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diff_rrt.py
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diff_rrt.py
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import numpy as np
import matplotlib.pyplot as plt
import matplotlib.animation as animation
import matplotlib as mpl
import networkx as nx
import itertools
from lqr_tools import LQR_QP, dtfh_lqr, simulate_lti_fb_dt, AQR, final_value_LQR, lqr_dim
class Diff_RRT():
def __init__(self,linA,linB,dyn_f,cost_0,cost_1,cost_2,n,m,max_time_horizon):
(self.n,self.m) = (n,m) #n+1 for affine term
self.max_time_horizon = max_time_horizon
self.linA = linA
self.linB = linB
self.dyn_f = dyn_f
self.cost_zeroth_ord = cost_0
self.cost_first_ord = cost_1
self.cost_second_ord = cost_2
def action_state_valid(x,u):
return True
self.action_state_valid = action_state_valid
self.max_nodes_per_extension = None
self.max_steer_cost = np.inf
self.u0 = np.zeros(self.m)
def get_ABc(self,x,u):
"""
x[k+1] = A(x[k] - x0) + B(u[k] - u0) + f(x0,u)
x[k+1] = A*x[k] + B*u[k] + -A*x0 -B*u0 + f(x0,u)
"""
A = self.linA(x,u)
B = self.linB(x,u)
c = self.dyn_f(x,u).reshape(self.n,1) - np.dot(A,x.reshape(self.n,1)) - np.dot(B,u.reshape(self.m,1))
A = np.matrix(A)
B = np.matrix(B)
c = np.matrix(c)
assert A.shape == (self.n,self.n)
assert B.shape == (self.n,self.m)
assert c.shape == (self.n,1)
return A,B,c
def get_QRqrd(self,x,u):
n,m = self.n, self.m
d = self.cost_zeroth_ord(x,u)
_qr = self.cost_first_ord(x,u)
q = _qr[0,0:n].reshape(n,1)
r = _qr[0,n:n+m].reshape(m,1)
_QR = self.cost_second_ord(x,u)
#ignoring cross terms between x and u #FIXME
Q = _QR[0:n,0:n]
R = _QR[n:n+m,n:n+m]
assert Q.shape == (n,n)
assert R.shape == (m,m)
return Q,R,q,r,d
def run_forward(self,x0,us):
n = self.n+1 #+1 for time dimension
m = self.m
#us = np.reshape(us,newshape=(-1,m))
assert len(x0) == n
T = us.shape[0]
xs = np.zeros(shape=(T+1,n))
xs[0] = np.squeeze(x0)
for i in range(1,T+1):
xs[i,0:self.n] = self.dyn_f(xs[i-1,0:self.n].T,us[i-1].T).reshape((1,self.n))
xs[i,self.n] = xs[i-1,self.n] + 1 #increase time
return xs[1:]
def run_forward_fb(self,x0,gain_schedule):
raise NotImplementedError
A = self.A
B = self.B
n = self.n+1 #+1 for time dimension
m = self.m
assert len(x0) == n
T = gain_schedule.shape[0]
assert gain_schedule.shape[2] == n-1 #no time in the gain
assert gain_schedule.shape[1] == m
xs = np.zeros(shape=(T+1,n))
us = np.zeros(shape=(T,m))
xs[0] = np.squeeze(x0)
for i in range(1,T+1):
us[i-1] = -1* np.dot(gain_schedule[i-1],xs[i-1,0:self.n])
xs[i,0:self.n] = np.dot(A,xs[i-1,0:self.n].T) + np.dot(B,us[i-1].T)
xs[i,self.n] = xs[i-1,self.n] + 1
return xs[1:],us
def collision_free(self,from_node,action):
"""
check that taking action from from_node produces a collision free trajectory
if not, return a partial trajectory for the state (x_path) and control (u_path)
u_path is a list of actions -- it partitions the actions.
"""
x0 = from_node['state']
x_path = []
u_path = []
all_the_way = True
if len(action) > 0:
if self.action_state_valid(x0,action[0]):
x_path_np = self.run_forward(x0,action)
for i in range(len(x_path_np)):
if not self.action_state_valid(x_path_np[i],action[i]):
all_the_way = False
break
x_path.append(x_path_np[i])
u_path.append(action[[i]])
else:
all_the_way = False
u_path_all = np.zeros(shape=(len(u_path),self.m))
#don't return any intermediate points
if False:
for i in range(len(u_path)):
u_path_all[i] = u_path[i][0]
if len(x_path)>0:
return [x_path[-1]], [u_path_all], all_the_way
else:
return [], [], all_the_way
#downsample
if self.max_nodes_per_extension is not None and len(x_path) > self.max_nodes_per_extension:
x_path_ds = []
u_path_ds = []
#getting indices correct here was a pain.
inds = np.array(np.round(np.linspace(0,len(x_path),self.max_nodes_per_extension)),dtype=np.int)
for (i,j) in zip(inds[0:-1],inds[1:]):
x_path_ds.append(x_path[j-1])
l = j-i
action_part = np.zeros(shape=(l,self.m))
for k in range(l):
action_part[k] = u_path[i+k][0]
u_path_ds.append(action_part)
assert len(u_path) == sum( [len(action_part) for action_part in u_path] )
x_path = x_path_ds
u_path = u_path_ds
#u_path = np.array(u_path)
return x_path, u_path, all_the_way
def same_state(self,a,b):
return a[self.n] == b[self.n] and np.allclose(a,b,atol=1e-4) #time has to be identical and phase-space state has to be approximate
def cost(self,x_from,action):
#this does not include the cost of being in at the state arrived by taking the last action
#that is, with x_(k+1) = f(x_k,u_k) this sums cost(x_k,u_k) for k in range(len(action))
assert len(x_from) == self.n+1
if len(action) == 0:
#null action
return 0 #is captured in dynamics below, but if the action is actually null, then the next assertion fails
assert action.shape[1] == self.m
x_path = self.run_forward(x_from,action)
cost = 0
for i in range(action.shape[0]):
#x_path does not include x_from
x = x_path[[i-1],0:self.n].T if i>0 else x_from[0:self.n] #don't include time
u = action[[i],:].T
cost += np.squeeze( self.cost_zeroth_ord(x,u) )
return cost
def node_cache_ctg(self,node):
A,B,c = self.get_ABc(node['state'][0:self.n],self.u0)
Q,R,q,r,d = self.get_QRqrd(node['state'][0:self.n],self.u0)
max_time_horizon = self.max_time_horizon
#print 'calculate ctg for', node
x = node['state']
#reverse system
Ar = A.I
Br = -A.I * B
cr = -A.I * c
kmax = max_time_horizon - x[self.n] +1
assert kmax > 0
#ctg[0] is cost-to-go zero steps -- very sharp quadratic
#so ctg[k] with k = max_time_horizon - from_node['state'][self.n] is time to go
Fs, Ps = final_value_LQR(Ar,Br,Q,R,x[0:self.n],kmax,c=cr,q=q,r=r,d=d)
#storing in reverse order is easier to think about.
#node['gain'][i] is what you should do with i steps left to go.
node['ctg'] = Ps[::-1]
node['gain'] = Fs[::-1]
node['dynamics'] = (A,B,c)
node['reverse_dynamics'] = (Ar,Br,cr)
def steer(self,x_from_node,x_toward,cost_limit=True):
A,B,c = self.get_ABc(x_from_node['state'][0:self.n],self.u0)
Q,R,q,r,d = self.get_QRqrd(x_from_node['state'][0:self.n],self.u0)
x_from = x_from_node['state']
assert len(x_from) == self.n+1
T = x_toward[self.n] - x_from[self.n] #how much time to do the steering
assert T-int(T) == 0 #discrete time
T=int(T)
if T<=0:
return (x_from,np.zeros(shape=(0,self.m))) #stay here
desired = np.matrix(x_toward[0:self.n].reshape(self.n,1))
Qf = np.eye(self.n) * 1e8
qf = -np.dot(Qf,desired)
Qhf = np.zeros(shape=(self.n+1,self.n+1))
Qhf[0:self.n,0:self.n] = Qf
Qhf[0:self.n,[self.n]] = qf
Qhf[[self.n],0:self.n] = qf.T
Qhf[self.n,self.n] = np.dot(desired.T,np.dot(Qf,desired))
(Ah,Bh,Qh,Rh,pk) = AQR( A=A,
B=B,
c=c,
Q=Q,
R=R,
q=q,
r=r,
d=d,
ctdt='dt')
Fs, Ps = dtfh_lqr(A=Ah,B=Bh,Q=Qh,R=R,N=T,Q_terminal=Qhf)
#print Fs
xs = np.zeros(shape=(T+1,self.n+1))
us = np.zeros(shape=(T,self.m))
xs[0] = x_from
cost = 0
for i in range(T):
us[i] = -1 * (np.dot(Fs[i,:,0:self.n],xs[i,0:self.n]) + Fs[i,:,self.n]) + pk #FIXME
#xs[i+1,0:self.n] = np.dot(A,xs[i,0:self.n].T) + np.dot(B,us[i].T) + c #pretend the system is linear
#xs[i+1,self.n] = xs[i,self.n] + 1
xs[i+1] = self.run_forward(xs[i], us[[i]] )
cost += self.cost(xs[i].T,us[i].reshape(1,self.m))
if cost_limit and cost > self.max_steer_cost:
break
if i < T-1:
us = us[:i]
x_final = xs[i+1]
else:
x_final = xs[T]
return (x_final, us)
def steer_cache(self,x_from_node,x_toward,cost_limit=True):
A,B,c = self.get_ABc(x_from_node['state'][0:self.n],self.u0)
x_from = x_from_node['state']
assert len(x_from) == self.n+1
T = x_toward[self.n] - x_from[self.n] #how much time to do the steering
assert T-int(T) == 0 #discrete time
T=int(T)
if T <= 0:
return (x_from,np.zeros(shape=(0,self.m))) #stay here
if T < 10:
#this technique isn't too accurate for short times due to slack in the final-value constraint
#so do something else.
return self.steer(x_from_node,x_toward,cost_limit)
desired = np.matrix(x_toward[0:self.n].reshape(self.n,1))
if 'gain' not in x_from_node:
self.node_cache_ctg(x_from_node)
Fs = x_from_node['gain']
if T >len(Fs):
print "requested uncached steer!!!"
return self.steer(x_from_node,x_toward) #fixme should cache more
#reverse system
Ar = A.I
Br = -A.I * B
cr = A.I * c
xs_r = np.zeros(shape=(T+1,self.n+1)) #r for reverse
us = np.zeros(shape=(T,self.m))
#we're driving backwards. start at x_toward.
xs_r[0] = x_toward
for i in range(T):
j = T - i-1 #gain matrices Fs[j] is what you should do with j steps remaining
us[i] = -1 * (np.dot(Fs[j,:,0:self.n],xs_r[i,0:self.n]) + Fs[j,:,self.n])
xs_r[i+1,0:self.n] = np.dot(Ar,xs_r[i,0:self.n].T) + np.dot(Br,us[i].T) + np.array(cr).T[0] #sloppy dimensions
xs_r[i+1,self.n] = xs_r[i,self.n] - 1 #reverse time
us_r = us
us = us_r[::-1]
xs = np.zeros(shape=(T+1,self.n+1))
cost = 0
for i in range(T):
xs[i+1] = self.run_forward(xs[i], us[[i]] )
assert xs[i+1,self.n] == xs[i,self.n] + 1
cost += self.cost(xs[i].T,us[i].reshape(1,self.m))
if cost_limit and cost > self.max_steer_cost:
break
return (xs[i+1], us[:i])
def steer_QP(self,x_from_node,x_toward):
raise NotImplemented
A,B,c = self.get_ABc(x_from_node['state'][0:self.n],self.u0)
Q,R,q,r,d = self.get_QRqrd(node['state'][0:self.n],self.u0)
x_from = x_from_node['state']
assert len(x_from) == self.n+1
T = x_toward[self.n] - x_from[self.n] #how much time to do the steering
assert T-int(T) == 0 #discrete time
T=int(T)
if T<=0:
return (x_from,np.zeros(shape=(0,self.m))) #stay here
try:
qpsol, qpmats ,xs,us = LQR_QP(A,B,Q,R,(T+1),
x0=x_from[0:self.n],
xT=x_toward[0:self.n])
except ValueError as e:
#quadratic program is probably infeasible. This can happen if the time horizon is too short and the system's reachability doesn't include the final-value constraint
return (x_from,np.zeros(shape=(0,self.m))) #stay here (could do something smarter)
(QP_P,QP_q,QP_A,QP_B) = qpmats
xs = xs.T
us = us.T
x_actual = np.concatenate(( xs[-1],
[x_from[self.n]+us.shape[0]]
))
return (x_actual, us)
def distance_direct(self,from_node,to_point):
#print from_node['state'], to_point
#to_point is an array and from_point is a node
assert len(to_point)==self.n+1
x_actual,action = self.steer(from_node,to_point)
if self.same_state(x_actual,to_point): #if actually drove there:
return self.cost(from_node['state'],action)
else:
return np.inf
def distance_direct_qp(self,from_node,to_point):
#print from_node['state'], to_point
#to_point is an array and from_point is a node
assert len(to_point)==self.n
x_actual,action = self.steer_QP(from_node,to_point)
if self.same_state(x_actual,to_point): #if actually drove there:
return self.cost(from_node['state'],action)
else:
return np.inf
def distance(self,from_node,to_point):
#to_point is an array and from_point is a node
A,B,c = self.get_ABc(from_node['state'][0:self.n],self.u0)
Q,R,q,r,d = self.get_QRqrd(from_node['state'][0:self.n],self.u0)
x_from = from_node['state']
x_toward = to_point
assert len(x_toward)==self.n+1
T = x_toward[self.n] - x_from[self.n] #how much time to do the steering
assert T-int(T) == 0 #discrete time
T=int(T)
if T<0:
return np.inf
elif T==0:
return 0 if self.same_state(x_from,x_toward) else np.inf
desired = np.matrix(x_toward[0:self.n].reshape(self.n,1))
#we want the final bowl to be centered at desired:
#(x-x_d)^T * Qf * (x-x_d)
#xT*Qf*x -x_dT * Qf * x - xT *Qf *x_d * x_dT * Qf * x_d
Qf = np.eye(self.n) * 1e8
qf = -np.dot(Qf,desired)
Qhf = np.zeros(shape=(self.n+1,self.n+1))
Qhf[0:self.n,0:self.n] = Qf
Qhf[0:self.n,[self.n]] = qf
Qhf[[self.n],0:self.n] = qf.T
Qhf[self.n,self.n] = np.dot(desired.T,np.dot(Qf,desired))
(Ah,Bh,Qh,Rh,pk) = AQR( A=A,
B=B,
Q=Q,
R=R,
c=c,
q=q,
r=r,
d=d,
ctdt='dt')
T = T+1
Fs, Ps = dtfh_lqr(A=Ah,B=Bh,Q=Qh,R=R,N=T,Q_terminal=Qhf)
x_from_homo = np.zeros(self.n+1)
x_from_homo[0:self.n] = x_from[0:self.n]
x_from_homo[self.n] = 1
#assert False
return np.dot(x_from_homo.T,np.dot(Ps[0],x_from_homo))
def distance_cache(self,from_node,to_point):
#to_point is an array and from_point is a node
max_time_horizon = self.max_time_horizon
x_from = from_node['state']
x_toward = to_point
assert len(x_toward)==self.n+1
T = x_toward[self.n] - x_from[self.n] #how much time to do the steering
assert T-int(T) == 0 #discrete time
T=int(T)
if T<0:
return np.inf
elif T==0:
return 0 if self.same_state(x_from,x_toward) else np.inf
if T < 5:
#this technique isn't too accurate for short times due to slack in the final-value constraint
#so do something else.
return self.distance(from_node,to_point)
if 'ctg' not in from_node:
self.node_cache_ctg(from_node)
if T >= len(from_node['ctg']):
print 'requested uncached distance!!!'
distance(from_node,to_point)
ctg = from_node['ctg'][T]
x_to_homo = np.zeros(self.n+1)
x_to_homo[0:self.n] = x_toward[0:self.n]
x_to_homo[self.n] = 1
return np.dot(x_to_homo,np.dot(ctg,x_to_homo.T))
def check_cache_distances(self,rrt,to_point):
for node in rrt.tree.nodes():
from_node = rrt.tree.node[node]
d1 = self.distance(from_node,to_point)
d2 = self.distance_cache(from_node,to_point)
d3 = self.distance_direct(from_node,to_point)
d4 = self.distance_direct_qp(from_node,to_point)
d5 = self.distance_direct_steer_cache(from_node,to_point)
T = to_point[self.n] - from_node['state'][self.n]
print d1,d2,d3,d4,d5,T
# if not (a == np.inf and b == np.inf):
# print a,b,abs(a-b)/(abs(a)+abs(b))