p高位攻击
文章目录
- 题一
-
- 题目描述(p>>128<<128)：
- 题目分析：
-
- 方法一（后方有0填充）：
- 方法二（后方无0填充）：
- 方法三（后方无0填充）：
- 方法四（后方有0填充）：
- 题二
-
- 题目描述：
- 题三
-
- 题目描述：
题一
题目描述(p>>128<<128)：
e = 0x10001
p>>128<<128 = 0xd1c520d9798f811e87f4ff406941958bab8fc24b19a32c3ad89b0b73258ed3541e9ca696fd98ce15255264c39ae8c6e8db5ee89993fa44459410d30a0a8af700ae3aee8a9a1d6094f8c757d3b79a8d1147e85be34fb260a970a52826c0a92b46cefb5dfaf2b5a31edf867f8d34d2222900000000000000000000000000000000
n = 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
c = 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
题目分析：
（注：p>>128<<128------低位数据缺失，丢失了p的后128位，p抹除掉低128位数据以后的值。）
- sage中求p的多种解法
方法一（后方有0填充）：
# sage
p_high = 0xd1c520d9798f811e87f4ff406941958bab8fc24b19a32c3ad89b0b73258ed3541e9ca696fd98ce15255264c39ae8c6e8db5ee89993fa44459410d30a0a8af700ae3aee8a9a1d6094f8c757d3b79a8d1147e85be34fb260a970a52826c0a92b46cefb5dfaf2b5a31edf867f8d34d2222900000000000000000000000000000000
n = 0x79e0bf9b916e59286163a1006f8cefd4c1b080387a6ddb98a3f3984569a4ebb48b22ac36dff7c98e4ebb90ffdd9c07f53a20946f57634fb01f4489fcfc8e402865e152820f3e2989d4f0b5ef1fb366f212e238881ea1da017f754d7840fc38236edba144674464b661d36cdaf52d1e5e7c3c21770c5461a7c1bc2db712a61d992ebc407738fc095cd8b6b64e7e532187b11bf78a8d3ddf52da6f6a67c7e88bef5563cac1e5ce115f3282d5ff9db02278859f63049d1b934d918f46353fea1651d96b2ddd874ec8f1e4b9d487d8849896d1c21fb64029f0d6f47e560555b009b96bfd558228929a6cdf3fb6d47a956829fb1e638fcc1bdfad4ec2c3590dea1ed3
c = 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
def phase3(p_high,n,c):
R.<x> = PolynomialRing(Zmod(n),implementation = 'NTL')
p = p_high + x
x0 = p.small_roots(X = 2 ^ 128,beta = 0.1)[0]
print(int(p(x0))) # 或 print(p_high + x0)
phase3(p_high,n,c)
'''
p = 147305526294483975294006704928271118039370615054437206404408410848858740256154476278591035455064149531353089038270283281541411458250950936656537283482331598521457077465891874559349872035197398406708610440618635013091489698011474611145014167945729411970665381793142591665313979405475889978830728651549052207969
'''
方法二（后方无0填充）：
# sage
p_high = 0xd1c520d9798f811e87f4ff406941958bab8fc24b19a32c3ad89b0b73258ed3541e9ca696fd98ce15255264c39ae8c6e8db5ee89993fa44459410d30a0a8af700ae3aee8a9a1d6094f8c757d3b79a8d1147e85be34fb260a970a52826c0a92b46cefb5dfaf2b5a31edf867f8d34d2222900000000000000000000000000000000
n = 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
c = 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
def phase3(p_high,n,c):
R.<x> = PolynomialRing(Zmod(n),implementation = 'NTL')
p_high = p_high << 128
p = p_high + x
x0 = p.small_roots(X = 2 ^ 128,beta = 0.1)[0]
print(int(p(x0))) # 或 print(p_high + x0)
phase3(p_high,n,c)
'''
p = 147305526294483975294006704928271118039370615054437206404408410848858740256154476278591035455064149531353089038270283281541411458250950936656537283482331598521457077465891874559349872035197398406708610440618635013091489698011474611145014167945729411970665381793142591665313979405475889978830728651549052207969
'''
方法三（后方无0填充）：
# sage
(此处解法中p4为p去除0的剩余位)
p_high = 0xd1c520d9798f811e87f4ff406941958bab8fc24b19a32c3ad89b0b73258ed3541e9ca696fd98ce15255264c39ae8c6e8db5ee89993fa44459410d30a0a8af700ae3aee8a9a1d6094f8c757d3b79a8d1147e85be34fb260a970a52826c0a92b46cefb5dfaf2b5a31edf867f8d34d22229
n = 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
c = 0x1b2b4f9afed5fb5f9876757e959c183c2381ca73514b1918d2f123e386bebe9832835350f17ac439ac570c9b2738f924ef49afea02922981fad702012d69ea3a3c7d1fc8efc80e541ca2622d7741090b9ccd590906ac273ffcc66a7b8c0d48b7d62d6cd6dd4cd75747c55aac28f8be3249eb255d8750482ebf492692121ab4b27b275a0f69b15baef20bf812f3cbf581786128b51694331be76f80d6fb1314d8b280eaa16c767821b9c2ba05dfde5451feef22ac3cb3dfbc88bc1501765506f0c05045184292a75c475486b680f726f44ef8ddfe3c48f75bb03c8d44198ac70e6b7c885f53000654db22c8cee8eb4f65eaeea2da13887aaf53d8c254d2945691
pbits = 1024 # p原本位数
kbits = pbits - p_high.nbits() # p丢失位数
p_high = p_high << kbits
PR.<x> = PolynomialRing(Zmod(n))
f = x + p_high
p0 = f.small_roots(X = 2 ^ kbits,beta = 0.4)[0]
print(p4 + p0)
'''
p = 147305526294483975294006704928271118039370615054437206404408410848858740256154476278591035455064149531353089038270283281541411458250950936656537283482331598521457077465891874559349872035197398406708610440618635013091489698011474611145014167945729411970665381793142591665313979405475889978830728651549052207969
'''
方法四（后方有0填充）：
# sage
p_high = 0xd1c520d9798f811e87f4ff406941958bab8fc24b19a32c3ad89b0b73258ed3541e9ca696fd98ce15255264c39ae8c6e8db5ee89993fa44459410d30a0a8af700ae3aee8a9a1d6094f8c757d3b79a8d1147e85be34fb260a970a52826c0a92b46cefb5dfaf2b5a31edf867f8d34d2222900000000000000000000000000000000
n = 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
c = 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
pbits = 1024
kbits = 128
PR.<x> = PolynomialRing(Zmod(n))
f = x + p_high
p0 = f.small_roots(X = 2 ^ kbits,beta = 0.4)[0]
print(p_high + p0)
'''
p = 147305526294483975294006704928271118039370615054437206404408410848858740256154476278591035455064149531353089038270283281541411458250950936656537283482331598521457077465891874559349872035197398406708610440618635013091489698011474611145014167945729411970665381793142591665313979405475889978830728651549052207969
'''
- 其实以上解法都大同小异，放在一起只是为了更好理解
- 求出p后，接下来便是平平无奇的rsa解明文，代码如下：
import gmpy2
from Crypto.Util.number import *
n = 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
c = 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
e = 65537
p = 147305526294483975294006704928271118039370615054437206404408410848858740256154476278591035455064149531353089038270283281541411458250950936656537283482331598521457077465891874559349872035197398406708610440618635013091489698011474611145014167945729411970665381793142591665313979405475889978830728651549052207969
q = n // p
phi = (p-1)*(q-1)
print(gmpy2.gcd(e,phi))
d = gmpy2.invert(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))
- 得到：flag{Kn0wn_Hi9h_Bit5}
题二
题目描述：
from Crypto.Util.number import *
from secret import flag
flag = b'flag{*********}'
m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
n = p * q
e = 114
c = pow(m,e,n)
print(c)
print(p >> 200)
print(n)
# c = 4981370648841772812759645290740849305394680703208798679296466901875830602835273402860232301263281323578956193947979697234640828088984992529165349436050379602381023059635247562226192384089521639938396211636613132291696135696985578958227320544060232615333466684704244997055833821133086665356126147182204658744167431612986909752009485714137028204041440181653812250548914729617593568901044728464293232061709058144788756823288190386071071979728390993033661221130338943191220680445314588574185565138844949934691183548291792150029676489045342419826189506616272247940278820931530398810621850374268800818970515221497093852109
# p_high = 62037304914409314363888940906845820031382619388386590204815535497699521033644001814874589864676342418539729790446530529473631795496696578029445470682035483391568820927435567100377626022924900710513454770616746573110984342344183967600234091673261776
# n = 10315159385090642346129000730749042701431892949303034712476198921384639021767097119992198421632142955005047146294210952031882321038272269972695714084199530336742619691272883151455898061330316812891004827724782855036289498818157782936179413509824274682055131552093071749522986951202502017564120645520386407170556413591537187759567563157956331577316042296031033014710853038209000676314440817362756989634719336973373719581572614119144998829076893422175956726616346716072744575347893245428145235967836165207095908913238287634122873060994828380614739915448587956681845973466847711337763120292734433687845920176310499582951
- p的求法和上面一样，只需修改些许参数即可，大家可自行尝试，这里我便不再说了，最终得到p:
p = 99690105430259549732952386298363416480730988331578091065948950836198178325904426675017504756348563688521763268566954512895974110780822714951824351709232320913381679046309934991336770483285399157355308073567950907088479972767984569322594411195698421500521401221792581871025328456951904596576566123729811756413
- 值得注意的是此处 gcd(e, phi) = 6, 已经学到高位攻击了，想必不互素的情况大家也不陌生了，以下是解题代码：
import gmpy2
from Crypto.Util.number import *
p = 99690105430259549732952386298363416480730988331578091065948950836198178325904426675017504756348563688521763268566954512895974110780822714951824351709232320913381679046309934991336770483285399157355308073567950907088479972767984569322594411195698421500521401221792581871025328456951904596576566123729811756413
c = 4981370648841772812759645290740849305394680703208798679296466901875830602835273402860232301263281323578956193947979697234640828088984992529165349436050379602381023059635247562226192384089521639938396211636613132291696135696985578958227320544060232615333466684704244997055833821133086665356126147182204658744167431612986909752009485714137028204041440181653812250548914729617593568901044728464293232061709058144788756823288190386071071979728390993033661221130338943191220680445314588574185565138844949934691183548291792150029676489045342419826189506616272247940278820931530398810621850374268800818970515221497093852109
n = 10315159385090642346129000730749042701431892949303034712476198921384639021767097119992198421632142955005047146294210952031882321038272269972695714084199530336742619691272883151455898061330316812891004827724782855036289498818157782936179413509824274682055131552093071749522986951202502017564120645520386407170556413591537187759567563157956331577316042296031033014710853038209000676314440817362756989634719336973373719581572614119144998829076893422175956726616346716072744575347893245428145235967836165207095908913238287634122873060994828380614739915448587956681845973466847711337763120292734433687845920176310499582951
e = 114
q = n // p
phi = (p-1)*(q-1)
print(gmpy2.gcd(e,phi))
dt = gmpy2.invert(e//6,phi)
# gcd = 6
# c = (m ** 6) ** (e // 6) mod n
# 这里便是将m**6当作一个新的m,e//6当作一个新的e
# c = pow(m_6,e//6,n)
m_6 = pow(c,dt,n)
m = gmpy2.iroot(m_6,6)[0]
print(long_to_bytes(m))
- 得到：flag{Wish_you_good_luck!}
题三
题目描述：
from Crypto.Util.number import getPrime, bytes_to_long
FLAG = b"flag{}"
def enc(m):
return pow(m, e, N)
if __name__ == "__main__":
l = 256
p = getPrime(1024)
N = p * getPrime(1024)
e = 65537
a = (p >> l) << l
print("N:", N)
print("Known part of p:", hex(a))
print("Length of the unknown part:", l)
print("enc:", enc(bytes_to_long(FLAG)))
'''
N: 13139369168613206469808493070119137888363636548621629780897948879328793540933675072448361493321304924953815474270401406259487525517560528123707016104942485164559271692275987380567766009184969340122208041180122234792566147648471202470677782205185423853314467362074540818483729953544353584322270414479260852672948012862257167187569701381652931473637503302338392147780573148724508117699531886205586281824118899931516823621049590863613262210219765105389989391065557707559113268724368695051264276619633555407916385088611885715568165460641318205321508100969473959719364829756492542217470309748646183210141490634293731384313
Known part of p: 0xce2f93251a3a97404a11c1fe88cf15c7aaf26ffd508ff006933bff2e9ea0c6197a98f1188f03b74b16d564e958a84c877fc0e21faf00f0ae42f26bde226ebf7c9732f17d860b81d139799832d510b91001967fc33ff2d9fbd4c4767fa2438e480000000000000000000000000000000000000000000000000000000000000000
Length of the unknown part: 256
enc: 10994040238462701583659972453101990790105813765012990749608185072718600854440721383239626137413394180017327728436373282101414910961527694291269438429124480262436140242346606299693570528165786128342639697068827177018329599990851910019879374047005450085901527704075107943071874274180276690059744636245418370546246757239666384320024728428669095536636012185174919921512858446990367609436305909104080544807406108497002947094974486271586520410674562299953931051491793506395053049529401503138822284281986730639405804678960905130223082038659899687410133538282139577181626031301078543719687909557541703207389570582210505090095
'''
- 解法和题1一模一样，只需修改些许参数即可，就不过多说了
- 最后得到：flag{b082411e0d6e296d4598b69825fc0cc5}