【Python】Coding the Matrix:Week 5: Dimension Homework 5

这一周的作业，刚压线写完。Problem3 没有写，不想证明了。从Problem 9 开始一直到最后难度都挺大的，我是在论坛上看过了别人的讨论才写出来的，挣扎了很久。

Problem 9在给定的基上分解向量，里面调用了hw4的一些函数，通过solve函数获得矩阵方程的解

Problem 10判断矩阵是不是可逆的，注意判断矩阵是不是square的

Problem 11和Problem 12 都是求逆，也是解方程，只是函数的参数需要参考一下源码

发现一个有趣的事情，Coding the Matrix的论坛上有个老头胡子都白了也在学这个课程，好励志的感觉，不过人家貌似是教授来着。

# version code 941
# Please fill out this stencil and submit using the provided submission script.

from vecutil import list2vec
from solver import solve
from matutil import *
from mat import Mat
from GF2 import one
from vec import Vec
from independence import *
from triangular import *
from hw4 import *

## Problem 1
w0 = list2vec([1,0,0])
w1 = list2vec([0,1,0])
w2 = list2vec([0,0,1])

v0 = list2vec([1,2,3])
v1 = list2vec([1,3,3])
v2 = list2vec([0,3,3])

# Fill in exchange_S1 and exchange_S2
# with appropriate lists of 3 vectors

exchange_S0 = [w0,w1,w2]
exchange_S1 = [v0,w1,w2]
exchange_S2 = [v0,v1,w2]
exchange_S3 = [v0,v1,v2]

## Problem 2
w0 = list2vec([0,one,0])
w1 = list2vec([0,0,one])
w2 = list2vec([one,one,one])

v0 = list2vec([one,0,one])
v1 = list2vec([one,0,0])
v2 = list2vec([one,one,0])

exchange_2_S0 = [w0, w1, w2]
exchange_2_S1 = [w0,w1,v1]
exchange_2_S2 = [w0,v0,v1]
exchange_2_S3 = [v0, v1, v2]

## Problem 3
def morph(S, B):
    '''
    Input:
        - S: a list of distinct Vec instances
        - B: a list of linearly independent Vec instances
        - Span S == Span B
    Output: a list of pairs of vectors to inject and eject
    Example:
        >>> #This is how our morph works.  Yours may yield different results.
        >>> S = [list2vec(v) for v in [[1,0,0],[0,1,0],[0,0,1]]]
        >>> B = [list2vec(v) for v in [[1,1,0],[0,1,1],[1,0,1]]]
        >>> morph(S, B)
        [(Vec({0, 1, 2},{0: 1, 1: 1, 2: 0}), Vec({0, 1, 2},{0: 1, 1: 0, 2: 0})), (Vec({0, 1, 2},{0: 0, 1: 1, 2: 1}), Vec({0, 1, 2},{0: 0, 1: 1, 2: 0})), (Vec({0, 1, 2},{0: 1, 1: 0, 2: 1}), Vec({0, 1, 2},{0: 0, 1: 0, 2: 1}))]

    '''
    pass

## Problem 4
# Please express each solution as a list of vectors (Vec instances)

row_space_1 = [list2vec([1,2,0]),list2vec([0,2,1])]
col_space_1 = [list2vec([1,0]),list2vec([0,1])]

row_space_2 = [list2vec([1,4,0,0]),list2vec([0,2,2,0]),list2vec([0,0,1,1])]
col_space_2 = [list2vec([1,0,0]),list2vec([4,2,0]),list2vec([0,0,1])]

row_space_3 = [list2vec([1])]
col_space_3 = [list2vec([1,2,3])]

row_space_4 = [list2vec([1,0]),list2vec([2,1])]
col_space_4 = [list2vec([1,2,3]),list2vec([0,1,4])]

## Problem 5
def my_is_independent(L):
    '''
    input:  A list, L, of Vecs
    output: A boolean indicating if the list is linearly independent

    >>> L = [Vec({0, 1, 2},{0: 1, 1: 0, 2: 0}), Vec({0, 1, 2},{0: 0, 1: 1, 2: 0}), Vec({0, 1, 2},{0: 0, 1: 0, 2: 1}), Vec({0, 1, 2},{0: 1, 1: 1, 2: 1}), Vec({0, 1, 2},{0: 1, 1: 1, 2: 0}), Vec({0, 1, 2},{0: 0, 1: 1, 2: 1})]
    >>> my_is_independent(L)
    False
    >>> my_is_independent(L[:2])
    True
    >>> my_is_independent(L[:3])
    True
    >>> my_is_independent(L[1:4])
    True
    >>> my_is_independent(L[0:4])
    False
    >>> my_is_independent(L[2:])
    False
    >>> my_is_independent(L[2:5])
    False
    '''
    if rank(L)==len(L):return True
    else:return False

## Problem 6
def subset_basis(T):
    '''
    input: A list, T, of Vecs
    output: A list, S, containing Vecs from T, that is a basis for the
    space spanned by T.

    >>> a0 = Vec({'a','b','c','d'}, {'a':1})
    >>> a1 = Vec({'a','b','c','d'}, {'b':1})
    >>> a2 = Vec({'a','b','c','d'}, {'c':1})
    >>> a3 = Vec({'a','b','c','d'}, {'a':1,'c':3})
    >>> subset_basis([a0,a1,a2,a3]) == [Vec({'c', 'b', 'a', 'd'},{'a': 1}), Vec({'c', 'b', 'a', 'd'},{'b': 1}), Vec({'c', 'b', 'a', 'd'},{'c': 1})]
    True
    '''
    r=[]
    for x in T:
        r.append(x)
        #print(x)
        if my_is_independent(r)==False:r.remove(x)
    #print(r)
    return r     

## Problem 7
def my_rank(L):
    '''
    input: A list, L, of Vecs
    output: The rank of the list of Vecs

    >>> my_rank([list2vec(v) for v in [[1,2,3],[4,5,6],[1.1,1.1,1.1]]])
    2
    '''
    return len(subset_basis(L))

## Problem 8
# Please give each answer as a boolean

only_share_the_zero_vector_1 = True
only_share_the_zero_vector_2 = True
only_share_the_zero_vector_3 = True

## Problem 9
def direct_sum_decompose(U_basis, V_basis, w):
    '''
    input:  A list of Vecs, U_basis, containing a basis for a vector space, U.
    A list of Vecs, V_basis, containing a basis for a vector space, V.
    A Vec, w, that belongs to the direct sum of these spaces.
    output: A pair, (u, v), such that u+v=w and u is an element of U and
    v is an element of V.

    >>> U_basis = [Vec({0, 1, 2, 3, 4, 5},{0: 2, 1: 1, 2: 0, 3: 0, 4: 6, 5: 0}), Vec({0, 1, 2, 3, 4, 5},{0: 11, 1: 5, 2: 0, 3: 0, 4: 1, 5: 0}), Vec({0, 1, 2, 3, 4, 5},{0: 3, 1: 1.5, 2: 0, 3: 0, 4: 7.5, 5: 0})]
    >>> V_basis = [Vec({0, 1, 2, 3, 4, 5},{0: 0, 1: 0, 2: 7, 3: 0, 4: 0, 5: 1}), Vec({0, 1, 2, 3, 4, 5},{0: 0, 1: 0, 2: 15, 3: 0, 4: 0, 5: 2})]
    >>> w = Vec({0, 1, 2, 3, 4, 5},{0: 2, 1: 5, 2: 0, 3: 0, 4: 1, 5: 0})
    >>> direct_sum_decompose(U_basis, V_basis, w) == (Vec({0, 1, 2, 3, 4, 5},{0: 2.0, 1: 4.999999999999972, 2: 0.0, 3: 0.0, 4: 1.0, 5: 0.0}), Vec({0, 1, 2, 3, 4, 5},{0: 0.0, 1: 0.0, 2: 0.0, 3: 0.0, 4: 0.0, 5: 0.0}))
    True
    '''
    T=U_basis + V_basis
    x=vec2rep(T,w)
    #print(T,w,x)
    rep=list(x.f.values())
    u1=list2vec(rep[0:len(U_basis)])
    v1=list2vec(rep[len(U_basis):len(T)])
    u=rep2vec(u1,U_basis)
    v=rep2vec(v1,V_basis)
    return (u,v)

## Problem 10
def is_invertible(M):
    '''
    input: A matrix, M
    outpit: A boolean indicating if M is invertible.

    >>> M = Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): 0, (1, 2): 1, (3, 2): 0, (0, 0): 1, (3, 3): 4, (3, 0): 0, (3, 1): 0, (1, 1): 2, (2, 1): 0, (0, 2): 1, (2, 0): 0, (1, 3): 0, (2, 3): 1, (2, 2): 3, (1, 0): 0, (0, 3): 0})
    >>> is_invertible(M)
    True
    '''
    L=mat2coldict(M)
    L=list(L.values())
    if len(L)!=len(L[0].D):return False
    else:return rank(L)==len(L)

## Problem 11
def find_matrix_inverse(A):
    '''
    input: An invertible matrix, A, over GF(2)
    output: Inverse of A

    >>> M = Mat(({0, 1, 2}, {0, 1, 2}), {(0, 1): one, (1, 2): 0, (0, 0): 0, (2, 0): 0, (1, 0): one, (2, 2): one, (0, 2): 0, (2, 1): 0, (1, 1): 0})
    >>> find_matrix_inverse(M) == Mat(({0, 1, 2}, {0, 1, 2}), {(0, 1): one, (2, 0): 0, (0, 0): 0, (2, 2): one, (1, 0): one, (1, 2): 0, (1, 1): 0, (2, 1): 0, (0, 2): 0})
    True
    '''
    B=[]
    for i in range(len(A.D[0])):
        b=Vec(A.D[0],{})
        b[i]=one
        t=solve(A,b)
        B.append(t)
    B=coldict2mat(B)
    return B

## Problem 12
def find_triangular_matrix_inverse(A):
    '''
    input: An upper triangular Mat, A, with nonzero diagonal elements
    output: Inverse of A
    >>> A = listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
    >>> find_triangular_matrix_inverse(A) == Mat(({0, 1, 2, 3}, {0, 1, 2, 3}), {(0, 1): -0.5, (1, 2): -0.3, (3, 2): 0.0, (0, 0): 1.0, (3, 3): 1.0, (3, 0): 0.0, (3, 1): 0.0, (2, 1): 0.0, (0, 2): -0.05000000000000002, (2, 0): 0.0, (1, 3): -0.87, (2, 3): -0.1, (2, 2): 1.0, (1, 0): 0.0, (0, 3): -3.545, (1, 1): 1.0})
    True
    '''
    B=[]
    C=mat2rowdict(A)
    for i in range(len(A.D[0])):
        b=Vec(A.D[0],{})
        b[i]=1
        t=triangular_solve(C,range(len(C)),b)
        B.append(t)
    B=coldict2mat(B)
    return B