def labelMobileDet( M, odd, parent ):
'''
labelMobileDet( M, odd, parent )
In: M is a mobile.
odd and parent are vertices in M.
{odd,parent} is an edge in M.
odd is of an odd generation (and thus,
parent is of even generation).
parent has already been given a label.
Out: All descendants of odd of even generation
have been given a label by the (-1)-rule.
'''
# The labels are determined by the label of odd's parent.
parentLabel = M.node[parent][ 'label' ]
# The children to be labeled.
children = graphUtil.childrenOf( odd, M, parent )
# Make sure that the children get the labels in the correct order.
children.sort()
# Label odd's children.
for i in range(len( children )):
v = children[i]
# The next vertex gets a label that is
# one less than the preceding vertex.
M.node[v][ 'label' ] = parentLabel - i - 1
# Find all children of v (they are of an odd generation)
# and label them.
childrenOf_v = graphUtil.childrenOf( v, M, odd )
for child_v in childrenOf_v:
labelMobileDet( M, child_v, v )
def relabelTree( T, root ): ''' S = relabelTree( T, root ) In: T is a tree with vertices 0, 1, ..., n-1 and root is a vertex in T. Each vertex in T has an integer 'age' attribute such that siblings have different ages. We take siblings to be drawn in ascending order by age from left to right in the plane. Out: S is a new tree that is just like T, except the vertices have been relabeled to 0, 1, ..., n-1 such that the root is 0 and for each vertex its children are in ascending order from left to right, all vertices in generation <=k are smaller than all vertices in generation >k (for all k). ''' n = nx.number_of_nodes( T ) # Dictionary for the new labels. newLabels = dict(zip( range(n), range(n) )) nextLabel = 0 Q = Queue.Queue( maxsize = n ) Q.put( ( root, None ) ) while not Q.empty(): # We have found labels for all the nextLabel vertices # that are not descendants of the vertices in the queue Q. # The next vertex, node, gets the next number. node, parent = Q.get() newLabels[node] = nextLabel nextLabel = nextLabel + 1 # Put each child of node on the queue in ascending order by age. children = graphUtil.childrenOf( node, T, parent ) children.sort( key = lambda child: T.node[child][ 'age' ] ) for child in children: Q.put( ( child, node ) ) # Relabel the vertices according to the dictionary. return nx.relabel_nodes( T, newLabels )
def findContourSequence( contSeq, T, node, parent ): ''' findContourSequence( contSeq, T, node, parent ) In: contSeq is a list T is a tree such that for each node, its children appear in ascending order from left to right in the plane. node and parent are nodes in T, except if parent == None, in which case node is the root of T. Out: node and all its descendants (assuming parent is the parent of node) have been added to contSeq (with repetition) in the order they appear in a clockwise walk around the subtree induced by node and its descendants. ''' contSeq.append( node ) children = graphUtil.childrenOf( node, T, parent ) # Ensure that the children are visited in the correct order. children.sort() # Recursively finish the objective for each child of node. for child in children: findContourSequence( contSeq, T, child, node ) # If node is not the root, walk back to its parent. if not parent is None: contSeq.append( parent )
def findContourSequence(contSeq, T, node, parent):
'''
findContourSequence( contSeq, T, node, parent )
In: contSeq is a list
T is a tree such that for each node, its
children appear in ascending order from left
to right in the plane.
node and parent are nodes in T,
except if parent == None, in which case node
is the root of T.
Out: node and all its descendants (assuming parent is
the parent of node) have been added to contSeq
(with repetition) in the order they appear in a
clockwise walk around the subtree induced by node
and its descendants.
'''
contSeq.append(node)
children = graphUtil.childrenOf(node, T, parent)
# Ensure that the children are visited in the correct order.
children.sort()
# Recursively finish the objective for each child of node.
for child in children:
findContourSequence(contSeq, T, child, node)
# If node is not the root, walk back to its parent.
if not parent is None:
contSeq.append(parent)
def labelMobileRand( M, odd, parent ):
'''
labelMobileRand( M, odd, parent )
In: M is a mobile.
odd and parent are vertices in M.
{odd,parent} is an edge in M.
odd is of an odd generation (and thus,
parent is of even generation).
parent has already been given a label.
Out: All descendants of odd of even generation
have been given a random label.
'''
# The children to be labeled.
children = graphUtil.childrenOf( odd, M, parent )
# Make sure that the children get the labels in the correct order.
children.sort()
n = len(children)
if n == 0:
return
# Make use of the multinomial method to get labels
# such that the differences of nearby labels are
# i.i.d.
# Maximum number of trials:
maxiter = 999
# Probability distribution of jumps + 1:
xi = lambda k: 0.5 ** ( k + 1. )
trials = 0
while True:
N = graphUtil.multinomial( n + 1, xi )
K = len(N)
trials += 1
# If sum( (j-1)*N[j] ) == 0, then we are done.
# If not, try again.
if ( ( np.arange(K) - 1 ) * N ).sum() == 0:
break
if trials >= maxiter:
raise RuntimeError('Maximum number of trials reached' +\
' (%d). ' % maxiter +\
'Try again or choose another' +\
'probability distribution.')
# Create a vector with N[j] copies of j
# for j = 0, ..., K-1 and permute it randomly.
Jumps = np.concatenate([ ( j - 1 ) * np.ones( N[j] ) for j in range(K) ])
Jumps = np.random.permutation( Jumps ).astype( int )
# Now, Jumps is a vector of n + 1 independent random variables
# distributed by f(k) = 0.5^(k+2) for k >= -1 and sum(Jumps) == 0.
# The labels are determined by the label of odd's parent.
currentLabel = M.node[parent][ 'label' ]
# Label odd's children.
for i in range(len( children )):
v = children[i]
currentLabel = currentLabel + Jumps[i]
M.node[v][ 'label' ] = currentLabel
# Find all children of v (they are of an odd generation)
# and label them.
childrenOf_v = graphUtil.childrenOf( v, M, odd )
for child_v in childrenOf_v:
labelMobileRand( M, child_v, v )