def test_gaussel3(self) :
# matrix dari vektor baris
X = array([ # kasus j = k : solusi unik
[1,2,3],
[4,5,6],
[7,8,9]
])
Y = array([
[10,11,12]
])
print "kasus unik normal"
print solve_linear( (X.astype(float)).T, Y.astype(float) )
def test_gaussel0(self) :
# matrix dari vektor baris
X = array([ # kasus j = k : solusi unik
[1,2,3],
[4,5,6]
])
Y = array([
[5,7,10]
])
print "linear independen (1)"
print "================="
print solve_linear( (X.astype(float)).T, Y.astype(float) )
def test_gaussel2(self) :
# matrix dari vektor baris
X = array([ # kasus j = k : solusi unik
[1,2,3],
[4,5,6],
[5,7,10]
])
Y = array([
[0,0,0]
])
print "sistem homogen"
print "=============="
print solve_linear( (X.astype(float)).T, Y.astype(float) )
def shortest(B) :
(j,k) = B.shape
if j==1 :
return B
b = LLL(B).getBasis() # representasi basis dalam array 2d
while True :
# aproximasi reduced basis
b_ = orthoproject(b[0], b[1:])
# rekursif
b__ = shortest(b_)
# lifting dg transformasi linear, catatan 13/12
T = solve_linear( b_.T,b__ ) # b_ perlu ditransform, liat spek
# transpose T karena msh kolom
v = matrix( T.T )*matrix( b[1:] ) # ubah ke matrix dulu
b[1:] = np.asarray(v) # override array lagi
b0n = norm(b[0])
b1n = norm(b[1])
if b1n**2 >= (3.0/4)*b0n**2 :
break
# swap
t = np.copy(b[0])
b[0] = b[1]
b[1] = t
# temukan nilai batas j0
g = Gram(b)
# cari vektor yg norm-nya lebih besar dari b0
i = 1
while i < j :
v = g.getBstar(i)
if norm(v) >= b0n :
break
i += 1
# enumerate
v1 = enumerate(b[:i], g)
# select-basis
B = select_basis( np.append([v1], b, axis=0) )
# ulangi lagi step spt yg didalam loop
B_ = orthoproject( B[0], B[1:] )
B__ = shortest(B_) # rekursif
T = solve_linear( B_.T, B__ )
# tangani kondisi ketika pengenolan
# apakah dalam enumerate alpha diubah2 secara share
V = matrix( T.T )*matrix( B[1:] ) # perlu bentuk matrix
B[1:] = np.asarray(V)
return B
def test_gaussel0(self) :
X = array( [[1,2,3]] )
Y = array( [[1,2,3]] )
print "kasus paling dasar"
print "=================="
T = solve_linear( (X.astype(float)).T, Y.astype(float) ) # input kolom perlu transp
print T
def test_gaussel7(self) :
X = array([ # kasus j = k : solusi unik
[0,-1,-1],
[4,7,8],
[-3,-5,-6]
])
Y = array([
[3, 4, 5]
])
print "kasus unik full rank"
T = solve_linear( (X.astype(float)).T, Y.astype(float) ) # input kolom perlu transp
print T
def test_gaussel6(self) :
X = array([ # kasus j = k : solusi unik
[2,-1,0],
[-1,2,-1],
[0,-1,1]
])
Y = array([
[0, 0, 1]
])
print "kasus unik full rank"
T = solve_linear( (X.astype(float)).T, Y.astype(float) ) # input kolom perlu transp
print T
def test_gaussel5(self) :
X = array([ # kasus j = k : solusi unik
[1,1,1],
[1,2,2],
[1,2,3]
])
Y = array([
[1, 1, 1]
])
print "kasus unik full rank"
T = solve_linear( (X.astype(float)).T, Y.astype(float) ) # input kolom perlu transp
print T
def test_gaussel4(self) :
X = array([ # kasus j = k : solusi unik
[0,-2,-2],
[1,4,5],
[-1,-1,-4]
])
Y = array([
[3, 1, -2]
])
print "kasus unik full rank"
T = solve_linear( (X.astype(float)).T, Y.astype(float) ) # input kolom perlu transp
print T
def test_gaussel2(self):
X = array([ # kasus j = k : solusi unik
[1,2,3],
[7,8,9]
])
Y = array([
[4,5,6],
[10, 11, 12]
])
print "kasus unik not full rank"
T = solve_linear( (X.astype(float)).T, Y.astype(float) ) # input kolom perlu transp
print T
def select_basis(b) : # b is numpy array
(j,k) = b.shape # j = k+1
global Z
if Z is None :
Z = np.zeros(k) # TODO : cek apakah rekursif berpengaruh
b[0] = map(lambda x: round(x,14), b[0]) # cek pembulatan nol pd b0
if all( b[0]==Z ) :
if len(b) > 1 :
basis = b[1:]
else :
basis = []
else :
# tes apakah b[0]= kombinasi linear dari b[1]..b[j-1]
# menggunakan eliminasi Gauss
c = solve_linear( (b[1:j].astype(float)).T , [ b[0] ] )
a = np.empty([j-1,k], dtype=object)
if c==[] : # b[0] bebas linear thd b lain
a[0] = np.copy(b[0])
else :
xs = c.T[0] # c msh dalam represnt kolom
# khusus bagian ini lihat buku Bremner p.185
# kita tidak perlu y dan z karena sdh ditangani oleh Fraction
# tidak perlu menangani x_i = 0
i = findNZ1( xs )
f = Fraction( str(xs[i]) )
m = f.denominator
i2 = i+1
while i2 < len(xs) :
if xs[i2] :
f = Fraction( str(xs[i2]) )
m = lcm( f.denominator, m )
i2 += 1
d = long( round( m*xs[i] ) )
i += 1
while i < len(xs) :
if xs[i] :
x = long( round(m*xs[i]) )
d = gcd( x,d )
i += 1
# python perlu denominator tetap positif
if d < 0 :
m *= -1
d *= -1
# f adalah bentuk rasional m/d yg jika relative prima = p/q
# yg berarti faktor bersama nya dikeluarkan dan diambil q nya
f = Fraction( '%d/%d' % (m,d) )
if m < 0 :
a[0] = ( -1.0/f.denominator )*b[0]
else :
a[0] = ( 1.0/f.denominator )*b[0]
# proyeksi b ke a[0]. karena b[0] sdh diproses, maka
# b yg diproyeksi berikutnya adalah 1,2..j-1
b_ = orthoproject( a[0], b[1:] )
# rekursif
v = select_basis(b_)
# hasil dari rekursif digunakan utk lifting dari c ke a
i = 1 # a[0] sudah diproses
for c in v :
a[i] = lifting(b, c)
i += 1
basis = a
return basis
def test_step_4dim(self) :
b_ = np.array([
[2.46632124352332, 0.25906735751295, -2.11917098445596, -3.59067357512953], [3.4786437692646395, -3.6177895200352275, 4.313518273888157, -0.41743725231175377]], dtype=object)
b = np.array([[ 2.46632124, 0.25906736, -2.11917098, -3.59067358],
[ 3.24352332, -3.64248705, 4.51554404, -0.07512953]])
T = solve_linear(b.T,b_)
