Realworld CTF 2023 Ferris_proxy

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Realworld CTF 2023 Ferris_proxy

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看雪论坛作者ID:1mmortal


题目含有两个程序client和server以及一个包含flag传输的pcapng流量文件。打开题目一看,好消息是binary没去符号,坏消息是这是个rust。 用strace看了一下server端的监听端口是8888, 再打开pcapng看了一下,协议都是tcp,再上面一层的协议就只显示data了。 所以题目应该是要求我们逆向协议。

client::main


因为是有符号的,所以可以很轻松的找到main的函数。因为是协议,所以server和client的实现应该是对称的,这边以client端的为主进行分析。一打开client可以看到程序通过serde_yaml::de::from_str::h51d2279d81ebe645 读入了硬编码的rsa 密钥, client程序里硬编码的publickey实际上是server端的publickey,privatekey则是自己保存,和RSA公钥加密私钥解密的流程相符合。 所以协议某处肯定是用到了RSA加密方案。

继续往下看,程序在完成connect以后,有个lib::protocol::rc4::Rc4$LT$T$C$V$GT$::new::h22d438f41ed76e26 的调用,而接下来跟着就是一个tokio::task::spawn的操作, 所以怀疑流量是经过了rc4加密,调试了一下发现加密密钥是explorer。因此可以从pcapng中把所有data取出来,分为server2client和client2server两个方向,各自使用rc4进行解密。可以得到如下的流量,含有明显是包头的12字节:
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lib::protocol::crypto::Crypto$LT$T$C$V$GT$::key_exchange


因为之前看到了RSA密钥,所以就在函数列表里搜了RSA,然后顺藤摸瓜到了key_exchange这个函数。除了rsa以外,还在函数列表里搜到了aes_128_cbc的字样。那么按照一般理解来说,这里应该是用rsa加密了aes的密钥作为key进行了key_exchange, 之后的流量就用aes进行对称加密了。

所以我们找一下两个关键的rsa_encrypt 和 rsa_decrypt 来看看key_exchange部分中,发出的数据和接收的数据怎么产生aes的密钥。

Realworld CTF 2023 Ferris_proxy

Realworld CTF 2023 Ferris_proxy

可以看到发出给对方的rsa加密数据是自己这边产生的随机数,而rsa解密出来的数据也是对方产生的随机数。之后的一系列zip,map等操作应该是执行了类似于python的map(lambda x, y: x + y, [1, 3, 5, 7, 9], [2, 4, 6, 8, 10]) 这种操作。但是因为不熟悉rust,所以找不到对应的操作函数。之后抄了一份官方样例自己编译了一份rust的binary,它的源码和binary如下:
fn main() {    let a = vec![1,2,3];    let b = vec![2,3,4];    let c = a.into_iter().zip(b)        .map(|(a,b)|a^b)        .collect::<Vec<_>>();}

Realworld CTF 2023 Ferris_proxy

我发现rust同一个域下会生成一个map内的匿名函数,所以按着这个找题目程序里面应该也有。
Realworld CTF 2023 Ferris_proxy
因此可以知道,密钥是通过双方生成的随机数异或而成。

最后,还可以注意到decrypt之后,算出des密钥后有个sha256并进行发送的过程,在往下找代码的话还能看到:
Realworld CTF 2023 Ferris_proxy

说明应该是发送des密钥的sha256,让传输双方进行校验。

流量分析(lib::protocol::crypto::Crypto$LT$T$C$V$GT$::process_data_dec_in)


说句实话,包头的解析这部分猜的成分很大hhh, 当然其实小标题里那个函数里面应该都包括了程序解析数据包的逻辑,慢慢看也行。

Realworld CTF 2023 Ferris_proxy

总的来说就是包头由三个4byte的数据组成,分别代表length,action和streamid。其中action里0代表开启一个stream,1是传输,2对应关闭。
每个stream都要进行key exchange,也就是接下来的流量是rsa加密的aes密钥,加密后应该有256字节长,但是流量里有258字节,发现前两个字节恰好是0100,估计也是长度。再后面跟了一个aes密钥的sha256。最后就是aes加密的流量了。

aes加密的流量也有head,是一个4字节的代表数据长度的值,其后跟着一个16字节的当前加密所用的iv。因此可以直接进行aes_128_cbc解密。

solve


附件因为大小限制只传了client,由于两边对称所以其实看一个也行。两个text文件为rc4解密后的hex数据。
from Crypto.Cipher import ARC4from Crypto.PublicKey import RSAfrom Crypto.Cipher import PKCS1_v1_5 as PKCS1_cipherimport base64import binasciiimport pickledef get_key(key_file):    with open(key_file) as f:        data = f.read()        key = RSA.importKey(data)    return key def decrypt_data(encrypt_msg:bytes, private_key):    cipher = PKCS1_cipher.new(private_key)    back_text = cipher.decrypt(encrypt_msg, "connot decrypt.")    return back_text c2s_key = get_key('s_private.pem')with open('dec_c2s.txt','r') as f:    lines =f.readlines()d_lst = [b'',b'',b'',b'']for line in lines:    i = 0    content = binascii.unhexlify(line.strip())    while i<len(content):        legth = int.from_bytes(content[i:i+4],'big')        dontknow =  int.from_bytes(content[i+4:i+8],'big')        streamid =  int.from_bytes(content[i+8:i+12],'big')        data = content[i+12:i+4+legth]        d_lst[streamid] += data        i += 4+legthprint(d_lst)dec = [[],[],[],[]]for j in range(len(d_lst)):    enc = d_lst[j]    i = 0    tmp = b''    while i<len(enc):        if(enc[i:i+2]==b'x01x00'):            if(tmp != b''):                dec[j].append(tmp)             tmp = decrypt_data(d_lst[j][i+2:i+258],c2s_key)            dec[j].append(tmp)            tmp = b''            i += 258        else:            tmp += bytes([enc[i]])            i += 1    if(tmp != b''):        dec[j].append(tmp)print()with open('c2s_aesenc.list','wb') as f: # [[stream0_aes_key1,keyhash+enced_data],...]    pickle.dump(dec, f)

from Crypto.Cipher import AESfrom Crypto.Hash import SHA256import pickle with open('c2s_aesenc.list','rb') as f:    c2s_data = pickle.load(f)with open('s2c_aesenc.list','rb') as f:    s2c_data = pickle.load(f)  def decrypt(data,key,iv):    bs = AES.block_size    # unpad = lambda s : s[0:-ord(s[-1])]    cipher = AES.new(key, AES.MODE_CBC, iv)    data  = (cipher.decrypt(data))    return (data) for streamid in range(0,4):    c2s_key = c2s_data[streamid][0]    s2c_key = s2c_data[streamid][0]    key = bytes([s2c_key[i] ^ c2s_key[i] for i in range(16)])    data = s2c_data[streamid][1]    assert(data[0:32] == SHA256.new(key).digest())    i=32    while i < len(data):        legth = int.from_bytes(data[i:i+4],'big')        enc = data[i+4:i+4+legth]        iv = data[i+4+legth:i+4+legth+16]        print(streamid, decrypt(enc,key,iv))        i+= 4+legth+16    print(i,len(data)) print('--------------------------------------')for streamid in range(4):    c2s_key = c2s_data[streamid][0]    s2c_key = s2c_data[streamid][0]    key = bytes([s2c_key[i] ^ c2s_key[i] for i in range(16)])    data = c2s_data[streamid][1]    assert(data[0:32] == SHA256.new(key).digest())    i=32    while i < len(data):        legth = int.from_bytes(data[i:i+4],'big')        enc = data[i+4:i+4+legth]        iv = data[i+4+legth:i+4+legth+16]        print(streamid, decrypt(enc,key,iv))        i+= 4+legth+16
最后可以在解密结果中看到flag, 不过我感觉这个pcapng也是不全的,比较server2client部分没有截取到streamid==3时action==2的关闭请求。
0 b'x15xbf@;xe8Kx9exb0xd6x9fZ-xe4x9a}xc8'0 b'xe3x9ax0c/xf3x11.x05(Rxb6!xadxc6x8dx80'0 b'xe5x9cqfW?6xf2qx0exa5x84x0ex80(,HTTP/1.0 200 OKrnServer: SimpleHTTP/0.6 Python/3.6.9rnDate: Fri, 06 Jan 2023 18:16:47 GMTrnContent-type: text/html; charset=utf-8rnContent-Length: 338rnrn<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">n<html>n<head>n<meta http-equiv="Content-Type" content="text/html; charset=utf-8">n<title>Directory listing for /</title>n</head>n<body>n<h1>Directory listing for /</h1>n<hr>n<ul>n<li><a href="flag.txt">flag.txt</a></li>n</ul>n<hr>n</bo'620 6201 b'xe98x8b;x92x13xddxa8vxe7`xbc2xa4xfexa6'1 b'xb6xf7xe5xa7x19{xLxf5tx9axa0xcejSxa7'1 b'xd0xa9XP/x02pxbbDx05xf6ex12xcbnxf9HTTP/1.0 404 File not foundrnServer: SimpleHTTP/0.6 Python/3.6.9rnDate: Fri, 06 Jan 2023 18:16:47 GMTrnConnection: closernContent-Type: text/html;charset=utf-8rnContent-Length: 469rnrn<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"n        "http://www.w3.org/TR/html4/strict.dtd">n<html>n    <head>n        <meta http-equiv="Content-Type" content="text/html;charset=utf-8">n        <title>Error response</title>n    </head>n    <body>n        <h1>Error response</h1>n        <p>Error code: 404</p>n        <p>Message: File not found.</p>n        <p>Error code explanation: HTTPStatus.NOT_FOUND - Nothing matches the given URI.</p>n    </b'780 7802 b'evxe5x1a=Gxb3bxaexa5xda2x9e5xb0x9c'2 b'x02jx8c$xbexc53mx0f#xce+xbax11xcbx03'104 1043 b'xa1{qxb2;xe4{bd)xe4(x81xd0xf01'3 b'xbaxc9xc6QGxa5;xf8xdc&xd6x1cx91I-L'3 b'%gxb8xf6xe7x13Px985x95#xc0(xdfxdex13HTTP/1.0 200 OKrnServer: SimpleHTTP/0.6 Python/3.6.9rnDate: Fri, 06 Jan 2023 18:16:49 GMTrnContent-type: text/plainrnContent-Length: 38rnLast-Modified: Fri, 06 Jan 2023 18:15:52 GMTrnrnrwctf{l1fe_1s_sh0rt_DO0'348 348



Realworld CTF 2023 Ferris_proxy


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Realworld CTF 2023 Ferris_proxy

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