```markdown
**Complexity per iteration of three methods is identical**

1. **Use block elimination to solve KKT system:**

   \[
   \begin{bmatrix}
   \text{diag}(x)^{-2} & A^T \\
   A & 0
   \end{bmatrix}
   \begin{bmatrix}
   \Delta x \\
   w
   \end{bmatrix}
   =
   \begin{bmatrix}
   \text{diag}(x)^{-1}1 \\
   0
   \end{bmatrix}
   \]

   Reduces to solving \( A \text{diag}(x)^2 A^T w = b \).

2. **Solve Newton system:**

   \[
   A \text{diag}(A^T \nu)^{-2} A^T \Delta \nu = -b + A \text{diag}(A^T \nu)^{-1} 1
   \]

3. **Use block elimination to solve KKT system:**

   \[
   \begin{bmatrix}
   \text{diag}(x)^{-2} & A^T \\
   A & 0
   \end{bmatrix}
   \begin{bmatrix}
   \Delta x \\
   \Delta \nu
   \end{bmatrix}
   =
   \begin{bmatrix}
   \text{diag}(x)^{-1}1 \\
   Ax - b
   \end{bmatrix}
   \]

   Reduces to solving \( A \text{diag}(x)^2 A^T w = 2Ax - b \).

**Conclusion:** In each case, solve \( A D A^T w = h \) with \( D \) positive diagonal.
```
