hw6.py

# version code 988
# Please fill out this stencil and submit using the provided submission script.

from matutil import *
from GF2 import one
from vecutil import zero_vec


## Problem 1
# Write each matrix as a list of row lists

echelon_form_1 = [[1, 2, 0, 2, 0],
                  [0, 1, 0, 3, 4],
                  [0, 0, 2, 3, 4],
                  [0, 0, 0, 2, 0],
                  [0, 0, 0, 0, 4]]

echelon_form_2 = [[0, 4, 3, 4, 4],
                  [0, 0, 4, 2, 0],
                  [0, 0, 0, 0, 1],
                  [0, 0, 0, 0, 0]]

echelon_form_3 = [[1, 0, 0, 1],
                  [0, 0, 0, 1],
                  [0, 0, 0, 0]]

echelon_form_4 = [[1, 0, 0, 0],
                  [0, 1, 0, 0],
                  [0, 0, 0, 0],
                  [0, 0, 0, 0]]



## Problem 2

def is_echelon(A):
    '''
    Input:
        - A: a list of row lists
    Output:
        - True if A is in echelon form
        - False otherwise
    Examples:
        >>> is_echelon([[1,1,1],[0,1,1],[0,0,1]])
        True
        >>> is_echelon([[0,1,1],[0,1,0],[0,0,1]])
        False
    '''
    i = -1 # initializing lowest index of first non-zero entry
    ncols = len(A[0])
    
    for row in A:
        j = 0 #initializing current row index
        while j < ncols and row[j] == 0:
            j += 1
        if j <= i and j < ncols-1: 
            return False
        i = j
    return True

## Problem 3
# Give each answer as a list

echelon_form_vec_a = [1, 0, 3, 0]
echelon_form_vec_b = [-3, 0, -2, 3]
echelon_form_vec_c = [-5, 0, 2, 0, 2]



## Problem 4
# If a solution exists, give it as a list vector.
# If no solution exists, provide "None".

solving_with_echelon_form_a = None
solving_with_echelon_form_b = [21, 0, 2, 0, 0]



## Problem 5
def echelon_solve(rowlist, label_list, b):
    '''
    Input:
        - rowlist: a list of Vecs
        - label_list: a list of labels establishing an order on the domain of
                      Vecs in rowlist
        - b: a vector (represented  as a list)
    Output:
        - Vec x such that rowlist * x is b
    >>> D = {'A','B','C','D','E'}
    >>> U_rows = [Vec(D, {'A':one, 'E':one}), Vec(D, {'B':one, 'E':one}), Vec(D,{'C':one})] 
    >>> b_list = [one,0,one]>>> cols = ['A', 'B', 'C', 'D', 'E']
    >>> echelon_solve(U_rows, cols, b_list)
    Vec({'B', 'C', 'A', 'D', 'E'},{'B': 0, 'C': one, 'A': one})
    '''
    D = rowlist[0].D
    ncols = len(D)
    x = zero_vec(D)
    
    # remove all zero rows at the bottom
    i = len(rowlist)
    while rowlist[i-1] == zero_vec(D):
        i -= 1
    for j in reversed(range(i)):
        row = rowlist[j]
        pivot_index = 0 # lowest possible pivot index (should be set to j)
        c = label_list[pivot_index]
        while row[c] == 0:
            pivot_index += 1
            c = label_list[pivot_index]
        x[c] = (b[j] - x*row)/row[c]
    return x

## Problem 6
D = {'A', 'B', 'C', 'D'}
rowlist = [Vec(D, {'A':one, 'B':one, 'D':one}), 
           Vec(D, {'B':one}),
           Vec(D, {'C':one}),
           Vec(D, {'D':one})]    # Provide as a list of Vec instances
label_list = ['A', 'B', 'C', 'D'] # Provide as a list
b = [one, one, 0, 0]          # Provide as a list



## Problem 7
null_space_rows_a = {3, 4} # Put the row numbers of M from the PDF



## Problem 8
null_space_rows_b = {4}



## Problem 9
# Write each vector as a list
closest_vector_1 = [8/5, 16/5]
closest_vector_2 = [0, 1, 0]
closest_vector_3 = [3, 2, 1, -4]



## Problem 10
# Write each vector as a list

project_onto_1 = [2, 0]
projection_orthogonal_1 = [0, 1]

project_onto_2 = [-1/6, -1/3, 1/6]
projection_orthogonal_2 = [7/6, 4/3, 23/6]

project_onto_3 = [1, 1, 4]
projection_orthogonal_3 = [0, 0, 0]



## Problem 11
norm1 = 3
norm2 = 4
norm3 = 1