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
    0  7.40e+00  0.00e+00  0.00e+00  0.00e+00
    1  2.90e-01  2.73e+00  2.71e+00  2.71e+00   2.7e+00  1.1e-01
    2  1.41e-01  2.76e+00  2.83e+00  4.00e+00   7.3e-01  3.5e-01
    3  3.33e-02  2.78e+00  2.88e+00  5.07e+00   5.2e-01  1.2e-01
    4  2.90e-02  2.78e+00  3.00e+00  6.11e+00   6.5e-01  5.6e-01
    5  1.41e-10  2.78e+00  3.14e+00  7.37e+00   9.3e-01  4.5e-09

Simple stats
 niter: 5
 solved: true
 inconsistent: false
 residuals: []
 Aresiduals: []
 κ₂(A): []
 status: solution good enough for the tolerances given
Primal feasibility: 1.9e-11
Dual   feasibility: 1.7e-16
Error in x: 1.9e-11
Error in y: 1.3e-11