#=======================================================================
# Author: Isai Damier
# Title: Dijkstra's Single Source Shortest Path
# Project: geekviewpoint
# Package: algorithm.graph
#
# Statement:
# I live in the Bay Area (a.k.a. Silicon Valley). But my parents live
# in Boston. Suppose I own a small airplane and know the distance
# between all pairs of airports that my plane could reach in one
# nonstop flight. How do I find the shortest path from my airport
# to my parent's 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.
#
# Greed as a Virtue:
#
# Oftentimes, the worse vices sublimate into the greatest virtues.
# As shown by the late Professor Edsger Wybe Dijkstra, the best known
# solution to this problem is a greedy algorithm. If we call my
# starting airport s and my ending airport e, then the intuition
# governing Dijkstra's "Single Source Shortest Path" algorithm
# goes like this:
#
# 0] To begin with, my probable itinerary T only contains one airport,
# s, since that's where I am starting from: T={s}
# 1] Then among all the airports that I could reach from s thru a
# nonstop flight, I pick the nearest one (the greedy choice), say
# v2. So now T={s,v2}. I also write down some notes next to v2:
# the distance from s to v2 and that I went directly from s to v2.
# 2] Then among all the airports that I could fly directly into from
# either s or v2, I pick the one nearest to s, say w3. Again, I
# take some notes next to w3: the distance from s to w3 and whether
# I reached w3 from v2 or from s. So now T={s,v2,w3}
# 3] Then I repeat the process until my probable itinerary contains e
# (i.e. Logan Airport in Boston). As soon as I reach e I can stop
# computing. Say T={s,v2,w3,w4,x5,y5,y6,z6,e}.
# 4] At this point, to trace out the actual path I am to take, I
# backtrack from e to s. Recall that I always took note of how I
# reached an airport. So if I reached e from z6, z6 from y5, y5
# from w4, w4 from w3, w3 from s, then to go from s to e I will
# travel s -> w3 -> w4 -> y5 -> z6 -> e as my shortest path.
#
# You probably noticed that v2 and x5 are not on the final path.
# That's okay. That's the price for being greedy. But on the bright
# side (if you can see it), I know the shortest distance from s to all
# the airports that's in T. Hence, not only do I now know the shortest
# distance from s to e, I also know the shortest distance from s to z6,
# from s to y5, from s to x5, etc.
#
# Hence in solving for the shortest distance from s to e,
# I inadvertently may have solved for the shortest distance to all
# the airports in America without any additional effort. I say "may"
# because you may choose to halt the search after reaching e. But in
# reality Dijkstra's algorithm tend to not supply e as a parameter;
# consequently, you normally solve for the distance from s to all
# the other vertices. Then you use the result as you wish; say to find
# find the shortest path to e. Again, there is no additional cost.
# Either way, time complexity is O((V+E) log V).
#
#
# Actual Algorithm:
#
# 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]=100 would mean the distance between s and v is 100.
#
# 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. Therefore set minDist[s]=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, recalculate the distance from s to x. Note that as long as
# a vertex is still in the queue it is effectively not in T
# (the probable itinerary), and that by removing a vertex v from the
# queue we are essentially adding it to T.
#
# 6] Re-prioritize the queue based on the recalculated values of minDist.
#
# 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.
#=======================================================================
from sys import maxint
from Queue import PriorityQueue
class DijkstraSingleSourceShortestPath( 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
pq = self.createPriorityQueue( minDist )
while not pq.empty():
self.updatePriorityQueue( pq )
v = pq.get()[1]
for XD in self.adjacency( weight, pq, v ):
x = XD[1]
# triangle inequality
if x is not None and minDist[x] > minDist[v] + weight[v][x]:
minDist[x] = minDist[v] + weight[v][x]
pred[x] = v
XD[0] = minDist[x] # update pq.
return [pred, minDist]
#=====================================================================
# Create a priority queue and load it with the vertices sorted by
# minDist.
#=====================================================================
def createPriorityQueue( self, dist ):
pq = PriorityQueue( len( dist ) )
for v in range( len( dist ) ):
pq.put( [dist[v], 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]
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 only checks its structure
# when it is adding or removing elements. It 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.DijkstraSingleSourceShortestPath import DijkstraSingleSourceShortestPath
class Test( unittest.TestCase ):
def testSingleSourceShortestPath( self ):
dijkstra = DijkstraSingleSourceShortestPath()
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 = dijkstra.singleSourceShortestPath( weight, source )
self.assertEquals( expResult, result )
weight = [
[None, 6, 8, 18, None, None],
[None, None, None, None, 11, None],
[None, None, None, 9, None, None],
[None, None, None, None, None, None],
[None, None, None, None, None, 3],
[None, None, 7, 4, None, None], ]
expResult = [[-1, 0, 0, 2, 1, 4], [0, 6, 8, 17, 17, 20]]
result = dijkstra.singleSourceShortestPath( weight, source )
self.assertEquals( expResult, result )
