Crypto
老王的秘密
根据题目提示区块链与数据格式，本题为秘密共享，使用github下现成库可解决：
https://github.com/blockstack/secret-sharing
解题代码如下：
shares=['1-fddc7d57594928fb74a507ab9cba0b28b92bb6e7b36a9925a105eeddac020e64',
'3-84f82314003c9690eeacd823b22680ccbe93ac098cabdd0a992c095dde0031cf',
'5-b0e2e8d2cadc91f8f2f357a42e26aeabaccbfa7731437298ca23d8a4a5424ce4',
'7-810e7545213971a3c7c2dce3d0998764d0bc1e3b866b15ad0deebaa7abcf64c5',
'9-b4da0bd03394e4bdfef92f16365e8811d9614f11b99111bcf8a4e68ba79626a2',
'b-661069e7d491719759a3199be1f65ffb6db92d1b014abb4e33ca7e32f85ee276',
'd-1f84ab9b467a4ec4de4451ed187987785b567bbdde0126d0722e3335a5307d68',
'f-9001dc36dd28c5c5dd7333968e7263986f55dd79cd9be286d21f45e46f53c399']
from secretsharing import PlaintextToHexSecretSharer
PlaintextToHexSecretSharer.recover_secret(shares[0:8])
遗失的秘密
本题为PCTF原题改编，解题原理如下：
* given a candidate for (p mod 16**(t - 1)), generate all possible candidates for (p mod 16**t) (check against mask for prime1)
* calculate q = n * invmod(p, 16**t) (and check against mask for prime2)
* calculate d = invmod(e, 16**t) * (1 + k * (N - p - q + 1)) (and check against mask for private exponent)
* calculate d_p = invmod(e, 16**t) * (1 + k_p * (p - 1)) (and check against mask for exponent1)
* calculate d_q = invmod(e, 16**t) * (1 + k_q * (q - 1)) (and check against mask for exponent2)
* if any of checks failed - check next candidate
利用公钥和私钥内的信息可以拼出n，但是这个n是不完整的，其后8位为空。幸好我们知道p和q的后8位，因此可以计算出完整的n。解题代码如下：
#!/usr/bin/env python
#-*- encoding: utf8 -*-
import re
import pickle
from itertools import product
from libnum import invmod, gcd
def solve_linear(a, b, mod):
if a & 1 == 0 or b & 1 == 0:
return None
return (b * invmod(a, mod)) & (mod - 1) # hack for mod = power of 2
def to_n(s):
s = re.sub(r"[^0-9a-f]", "", s)
return int(s, 16)
def msk(s):
cleaned = "".join(map(lambda x: x[-2:], s.split(":")))
return msk_ranges(cleaned), msk_mask(cleaned), msk_val(cleaned)
def msk_ranges(s):
return [range(16) if c == " " else [int(c, 16)] for c in s]
def msk_mask(s):
return int("".join("0" if c == " " else "f" for c in s), 16)
def msk_val(s):
return int("".join("0" if c == " " else c for c in s), 16)
E = 65537
N = to_n("""00:c4:9d:36:a4:77:76:12:12:85:24:6c:74:1d:7d:
b3:ce:f4:c3:a4:69:cd:0b:2e:8f:d6:75:e3:80:b8:
e8:1c:ce:e8:60:90:45:56:73:ab:32:32:00:7f:6a:
76:3e:b6:10:d3:a2:74:da:f9:4e:a5:7e:ae:ef:f4:
da:82:57:6d:68:82:50:d8:b1:fc:92:b1:5c:7d:54:
f5:7e:d0:06:8a:60:ff:82:70:72:20:68:4b:71:ba:
87:44:57:c1:97:a0:8a:2d:53:93:f3:0a:60:87:a3:
85:c8:45:e6:0a:88:85:b5:ff:c7:09:9a:76:03:fe:
99:b6:fb:8a:1e:9f:a8:42:3a:0a:c9:a9:bf:1c:87:
2c:c4:99:10:db:46:e3:a9:a5:79:93:8c:75:71:ec:
c6:3b:af:44:dc:60:c4:53:f6:3c:e8:73:2f:50:10:
38:e7:6f:d0:a5:4b:ae:e3:1e:43:11:42:2c:a2:38:
e6:3f:0b:13:54:63:e8:2f:9e:61:ab:08:65:97:e0:
27:30:19:fd:a7:fe:5c:d8:11:b8:34:87:ad:02:c2:
bc:cd:73:d3:86:be:fd:2a:b4:fe:7d:7e:d3:64:bb:
6f:63:ed:a6:1d:ee:f2:80:da:9d:7a:23:7f:c1:39:
b0:98:0c:85:8f:d0:4b:9f:e4:1a:26:fc:44:d1:67:
03:32:03:0c:91:61:23:4c:81:6f:42:18:88:41:dc:
27:55:a3:07:7c:a1:ad:f3:58:4d:91:07:65:f1:63:
f2:34:d5:17:0e:59:c6:bb:b6:6d:7d:0c:d2:64:4b:
b9:9c:52:59:03:8e:2a:43:23:76:33:c3:e8:72:3b:
1c:e0:40:97:36:5f:ae:00:d7:e3:09:eb:df:55:44:
22:b4:09:00:b5:09:41:70:6c:5c:3b:98:d3:34:7e:
60:a2:b8:93:bd:af:32:77:48:48:8a:a5:9c:0e:6a:
a1:79:36:86:8c:e9:3f:b1:a2:a7:4a:3a:d8:d6:f6:
dd:62:d8:ae:9e:13:bb:0c:6b:b1:65:68:0d:7e:58:
3f:68:1e:91:49:13:19:68:2b:fd:3c:5e:52:fa:76:
b0:57:fc:0e:35:d8:71:56:41:06:ef:50:99:56:dd:
d4:9a:1f:d3:46:26:12:9c:15:4b:43:fc:1b:de:c9:
06:ad:82:56:63:c8:a4:83:32:d2:35:05:23:15:52:
d9:0a:73:85:5e:c9:c2:56:af:69:d2:5f:77:04:28:
c8:4c:b9:a6:d4:15:15:b5:15:99:13:ef:a9:a5:de:
5a:74:b1:03:cf:32:a5:03:69:f8:e9:bb:7e:16:31:
5e:43:e7:02:51:ac:c5:f6:bf:ef:1c:74:f7:13:0c:
19:ad:b7""")
# prime1 p
p_ranges, pmask_msk, pmask_val = msk("""00: :05:89: :bd:35: : :23: : : : :84:
: :ed: :70:14: : : :10: : :87: :51:
ea: :97:69: :52: : : : : :ea: : :15:
: :34: :be:11:23: : :34:14: :94: :10:
: :74:87:37:ee:81:62:ee:95: : :dc:49:dd:
: :35: :81: :fa: : : :86: : : :fb:
:93: : :12: :14: :ab:76: :96: : :27:
:21: :04:01:41: :98: :ff: : :12:dc: :
cd: :39:95:30: :47: :fa:ff: :34: :ad: :
:52:02:fa:bc:14:22:22:48:61:62:bd:53: : :
72:08:cb:41:88: : : :63:91:30:fe: : :42:
87: :18:52: :39:dd: :68: :fe:06:88:81: :
: : :ae:fd: : :fb:21:37:59: :53: :fa:
:07:40:eb:33:77:51:64:10:dd: :73: :86:62:
:bf: :79: :34: :bb: :44:ff: :46:fe:90:
ef: :52:ad: : :fe: :69:18:89:bd:cd:09:46:
: :74:71: : : :41:66: : :11: :25: :
39:8b""")
# prime2 q
q_ranges, qmask_msk, qmask_val = msk("""00:ce:43:ef: :76:58:17:43:31: : :32:70: :
89: : :36:55:06: :79:66:78: : : : : :
:85: : : : : :33:bb: : :56: :66:cb:
:08: : :90:cb: : :24:fa:ca:47: : : :
:88: :83:01: :62: : : : : : :ad:ae:
: : :58: :ec: : : :09:04:86: :05:00:
:df:50:84:81:80: :ae: :24: :94:da: :04:
ce: :ef: : :ed:be:bf:43:78: : :05:93: :
08:52:05: : : : :ae: : : : :ab: : :
:76:ce: : : : :19:bd:22: :ef:dc:bf:ea:
ab:78:01: : :85: : : :ea: : :fb: : :
92:66:19: : :ab: : :82: : :31: : :da:
82: :13:82:43: : :94:13:41: : : :37: :
:04:56:02:87:dd: :58:27: : :24: : : :
28: : :09:14:89: : : :49:59: :16:eb:65:
:01:22: : :dd: :78: : :db:90: :ac: :
:fd: :03:74: : : : :92: :00:ba: : :
:05""")
# privateExponent d
_, dmask_msk, dmask_val = msk("""11: : :69:62:64: : : : :15: :13:de:de:
cf: : :17: : :75: :98:42:fc: :12:15:08:
: : : : :36: :be:25:48: : :19: : :
:47:11:19: :03: :49:fc:da: :96:45:eb: :
: : :91: :ea: : :55:ff: :37:58: : :
19: : :73:40: :91:15:01:da:91:22:fd:32: :
: :50: : :66: : : :42: : :ef: : :
df:42: :97:30: :39: : : : : : :dc: :
: : : : :38: : : :88:28: :05: : :
78:59:fa: :86: :19:24: : : : :da:cf:15:
39: : : : :ef:55: :ce:47: :58:89: :fb:
:24: : : :92: : :ee: : :db:67:31:ce:
:28: :72:ec:89: :04: : :50: : : : :
:37: :44: : : : :56: :38: :bb:47:bb:
66:83:99:22:07:72: : :48:52:02: : : :29:
:82:56: :67: :95: : :56:94: : :71: :
bf:27:98: : :54:98:26:06:87: :ae: :53:be:
: :80:37:60:61:ea:ef:de: : :df:90:81: :
70: :06:33:26: :75:fe:95: :92: :78:cd:05:
64:cc:68: : :36:54: :bd:16:90:ee:60: : :
: :41: : :91: :79:58:06:50: :46: : :
45: :09:ca:ac:16: :27:98: : :ba:82: :77:
93:98:ad: :15: :67:53:97:ad:ee:50:44: :31:
07: :ff:01: :09: : : : : :46: : :42:
15: :db:df:42:be: : : :78: :41: : : :
:14: : :25:fc: :84: : : : : : :20:
da:46:01:eb:87: :12:57: : :56:af: :87:93:
60: :02: :18:89:63:72:ad: :ed:cf: : :84:
:22: :13: : :dd: :ff: : : :de:62:37:
:19:66: : :86:02: :38: : : : :ec:14:
12: :43:93:19:65:98: : :03: : : :ef: :
: :ca:07:92:22: : :bb:15:eb: : : :35:
:72:29:cd: : :99: : : : :41:06: : :
:43:33: :32: : :54:be:92:62: :78:59:42:
79:89""")
# exponent1
_, dpmask_msk, dpmask_val = msk(""" :39: :28:16:02:89:ce:11:fe: : : : :af:
: : :ed:97: : :11:20:ba:ae:98:ad: : :
:10:87:ac:07: : : : :50: : :70:50:52:
df:89:eb:02: : : : :93:11: : :12: :56:
:08: : :ea: :10:fa:19: : : :54:45:07:
: :bc:ff:33: :db:63:49:fe:52: :33: : :
bf:cd:45:91: :10: : :92:81:40:03: :80: :
29: :30: :ed:43:64:ca: :bf:64: : :bf: :
: : :24:72:84: : :ff: : :24: :81:27:
db:23: :64: :67: :ba: : :bc: : : : :
:ae:88: : : : : :91: : :14: :ba:ef:
:89: : : : : : : : :05: :75:52: :
: : :be:ad:df: :02:88:00: : :15:45: :
cf:32: :ca: :93: :32: :40: :27:dd: :19:
73:dc: : : : : :cf: : :dd: : :ca: :
ee: :ca: : : :49: :27: :58:53: :64:25:
:22:06:16:ff:62:bc: : : : :24:fc: : :
df""")
# exponent2
_, dqmask_msk, dqmask_val = msk("""02: :bd: :19:25:98:75: :65: :55:28:33:bc:
34:84:91:01:96: : :08: :32:45: :27: : :
:fe: :bb:63:32:68: :51:bd:75:40: :52:52:
: : :78:85:fc:94: :07: :14: : : : :
15:dd: : :93: :01: : :77:ca: :40: :da:
:89:bc:87:62:dc:ac:61:88: : :70: :69: :
:36: : :21:08: :dc:73: :ad:da:ee:fe: :
96: :58: : :46: :29:ff:97:ce: : : :cb:
51: : :81: :22: : :19: :10:69:41:36:ca:
:22:49: :cc:cf:06: : :08: :76: : :45:
98: : :45: : : :69:13:65: : :da:54: :
19: :ee:24: :73: : : : : : :18:53:40:
21:25: : :84:52:cd: :49:33:78: : :ed: :
25:27: : : :ca: : : :ca: : :bc: :02:
31:70: :10:ca:84:59: : : :52: :27:76: :
47: :66:bf:ff: :03: :99:ff: :df: : : :
:46:27:45: :65:07: :48:da:dc: :80: : :
f9""")
def search(K, Kp, Kq, check_level, break_step):
max_step = 0
cands = [0]
for step in range(1, break_step + 1):
#print " ", step, "( max =", max_step, ")"
max_step = max(step, max_step)
mod = 1 << (4 * step)
mask = mod - 1
cands_next = []
for p, new_digit in product(cands, p_ranges[-step]):
pval = (new_digit << ((step - 1) * 4)) | p
if check_level >= 1:
qval = solve_linear(pval, N & mask, mod)
if qval is None or not check_val(qval, mask, qmask_msk, qmask_val):
continue
if check_level >= 2:
val = solve_linear(E, 1 + K * (N - pval - qval + 1), mod)
if val is None or not check_val(val, mask, dmask_msk, dmask_val):
continue
if check_level >= 3:
val = solve_linear(E, 1 + Kp * (pval - 1), mod)
if val is None or not check_val(val, mask, dpmask_msk, dpmask_val):
continue
if check_level >= 4:
val = solve_linear(E, 1 + Kq * (qval - 1), mod)
if val is None or not check_val(val, mask, dqmask_msk, dqmask_val):
continue
if pval * qval == N:
print "Kq =", Kq
print "pwned"
print "p =", pval
print "q =", qval
p = pval
q = qval
d = invmod(E, (p - 1) * (q - 1))
coef = invmod(p, q)
from Crypto.PublicKey import RSA
print RSA.construct(map(long, (N, E, d, p, q, coef))).exportKey()
quit()
cands_next.append(pval)
if not cands_next:
return False
cands = cands_next
return True
def check_val(val, mask, mask_msk, mask_val):
test_mask = mask_msk & mask
test_val = mask_val & mask
return val & test_mask == test_val
# K = 4695
# Kp = 15700
# Kq = 5155
for K in range(1, E):
if K % 100 == 0:
print "checking", K
if search(K, 0, 0, check_level=2, break_step=20):
print "K =", K
break
for Kp in range(1, E):
if Kp % 1000 == 0:
print "checking", Kp
if search(K, Kp, 0, check_level=3, break_step=30):
print "Kp =", Kp
break
for Kq in range(1, E):
if Kq % 100 == 0:
print "checking", Kq
if search(K, Kp, Kq, check_level=4, break_step=9999):
print "Kq =", Kq
break
美好的回忆
设c[i]为密文按8位分割的段，m[i]为明文按8位分割的段，根据加密算法，有
c[0] = m[0] ^ iv ^ key
c[1] = m[1] ^ c[0] ^ key
…
c[n] = m[n] ^ c[n-1] ^ key
我们已知c[i]和m[0]、m[1]，根据c[1] = m[1] ^ c[0] ^ key可求出key，再根据c[0] = m[0] ^ iv ^ key求出iv，即可依次求出明文。解题代码如下：
#!/usr/bin/python
#encoding=utf-8
def strxor(a, b):
c = ''
for i in xrange(len(a)):
c += chr(ord(a[i])^ord(b[i%len(b)]))
return c
start = 'have a good time.'
ciphertext = open('flag.txt.encrypted','rb').read()
c = []
for i in xrange(8):
c.append(ciphertext[i*8:i*8+8])
m = []
for i in xrange(8):
m.append('1')
m[0] = start[0:8]
m[1] = start[8:16]
key = strxor(strxor(c[1],m[1]),c[0])
iv = strxor(strxor(c[0],key),m[0])
result = ''
for i in xrange(8):
m[i] = strxor(strxor(c[i],key),iv)
iv = c[i]
result += m[i]
print result
悲伤的结局
本题加密算法与上题相同，但是仅知道明文尾部部分信息，因此需要判断填充位。填充位有8种可能，分别对应8个不同的m[n]和m[n-1]，由
c[n] = m[n] ^ c[n-1] ^ key
c[n-1] = m[n-1] ^ c[n-2] ^ key
两式可以算出两个key。对于8种不同的填充，如果两个key相同，说明此时填充位正确且key正确。
但是本题实际上有个花招，即最后一部分加密结果被篡改，所以如果要求完全相同是求不出解的。这时可以输出每种填充时求得的两个key进行观察，发现当填充位为4时两个key比较相像。将此时的key当作真key，即可解出除第一段外的所有明文。
解题代码如下：
#!/usr/bin/python
#encoding=utf-8
def strxor(a, b):
c = ''
for i in xrange(len(a)):
c += chr(ord(a[i])^ord(b[i%len(b)]))
return c
last = 'keep away from xiaocui!'
ciphertext = open('flag.txt.encrypted','rb').read()
pad = -1
key = ''
for i in xrange(8):
tmpm = last[i-23:]+chr(i+1)*(i+1)
k1 = strxor(strxor(ciphertext[-16:-8],tmpm[8:16]),ciphertext[-24:-16])
k2 = strxor(strxor(ciphertext[-8:],tmpm[16:24]),ciphertext[-16:-8])
print i
print k1.encode('hex')
print k2.encode('hex')
if(k1[:3]==k2[:3]):
pad = i+1
key = k1
break
plaintext = ''
count = len(ciphertext)/8
for i in xrange(count-1):
plaintext += strxor(strxor(ciphertext[i*8+8:i*8+16],ciphertext[i*8:i*8+8]),key)
print plaintext