDist.
# For python it does not suffice to simply pass new values to
# the array objects that constitute the queue. The
# PriorityQueue data structure in python is unaware of any
# direct changes to the objects it comprises. Therefore to force
# the queue to re-prioritize, an element is removed and then
# immediately added back.
#=====================================================================
def updatePriorityQueue( self, pq ) :
pq.put( pq.get() )
import unittest
from graph.ConstrainedShortestPath import ConstrainedShortestPath
class Test( unittest.TestCase ):
def testConstrainedShortestPath( self ):
dijkstra = ConstrainedShortestPath()
D = [
[None, 2, 2, None, None, None, None, None, None, None],
[None, None, None, 4, None, 16, None, None, None, None],
[None, None, None, None, 3, None, None, 8, None, None],
[None, None, None, None, None, None, 4, 4, None, None],
[None, None, None, None, None, None, None, 4, None, None],
[None, None, None, None, None, None, None, None, 13, None],
[None, None, None, None, None, None, None, None, None, 2],
[None, None, None, None, None, None, None, None, None, 6],
[None, None, None, None, None, None, None, None, None, 4],
[None, None, None, None, None, None, None, None, 5, None]]
S = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
money = 199
expected = [0, 1, 5, 8]
result = dijkstra.constrainedShortestPath( D, S, money, 0, 8 )
self.assertEquals( expected, result )
