Coding the Matrix Week 3 The Matrix 矩陣

  本周共有三次作業。所花費的時間為一天左右，還算可以，需要注意的是考慮一些特殊情況，寫出能夠通用的程式，這就行了。

  體會

  set()和{}結果相同，可以通用，可以採取後者。

>>> type({})<class 'dict'>>>> type(set())<class 'set'>>>> type(dict())<class 'dict'>>>> {}==set()False>>> {}==dict()True

  作業1 hw3

  這一節講的是矩陣的運算。值得注意的是左乘和右乘疏鬆陣列的意義，和矩陣求逆的方法和線性方程組的通解。

  左乘矩陣，對於這個疏鬆陣列的任意點（i,j），相當於把矩陣第j行加到結果的i行。右乘矩陣，對於點（i,j），相當於把矩陣第i列加到結果的j列。用來計算包含疏鬆陣列的乘法十分方便。

  代碼如下

# version code 893# Please fill out this stencil and submit using the provided submission script.from mat import Matfrom vec import Vecfrom matutil import *## Problem 1# Please represent your solutions as lists.vector_matrix_product_1 = [1,0]vector_matrix_product_2 = [0,4.44]vector_matrix_product_3 = [14,20,26]## Problem 2# Represent your solution as a list of rows.# For example, the identity matrix would be [[1,0],[0,1]].M_swap_two_vector = [[0,1],[1,0]]## Problem 3three_by_three_matrix = [[1,0,1],[0,1,0],[1,0,0]] # Represent with a list of rows lists.## Problem 4multiplied_matrix = [[2,0,0],[0,4,0],[0,0,3]] # Represent with a list of row lists.## Problem 5# Please enter a boolean representing if the multiplication is valid.# If it is not valid, please enter None for the dimensions.part_1_valid = False # True or Falsepart_1_number_rows = None # Integer or Nonepart_1_number_cols = None # Integer or Nonepart_2_valid = Falsepart_2_number_rows = Nonepart_2_number_cols = Nonepart_3_valid = Truepart_3_number_rows = 1part_3_number_cols = 2part_4_valid = Truepart_4_number_rows = 2part_4_number_cols = 1part_5_valid = Falsepart_5_number_rows = Nonepart_5_number_cols = Nonepart_6_valid = Truepart_6_number_rows = 1part_6_number_cols = 1part_7_valid = Truepart_7_number_rows = 3part_7_number_cols = 3## Problem 6# Please represent your answer as a list of row lists.small_mat_mult_1 = [[8,13],[8,14]]small_mat_mult_2 = [[24,11,4],[1,3,0]]small_mat_mult_3 = [[3,13]]small_mat_mult_4 = [[14]]small_mat_mult_5 = [[1,2,3],[2,4,6],[3,6,9]]small_mat_mult_6 = [[-2,4],[1,1],[1,-3]]## Problem 7# Please represent your solution as a list of row lists.part_1_AB = [[5,2,0,1],[2,1,-4,6],[2,3,0,-4],[-2,3,4,0]]part_1_BA = [[1,-4,6,2],[3,0,-4,2],[3,4,0,-2],[2,0,1,5]]part_2_AB = [[5,1,0,2],[2,6,-4,1],[2,-4,0,3],[-2,0,4,3]]part_2_BA = [[3,4,0,-2],[3,0,-4,2],[1,-4,6,2],[2,0,1,5]]part_3_AB = [[1,0,5,2],[6,-4,2,1],[-4,0,2,3],[0,4,-2,3]]part_3_BA = [[3,4,0,-2],[1,-4,6,2],[2,0,1,5],[3,0,-4,2]]## Problem 8# Please represent your answer as a list of row lists.# Please represent the variables a and b as strings.# Represent multiplication of the variables, make them one string.# For example, the sum of 'a' and 'b' would be 'a+b'.matrix_matrix_mult_1    = [[1,'a+b'],[0,1]]matrix_matrix_mult_2_A2 = [[1,2],[0,1]]matrix_matrix_mult_2_A3 = [[1,3],[0,1]]# Use the string 'n' to represent variable the n in A^n.matrix_matrix_mult_2_An = [[1,'n'],[0,1]]## Problem 9# Please represent your answer as a list of row lists.your_answer_a_AB = [[0,0,2,0],[0,0,5,0],[0,0,4,0],[0,0,6,0]]your_answer_a_BA = [[0,0,0,0],[4,4,4,0],[0,0,0,0],[0,0,0,0]]your_answer_b_AB = [[0,2,-1,0],[0,5,3,0],[0,4,0,0],[0,6,-5,0]]your_answer_b_BA = [[0,0,0,0],[1,5,-2,3],[0,0,0,0],[4,4,4,0]]your_answer_c_AB = [[6,0,0,0],[6,0,0,0],[8,0,0,0],[5,0,0,0]]your_answer_c_BA = [[4,2,1,-1],[4,2,1,-1],[0,0,0,0],[0,0,0,0]]your_answer_d_AB = [[0,3,0,4],[0,4,0,1],[0,4,0,4],[0,-6,0,-1]]your_answer_d_BA = [[0,11,0,-2],[0,0,0,0],[0,0,0,0],[1,5,-2,3]]your_answer_e_AB = [[0,3,0,8],[0,-9,0,2],[0,0,0,8],[0,15,0,-2]]your_answer_e_BA = [[-2,12,4,-10],[0,0,0,0],[0,0,0,0],[-3,-15,6,-9]]your_answer_f_AB = [[-4,4,2,-3],[-1,10,-4,9],[-4,8,8,0],[1,12,4,-15]]your_answer_f_BA = [[-4,-2,-1,1],[2,10,-4,6],[8,8,8,0],[-3,18,6,-15]]## Problem 10column_row_vector_multiplication1 = Vec({0, 1}, {0:13,1:20})column_row_vector_multiplication2 = Vec({0, 1, 2}, {0:24,1:11,2:4})column_row_vector_multiplication3 = Vec({0, 1, 2, 3}, {0:4,1:8,2:11,3:3})column_row_vector_multiplication4 = Vec({0,1}, {0:30,1:16})column_row_vector_multiplication5 = Vec({0, 1, 2}, {0:-3,1:1,2:9})## Problem 11def lin_comb_mat_vec_mult(M, v):    assert(M.D[1] == v.D)    nm=mat2coldict(M)    return sum([v[x]*nm[x] for x in M.D[1]])## Problem 12def lin_comb_vec_mat_mult(v, M):    assert(v.D == M.D[0])    nm=mat2rowdict(M)    return sum([v[x]*nm[x] for x in M.D[0]])## Problem 13def dot_product_mat_vec_mult(M, v):    assert(M.D[1] == v.D)    nm=mat2rowdict(M)    return Vec(M.D[0],{x:v*nm[x] for x in M.D[0]})## Problem 14def dot_product_vec_mat_mult(v, M):    assert(v.D == M.D[0])    nm=mat2coldict(M)    return Vec(M.D[1],{x:v*nm[x] for x in M.D[1]})## Problem 15def Mv_mat_mat_mult(A, B):    assert A.D[1] == B.D[0]    nm=mat2coldict(B)    return coldict2mat({x:A*nm[x] for x in B.D[1]})## Problem 16def vM_mat_mat_mult(A, B):    assert A.D[1] == B.D[0]    na=mat2rowdict(A)    return rowdict2mat({x:na[x]*B for x in A.D[0]})## Problem 17def dot_prod_mat_mat_mult(A, B):    assert A.D[1] == B.D[0]    nb=mat2coldict(B)    na=mat2rowdict(A)    return Mat((A.D[0],B.D[1]),{(x,y):na[x]*nb[y] for x in A.D[0] for y in B.D[1]})## Problem 18solving_systems_x1 = -0.2solving_systems_x2 = 0.4solving_systems_y1 = 0.8solving_systems_y2 = -0.6solving_systems_m = Mat(({0, 1}, {0, 1}), {(0,0):-0.2,(0,1):0.8,(1,0):0.4,(1,1):-0.6})solving_systems_a = Mat(({0, 1}, {0, 1}), {(0,0):3,(0,1):4,(1,0):2,(1,1):1})solving_systems_a_times_m = Mat(({0, 1}, {0, 1}), {(0,0):1,(0,1):0,(1,0):0,(1,1):1})solving_systems_m_times_a = Mat(({0, 1}, {0, 1}), {(0,0):1,(0,1):0,(1,0):0,(1,1):1})## Problem 19# Please write your solutions as booleans (True or False)are_inverses1 = Trueare_inverses2 = Trueare_inverses3 = Falseare_inverses4 = False

  作業2 mat

  這個作業是完成mat class，重載了很多運算子。值得注意的是，getitem對於程式中的稀疏儲存方式，在dict中如無法找到對應項應該返回0.做的時候注意細節。

  代碼如下：

from vec import Vecfrom GF2 import onedef getitem(M, k):    "Returns the value of entry k in M.  The value of k should be a pair."    assert k[0] in M.D[0] and k[1] in M.D[1]    return M.f[k] if k in M.f.keys() else 0def setitem(M, k, val):    "Sets the element of v with label k to be val.  The value of k should be a pair"    assert k[0] in M.D[0] and k[1] in M.D[1]    M.f[k]=valdef add(A, B):    "Returns the sum of A and B"    assert A.D == B.D    return Mat(A.D,{(x,y):A[x,y]+B[x,y] for x in A.D[0] for y in A.D[1]})def scalar_mul(M, alpha):    "Returns the product of scalar alpha with M"     return Mat(M.D,{x:y*alpha for x,y in M.f.items()})def equal(A, B):    "Returns true iff A is equal to B"    assert A.D == B.D    for x in A.D[0]:        for y in A.D[1]:            if A[x,y]!=B[x,y]:return False    return Truedef transpose(M):    "Returns the transpose of M"    return Mat((M.D[1],M.D[0]),{(b,a):y for (a,b),y in M.f.items()})def vector_matrix_mul(v, M):    "Returns the product of vector v and matrix M"    assert M.D[0] == v.D    return Vec(M.D[1],{x:sum([v[y]*M[y,x] for y in M.D[0]]) for x in M.D[1]})def matrix_vector_mul(M, v):    "Returns the product of matrix M and vector v"    assert M.D[1] == v.D    return Vec(M.D[0],{x:sum([v[y]*M[x,y] for y in M.D[1]]) for x in M.D[0]})def matrix_matrix_mul(A, B):    "Returns the product of A and B"    assert A.D[1] == B.D[0]    return Mat((A.D[0],B.D[1]),{(x,y):sum([A[x,t]*B[t,y] for t in A.D[1]]) for x in A.D[0] for y in B.D[1]})################################################################################class Mat:    def __init__(self, labels, function):        self.D = labels        self.f = function    __getitem__ = getitem    __setitem__ = setitem    transpose = transpose    def __neg__(self):        return (-1)*self    def __mul__(self,other):        if Mat == type(other):            return matrix_matrix_mul(self,other)        elif Vec == type(other):            return matrix_vector_mul(self,other)        else:            return scalar_mul(self,other)            #this will only be used if other is scalar (or not-supported). mat and vec both have __mul__ implemented    def __rmul__(self, other):        if Vec == type(other):            return vector_matrix_mul(other, self)        else:  # Assume scalar            return scalar_mul(self, other)    __add__ = add    def __sub__(a,b):        return a+(-b)    __eq__ = equal    def copy(self):        return Mat(self.D, self.f.copy())    def __str__(M, rows=None, cols=None):        "string representation for print()"        if rows == None:            try:                rows = sorted(M.D[0])            except TypeError:                rows = sorted(M.D[0], key=hash)        if cols == None:            try:                cols = sorted(M.D[1])            except TypeError:                cols = sorted(M.D[1], key=hash)        separator = ' | '        numdec = 3        pre = 1+max([len(str(r)) for r in rows])        colw = {col:(1+max([len(str(col))] + [len('{0:.{1}G}'.format(M[row,col],numdec)) if isinstance(M[row,col], int) or isinstance(M[row,col], float) else len(str(M[row,col])) for row in rows])) for col in cols}        s1 = ' '*(1+ pre + len(separator))        s2 = ''.join(['{0:>{1}}'.format(c,colw[c]) for c in cols])        s3 = ' '*(pre+len(separator)) + '-'*(sum(list(colw.values())) + 1)        s4 = ''.join(['{0:>{1}} {2}'.format(r, pre,separator)+''.join(['{0:>{1}.{2}G}'.format(M[r,c],colw[c],numdec) if isinstance(M[r,c], int) or isinstance(M[r,c], float) else '{0:>{1}}'.format(M[r,c], colw[c]) for c in cols])+'\n' for r in rows])        return '\n' + s1 + s2 + '\n' + s3 + '\n' + s4    def pp(self, rows, cols):        print(self.__str__(rows, cols))    def __repr__(self):        "evaluatable representation"        return "Mat(" + str(self.D) +", " + str(self.f) + ")"

  作業3 ecc_lab

  這個是相對來說比較難一點，因為是關於漢明碼編碼解碼的。做的時候花了不少時間，不過把編解碼的過程弄清楚就沒有問題了，回頭來看還是比較簡單的。

  代碼如下：

from vec import Vecfrom mat import Matfrom bitutil import *from GF2 import onefrom matutil import *## Task 1 part 1""" Create an instance of Mat representing the generator matrix G. You can usethe procedure listlist2mat in the matutil module (be sure to import first).Since we are working over GF (2), you should use the value one from theGF2 module to represent 1"""G = listlist2mat([[one,0,one,one],[one,one,0,one],[0,0,0,one],[one,one,one,0],[0,0,one,0],[0,one,0,0],[one,0,0,0]])## Task 1 part 2# Please write your answer as a list. Use one from GF2 and 0 as the elements.encoding_1001 = [0,0,one,one,0,0,one]## Task 2# Express your answer as an instance of the Mat class.R = listlist2mat([[0,0,0,0,0,0,one],[0,0,0,0,0,one,0],[0,0,0,0,one,0,0],[0,0,one,0,0,0,0]])## Task 3# Create an instance of Mat representing the check matrix H.H = listlist2mat([[0,0,0,one,one,one,one],[0,one,one,0,0,one,one],[one,0,one,0,one,0,one]])## Task 4 part 1def find_error(e):    """    Input: an error syndrome as an instance of Vec    Output: the corresponding error vector e    Examples:        >>> find_error(Vec({0,1,2}, {0:one}))        Vec({0, 1, 2, 3, 4, 5, 6},{3: one})        >>> find_error(Vec({0,1,2}, {2:one}))        Vec({0, 1, 2, 3, 4, 5, 6},{0: one})        >>> find_error(Vec({0,1,2}, {1:one, 2:one}))        Vec({0, 1, 2, 3, 4, 5, 6},{2: one})        """    nh=mat2coldict(H)    for t in H.D[1]:        if nh[t]==e:            return Vec(H.D[1],{t:one})    return Vec(H.D[1],{})## Task 4 part 2# Use the Vec class for your answers.non_codeword = Vec({0,1,2,3,4,5,6}, {0: one, 1:0, 2:one, 3:one, 4:0, 5:one, 6:one})error_vector = Vec({0, 1, 2, 3, 4, 5, 6},{6: one})code_word = Vec({0, 1, 2, 3, 4, 5, 6},{0: one, 1: 0, 2: one, 3: one, 4: 0, 5: one, 6: 0})original = Vec({0, 1, 2, 3},{0: 0, 1: one, 2: 0, 3: one}) # R * code_word## Task 5def find_error_matrix(S):    """    Input: a matrix S whose columns are error syndromes    Output: a matrix whose cth column is the error corresponding to the cth column of S.    Example:        >>> S = listlist2mat([[0,one,one,one],[0,one,0,0],[0,0,0,one]])        >>> find_error_matrix(S)        Mat(({0, 1, 2, 3, 4, 5, 6}, {0, 1, 2, 3}), {(1, 2): 0, (3, 2): one, (0, 0): 0, (4, 3): one, (3, 0): 0, (6, 0): 0, (2, 1): 0, (6, 2): 0, (2, 3): 0, (5, 1): one, (4, 2): 0, (1, 0): 0, (0, 3): 0, (4, 0): 0, (0, 1): 0, (3, 3): 0, (4, 1): 0, (6, 1): 0, (3, 1): 0, (1, 1): 0, (6, 3): 0, (2, 0): 0, (5, 0): 0, (2, 2): 0, (1, 3): 0, (5, 3): 0, (5, 2): 0, (0, 2): 0})    """    return coldict2mat({x:find_error(mat2coldict(S)[x]) for x in S.D[1]})## Task 6s = "I'm trying to free your mind, Neo. But I can only show you the door. You’re the one that has to walk through it."P = bits2mat(str2bits(s))## Task 7C = G*Pbits_before = 896bits_after = 1568## Ungraded TaskCTILDE = None## Task 8def correct(A):    """    Input: a matrix A each column of which differs from a codeword in at most one bit    Output: a matrix whose columns are the corresponding valid codewords.    Example:        >>> A = Mat(({0,1,2,3,4,5,6}, {1,2,3}), {(0,3):one, (2, 1): one, (5, 2):one, (5,3):one, (0,2): one})        >>> correct(A)        Mat(({0, 1, 2, 3, 4, 5, 6}, {1, 2, 3}), {(0, 1): 0, (1, 2): 0, (3, 2): 0, (1, 3): 0, (3, 3): 0, (5, 2): one, (6, 1): 0, (3, 1): 0, (2, 1): 0, (0, 2): one, (6, 3): one, (4, 2): 0, (6, 2): one, (2, 3): 0, (4, 3): 0, (2, 2): 0, (5, 1): 0, (0, 3): one, (4, 1): 0, (1, 1): 0, (5, 3): one})    """    return find_error_matrix(H*A)+A