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  7.70e+00  0.00e+00  0.00e+00  0.00e+00   ✗ ✗ ✗ ✗  ✗ ✗ ✗ ✗  0.49s
    1  1.39e-01  2.80e+00  2.76e+00  2.76e+00   2.8e+00  5.0e-02  0.92s
    2  4.61e-02  2.80e+00  2.98e+00  4.22e+00   1.1e+00  3.6e-01  1.09s
    3  4.14e-02  2.80e+00  3.13e+00  5.46e+00   6.9e-01  6.2e-01  1.09s
    4  9.85e-03  2.80e+00  3.17e+00  6.67e+00   5.1e-01  1.2e-01  1.09s
    5  1.49e-11  2.80e+00  3.20e+00  7.49e+00   4.3e-01  6.4e-10  1.09s

SimpleStats
 niter: 5
 solved: true
 inconsistent: false
 residuals: []
 Aresiduals: []
 κ₂(A): []
 timer: 1.09s
 status: solution good enough for the tolerances given
Primal feasibility: 1.9e-12
Dual   feasibility: 1.6e-16
Error in x: 1.9e-12
Error in y: 1.8e-12