CRYPTOHACK BLOCK CIPHERS
admin
2024-04-14 03:33:29
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一、ECB CBC WTF

题目:

from Crypto.Cipher import AESKEY = ?
FLAG = ?@chal.route('/ecbcbcwtf/decrypt//')
def decrypt(ciphertext):ciphertext = bytes.fromhex(ciphertext)cipher = AES.new(KEY, AES.MODE_ECB)try:decrypted = cipher.decrypt(ciphertext)except ValueError as e:return {"error": str(e)}return {"plaintext": decrypted.hex()}@chal.route('/ecbcbcwtf/encrypt_flag/')
def encrypt_flag():iv = os.urandom(16)cipher = AES.new(KEY, AES.MODE_CBC, iv)encrypted = cipher.encrypt(FLAG.encode())ciphertext = iv.hex() + encrypted.hex()return {"ciphertext": ciphertext}

hint:

Here you can encrypt in CBC but only decrypt in ECB. That shouldn't be a weakness because they're different modes... right?

译文:

在这里,您可以在CBC中加密,但只能在ECB中解密。这不应该是一个弱点,因为它们是不同的模式……对吧?

思路:

本题告诉了我们用CBC模式进行加密,但使用ECB进行解密,其中,我们已经知道了CBC模式中加密的IV值的大小,我们又已经知道了将密文用ECB解密的值,通过CBC模式的解密结构,我们知道将第一段密文解密后于IV进行异或运算既是原明文,而第一段密文则是第二段密文加密的IV值,所以解密时将其反过来用,作为IV解密即可:

wp:

import requests
from Crypto.Util.number import *
result=requests.get('http://aes.cryptohack.org/ecbcbcwtf/encrypt_flag')
ciphertext=result.json()["ciphertext"]
# print(result.text)# print(len(bytes.fromhex(ciphertext)))
ciphertext=bytes.fromhex(ciphertext)
c1=hex(bytes_to_long(ciphertext[0:16]))[2:]
c2=hex(bytes_to_long(ciphertext[16:32]))[2:]
c3=hex(bytes_to_long(ciphertext[32:48]))[2:]
print(c1)
print(c2)
print(c3)# C=hex(bytes_to_long(ciphertext))
# print(c1)
# print(c2)
# print(c3)
# c1=0x9d2691928a24ed9215416815e80164d1
# c2=0xf77ea0b2c315db6b56a48c8250b0c9b8
# c3=0x6a61b98b2e5d26af3976d7c36a76a847
# k1=int(c1) ^ int(c2)
# k3=int(c2) ^ int(c3)
a1=requests.get(f'http://aes.cryptohack.org/ecbcbcwtf/decrypt/{c2}')
a2=requests.get(f'http://aes.cryptohack.org/ecbcbcwtf/decrypt/{c3}')
m1=a1.json()["plaintext"]
m2=a2.json()["plaintext"]
M1=int(m1,16)^int(c1,16)
M2=int(m2,16)^int(c2,16)
print(long_to_bytes(M1)+long_to_bytes(M2))

 二、ECB Oracle

题目:

from Crypto.Cipher import AES
from Crypto.Util.Padding import pad, unpadKEY = ?
FLAG = ?@chal.route('/ecb_oracle/encrypt//')
def encrypt(plaintext):plaintext &#61; bytes.fromhex(plaintext)padded &#61; pad(plaintext &#43; FLAG.encode(), 16)cipher &#61; AES.new(KEY, AES.MODE_ECB)try:encrypted &#61; cipher.encrypt(padded)except ValueError as e:return {"error": str(e)}return {"ciphertext": encrypted.hex()}</code></pre> 
<h3>hint&#xff1a;</h3> 
<blockquote> <p>ECB is the most simple mode, with each plaintext block encrypted entirely independently. In this case, your input is prepended to the secret flag and encrypted and that&#39;s it. We don&#39;t even provide a decrypt function. Perhaps you don&#39;t need a padding oracle when you have an "ECB oracle"?</p> <p><em>译文&#xff1a;</em></p> <p>ECB是最简单的模式&#xff0c;每个明文块完全独立加密。在本例中&#xff0c;您的输入被预先添加到秘密标志并加密&#xff0c;仅此而已。我们甚至不提供解密函数。或许当你有了“欧洲央行的预言者”&#xff0c;你就不需要填充预言了?</p> 
</blockquote> 
<h3>思路&#xff1a;</h3> 
<p> 本题为一道典型的ECB加密解密问题&#xff0c;首先&#xff0c;我们即不知道明文也不知道密文。我们可以发现题目中给与了我们加密算法&#xff0c;所以我们可以自己设置明文代入进行尝试&#xff1a;</p> 
<pre><code class="language-python">import requests
from tqdm import tqdm
from Crypto.Util.number import *
# m &#61; &#39;&#39;
for i in range(30):m &#61; i*&#39;61&#39;result &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)try:a &#61; result.json()["ciphertext"]print(i,len(a)//2)except:pass</code></pre> 
<p>从输出的值中&#xff0c;我们可以发现代码在第6和第22时发生改变。于是&#xff0c;我们可以推断明文为26bits长&#xff08;32 - 6、48 - 22&#xff09;&#xff0c;而加密算法中是以16bits为块进行加密和解密&#xff08;22 - 6&#xff09;。</p> 
<p>所以&#xff0c;我们可以代入15*a进行加密&#xff0c;这样前16为就为15 * a加上明文第一位&#xff0c;接着&#xff0c;我们对十六位进行暴力破解即可得到我们所需的明文第一位&#xff0c;第一到第十五位都为这个思路&#xff1a;</p> 
<pre><code class="language-python">k&#61;&#39;&#39;
for i in range(15):m &#61; &#39;61&#39;*(15-i)print(bytes.fromhex(m))result &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)m &#61; result.json()["ciphertext"]m1 &#61; m[:32]# print(m1)for j in range(32,127):m &#61; &#39;61&#39; * (15-i) &#43; km &#43;&#61; str(hex(j)[2:])result1 &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)n &#61; result1.json()["ciphertext"]n1 &#61; n[:32]# print(n1)try:if n1 &#61;&#61; m1:k &#43;&#61; str(hex(j)[2:])print(bytes.fromhex(k))breakexcept:pass# b&#39;crypto{p3n6u1n5&#39;</code></pre> 
<p>后11位思路类似&#xff0c;仅需要我们将代码扩张到前32位进行暴力破解运算&#xff1a;</p> 
<pre><code class="language-python">k&#61;&#39;63727970746f7b70336e3675316e35&#39;
for i in range(11):m &#61; &#39;61&#39;*(16-i)print(bytes.fromhex(m))result &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)m &#61; result.json()["ciphertext"]m1 &#61; m[32:64]# print(m1)for j in range(32,127):m &#61; &#39;61&#39; * (16-i) &#43; km &#43;&#61; str(hex(j)[2:])result1 &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)n &#61; result1.json()["ciphertext"]n1 &#61; n[32:64]# print(n1)try:if n1 &#61;&#61; m1:k &#43;&#61; str(hex(j)[2:])print(bytes.fromhex(k))breakexcept:pass</code></pre> 
<h3>wp&#xff1a;</h3> 
<pre><code class="language-python">import requests
from tqdm import tqdm
from Crypto.Util.number import *
# m &#61; &#39;&#39;
# for i in range(30):
#     m &#61; i*&#39;61&#39;
#     result &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)
#     try:
#         a &#61; result.json()["ciphertext"]
#         print(i,len(a)//2)
#     except:
#         passk&#61;&#39;&#39;
for i in range(15):m &#61; &#39;61&#39;*(15-i)print(bytes.fromhex(m))result &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)m &#61; result.json()["ciphertext"]m1 &#61; m[:32]# print(m1)for j in range(32,127):m &#61; &#39;61&#39; * (15-i) &#43; km &#43;&#61; str(hex(j)[2:])result1 &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)n &#61; result1.json()["ciphertext"]n1 &#61; n[:32]# print(n1)try:if n1 &#61;&#61; m1:k &#43;&#61; str(hex(j)[2:])print(bytes.fromhex(k))breakexcept:pass# b&#39;crypto{p3n6u1n5&#39;k&#61;&#39;63727970746f7b70336e3675316e35&#39;
for i in range(11):m &#61; &#39;61&#39;*(16-i)print(bytes.fromhex(m))result &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)m &#61; result.json()["ciphertext"]m1 &#61; m[32:64]# print(m1)for j in range(32,127):m &#61; &#39;61&#39; * (16-i) &#43; km &#43;&#61; str(hex(j)[2:])result1 &#61; requests.get(f&#39;http://aes.cryptohack.org/ecb_oracle/encrypt/{m}&#39;)n &#61; result1.json()["ciphertext"]n1 &#61; n[32:64]# print(n1)try:if n1 &#61;&#61; m1:k &#43;&#61; str(hex(j)[2:])print(bytes.fromhex(k))breakexcept:pass</code></pre> 
<h1>三、FLIPPING COOKIE</h1> 
<h3>题目&#xff1a;</h3> 
<pre><code class="language-python">from Crypto.Cipher import AES
import os
from Crypto.Util.Padding import pad, unpad
from datetime import datetime, timedeltaKEY &#61; ?
FLAG &#61; ?&#64;chal.route(&#39;/flipping_cookie/check_admin/<cookie>/<iv>/&#39;)
def check_admin(cookie, iv):cookie &#61; bytes.fromhex(cookie)iv &#61; bytes.fromhex(iv)try:cipher &#61; AES.new(KEY, AES.MODE_CBC, iv)decrypted &#61; cipher.decrypt(cookie)unpadded &#61; unpad(decrypted, 16)except ValueError as e:return {"error": str(e)}if b"admin&#61;True" in unpadded.split(b";"):return {"flag": FLAG}else:return {"error": "Only admin can read the flag"}&#64;chal.route(&#39;/flipping_cookie/get_cookie/&#39;)
def get_cookie():expires_at &#61; (datetime.today() &#43; timedelta(days&#61;1)).strftime("%s")cookie &#61; f"admin&#61;False;expiry&#61;{expires_at}".encode()iv &#61; os.urandom(16)padded &#61; pad(cookie, 16)cipher &#61; AES.new(KEY, AES.MODE_CBC, iv)encrypted &#61; cipher.encrypt(padded)ciphertext &#61; iv.hex() &#43; encrypted.hex()return {"cookie": ciphertext}</code></pre> 
<h3>hint&#xff1a;</h3> 
<blockquote> <p>You can get a cookie for my website, but it won&#39;t help you read the flag... I think.</p> <p><em>译文&#xff1a;</em></p> <p>你可以为我的网站获得一块cookie&#xff0c;但它不会帮助你读flag…我认为。</p> 
</blockquote> 
<h3>思路&#xff1a; </h3> 
<p>这是一到典型的CBC字节翻转问题&#xff0c;我们已知CBC的加密模式&#xff0c;其中&#xff0c;第一段密文为第一段明文于IV进行异或运算再进行AES加密&#xff0c;最终得到我们所需的第一段密文&#xff0c;第二段相比第一段既是将输入的IV改为我们求得的第一段密文。</p> 
<p>我们已知第一段明文为&#xff1a;</p> 
<pre>a1 &#61; b&#39;admin&#61;False;expi&#39;</pre> 
<p>我们需要将admin改为True&#xff0c;从而让网站以为我们拥有网站的控制权&#xff0c;因为我们通过代码可知&#xff0c;当admin&#61;True&#xff1b;时网站会返回我们flag值。所以我们需要构建一个新的IV使密文被加密后为True&#xff0c;从而骗过网站。</p> 
<p>Dec (m1) &#61; c1 ^ IV</p> 
<p>c1_new  &#61; Dec (m1) ^ IV_new</p> 
<p>其中&#xff0c;各个条件我们都已知&#xff0c;除了IV_new&#xff0c;于是&#xff0c;我们可以构建出IV_new&#xff0c;并将新的IV返回到网站中输入&#xff0c;最终求得flag。</p> 
<h3>wp&#xff1a;</h3> 
<pre><code class="language-python">import requests
from Crypto.Util.number import *result &#61; requests.get(&#39;http://aes.cryptohack.org/flipping_cookie/get_cookie&#39;)
m &#61; result.json()["cookie"]
M &#61; bytes.fromhex(m)
# print(m)
IV &#61; m[:32]
k &#61; m[32:]a1 &#61; b&#39;admin&#61;False;expi&#39;
a2 &#61; b&#39;admin&#61;True;00000&#39;
m1&#61;hex(bytes_to_long(a1))
m2&#61;hex(bytes_to_long(a2))
# print(m1)
IV_new &#61; int(m1,16)^int(m2,16)^int(IV,16)
# print(IV_new)
IV_new &#61; hex(IV_new)[2:]
print(IV_new)
result2 &#61; requests.get(f&#39;http://aes.cryptohack.org/flipping_cookie/check_admin/{k}/{IV_new}&#39;)
l &#61; result2.json()
print(l)
</code></pre> 
<h1>四、LAZY CBC</h1> 
<h3>题目&#xff1a;</h3> 
<pre><code class="language-python">from Crypto.Cipher import AESKEY &#61; ?
FLAG &#61; ?&#64;chal.route(&#39;/lazy_cbc/encrypt/<plaintext>/&#39;)
def encrypt(plaintext):plaintext &#61; bytes.fromhex(plaintext)if len(plaintext) % 16 !&#61; 0:return {"error": "Data length must be multiple of 16"}cipher &#61; AES.new(KEY, AES.MODE_CBC, KEY)encrypted &#61; cipher.encrypt(plaintext)return {"ciphertext": encrypted.hex()}&#64;chal.route(&#39;/lazy_cbc/get_flag/<key>/&#39;)
def get_flag(key):key &#61; bytes.fromhex(key)if key &#61;&#61; KEY:return {"plaintext": FLAG.encode().hex()}else:return {"error": "invalid key"}&#64;chal.route(&#39;/lazy_cbc/receive/<ciphertext>/&#39;)
def receive(ciphertext):ciphertext &#61; bytes.fromhex(ciphertext)if len(ciphertext) % 16 !&#61; 0:return {"error": "Data length must be multiple of 16"}cipher &#61; AES.new(KEY, AES.MODE_CBC, KEY)decrypted &#61; cipher.decrypt(ciphertext)try:decrypted.decode() # ensure plaintext is valid asciiexcept UnicodeDecodeError:return {"error": "Invalid plaintext: " &#43; decrypted.hex()}return {"success": "Your message has been received"}</code></pre> 
<h3>hint&#xff1a;</h3> 
<blockquote> <p>I&#39;m just a lazy dev and want my CBC encryption to work. What&#39;s all this talk about initialisations vectors? Doesn&#39;t sound important.</p> <p><em>译文&#xff1a;</em></p> <p>我只是一个懒惰的开发人员&#xff0c;希望我的CBC加密工作。这些关于初始化向量的讨论是怎么回事?听起来不重要。</p> 
</blockquote> 
<h3>思路&#xff1a;</h3> 
<p>在这道题中&#xff0c;我们有密文的生成代码&#xff0c;和密文的解密代码&#xff0c;以及flag的获得代码</p> 
<p>CBC的生成思路&#xff1a;</p> 
<p>c1 &#61; Dec(m1) ^ IV  ——&#xff08;1&#xff09;</p> 
<p>c2 &#61; Dec(m2) ^ m1  ——&#xff08;2&#xff09;</p> 
<p>c3 &#61; Dec(m3) ^ m2  ——&#xff08;3&#xff09;</p> 
<p>由于密文明文我们都可以自己设置&#xff0c;于是我们不妨让m2&#61;&#61;0&#xff1b;m1&#61;m3&#xff0c;于是让&#xff08;1&#xff09;和&#xff08;3&#xff09;进行异或运算&#xff0c;可以求得我们所需的IV值&#xff0c;将其返回网站即可&#xff1a;</p> 
<h3>wp&#xff1a;</h3> 
<pre><code class="language-python">import requests
from Crypto.Util.number import *
# a &#61; b&#39;0&#39;*16
# A &#61; hex(bytes_to_long(a))
# print(A)b &#61; &#39;61616161616161616161616161616161&#39;*3
result &#61; requests.get(f&#39;http://aes.cryptohack.org/lazy_cbc/encrypt/{b}&#39;)
m &#61; result.json()["ciphertext"]
print(m)
k1 &#61; m[:32] &#43; &#39;0&#39;*32 &#43; m[:32]
k &#61; requests.get(f&#39;http://aes.cryptohack.org/lazy_cbc/receive/{k1}&#39;)
k&#61;k.json()
print(k)b1 &#61; &#39;6161616161616161616161616161616116874d46ae7c02bcd5fe3d27fcba5b60966f8a5cc032ff235c45da24d505e288&#39;
# result1 &#61; requests.get(f&#39;http://aes.cryptohack.org/lazy_cbc/encrypt/{b1}&#39;)
# result1 &#61; result1.json()
# print(result1)key &#61; hex(int(b1[:32],16)^int(b1[64:96],16))[2:]
print(key)
m &#61; requests.get(f&#39;http://aes.cryptohack.org/lazy_cbc/get_flag/{key}&#39;)
m &#61; m.json()
print(m)M &#61; 0x63727970746f7b35306d335f703330706c335f64306e375f3768316e6b5f49565f31355f316d70307237346e375f3f7d
print(long_to_bytes(int(M)))</code></pre> 
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