# ## MIT License

# Permission is hereby granted, free of charge, to any person obtaining a copy
# of this software and associated documentation files (the "Software"), to deal
# in the Software without restriction, including without limitation the rights
# to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
# copies of the Software, and to permit persons to whom the Software is
# furnished to do so, subject to the following conditions:
#
# The above copyright notice and this permission notice shall be included in
# all copies or substantial portions of the Software.
#
# THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
# IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
# FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
# AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
# LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
# OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
# THE SOFTWARE.
#

# In[54]:
import numpy as np
import numpy.linalg as la
np.set_printoptions(linewidth=150, suppress=True, precision=3)
# In[55]:
A = np.random.randn(4, 4)
A
# Initialize `L` and `U`:
# In[56]:
L = np.eye(len(A))
U = np.zeros_like(A)
# Recall the "recipe" for LU factorization:
#
# $$\let\B=\boldsymbol \begin{array}{cc}
# & \left[\begin{array}{cc}
# u_{00} & \B{u}_{01}^T\\
# & U_{11}
# \end{array}\right]\\
# \left[\begin{array}{cc}
# 1 & \\
# \B{l}_{10} & L_{11}
# \end{array}\right] & \left[\begin{array}{cc}
# a_{00} & \B{a}_{01}\\
# \B{a}_{10} & A_{11}
# \end{array}\right]
# \end{array}$$
#
# Find $u_{00}$ and $u_{01}$. Check `A - L@U`.
# In[57]:
U[0] = A[0]
A - L@U
# Find $l_{10}$. Check `A - L@U`.
# In[58]:
L[1:,0] = A[1:,0]/U[0,0]
A - L@U
# Recall $A_{22} =\B{l}_{21} \B{u}_{12}^T + L_{22} U_{22}$. Write the next step generic in terms of `i`.
#
# After the step, print `A-L@U` and `remaining`.
# In[59]:
i = 1
remaining = A - L@U
# In[62]:
U[i, i:] = remaining[i, i:]
L[i+1:,i] = remaining[i+1:,i]/U[i,i]
remaining[i+1:, i+1:] -= np.outer(L[i+1:,i], U[i, i+1:])
i = i + 1
print(remaining)
print(A-L@U)
# In[ ]: