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  7.62e+00  0.00e+00  0.00e+00  0.00e+00   ✗ ✗ ✗ ✗  ✗ ✗ ✗ ✗  0.00s
    1  3.92e-01  2.77e+00  2.75e+00  2.75e+00   2.8e+00  1.4e-01  0.00s
    2  1.26e-01  2.80e+00  2.91e+00  4.12e+00   9.0e-01  2.9e-01  0.00s
    3  2.51e-02  2.81e+00  3.04e+00  5.31e+00   8.5e-01  1.7e-01  0.00s
    4  4.27e-03  2.81e+00  3.10e+00  6.26e+00   6.1e-01  1.0e-01  0.00s
    5  2.88e-11  2.81e+00  3.13e+00  7.07e+00   4.5e-01  3.0e-09  0.00s

SimpleStats
 niter: 5
 solved: true
 inconsistent: false
 indefinite: false
 npcCount: 0
 residuals: []
 Aresiduals: []
 qvals: []
 κ₂(A): []
 allocation timer: 2.15μs
 timer: 60.65μs
 status: solution good enough for the tolerances given
Primal feasibility: 3.8e-12
Dual   feasibility: 1.4e-16
Error in x: 3.7e-12
Error in y: 3.2e-12