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```''' This program uses the stepping stone algorithum to solve
the transhipment problem. That is how to transport various quuantities
of material to various destinations minimising overall cost, given
the various costs of sending a unit from each source to each destination.
The sum of supply and demand must equal.'''

def PrintOut():
GetDual()
nCost = 0
print
print '    DEMAND' + ' ' * ( m * 10) + 'SUPPLY'
print '%10i' % y,
print
for x in range( n):
for y in range( m):
nCost += aCost[ x][ y] * aRoute[ x][ y]
if aRoute[ x][ y] == 0:
print '[<%2i>%4i]' %( aCost[ x][ y], aDual[ x][ y]),
else:
print '[<%2i>(%2i)]' %( aCost[ x][ y], aRoute[ x][ y] + 0.5),
print ' : %i' % aSupply[ x]
print 'Cost: ', nCost
print 'Press ENTER to continue'
raw_input()

def NorthWest():
''' The simplest method to get an initial solution.
Not the most efficient'''
global aRoute
u  = 0
v  = 0
aS = [ 0] * m
aD = [ 0] * n
while u <= n - 1 and v <= m - 1:
if aDemand[ v] - aS[ v] < aSupply[ u] - aD[ u]:
z              = aDemand[ v] - aS[ v]
aRoute[ u][ v] = z
aS[ v]        += z
v             += 1
else:
z              = aSupply[ u] - aD[ u]
aRoute[ u][ v] = z
aS[ v]        += z
u             += 1

def NotOptimal():
global PivotN
global PivotM
nMax = -nVeryLargeNumber
GetDual()
for u in range( 0, n):
for v in range( 0, m):
if x > nMax:
nMax = x
PivotN = u
PivotM = v
return ( nMax > 0)

def GetDual():
for u in range( 0, n):
for v in range( 0, m):
aDual[ u][ v] = -0.5 # null value
if aRoute[ u][ v] == 0:
aPath = FindPath( u, v)
z     = -1
x     = 0
for w in aPath:
x += z * aCost[ w[ 0]][ w[ 1]]
z *= -1

def FindPath( u, v):
aPath = [[ u, v]]
if not LookHorizontaly( aPath, u, v, u, v):
print 'Path error, press key', u, v
raw_input()
return aPath

def LookHorizontaly( aPath, u, v, u1, v1):
for i in range( 0, m):
if i != v and aRoute[ u][ i] != 0:
if i == v1:
aPath.append( [ u, i])
return True # complete circuit
if LookVerticaly( aPath, u, i, u1, v1):
aPath.append( [ u, i])
return True

def LookVerticaly( aPath, u, v, u1, v1):
for i in range( 0, n):
if i != u and aRoute[ i][ v] != 0:
if LookHorizontaly( aPath, i, v, u1, v1):
aPath.append([ i, v])
return True

def BetterOptimal():
global aRoute
aPath = FindPath( PivotN, PivotM)
nMin  = nVeryLargeNumber
for w in range( 1, len( aPath), 2):
t = aRoute[ aPath[ w][ 0]][ aPath[ w][ 1]]
if t < nMin:
nMin = t
for w in range( 1 , len( aPath), 2):
aRoute[ aPath[ w][ 0]][ aPath[ w][ 1]]         -= nMin
aRoute[ aPath[ w - 1][ 0]][ aPath[ w - 1][ 1]] += nMin

# example 1
aCost = [[ 2, 1, 3, 3, 2, 5]
,[ 3, 2, 2, 4, 3, 4]
,[ 3, 5, 4, 2, 4, 1]
,[ 4, 2, 2, 1, 2, 2]]

aDemand = [ 30, 50, 20, 40, 30, 11]
aSupply = [ 50, 40, 60, 31]

''' example 2
aCost = [[ 1, 2, 1, 4, 5, 2]
,[ 3, 3, 2, 1, 4, 3]
,[ 4, 2, 5, 9, 6, 2]
,[ 3, 1, 7, 3, 4, 6]]
aDemand = [ 20, 40, 30, 10, 50, 25]
aSupply = [ 30, 50, 75, 20]
'''
''' example3
aCost = [[ 5, 3, 6, 2]
,[ 4, 7, 9, 1]
,[ 3, 4, 7, 5]]
aDemand = [ 16, 18, 30, 25]
aSupply = [ 19, 37, 34]
'''
n = len( aSupply)
nVeryLargeNumber = 99999999999
# add a small amount to prevent degeneracy
# degeneracy can occur when the sums of subsets of supply and demand equal
elipsis = 0.001
k += elipsis / len( aDemand)
aSupply[ 1] += elipsis
# initialisation
aRoute = []
for x in range( n):
aRoute.append( [ 0] * m)
for x in range( n):
NorthWest()
PivotN = -1
PivotM = -1
PrintOut()
# MAIN
while NotOptimal():
print 'PIVOTING ON', PivotN, PivotM
BetterOptimal()
PrintOut()
print "FINISHED"
```