【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