# coding: utf-8
# ### Conjugate Gradient: the Krylov subspace method for Quadratic Optimization $\def\b#1{\boldsymbol{ #1}}\def\B#1{\boldsymbol{ #1}}$
#
# Conjugate gradient can be thought of as a Krylov subspace method for the quadratic optimization problem $\min(f(\b x)$ where
# $$f(\b x)= \frac{1}{2}\b x^T\b A \b x + \b c^T \b x$$
# and $\b A$ is symmetric positive definite (this makes the problem convex).
#
# In this demo, we compare the result of conjugate gradient to an explicitly constructed Krylov subspace.
#
# We start by picking a random $\b A$ and $\b c$:
# In[1]:
import numpy as np
import numpy.linalg as la
import scipy.optimize as sopt
n = 32
# make A a random SPD matrix
Q = la.qr(np.random.randn(n, n))[0]
A = Q @ (np.diag(np.random.rand(n)) @ Q.T)
c = np.random.randn(n)
# pick number of steps of CG to execute
k = 10
# define quadratic objective function
def f(x):
return .5*np.inner(x, A @ x) + np.inner(c,x)
# ### Quadratic Programming using Arnoldi
#
# First, we form a Krylov subspace using Arnoldi iteration on $\b A$ with starting guess $\b c$, then select the minima via
# $$\B x_k = -||\B c||_2\B Q_k \B T_k^{-1} \B e_1 ,$$
# which is a minimizer within this Krylov subspace since
# \begin{align*}
# \min_{\B x \in \mathcal{K}_k(\B A, \B c)} f(\B x) &= \min_{\B y \in \mathbb{R}^k} f(\B Q_k \B y) \\
# &=\min_{\B y \in \mathbb{R}^k} \B y^T \B Q_k^T \B A \B Q_k \B y + \B c^T \B Q_k \B y \\
# &= \min_{\B y \in \mathbb{R}^k} \B y^T \B T_k \B y + ||\B c||_2\B e_1^T \B y.
# \end{align*}
#
# In[2]:
# initialize Krylov subspace, taking first column as c/||c||_2
Q = np.zeros((n,k))
T = np.zeros((k,k))
x_kr = np.zeros((n,k))
Q[:,0] = c / la.norm(c)
for i in range(1,k+1):
# perform Lanczos iteration
u = A @ Q[:,i-1]
for j in range(0,i):
T[j,i-1] = np.inner(Q[:,j],u)
u = u - T[j,i-1]*Q[:,j]
if i0:
# compute approximate minima x_kr as -||\b c||_2\b Q \b T^{-1} \b e_1
e1 = np.zeros(i)
e1[0] = la.norm(c)
x_kr[:,i-1] = - Q[:,:i] @ la.solve(T[:i,:i],e1)
print(f(x_kr[:,i-1]), "= min_{ x_kr in Krylov subspace of dimension ", i,"} f(x_kr)")
# ### Nonlinear Programming using Parallel Tangents for Quadratic Objective
#
# Next, we implement Conjugate Gradient using the parallel tangents method, which is general to arbitrary nonlinear objective functions (but the convergence and optimality properties are only true for a convex quadratic objective function).
#
# The parallel tangent method performs two line minimizations at each iteration:
# * Perform a step of steepest descent to generate $\B{\hat{x}}_{k}$ from $\B x_k$.
# * Generate $\B x_{k+1}$ by minimizing over the line passing through $\B x_{k-1}$ and $\B{\hat{x}}_{k}$.
#
# To initialize the parallel tangents method, we use the two first approximate minima obtained from the Krylov subspace method above. Subsequent iterates are then reproduced exactly!
# In[3]:
# perform conjugate gradient by parallel tangents method
x_cg_pp = np.zeros((n, k))
x_cg_pp[:,0] = x_kr[:,0]
x_cg_pp[:,1] = x_kr[:,1]
# perform CG method via parallel tangents method
for i in range(2,k):
x = x_cg_pp[:,i-1]
# define function for steepest descent, based on current iterate x and grad_f(x)
# can use local variables within function
def f_1d(a):
return f(x + a*(A@x+c))
# solve for best steepest descent step using golden section search and form x_hat
a = sopt.golden(f_1d)
x_hat = x + a * (A@x+c)
# define function for extrapolation, based on last iterate x_cg_pp[:,i-2] and x_hat (the solution of steepest descent)
# can use local variables within function
def g_1d(a):
return f(x_hat + a*(x_hat - x_cg_pp[:,i-2]))
# solve for best extrapolation using golden section search and update x
a = sopt.golden(g_1d)
x = x_hat + a*(x_hat - x_cg_pp[:,i-2])
# store CG iterate as next column of x_cg_pp
x_cg_pp[:,i] = x
print(f(x), " = f( conjugate gradient parallel tangents iterate", i, ")")
# ### Quadratic Programming by Conjugate Gradient
#
# The 1D line minimizations in the parallel tangents method can be solved analytically given a quadratic objective, so the parallel tangents implementation can be transformed to use a couple of matrix vector products in place of the two golden section searches.
# In[4]:
# perform CG method via matrix-vector products
x_cg_mv = np.zeros((n, k))
x_cg_mv[:,0] = x_kr[:,0]
x_cg_mv[:,1] = x_kr[:,1]
# this implementation mimics parallel tangents implementation,
# one of the matrix vector products can in fact be avoided
for i in range(2,k):
x = x_cg_mv[:,i-1]
# compute gradient
g = A@x + c
# compute explicit equation for optimal step size for steepest descent
a = - np.inner(g,g)/np.inner(g, A@g)
# update x
x_hat = x + a * g
# determine new optimization direction
d = x_hat - x_cg_mv[:,i-2]
# compute explicit equation for optimal step size in direction d = x_hat - x_{k-1}
a = - np.inner(g,d)/np.inner(d, A@d)
#update x
x = x_hat + a * d
# store CG iterate as next column of x_cg_mv
x_cg_mv[:,i] = x
print(f(x), " = f( conjugate gradient matrix vector product iterate", i, ")")
# It can be shown that each step of Conjugate gradient causes an update $\b s_k = \b x_{k+1} - \b x_k$ that is $\b A$-conjugate ($\b A$-orthogonal) to the previous, i.e. $\b s_i^T\b A \b s_{i-1}$:
# In[5]:
for i in range(2,k):
print(np.inner(x_cg_mv[:,i]-x_cg_mv[:,i-1],A @ (x_cg_mv[:,i-1]-x_cg_mv[:,i-2])))
# In fact $\b s_i^T \b A \b s_j= 0$ if $i\neq j$.
# In[6]:
print((x_cg_mv[:,1:]-x_cg_mv[:,:-1]).T @ A @ (x_cg_mv[:,1:]-x_cg_mv[:,:-1]))
# This fact permits a further optimization of the conjugate gradient iteration, resulting in only 1 matrx-vector product (the resulting work-efficient method can be found in many textbooks, papers, and on wikipedia).