| Definitions |
| xk : | estimate of true image |
| Ik : | index set at xk |
| C : | given positive definite symmetric matrix |
| d : | given vector |
| Ω : | constraint set |
| PrΩ (·) : | projection function defined by Equation (III.5) |
| α : | step size |
| γ,m : | parameters used to update α |
| Set initial conditions: |
| Initialize xk ∈ Ω, k: = 0 |
| Set m=1, γ ∈ (0,1) |
| At iteration k:k=0,1,2,… |
| Construct the index set Ik at xk. |
| Ik=I(xk) = {i| x i k = 0,1 ≤ i ≤ n}. |
| Let x̅k be a solution of the problem |
| minf(x),x ∈ Ωk |
| Ωk = {x ∈ ℝn|xi = 0,i ∈ Ik} |
| that is solved by conjugate gradient method. |
| If x̅k ∈ Ω, then go to step 5. |
| Otherwise, construct point xk+1 |
| xk+1 = xk + λk (x̅k − xk), |
| where |
| λ k = min j ∈ I 1 k x j k x j k − x ¯ 攳 k , |
| I 1 k = { i | x i k − x ¯ i k 〉 0 , i ∈ I ( x k ) , 1 ≤ i ≤ n } , |
| and go to step 1 for k: = k+1. |
| Check optimality condition at point x̅k |
| { ( x ¯ k ) T C x ¯ k − d T x ¯ k = 0 C x ¯ k − d ≥ 0. |
| If this condition holds, then x̅k is the solution of the problem. |
| Otherwise, find the projection of v for α = γm on Ω : |
| v=PrΩ (x̅k − αf′(x̅k)). |
| If f(v) < f(x̅k), then go to step 1 for x̅k: = v. |
| Otherwise, go to step 6 for m:=m+1. |