using Krylov
using LinearAlgebra, Printf
m = 5
n = 8
λ = 1.0e-3
A = rand(m, n)
b = A * ones(n)
xy_exact = [A λ*I] \ b # In Julia, this is the min-norm solution!
(x, y, stats) = craig(A, b, λ=λ, atol=0.0, rtol=1.0e-20, verbose=1)
show(stats)
# Check that we have a minimum-norm solution.
# When λ > 0 we solve min ‖(x,s)‖ s.t. Ax + λs = b, and we get s = λy.
@printf("Primal feasibility: %7.1e\n", norm(b - A * x - λ^2 * y) / norm(b))
@printf("Dual feasibility: %7.1e\n", norm(x - A' * y) / norm(x))
@printf("Error in x: %7.1e\n", norm(x - xy_exact[1:n]) / norm(xy_exact[1:n]))
if λ > 0.0
@printf("Error in y: %7.1e\n", norm(λ * y - xy_exact[n+1:n+m]) / norm(xy_exact[n+1:n+m]))
end
CRAIG: system of 5 equations in 8 variables
k ‖r‖ ‖x‖ ‖A‖ κ(A) α β timer
0 7.94e+00 0.00e+00 0.00e+00 0.00e+00 ✗ ✗ ✗ ✗ ✗ ✗ ✗ ✗ 0.49s
1 1.77e-01 2.80e+00 2.84e+00 2.84e+00 2.8e+00 6.3e-02 0.93s
2 1.49e-01 2.80e+00 3.05e+00 4.31e+00 8.4e-01 7.1e-01 1.11s
3 6.85e-02 2.81e+00 3.16e+00 5.75e+00 7.7e-01 3.5e-01 1.11s
4 3.73e-03 2.81e+00 3.37e+00 7.12e+00 1.2e+00 6.3e-02 1.11s
5 6.48e-10 2.81e+00 3.37e+00 7.89e+00 1.6e-01 2.7e-08 1.11s
SimpleStats
niter: 5
solved: true
inconsistent: false
residuals: []
Aresiduals: []
κ₂(A): []
timer: 1.11s
status: solution good enough for the tolerances given
Primal feasibility: 8.2e-11
Dual feasibility: 2.4e-16
Error in x: 8.1e-11
Error in y: 7.0e-11