Dimension_problems.py

# version code 9913d98afecb+
coursera = 1
# Please fill out this stencil and submit using the provided submission script.

from vecutil import list2vec
from GF2 import one
from solver import solve
from matutil import listlist2mat, coldict2mat
from mat import Mat
from vec import Vec
from The_Basis_problems import vec2rep
from The_Basis_problems import exchange
from The_Basis_problems import is_superfluous
from The_Basis_problems import is_independent
from independence import rank
from The_Basis_problems import rep2vec
from matutil import mat2coldict
from matutil import mat2rowdict
from vecutil import zero_vec
from matutil import identity


## 1: (Problem 1) Iterative Exchange Lemma
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]


## 2: (Problem 2) Another Iterative Exchange Lemma
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, v0, w2]
exchange_2_S2 = [v1, v0, w2]
exchange_2_S3 = [v0, v1, v2]


## 3: (Problem 3) Morph Lemma Coding
def morph(S, B):
    '''
    Input:
        - S: a list of distinct Vecs
        - B: a list of linearly independent Vecs all in Span S
    Output: a list of pairs of vectors to inject and eject (see problem description)
    Example:
        >>> # This is how our morph works.  Yours may yield different results.
        >>> # Note: Make a copy of S to modify instead of modifying S itself.
        >>> from vecutil import list2vec
        >>> from vec import Vec
        >>> 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]]]
        >>> D = {0, 1, 2}
        >>> morph(S, B) == [(Vec(D,{0: 1, 1: 1, 2: 0}), Vec(D,{0: 1, 1: 0, 2: 0})), (Vec(D,{0: 0, 1: 1, 2: 1}), Vec(D,{0: 0, 1: 1, 2: 0})), (Vec(D,{0: 1, 1: 0, 2: 1}), Vec(D,{0: 0, 1: 0, 2: 1}))]
        True
        >>> S == [list2vec(v) for v in [[1,0,0],[0,1,0],[0,0,1]]]
        True
        >>> B == [list2vec(v) for v in [[1,1,0],[0,1,1],[1,0,1]]]
        True
        >>> from GF2 import one
        >>> D = {0, 1, 2, 3, 4, 5, 6, 7}
        >>> S = [Vec(D,{1: one, 2: one, 3: one, 4: one}), Vec(D,{1: one, 3: one}), Vec(D,{0: one, 1: one, 3: one, 5: one, 6: one}), Vec(D,{3: one, 4: one}), Vec(D,{3: one, 5: one, 6: one})]
        >>> B = [Vec(D,{2: one, 4: one}), Vec(D,{0: one, 1: one, 2: one, 3: one, 4: one, 5: one, 6: one}), Vec(D,{0: one, 1: one, 2: one, 5: one, 6: one})]
        >>> sol = morph(S, B)
        >>> sol == [(B[0],S[0]), (B[1],S[2]), (B[2],S[3])] or sol == [(B[0],S[1]), (B[1],S[2]), (B[2],S[3])]
        True
        >>> # Should work the same regardless of order of S
        >>> from random import random
        >>> sol = morph(sorted(S, key=lambda x:random()), B)
        >>> sol == [(B[0],S[0]), (B[1],S[2]), (B[2],S[3])] or sol == [(B[0],S[1]), (B[1],S[2]), (B[2],S[3])]
        True
    '''
    morphed_S = list(S)
    A = list(())
    ret = list()
    for b in B:
        w = exchange(morphed_S, A, b)
        morphed_S.remove(w)
        morphed_S.append(b)
        A.append(b)
        ret.append((b, w))
    return ret


## 4: (Problem 4) Row and Column Rank Practice
# Please express each solution as a list of Vecs

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([0,2,1]), 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) Subset Basis
def subset_basis(T):
    '''
    Input:
        - T: a set of Vecs
    Output: 
        - set S containing Vecs from T that is a basis for Span T.
    Examples:
        The following tests use the procedure is_independent, provided in module independence
        
        >>> from vec import Vec
        >>> from independence import is_independent
        >>> 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})
        >>> sb = subset_basis({a0, a1, a2, a3})
        >>> len(sb)
        3
        >>> all(v in [a0, a1, a2, a3] for v in sb)
        True
        >>> is_independent(sb)
        True

        >>> b0 = Vec({0,1,2,3},{0:2,1:2,3:4})
        >>> b1 = Vec({0,1,2,3},{0:1,1:1})
        >>> b2 = Vec({0,1,2,3},{2:3,3:4})
        >>> b3 = Vec({0,1,2,3},{3:3})
        >>> sb = subset_basis({b0, b1, b2, b3})
        >>> len(sb)
        3
        >>> all(v in [b0, b1, b2, b3] for v in sb)
        True
        >>> is_independent(sb)
        True

        >>> D = {'a','b','c','d'}
        >>> c0, c1, c2, c3, c4 = Vec(D,{'d': one, 'c': one}), Vec(D,{'d': one, 'a': one, 'c': one, 'b': one}), Vec(D,{'a': one}), Vec(D,{}), Vec(D,{'d': one, 'a': one, 'b': one})
        >>> subset_basis({c0,c1,c2,c3,c4}) == {c0,c1,c2,c4}
        True
    '''
    S = set(T)
    for s in S.copy():
        if is_superfluous(S, s):
            S.remove(s)
    return S


##: (Problem 6) Superset Basis Lemma in Python
def superset_basis(C, T):
    '''
    Input:
        - C: linearly independent set of Vecs
        - T: set of Vecs such that every Vec in C is in Span(T)
    Output:
        Linearly independent set S consisting of all Vecs in C and some in T
        such that the span of S is the span of T (i.e. S is a basis for the span
        of T).
    Example:
        >>> from vec import Vec
        >>> from independence import is_independent
        >>> 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})
        >>> sb = superset_basis({a0, a3}, {a0, a1, a2})
        >>> a0 in sb and a3 in sb
        True
        >>> is_independent(sb)
        True
        >>> all(x in [a0,a1,a2,a3] for x in sb)
        True
    '''
    S = set(C)
    S.union(set(T))
    for v in S:
        if is_superfluous(S, v):
            S.remove(v)
    return S


## 5: (Problem 7) My Is Independent Procedure
def my_is_independent(L):
    '''
    Input:
        - L: a list of Vecs
    Output:
        - boolean: true if the list is linearly independent
    Examples:
        >>> D = {0, 1, 2}
        >>> L = [Vec(D,{0: 1}), Vec(D,{1: 1}), Vec(D,{2: 1}), Vec(D,{0: 1, 1: 1, 2: 1}), Vec(D,{0: 1, 1: 1}), Vec(D,{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
        >>> L == [Vec(D,{0: 1}), Vec(D,{1: 1}), Vec(D,{2: 1}), Vec(D,{0: 1, 1: 1, 2: 1}), Vec(D,{0: 1, 1: 1}), Vec(D,{1: 1, 2: 1})]
        True
    '''
    return rank(L)==len(L)


## 6: (Problem 8) My Rank
def my_rank(L):
    '''
    Input: 
        - L: a list of Vecs
    Output: 
        - the rank of the list of Vecs
    Example:
        >>> L = [list2vec(v) for v in [[1,2,3],[4,5,6],[1.1,1.1,1.1]]]
        >>> my_rank(L)
        2
        >>> L == [list2vec(v) for v in [[1,2,3],[4,5,6],[1.1,1.1,1.1]]]
        True
        >>> my_rank([list2vec(v) for v in [[1,1,1],[2,2,2],[3,3,3],[4,4,4],[123,432,123]]])
        2
    '''
    return len(subset_basis(L))


## 8: (Problem 9) Direct Sum Unique Representation
def direct_sum_decompose(U_basis, V_basis, w):
    '''
    Input:
        - U_basis: a list of Vecs forming a basis for a vector space U
        - V_basis: a list of Vecs forming a basis for a vector space V
        - w: a Vec in the direct sum of U and V
    Output:
        - a pair (u, v) such that u + v = w, u is in U, v is in V
    Example:

        >>> D = {0,1,2,3,4,5}
        >>> U_basis = [Vec(D,{0: 2, 1: 1, 2: 0, 3: 0, 4: 6, 5: 0}), Vec(D,{0: 11, 1: 5, 2: 0, 3: 0, 4: 1, 5: 0}), Vec(D,{0: 3, 1: 1.5, 2: 0, 3: 0, 4: 7.5, 5: 0})]
        >>> V_basis = [Vec(D,{0: 0, 1: 0, 2: 7, 3: 0, 4: 0, 5: 1}), Vec(D,{0: 0, 1: 0, 2: 15, 3: 0, 4: 0, 5: 2})]
        >>> w = Vec(D,{0: 2, 1: 5, 2: 0, 3: 0, 4: 1, 5: 0})
        >>> (u, v) = direct_sum_decompose(U_basis, V_basis, w)
        >>> (u + v - w).is_almost_zero()
        True
        >>> U_matrix = coldict2mat(U_basis)
        >>> V_matrix = coldict2mat(V_basis)
        >>> (u - U_matrix*solve(U_matrix, u)).is_almost_zero()
        True
        >>> (v - V_matrix*solve(V_matrix, v)).is_almost_zero()
        True
        >>> ww = Vec(D,{0: 2, 1: 5, 2: 51, 4: 1, 5: 7})
        >>> (u, v) = direct_sum_decompose(U_basis, V_basis, ww)
        >>> (u + v - ww).is_almost_zero()
        True
        >>> (u - U_matrix*solve(U_matrix, u)).is_almost_zero()
        True
        >>> (v - V_matrix*solve(V_matrix, v)).is_almost_zero()
        True
        >>> U_basis == [Vec(D,{0: 2, 1: 1, 2: 0, 3: 0, 4: 6, 5: 0}), Vec(D,{0: 11, 1: 5, 2: 0, 3: 0, 4: 1, 5: 0}), Vec(D,{0: 3, 1: 1.5, 2: 0, 3: 0, 4: 7.5, 5: 0})]
        True
        >>> V_basis == [Vec(D,{0: 0, 1: 0, 2: 7, 3: 0, 4: 0, 5: 1}), Vec(D,{0: 0, 1: 0, 2: 15, 3: 0, 4: 0, 5: 2})]
        True
        >>> w == Vec(D,{0: 2, 1: 5, 2: 0, 3: 0, 4: 1, 5: 0})
        True
    '''
    U_V_basis = list(U_basis)
    U_V_basis.extend(V_basis)
    w_coordinates = vec2rep(U_V_basis, w)
    w_u_coord = list()
    w_v_coord = list()
    for n in range(len(U_V_basis)):
        if n < len(U_basis):
            w_u_coord.append(w_coordinates[n])
        else:
            w_v_coord.append(w_coordinates[n])

    u = rep2vec(list2vec(w_u_coord), U_basis)
    v = rep2vec(list2vec(w_v_coord), V_basis)
    return (u, v)


## 9: (Problem 10) Is Invertible Function
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

    >>> M1 = Mat(({0,1,2},{0,1,2}),{(0,0):1,(0,2):2,(1,2):3,(2,2):4})
    >>> is_invertible(M1)
    False
    '''
    coldict = mat2coldict(M)
    return (my_rank(coldict.values())==len(coldict.values()) and len(coldict.values())==len(mat2rowdict(M).values()))


## 10: (Problem 11) Inverse of a Matrix over GF(2)
def find_matrix_inverse(A):
    '''
    Input:
        - A: an invertible Mat over GF(2)
    Output:
        - A Mat that is the inverse of A
    Examples:
        >>> M1 = Mat(({0,1,2}, {0,1,2}), {(0, 1): one, (1, 0): one, (2, 2): one})
        >>> find_matrix_inverse(M1) == Mat(M1.D, {(0, 1): one, (1, 0): one, (2, 2): one})
        True
        >>> M2 = Mat(({0,1,2,3},{0,1,2,3}),{(0,1):one,(1,0):one,(2,2):one})
        >>> find_matrix_inverse(M2) == Mat(M2.D, {(0, 1): one, (1, 0): one, (2, 2): one})
        True
    '''
    cols = range(len(mat2coldict(A)))
    iden = mat2coldict(identity(A.D[0], one))
    B = list()
    for n in cols:
        ones = iden[n]
        B.append(solve(A, ones))
    return coldict2mat(B)

from triangular import triangular_solve


## 11: (Problem 12) Inverse of a Triangular Matrix
def find_triangular_matrix_inverse(A):
    '''
    Supporting GF2 is not required.

    Input:
        - A: an upper triangular Mat with nonzero diagonal elements
    Output:
        - Mat that is the inverse of A
    
    Example:
        >>> 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
    '''
    rows = range(len(mat2rowdict(A)))
    b = mat2coldict(identity(A.D[0], 1))
    rowlist = mat2rowdict(A)
    ret = list()
    for n in rows:
        ones = b[n]
        ret.append(triangular_solve(list(rowlist.values()), list(A.D[1]), ones))
    return coldict2mat(ret)