-
Notifications
You must be signed in to change notification settings - Fork 0
/
kannan.py
215 lines (192 loc) · 6.88 KB
/
kannan.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
from fractions import Fraction
import numpy as np
from numpy.linalg import norm
from numpy.matlib import dot, matrix
from util import solve_linear, lcm, gcd, findNZ1
# asumsi desain tanpa thread dan lock, gunakan glob variable
Z = None # vektor zero (nol)
def orthoproject(a,b) : # proyeksi orthogonal thd a
(j,k) = b.shape
b_ = np.empty( [j,k], )
for i in range( 0,j ) :
x = float( dot( b[i],a ))/dot( a,a )
b_[i] = b[i] - x*a
return b_
def lifting(a,b) : # proyeksi thd v in L'
x = float( dot(a, b) )/dot(a, a)
while True :
b = b - a
y = float( dot(a, b) )/dot(a, a)
if abs(x) < abs(y) :
break
x = y
# undo subtraction of b
return b + a
# spek dr Kannan menyatakan set b adalah linear dependent
# ref: Kannan p.
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
from L3 import Gram
alpha = [] # list of array of possible integer combination
# ref: Kannan p.
# input : k = basis ke, m = jml basis (indexing diawali dari 1/ aturan umum), g = Gram schmidt data
# return list of integer
def List(k,m,g) :
if k == 0 :
# list telah komplit, tambahkan ke alpha
global alpha
alpha = np.append( alpha, [ np.copy(alpha[0]) ], axis=0 )
return
#
# hitung beta0_k proposisi 2.13 ref Kannan p.
x = norm( g.getBstar(0) )/norm( g.getBstar(k-1) )
t = 0
for j in range( k+1, m+1 ) : # lihat catatan 18/12
t += g.getU( j-1,k-1 ) # lihat spek interface
beta0_k = long( round(- x - t) )
d = long( round(2*x) )
# tes utk tiap kombinasi nilai alpha
global alpha
for i in range( beta0_k , beta0_k + d + 1 ) :
alpha[0, k-1] = i
List(k-1,m,g) # rekursif
# ref: Kannan p.
# find shortest vector in lattice L(b1, b2, ... b_m)
def enumerate(b,g) :
(m,n) = b.shape
if m == 1 :
return b[0]
global alpha
alpha = np.array( [ np.zeros(m) ] , dtype=object)
# hitung bm(m) = b*_m
lim = norm(b[0])/norm( g.getBstar( m-1 ) )
for j in range( long( round(-lim )), long( round(lim )) + 1 ) :
alpha[0, m-1] = j
List(m-1,m,g)
# alpha sdh selesai, tinggal di iterasi cari SVP :)
B = matrix(b)
svp = np.ravel( alpha[1]*B )
sn = norm(svp)
k = len(alpha)
for i in range(2, k) :
x = np.ravel( alpha[i]*B )
xn = norm(x)
if sn > xn :
sn = xn
svp = x
return svp
from math import sqrt
from numpy.linalg import inv
from util import toLong
from L3 import LLL
# ref: Kannan p.
# main procedure of Kannan, find v1 and construct basis
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 init_global() : # will move to OO design
global Z, alpha
Z = None
alpha = []