Python triangular_solve Examples, triangular.triangular_solve Python Examples

Example #1

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def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    from triangular import triangular_solve
    from QR import factor
    from matutil import mat2rowdict

    Q, R = factor(A)
    rowlist = [mat2rowdict(R)[v] for v in mat2rowdict(R)]
    label_list = sorted(A.D[1], key=repr)

    return triangular_solve(rowlist, label_list, b * Q)

Example #2

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File: Dimension.py Project: omrigildor/cs53

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
    '''
    I = {i:Vec(A.D[0],{i:1}) for i in A.D[1]}
    rl = list()
    ll = list()
    cd = dict()
    rd = mat2rowdict(A)
    for k,i in rd.items():
        ll.append(k)
        rl.append(rd[k])
    for k in ll:
        cd[k]= triangular_solve(rl,ll,I[k])
    return coldict2mat(cd)

Example #3

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File: hw7.py Project: fperezlo/matrix

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    from triangular import triangular_solve
    from QR import factor
    from matutil import mat2rowdict
    
    Q,R = factor(A)
    rowlist = [mat2rowdict(R)[v] for v in mat2rowdict(R)]
    label_list = sorted(A.D[1], key = repr)
    
    return triangular_solve(rowlist, label_list, b*Q)

Example #4

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File: hw7.py Project: atomminder/Coursera_Brown_Coding_the_matrix

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q,R = factor(A)
    return triangular_solve(mat2rowdict(R),list(A.D[1]), Q.transpose()*b)

# test
#domain = ({'a','b','c'},{'A','B'})
#A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
#Q, R = factor(A)
#b = Vec(domain[0], {'a': 1, 'b': -1})
#x = QR_solve(A, b)
#result = A.transpose()*(b-A*x)
#print(result * result < 1E-10)

Example #5

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def QR_solve(A, b):
    '''
    Input:
        - A: a Mat with linearly independent columns
        - b: a Vec whose domain equals the set of row-labels of A
    Output:
        - vector x that minimizes norm(b - A*x)
    Note: This procedure uses the procedure QR_factor, which in turn uses dict2list and list2dict.
           You wrote these procedures long back in python_lab.  Make sure the completed python_lab.py
           is in your matrix directory.
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR_factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result.is_almost_zero()
        True
    '''
    Q, R = QR_factor(A)
    c = Q.transpose() * b
    rowlist = mat2rowdict(R)
    labelslist = sorted(A.D[1], key=repr)
    return triangular_solve(rowlist, labelslist, c)

Example #6

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File: Orthogonalization_problems.py Project: JohnBurchell/Learning

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat with linearly independent columns
        - b: a Vec whose domain equals the set of row-labels of A
    Output:
        - vector x that minimizes norm(b - A*x)
    Note: This procedure uses the procedure QR_factor, which in turn uses dict2list and list2dict.
           You wrote these procedures long back in python_lab.  Make sure the completed python_lab.py
           is in your matrix directory.
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR_factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result.is_almost_zero()
        True
    '''
    Q,R = QR_factor(A)
    keylist = sorted(R.D[1], key=repr)
    c = Q.transpose() * b
    res = triangular_solve(mat2rowdict(R), keylist, c)
    return res

Example #7

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File: hw7.py Project: LukeLu1263/Matrix-part-I

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    from triangular import triangular_solve
    from QR import factor
    from dictutil import dict2list
    from vecutil import vec2list

    Q,R = factor(A)
    b = Q.transpose()*b
    rowdict = mat2rowdict(R)
    rowlist = dict2list(rowdict, list(rowdict.keys()))
    x = triangular_solve(rowlist, list(rowlist[0].D), vec2list(b))
    return x

Example #8

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File: hw7.py Project: chrisKingsley/Coursera_Coding_the_Matrix

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q,R = factor(A)
    c = Q.transpose()*b
    return triangular_solve(mat2rowdict(R), sorted(A.D[1],key=repr), c)

# A=Mat(({'a','b','c'},{'A','B'}), {('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1, ('c','B'):-2})
# b = Vec({'a','b','c'}, {'a':1,'b':-1})
# x = QR_solve(A,b)
# print(x)
# print(A.transpose()*(b-A*x))

# print(QR_solve(least_squares_A1, least_squares_b1))
# print(QR_solve(least_squares_A2, least_squares_b2))

Example #9

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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
    '''
    I = {i: Vec(A.D[0], {i: 1}) for i in A.D[1]}
    rl = list()
    ll = list()
    cd = dict()
    rd = mat2rowdict(A)
    for k, i in rd.items():
        ll.append(k)
        rl.append(rd[k])
    for k in ll:
        cd[k] = triangular_solve(rl, ll, I[k])
    return coldict2mat(cd)

Example #10

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File: hw7.py Project: Shubashree/matrix_coding

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    from QR import factor
    from triangular import triangular_solve
    from mat import transpose
    from matutil import mat2rowdict
    Q, R = factor(A)
    qT = Q.transpose()
    c = qT * b
    rows = [rowVec for index, rowVec in mat2rowdict(R).items()]
    colLabels = sorted(A.D[1], key=repr)
    x_hat = triangular_solve(rows, colLabels, c)
    return x_hat

Example #11

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def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q, R = factor(A)
    c = Q.transpose() * b
    return triangular_solve(mat2rowdict(R), sorted(A.D[1], key=repr), c)


# A=Mat(({'a','b','c'},{'A','B'}), {('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1, ('c','B'):-2})
# b = Vec({'a','b','c'}, {'a':1,'b':-1})
# x = QR_solve(A,b)
# print(x)
# print(A.transpose()*(b-A*x))

# print(QR_solve(least_squares_A1, least_squares_b1))
# print(QR_solve(least_squares_A2, least_squares_b2))

Example #12

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File: hw5.py Project: pashanitw/coding-the-matrix

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
    """
    I_col_dict = {i: Vec(A.D[0], {i: 1}) for i in A.D[1]}
    # def triangular_solve(rowlist, label_list, b):
    rowlist = list()
    label_list = list()
    from matutil import mat2rowdict

    rowdict = mat2rowdict(A)
    for key in rowdict.keys():
        label_list.append(key)
        rowlist.append(rowdict[key])

    B_coldict = dict()
    from triangular import triangular_solve

    for key in label_list:
        B_coldict[key] = triangular_solve(rowlist, label_list, I_col_dict[key])
    return coldict2mat(B_coldict)

Example #13

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File: hw7.py Project: SherMM/matrix_coding

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    from QR import factor
    from triangular import triangular_solve
    from mat import transpose
    from matutil import mat2rowdict
    Q, R = factor(A)
    qT = Q.transpose()
    c = qT*b
    rows = [rowVec for index,rowVec in mat2rowdict(R).items()]
    colLabels = sorted(A.D[1], key=repr)
    x_hat = triangular_solve(rows,colLabels,c)
    return x_hat

Example #14

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File: hw5.py Project: zezei/linearAlgebra-coursera

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
    '''

    R = len(A.D[0])
    C = len(A.D[1])
    coldict = {}
    for i in range(C):
        b = [0] * R
        b[i] = 1
        s = triangular_solve(list(matutil.mat2rowdict(A).values()),
                             list(A.D[0]), list2vec(b))
        coldict[i] = s

    return coldict2mat(coldict)


# import hw5
# import hw4
# from mat import Mat
# import vecutil
# import matutil
# from vec import Vec
# from GF2 import one
# 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})]
# hw5.my_is_independent(L)
# hw5.my_is_independent(L[:2])
# S = [vecutil.list2vec(v) for v in [[1,0,0],[0,1,0],[0,0,1]]]
# B = [vecutil.list2vec(v) for v in [[1,1,0],[0,1,1],[1,0,1]]]
# ff = hw5.morph(S, B)
# ff == [(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}))]
# 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 = hw5.subset_basis([a0,a1,a2,a3])
# sb == [Vec({'c', 'b', 'a', 'd'},{'a': 1}), Vec({'c', 'b', 'a', 'd'},{'b': 1}), Vec({'c', 'b', 'a', 'd'},{'c': 1})]
# 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})
# ss = hw5.direct_sum_decompose(U_basis, V_basis, w)
# ss == (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}))
# 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})
# hw5.is_invertible(M)
# A = Mat(({0, 1}, {0, 1, 2}), {(0, 1): 2, (1, 2): 1, (0, 0): 1, (1, 0): 3, (0, 2): 3, (1, 1): 1})
# B = 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})
# C = Mat(({0, 1, 2}, {0, 1}), {(0, 1): 0, (2, 0): 2, (0, 0): 1, (1, 0): 0, (1, 1): 1, (2, 1): 1})
# D = Mat(({0, 1, 2}, {0, 1, 2}), {(0, 1): 5, (1, 2): 2, (0, 0): 1, (2, 0): 4, (1, 0): 2, (2, 2): 7, (0, 2): 8, (2, 1): 6, (1, 1): 5})
# E = Mat(({0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}), {(1, 2): 7, (3, 2): 7, (0, 0): 3, (3, 0): 1, (0, 4): 3, (1, 4): 2, (1, 3): 4, (2, 3): 0, (2, 1): 56, (2, 4): 5, (4, 2): 6, (1, 0): 2, (0, 3): 7, (4, 0): 2, (0, 1): 5, (3, 3): 4, (4, 1): 4, (3, 1): 23, (4, 4): 5, (0, 2): 7, (2, 0): 2, (4, 3): 8, (2, 2): 9, (3, 4): 2, (1, 1): 4})
# 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})
# hw5.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})
# A = matutil.listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
# hw5.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})

Example #15

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File: hw5.py Project: buptdjd/linearAlgebra-coursera

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
    '''

    R = len(A.D[0])
    C = len(A.D[1])
    coldict = {}
    for i in range(C):
        b = [0] * R
        b[i] = 1
        s = triangular_solve(list(matutil.mat2rowdict(A).values()), list(A.D[0]), list2vec(b))
        coldict[i] = s

    return coldict2mat(coldict)



# import hw5
# import hw4
# from mat import Mat
# import vecutil
# import matutil
# from vec import Vec
# from GF2 import one
# 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})]
# hw5.my_is_independent(L)
# hw5.my_is_independent(L[:2])
# S = [vecutil.list2vec(v) for v in [[1,0,0],[0,1,0],[0,0,1]]]
# B = [vecutil.list2vec(v) for v in [[1,1,0],[0,1,1],[1,0,1]]]
# ff = hw5.morph(S, B)
# ff == [(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}))]
# 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 = hw5.subset_basis([a0,a1,a2,a3])
# sb == [Vec({'c', 'b', 'a', 'd'},{'a': 1}), Vec({'c', 'b', 'a', 'd'},{'b': 1}), Vec({'c', 'b', 'a', 'd'},{'c': 1})]
# 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})
# ss = hw5.direct_sum_decompose(U_basis, V_basis, w)
# ss == (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}))
# 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})
# hw5.is_invertible(M)
# A = Mat(({0, 1}, {0, 1, 2}), {(0, 1): 2, (1, 2): 1, (0, 0): 1, (1, 0): 3, (0, 2): 3, (1, 1): 1})
# B = 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})
# C = Mat(({0, 1, 2}, {0, 1}), {(0, 1): 0, (2, 0): 2, (0, 0): 1, (1, 0): 0, (1, 1): 1, (2, 1): 1})
# D = Mat(({0, 1, 2}, {0, 1, 2}), {(0, 1): 5, (1, 2): 2, (0, 0): 1, (2, 0): 4, (1, 0): 2, (2, 2): 7, (0, 2): 8, (2, 1): 6, (1, 1): 5})
# E = Mat(({0, 1, 2, 3, 4}, {0, 1, 2, 3, 4}), {(1, 2): 7, (3, 2): 7, (0, 0): 3, (3, 0): 1, (0, 4): 3, (1, 4): 2, (1, 3): 4, (2, 3): 0, (2, 1): 56, (2, 4): 5, (4, 2): 6, (1, 0): 2, (0, 3): 7, (4, 0): 2, (0, 1): 5, (3, 3): 4, (4, 1): 4, (3, 1): 23, (4, 4): 5, (0, 2): 7, (2, 0): 2, (4, 3): 8, (2, 2): 9, (3, 4): 2, (1, 1): 4})
# 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})
# hw5.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})
# A = matutil.listlist2mat([[1, .5, .2, 4],[0, 1, .3, .9],[0,0,1,.1],[0,0,0,1]])
# hw5.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})

Example #16

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File: hw5.py Project: srikanth235/Matrix

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
    '''
    return coldict2mat([triangular_solve([row for row in mat2rowdict(A).values()], range(len(A.D[0])), Vec(A.D[0],{i:1})) for i in range(len(A.D[0]))])

Example #17

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File: hw5.py Project: pcp135/C-CtM

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
		'''
		(v,k)=(list(mat2rowdict(A).values()),list(mat2rowdict(A).keys()))
		return coldict2mat([triangular_solve(v,k,Vec(A.D[1], {e:1})) for e in A.D[1]])

Example #18

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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
		'''
    (v, k) = (list(mat2rowdict(A).values()), list(mat2rowdict(A).keys()))
    return coldict2mat(
        [triangular_solve(v, k, Vec(A.D[1], {e: 1})) for e in A.D[1]])

Example #19

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def find_triangular_matrix_inverse(A):
	L=[]
	label_list = []
	I = identity(A.D[0],1)
	R = mat2rowdict(A)
	w = mat2rowdict(I)
	label_list.extend(range(len(R)))
	for k,v in w.items():
		print(k,v,)
		s = triangular_solve(R, label_list, v)
		L.append(s)
	return coldict2mat(L)

Example #20

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File: hw5.py Project: manishmmulani/python_repo

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
    '''
    solution = identity(A.D[0],1)
    solutionrowdict = mat2rowdict(solution)
    Arowveclist = list(mat2rowdict(A).values())
    label_list = list(range(len(Arowveclist)))
    return coldict2mat({key:triangular_solve(Arowveclist, label_list, vec) for key,vec in solutionrowdict.items()})

Example #21

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File: hw5.py Project: airy-beam/online_courses

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
    '''
    cols = []
    i = identity(A.D[0], 1)
    i_cols = mat2coldict(i)
    for k in A.D[0]:
        cols.append( triangular_solve([ mat2rowdict(A)[k] for k in mat2rowdict(A) ], list(range(len(A.D[0]))), i_cols[k]) )
    return coldict2mat(cols)

Example #22

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File: hw5.py Project: johnmerm/matrix

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
    """
    cd = list()
    label_list = list(mat2rowdict(A).keys())
    rowlist = list(mat2rowdict(A).values())
    for j in label_list:
        c = triangular_solve(rowlist, label_list, list2vec([1 if i == j else 0 for i in label_list]))
        cd.append(c)

    return coldict2mat(cd)

Example #23

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File: hw5.py Project: niujie/CMLACSA

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
    '''
    from triangular import triangular_solve
    from matutil import mat2rowdict
    sol = []
    rowlist = mat2rowdict(A)
    label_list = list(rowlist[0].D)
    for i in A.D[0]:
        sol = sol + [triangular_solve(rowlist,label_list,Vec(A.D[0],{i:1}))]
    return coldict2mat(sol)

Example #24

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File: hw5.py Project: LukeLu1263/Matrix-part-I

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
    '''
    D = A.D[0]
    I = Mat((D,D), {(d,d):1 for d in D})
    col_I = mat2coldict(I)
    rowdict_A = mat2rowdict(A)
    rowlist_A = dict2list(rowdict_A, list(D))
    B = dict()
    for i in col_I.keys():
        B[i] = triangular_solve(rowlist_A, list(D), col_I[i])
    return coldict2mat(B)

Example #25

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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
    '''
    aa = [a for a in mat2rowdict(A).values()]
    dd = [d for d in A.D[0]]
    return coldict2mat([triangular_solve(aa, dd, i) for i in mat2coldict(identity(A.D[0], 1)).values()])

Example #26

0

Show file

File: Dimension_problems.py Project: mintzopg/CodingTheMatrix_code

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
    '''
    label_list = list(A.D[1])
    rowlist = list(mat2rowdict(A).values())
    return coldict2mat([triangular_solve(rowlist, label_list, Vec(A.D[1], {i:1})) for i in A.D[1]])

Example #27

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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
'''

    inverseColDict = dict()
    rowList = [val for val in mat2rowdict(A).values()]
    label_list = [d for d in A.D[0]]
    identVecList = [ val for val in mat2coldict(identity(A.D[0] , 1)).values()]
    for k in range(len(rowList)):
        inverseColDict[k] = triangular_solve(rowList, label_list, identVecList[k])
    
    return coldict2mat(inverseColDict)

Example #28

0

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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
    '''
    counter = 0
    resultantVec = Vec(A.D[0], {i : 0 for i in A.D[0]})
    matrix = list()
    while counter < len(A.D[0]):
        resultantVec.f[list(A.D[0])[counter]] = 1
        if counter > 0:
            resultantVec.f[list(A.D[0])[counter - 1]] = 0
        matrix.append(triangular_solve(list(mat2rowdict(A).values()), list(A.D[1]), resultantVec))
        counter += 1
    return coldict2mat(matrix)

Example #29

0

Show file

File: Dimension_problems.py Project: jeantardelli/coding_the_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
    '''
    rowlist = [v for _,v in mat2rowdict(A).items()]
    label_list = [label for label in A.D[0]]
    b = [[0 if j!= i else 1 for j in range(len(A.D[0]))] for i in range(len(A.D[1]))]
    veclist = {i: triangular_solve(rowlist, label_list, b[i]) for i in range(len(b))}
    return coldict2mat(veclist)

Example #30

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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
    '''
    label_list = list(A.D[1])
    rowlist = list(mat2rowdict(A).values())
    return coldict2mat([
        triangular_solve(rowlist, label_list, Vec(A.D[1], {i: 1}))
        for i in A.D[1]
    ])

Example #31

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File: Dimension_problems.py Project: CodemanNeil/CodingTheMatrix

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
    '''
    num_equations = len(A.D[0])
    b_vecs = [Vec(A.D[0], {i: 1}) for i in range(num_equations)]
    inverse_cols = [
        triangular_solve(mat2rowdict(A), list(A.D[1]), b_vecs[i])
        for i in range(num_equations)
    ]
    return coldict2mat(inverse_cols)

Example #32

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File: hw7.py Project: manishmmulani/python_repo

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q,R = factor(A)
    Qt = Q.transpose()
    Rx = Qt * b
    return triangular_solve(list(mat2rowdict(R).values()), sorted(R.D[1], key=repr), Rx)

Example #33

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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
    '''
    x = []
    n = len(A.D[0])
    rowlist = list(mat2rowdict(A).values())
    for i in range(n):
        bi = list2vec([1 if i == j else 0 for j in range(n)])
        x.append(triangular_solve(rowlist, range(n), bi))
    return coldict2mat(x)

Example #34

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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
    '''
    from triangular import triangular_solve
    A_rows = list(mat2rowdict(A).values())

    B = []  #initializing inverse matrix
    A_nrows = len(A.D[0])
    A_rowspace = set(range(A_nrows))

    for i in range(A_nrows):
        v = Vec(A_rowspace, {i: 1})  # corresponding col of I matrix
        b = triangular_solve(A_rows, range(A_nrows), v)
        B.append(b)  #add to list of inverse matrix columns

    return coldict2mat(B)

Example #35

0

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File: Dimension_problems.py Project: HatlessFox/SelfStudy

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
    '''
    inverse_cols = []
    dim = len(A.D[0])
    for i in range(dim):
        b = Vec(set(range(dim)), {i:1})
        inv_col = triangular_solve(list(mat2rowdict(A).values()), range(dim), b)
        inverse_cols.append(inv_col)
    return coldict2mat(inverse_cols)

Example #36

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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
    '''
    L = []
    label_list = []
    I = identity(A.D[0], 1)
    R = mat2rowdict(A)
    w = mat2rowdict(I)
    label_list.extend(range(len(R)))
    for k, v in w.items():
        s = triangular_solve(R, label_list, v)
        L.append(s)
    return coldict2mat(L)

Example #37

0

Show file

File: Dimension_problems.py Project: shubang93/code-the-matrix-hw

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)

Example #38

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def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q, R = factor(A)
    Qt = Q.transpose()
    Rx = Qt * b
    return triangular_solve(list(mat2rowdict(R).values()),
                            sorted(R.D[1], key=repr), Rx)

Example #39

0

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File: hw5.py Project: df1111/CodingTheMatrix

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
    '''
    L=[]
    label_list = []
    I = identity(A.D[0],1)
    R = mat2rowdict(A)
    w = mat2rowdict(I)
    label_list.extend(range(len(R)))
    for k,v in w.items():
        s = triangular_solve(R, label_list, v)
        L.append(s)
    return coldict2mat(L)

Example #40

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def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    import QR
    from triangular import triangular_solve

    (Q, R) = QR.factor(A)
    Rlist = [mat2rowdict(R)[i] for i in R.D[0]]
    return triangular_solve(Rlist, sorted(A.D[1], key=repr), b * Q)

Example #41

0

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File: hw7.py Project: tri2sing/CodingTheMatrix

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    Q, R = factor(A)
    c = Q.transpose()*b
    Rdict = mat2rowdict(R)
    Rrows = [Rdict[key] for key in Rdict]
    labels = sorted(A.D[1], key=repr)
    return triangular_solve(Rrows, labels, c)

Example #42

0

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File: hw7.py Project: fandres70/matrix

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    import QR
    from triangular import triangular_solve
    
    (Q,R) = QR.factor(A)
    Rlist = [mat2rowdict(R)[i] for i in R.D[0]]
    return triangular_solve(Rlist, sorted(A.D[1], key=repr), b*Q)

Example #43

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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
    '''
    I_col_dict = {i:Vec(A.D[0],{i:1}) for i in A.D[1]}
    #def triangular_solve(rowlist, label_list, b):
    rowlist = list()
    label_list = list()
    from matutil import mat2rowdict
    
    rowdict = mat2rowdict(A)
    for key in rowdict.keys():
        label_list.append(key)
        rowlist.append(rowdict[key])
    
    B_coldict = dict()
    from triangular import triangular_solve
    for key in label_list:
        B_coldict[key] = triangular_solve(rowlist,label_list,I_col_dict[key])
    return coldict2mat(B_coldict)

Example #44

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def QR_solve(A, b):
    '''
Input:
- A: a Mat
- b: a Vec
Output:
- vector x that minimizes norm(b - A*x)
Example:
>>> domain = ({'a','b','c'},{'A','B'})
>>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
>>> Q, R = QR.factor(A)
>>> b = Vec(domain[0], {'a': 1, 'b': -1})
>>> x = QR_solve(A, b)
>>> result = A.transpose()*(b-A*x)
>>> result * result < 1E-10
True
'''
    from triangular import triangular_solve
    from QR import factor
    from matutil import mat2rowdict
    Q, R = factor(A)
    c = Q.transpose() * b
    x = triangular_solve(list(mat2rowdict(R).values()), list(A.D[1]), c)
    return x

Example #45

0

Show file

File: hw7.py Project: niujie/CMLACSA

def QR_solve(A, b):
    '''
    Input:
        - A: a Mat
        - b: a Vec
    Output:
        - vector x that minimizes norm(b - A*x)
    Example:
        >>> domain = ({'a','b','c'},{'A','B'})
        >>> A = Mat(domain,{('a','A'):-1, ('a','B'):2,('b','A'):5, ('b','B'):3,('c','A'):1,('c','B'):-2})
        >>> Q, R = QR.factor(A)
        >>> b = Vec(domain[0], {'a': 1, 'b': -1})
        >>> x = QR_solve(A, b)
        >>> result = A.transpose()*(b-A*x)
        >>> result * result < 1E-10
        True
    '''
    from triangular import triangular_solve
    from QR import factor
    from matutil import mat2rowdict
    Q, R = factor(A)
    c = Q.transpose() * b
    x = triangular_solve(list(mat2rowdict(R).values()), list(A.D[1]), c)
    return x

Example #46

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File: HW.py Project: vasuharish/Coding_the_matrix

def QR_solve(A, b):
    Q, R = factor(A)
    R_rowlist = mat2rowdict(R)
    label_list = sorted(A.D[1], key=repr)
    return triangular_solve(R_rowlist, label_list, Q.transpose() * b)

Example #47

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File: HW.py Project: vasuharish/Coding_the_matrix

def new_tria_solve(rowlist, b):
    collist = remove_irr_col(rowlist)
    new_rowlist = [
        values for (keys, values) in mat2rowdict(coldict2mat(collist)).items()
    ]
    return triangular_solve(new_rowlist, list(new_rowlist[0].D), b)