from casadi import *

# look at: https://github.com/casadi/casadi/blob/master/docs/examples/python/vdp_indirect_multiple_shooting.py

class OpenLoopSolver:
    def __init__(self, N=60, T=6.0):
        self.T = T
        self.N = N

    def solve(self, x0):
        x = SX.sym('x')
        y = SX.sym('y')
        theta = SX.sym('theta')
        state = vertcat(x, y, theta)
        r = 0.03
        R = 0.05
        d = 0.02

        omegar = SX.sym('omegar')
        omegal = SX.sym('omegal')
        control = vertcat(omegar, omegal)

        # model equation
        f1 = (r / 2 * cos(theta) - r * d / (2 * R) * sin(theta)) * omegar + (r / 2 * cos(theta) + r * d / (2 * R) * sin(
            theta)) * omegal
        f2 = (r / 2 * sin(theta) + r * d / (2 * R) * cos(theta)) * omegar + (r / 2 * sin(theta) - r * d / (2 * R) * cos(
            theta)) * omegal
        f3 = r / (2 * R) * omegar - r / (2 * R) * omegal
        xdot = vertcat(f1, f2, f3)
        f = Function('f', [x, y, theta, omegar, omegal], [f1, f2, f3])
        print("f = {}".format(f))

        # cost functional
        L = x ** 2 + y ** 2 + 1e-2 * theta ** 2 + 1e-4 * (omegar ** 2 + omegal ** 2)

        # Fixed step Runge-Kutta 4 integrator
        M = 4  # RK4 steps per interval
        DT = self.T / self.N / M
        print("DT = {}".format(DT))
        f = Function('f', [state, control], [xdot, L])
        X0 = MX.sym('X0', 3)
        U = MX.sym('U', 2)
        X = X0
        Q = 0
        runge_kutta = True
        if runge_kutta:
            for j in range(M):
                k1, k1_q = f(X, U)
                k2, k2_q = f(X + DT / 2 * k1, U)
                k3, k3_q = f(X + DT / 2 * k2, U)
                k4, k4_q = f(X + DT * k3, U)
                X = X + DT / 6 * (k1 + 2 * k2 + 2 * k3 + k4)
                Q = Q + DT / 6 * (k1_q + 2 * k2_q + 2 * k3_q + k4_q)
        else:
            DT = self.T / self.N
            k1, k1_q = f(X, U)
            X = X + DT * k1
            Q = Q + DT * k1_q
        F = Function('F', [X0, U], [X, Q], ['x0', 'p'], ['xf', 'qf'])

        # F_euler = Function('F_euler', [X0, U], [Xeuler, Qeuler], ['x0', 'p'], ['xf', 'qf'])

        Fk = F(x0=[0.2, 0.3, 0.0], p=[-1.1, 1.1])
        print(Fk['xf'])
        print(Fk['qf'])

        # Start with an empty NLP
        w = []
        w0 = []
        lbw = []
        ubw = []
        J = 0
        g = []
        lbg = []
        ubg = []

        # Formulate the NLP
        Xk = MX(x0)
        for k in range(self.N):
            # New NLP variable for the control
            U1k = MX.sym('U1_' + str(k), 2)
            # U2k = MX.sym('U2_' + str(k))
            w += [U1k]
            lbw += [-10, -10]
            ubw += [10, 10]
            w0 += [0, 0]

            # Integrate till the end of the interval
            Fk = F(x0=Xk, p=U1k)
            Xk = Fk['xf']
            J = J + Fk['qf']

            # Add inequality constraint
            # g += [Xk[1]]
            # lbg += [-.0]
            # ubg += [inf]

        # Create an NLP solver
        prob = {'f': J, 'x': vertcat(*w), 'g': vertcat(*g)}
        self.solver = nlpsol('solver', 'ipopt', prob)

        # Solve the NLP
        sol = self.solver(x0=w0, lbx=lbw, ubx=ubw, lbg=lbg, ubg=ubg)
        w_opt = sol['x']

        # Plot the solution
        u_opt = w_opt
        x_opt = [x0]
        for k in range(self.N):
            Fk = F(x0=x_opt[-1], p=u_opt[2*k:2*k+2])
            x_opt += [Fk['xf'].full()]
        x1_opt = [r[0] for r in x_opt]
        x2_opt = [r[1] for r in x_opt]
        x3_opt = [r[2] for r in x_opt]

        tgrid = [self.T/self.N*k for k in range(self.N+1)]
        import matplotlib.pyplot as plt
        plt.figure(2)
        plt.clf()
        plt.plot(tgrid, x1_opt, '--')
        plt.plot(tgrid, x2_opt, '-')
        plt.plot(tgrid, x3_opt, '*')
        #plt.step(tgrid, vertcat(DM.nan(1), u_opt), '-.')
        plt.xlabel('t')
        plt.legend(['x1','x2','x3','u'])
        plt.grid()
        #plt.show()
        #return

        # alternative solution using multiple shooting (way faster!)
        opti = Opti() # Optimization problem

        # ---- decision variables ---------
        X = opti.variable(3,self.N+1) # state trajectory
        Q = opti.variable(1,self.N+1) # state trajectory
        posx   = X[0,:]
        posy   = X[1,:]
        angle = X[2,:]
        U = opti.variable(2,self.N)   # control trajectory (throttle)
        #T = opti.variable()      # final time

        # ---- objective          ---------
        #opti.minimize(T) # race in minimal time

        # ---- dynamic constraints --------
        #f = lambda x,u: vertcat(f1, f2, f3) # dx/dt = f(x,u)

        dt = self.T/self.N # length of a control interval
        for k in range(self.N): # loop over control intervals
           # Runge-Kutta 4 integration
           k1, k1_q = f(X[:,k],         U[:,k])
           k2, k2_q = f(X[:,k]+dt/2*k1, U[:,k])
           k3, k3_q = f(X[:,k]+dt/2*k2, U[:,k])
           k4, k4_q = f(X[:,k]+dt*k3,   U[:,k])
           x_next = X[:,k] + dt/6*(k1+2*k2+2*k3+k4)
           q_next = Q[:,k] + dt/6*(k1_q + 2 * k2_q + 2 * k3_q + k4_q)
           opti.subject_to(X[:,k+1]==x_next) # close the gaps
           opti.subject_to(Q[:,k+1]==q_next) # close the gaps
        opti.minimize(Q[:,self.N])

        # ---- path constraints -----------
        #limit = lambda pos: 1-sin(2*pi*pos)/2
        #opti.subject_to(speed<=limit(pos))   # track speed limit
        opti.subject_to(opti.bounded(-10,U,10)) # control is limited

        # ---- boundary conditions --------
        opti.subject_to(posx[0]==x0[0])   # start at position 0 ...
        opti.subject_to(posy[0]==x0[1]) # ... from stand-still
        opti.subject_to(angle[0]==x0[2])  # finish line at position 1
        #opti.subject_to(speed[-1]==0)  # .. with speed 0
        opti.subject_to(Q[:,0]==0.0)

        # ---- misc. constraints  ----------
        #opti.subject_to(X[1,:]>=0) # Time must be positive
        #opti.subject_to(X[2,:]<=4) # Time must be positive
        #opti.subject_to(X[2,:]>=-2) # Time must be positive

        # avoid obstacle
        #r = 0.25
        #p = (0.5, 0.5)
        #for k in range(self.N):
        #    opti.subject_to((X[0,k]-p[0])**2 + (X[1,k]-p[1])**2  > r**2)
        #    pass


        # ---- initial values for solver ---
        #opti.set_initial(speed, 1)
        #opti.set_initial(T, 1)

        # ---- solve NLP              ------
        opti.solver("ipopt") # set numerical backend
        sol = opti.solve()   # actual solve

        #x0 = sol.value(opti.x)
        #lam_g0 = sol.value(opti.lam_g)
        #opti.set_initial(opti.lam_g, lam_g0)
        #opti.set_initial(opti.x, x0)
        #opti.solve()

        from pylab import plot, step, figure, legend, show, spy

        plot(sol.value(posx),label="posx")
        plot(sol.value(posy),label="posy")
        plot(sol.value(angle),label="angle")

        plt.figure(3)
        plot(sol.value(posx), sol.value(posy))
        ax = plt.gca()
        circle = plt.Circle(p, r)
        ax.add_artist(circle)
        #plot(limit(sol.value(pos)),'r--',label="speed limit")
        #step(range(N),sol.value(U),'k',label="throttle")
        legend(loc="upper left")
        plt.show()
        pass
        # linearization
        # A = zeros(states.shape[0], states.shape[0])
        # for i in range(f.shape[0]):
        #     for j in range(states.shape[0]):
        #         A[i,j] = diff(f[i,0], states[j])
        # Alin = A.subs([(theta,0), (omegar,0), (omegal,0)])
        # print("A = {}".format(Alin))
        # B = zeros(f.shape[0], controls.shape[0])
        # for i in range(f.shape[0]):
        #     for j in range(controls.shape[0]):
        #         B[i,j] = diff(f[i,0], controls[j])
        # print("B = {}".format(B))
        # dfdtheta = diff(f, theta)
        #print(dfdtheta.doit())

        # takeaway: linearization is not helpful, because the linearized system is not stabilizable
        # -> alternative: use nonlinear control method