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 9.60e+00 0.00e+00 0.00e+00 0.00e+00
1 3.06e-01 2.74e+00 3.51e+00 3.51e+00 3.5e+00 1.1e-01
2 1.21e-01 2.77e+00 3.59e+00 5.07e+00 7.1e-01 2.8e-01
3 1.12e-01 2.78e+00 3.66e+00 6.42e+00 5.3e-01 4.9e-01
4 6.10e-02 2.79e+00 3.70e+00 7.83e+00 4.5e-01 2.5e-01
5 2.90e-09 2.79e+00 3.77e+00 9.05e+00 7.6e-01 3.6e-08
Simple stats
niter: 5
solved: true
inconsistent: false
residuals: []
Aresiduals: []
κ₂(A): []
status: solution good enough for the tolerances given
Primal feasibility: 3.0e-10
Dual feasibility: 1.9e-16
Error in x: 3.0e-10
Error in y: 1.4e-10