#!/usr/bin/env python
# coding: utf-8

# # Coding Back-Substitution
# 
# Copyright (C) 2020 Andreas Kloeckner
# 
# <details>
# <summary>MIT License</summary>
# 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
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# 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.
# </details>

# In[1]:


import numpy as np


# Here's an upper-triangular matrix $A$ and two vectors $x$ and $b$ so that $Ax=b$.
# 
# See if you can find $x$ by computation.

# In[2]:


n = 5

A = np.random.randn(n, n) * np.tri(n).T
print(A)

x = np.random.randn(n)
print(x)

b = A @ x


# In[3]:


xcomp = np.zeros(n)

for i in range(n-1, -1, -1):
    tmp = b[i]
    for j in range(n-1, i, -1):
        tmp -= xcomp[j]*A[i,j]
        
    xcomp[i] = tmp/A[i,i]


# Now compare the computed $x$ against the reference solution.

# In[4]:


print(x)
print(xcomp)


# Questions/comments:
# 
# * Can this fail?
# * What's the operation count?