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181 | def quadtree_gradient(C,W,CH,D,A):
"""Finite difference gradient matrix on a quadtree
Builds a finite difference gradient on a quadtree following a centered finite difference scheme, with the adjacency as suggested by Bickel et al. "Adaptative Simulation of Electrical Discharges".
Parameters
----------
C : numpy double array
Matrix of cell centers
W : numpy double array
Vector of cell half widths
CH : numpy int array
Matrix of child indeces (-1 if leaf node)
D : numpy int array
Vector of tree depths
A : scipy sparse.csr_matrix
Sparse node adjacency matrix, where a value of a in the (i,j) entry means that node j is to the a-th direction of i (a=1: left; a=2: right; a=3: bottom; a=4: top).
Returns
-------
G : scipy sparse.csr_matrix
sparse gradient matrix (first num_children rows are x derivatives, last are y derivatives)
stored_at : numpy double array
Matrix of child cell centers, where the values of G are stored
See also
--------
quadtree_laplacian, initialize_quadtree.
Notes
-----
This code is *purposefully* not optimized beyond asymptotics for simplicity in understanding its functionality and translating it to other programming languages beyond prototyping.
Examples
--------
```python
# Create a random point cloud
P = 2*np.random.rand(100,2) - 1
# Initialize the quadtree
C,W,CH,PAR,D,A = gpytoolbox.initialize_quadtree(P,graded=True,max_depth=8)
# Get the gradient matrix
G, stored_at = gpytoolbox.quadtree_gradient(C,W,CH,D,A)
```
"""
# Builds a finite difference gradient on a quadtree following a centered
# finite difference scheme, with the adjacency as suggested by
# Bickel et al. "Adaptative Simulation of Electrical
# Discharges". This code is *purposefully* not optimized beyond
# asymptotics for simplicity in understanding its functionality and
# translating it to other programming languages beyond prototyping.
#
# G = quadtree_gradient(C,W,CH,D,A)
# G,stored_at = quadtree_gradient(C,W,CH,D,A)
#
# Inputs:
# C #nodes by 3 matrix of cell centers
# W #nodes vector of cell widths (**not** half widths)
# CH #nodes by 4 matrix of child indeces (-1 if leaf node)
# D #nodes vector of tree depths
# A #nodes by #nodes sparse adjacency matrix, where a value of a in the
# (i,j) entry means that node j is to the a-th direction of i
# (a=1: left a=2: right a=3: bottom a=4: top).
#
# Outputs:
# G #2*num_children by #num_children sparse gradient matrix (first
# num_children rows are x derivatives, last are y derivatives)
# stored_at #num_children by 3 matrix of child cell centers, where the
# values of G are stored
# We will store Laplacian values at
# child cell indeces
children = np.nonzero(CH[:,1]==-1)[0]
# map from all cells to children
cell_to_children = -np.ones(W.shape[0],dtype=int)
cell_to_children[children] = np.linspace(0,children.shape[0]-1,children.shape[0],dtype=int)
# Vectors for constructing the Laplacian
I = []
J = []
vals = []
for i in range(children.shape[0]):
new_I = []
new_J = []
new_vals = []
l = [0,0,0,0,0]
new_dirs = []
child = children[i]
d = D[child]
num_dirs = 0
# Let's build d u(child)/dx^2 ~ u(child+W(child)*[1,0])/hr(hl+hr) -
# 2u(child)/hlhr + u(child-W(child)*[1,0])/hr(hl+hr)
# So, let's look for the value to the j direction. To do this, we seek the
# lowest-depth neighbor to the j direction. As a reminder the octree
# adjacency convention is i->j (1:left-2:right-3:bottom-4:top)
for j in range(1,5):
j_neighbors = (A[child,:]==j).nonzero()[1]
if len(j_neighbors)>0:
depths_j_neighbors = D[j_neighbors]
max_depth_j_neighbor = np.argmax(depths_j_neighbors)
max_depth_j = depths_j_neighbors[max_depth_j_neighbor]
max_depth_j_neighbor = j_neighbors[max_depth_j_neighbor]
# There are two options:
# One: the leaf node to our j direction has lower or equal depth to
# us
if max_depth_j<=d:
l[j] = (W[child] + W[max_depth_j_neighbor])/2.0
# then it's easy, just add this node
new_I.append(i)
# THIS HAS TO BE A CHILD !
assert(cell_to_children[max_depth_j_neighbor]>=0)
new_J.append(cell_to_children[max_depth_j_neighbor])
new_vals.append(1.0)
new_dirs.append(j)
else:
# In this case, assuming the grid is graded, there should
# be two j-neighbors at depth d+1
nn = j_neighbors[D[j_neighbors]==(d+1)]
assert len(nn)==2, "Are you sure you are inputting a graded quadtree?"
assert all(CH[nn,1]==-1)
# Then we simply average both
l[j] = (W[child] + W[nn[1]])/2.0
new_I.append(i)
new_I.append(i)
new_J.append(cell_to_children[nn[0]])
new_J.append(cell_to_children[nn[1]])
new_vals.append(0.5)
new_vals.append(0.5)
new_dirs.append(j)
new_dirs.append(j)
num_dirs = num_dirs + 1
# This is a cheeky way to identify corners and make the stencil
# backwards-forwards instead of centered in these cases
for j in range(1,5):
if l[j]==0:
new_I.append(i)
new_J.append(i)
new_vals.append(1.0)
new_dirs.append(j)
# print("Before")
# print(new_I)
# print(new_J)
# print(new_vals)
# print(new_dirs)
# At this point, we have to divide by the edge-lengths and add sign
for s in range(len(new_dirs)):
if new_dirs[s]==1:
new_vals[s] = -new_vals[s]/(l[1]+l[2])
elif new_dirs[s]==2:
new_vals[s] = new_vals[s]/(l[1]+l[2])
elif new_dirs[s]==3:
new_vals[s] = -new_vals[s]/(l[3]+l[4])
# These are the y derivatives so they go in the lower block
new_I[s] = new_I[s] + children.shape[0]
elif new_dirs[s]==4:
new_vals[s] = new_vals[s]/(l[3]+l[4])
# These are the y derivatives so they go in the lower block
new_I[s] = new_I[s] + children.shape[0]
# print("After")
# print(new_I)
# print(new_J)
# print(new_vals)
# print(new_dirs)
# And add them to the big sparse Laplacian construction vectors
I.extend(new_I)
J.extend(new_J)
vals.extend(new_vals)
G = csr_matrix((vals,(I,J)),(2*children.shape[0],children.shape[0]))
stored_at = C[children,:]
return G, stored_at
|