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  8.57e+00  0.00e+00  0.00e+00  0.00e+00   ✗ ✗ ✗ ✗  ✗ ✗ ✗ ✗  0.37s
1  2.58e-01  2.69e+00  3.19e+00  3.19e+00   3.2e+00  9.6e-02  0.78s
2  5.70e-02  2.70e+00  3.39e+00  4.79e+00   1.1e+00  2.5e-01  0.94s
3  3.23e-02  2.70e+00  3.46e+00  6.02e+00   6.2e-01  3.5e-01  0.94s
4  2.68e-03  2.70e+00  3.49e+00  7.14e+00   4.6e-01  3.8e-02  0.94s
5  1.41e-08  2.70e+00  3.49e+00  7.96e+00   1.2e-01  6.3e-07  0.94s

SimpleStats
niter: 5
solved: true
inconsistent: false
residuals: []
Aresiduals: []
κ₂(A): []
timer: 936.32ms
status: solution good enough for the tolerances given
Primal feasibility: 1.6e-09
Dual   feasibility: 2.1e-16
Error in x: 1.6e-09
Error in y: 1.5e-09