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Iris CTF 2025 Writeup

昨年に引き続き新年最初の CTF として Iris CTF に参加しました。

フル参加はしてないのですが、いくつか解いた問題の Writeup を簡単に書きました。

もくじ

Now this will run on my 486?(Rev)

It’s the future, which means tons of new CPU extensions! Unfortunately, we’re using such bleeding edge technology that your computer probably doesn’t support it.

Thankfully, with our proprietary JIT compiler, you can run it even on your 486!

問題バイナリとして与えられたプログラムを解析してみると、始めに sigaction で SIGILL(プロセッサがサポートしていない不正な命令の検出) 時のアクションとして sig4_func を登録しています。

int64_t (* act)(int64_t arg1, int64_t siginfo, void* arg3) = sig4_func
sigaction(sig: 4, &act, oact: nullptr)

また、その後は 0x378 バイト分のメモリ領域を確保してハードコードされたバイト列を書き込んだあと、そのメモリ領域に実行権限を割り当てて実行を試みています。

ただし、ここで書き込んだバイト列は X86-64 の実行要件を満たさないため、SIGILL が発火します。

rdi, rsi_1 = __builtin_memcpy(dest: &var_398, src: &ex_data, n: 0x378)
*rdi = *rsi_1
data_size = 0x379
// 3 = PROT_READ | PROT_WRITE
// 22h = MAP_PRIVATE | MAP_ANONYMOUS
int64_t rax_2 = mmap(addr: nullptr, len: 0x379, prot: 3, flags: 0x22, fd: 0xffffffff, offset: 0)
mem_data = rax_2
int32_t result = 1

if (rax_2 != -1)
  int64_t rax_3 = mmap(addr: nullptr, len: data_555555558010, prot: 3, flags: 0x22, fd: 0xffffffff, offset: 0)
  data_555555558020 = rax_3

  if (rax_3 != -1)
      memcpy(mem_data, &var_398, data_size)

      // Change protect into 5 = RX
      if (mprotect(mem_data, data_size, 5) == 0xffffffff)
          result = 1
      else if (mprotect(data_555555558020, data_555555558010, 3) == 0xffffffff)
          result = 1
      else
          int32_t result_1 = mem_data(0, 0, 0, 0, data_555555558020)
          result = result_1

          if (result_1 != 0)
              __printf_chk(flag: 1, format: "Program returned "incorrect" (%d…", zx.q(result_1))
          else
              puts(str: "Program returned "correct"!")

          munmap(mem_data, data_size)
          munmap(data_555555558020, data_555555558010)

if (rax == *(fsbase + 0x28))
  return result

SIGILL が発火した場合、sigaction で登録した sig4_func が呼び出されます。

image-20250106224444706

ここでは、先ほどメモリ領域に書き込んだバイトデータを解析して別のデータに置き換える操作を行っているようです。

このハードコードされたメモリデータを見ると、このデータは 0x17 から始まることがわかります。

image-20250106224542005

そこで、とりあえず 0x17 の解析部分を Python で置き換えてみました。

image-20250106230051684 

stack = bytearray()
for num in data:
    stack.extend(num.to_bytes(1, 'little'))

i = 0
op = stack[i]

if op == 0x17:
    tmp = stack[i+3:i+5]
    assert(len(tmp) == 2)
    struct.unpack("<H", tmp)[0]
    stack[i] = ((stack[i+1] & 7) | 0xb8)
    stack[i+1:i+3] = tmp
    stack[i+5] = 0x90
    stack[i+6] = 0x90
    stack[i+7] = 0x90

この結果を見てみると、どうやら 0x17 から始まる分岐の中では X86-64 で実行可能なコードをメモリ領域に書き込む操作をしていることがわかりました。(不要な部分は NOP で埋めているようです)

どうせプログラム自身で実行コードをメモリに書き込んでくれるならわざわざデコーダーを自作する必要もないよね、ということで gdb を使ってメモリに展開された実行コードを抽出することにしました。

以下のように、NOP 区切りでデコードされた実行コードがメモリに展開されていました。

0x7ffff7ffa000:      mov    eax,0x1000
0x7ffff7ffa005:      nop
0x7ffff7ffa006:      nop
0x7ffff7ffa007:      nop
0x7ffff7ffa008:      mov    ecx,0x67616c66
0x7ffff7ffa00d:      nop
0x7ffff7ffa00e:      nop
0x7ffff7ffa00f:      nop
0x7ffff7ffa010:      mov    DWORD PTR [rax+r8*1],ecx
・・・

以下が、メモリに展開された実行コードから NOP 部分を取り除いたものです。

このアセンブリコードを読むと、read で受け取った 0x20 バイトの文字列を 1 文字ずつ加算した結果が 0xcff になるかを検証しています。

0x7ffff7ffa000:      mov    eax,0x1000
0x7ffff7ffa008:      mov    ecx,0x67616c66
0x7ffff7ffa010:      mov    DWORD PTR [rax+r8*1],ecx
0x7ffff7ffa018:      mov    eax,0x1004
0x7ffff7ffa020:      mov    ecx,0x203f
0x7ffff7ffa028:      mov    DWORD PTR [rax+r8*1],ecx
0x7ffff7ffa030:      mov    ecx,0x1000
0x7ffff7ffa038:      mov    eax,0x1
0x7ffff7ffa040:      mov    edi,0x1
0x7ffff7ffa048:      lea    rsi,[rcx+r8*1]
0x7ffff7ffa050:      mov    edx,0x6
0x7ffff7ffa058:      syscall
0x7ffff7ffa060:      mov    ecx,0x2000
0x7ffff7ffa068:      xor    eax,eax
0x7ffff7ffa070:      xor    edi,edi
0x7ffff7ffa078:      lea    rsi,[rcx+r8*1]
0x7ffff7ffa080:      mov    edx,0x20
0x7ffff7ffa088:      syscall
0x7ffff7ffa090:      mov    edx,0x0
0x7ffff7ffa098:      mov    ebx,0x1f
0x7ffff7ffa0a0:      mov    eax,0x2000
0x7ffff7ffa0a8:      add    eax,ebx
0x7ffff7ffa0b0:      mov    eax,DWORD PTR [rax+r8*1]
0x7ffff7ffa0b8:      mov    ecx,0xff
0x7ffff7ffa0c0:      and    eax,ecx
0x7ffff7ffa0c8:      add    edx,eax
0x7ffff7ffa0d0:      mov    eax,0x0
0x7ffff7ffa0d8:      cmp    ebx,eax
0x7ffff7ffa0da:      setne  al
0x7ffff7ffa0dd:      movzx  eax,al
0x7ffff7ffa0e0:      test   eax,eax
0x7ffff7ffa0e2:      je     0x7ffff7ffa100
0x7ffff7ffa0e8:      mov    eax,0x1
0x7ffff7ffa0f0:      sub    ebx,eax
0x7ffff7ffa0f8:      jmp    0x7ffff7ffa0a0
0x7ffff7ffa100:      mov    eax,0xcff
0x7ffff7ffa108:      cmp    eax,edx
0x7ffff7ffa10a:      setne  al
0x7ffff7ffa10d:      movzx  eax,al
0x7ffff7ffa110:      test   eax,eax
0x7ffff7ffa112:      je     0x7ffff7ffa128
0x7ffff7ffa118:      mov    eax,0x1
0x7ffff7ffa120:      ret

当然、「すべての文字を加算すると 0xcff になる 0x20 文字の文字列」という情報からは Flag を 1 意に特定することはできませんので、適当にこの条件を満たす入力値を与えて再度デバッグを行います。

すると、先ほどはデコードされなかった新たな実行コードがメモリに書き込まれるようになります。

このコードを見てみると、入力値を 4 文字ずつ区切ってハードコードされた Key と XOR した結果を検証しており、この値を逆算することで正しい Flag を特定できることがわかります。

image-20250109005840051

ということで、これと同じ方法で先頭から 4 文字ずつ文字列を取得していくことで正解の Flag を特定することができました。

# gdb -x run.py
import gdb
import ctypes
from pprint import pprint

BINDIR = "./"
BIN = "main"
INPUT = "./in.txt"
# OUTPUT = "./out.txt"
BREAK1 = "*0x5555555553b5"
BREAK2 = "*0x55555555585b"

# 0xbf51b0d7  0xcc38c2be = siri

flag = "irisctf{wow_very_optimal_code!!}"
T = 0xCFF
A = 0

for c in flag:
    A += ord(c)

a = chr((T - A)//(32-len(flag)))
tmp = flag + a*(32-len(flag)-1)

B = 0
for c in tmp:
    B += ord(c)

tmp += chr(T - B)
with open("./in.txt", "w") as f:
    f.write(tmp)

gdb.execute('file {}/{}'.format(BINDIR, BIN))
gdb.execute('b {}'.format(BREAK1))
# gdb.execute('b {}'.format(BREAK2))
gdb.execute('run < ./in.txt')

for i in range(90):
    gdb.execute("continue")
    gdb.execute("continue")
    print(i)
    gdb.execute('x/i {}'.format(0x7ffff7ffa000 + i*8))

deldeldel(Forensic)

I managed to log more than just keys… perhaps it was too much data to capture?

問題バイナリとして USB キー入力操作をキャプチャした pcap ファイルが与えられます。

Cryptoverse CTF 2023 Writeup - The Cyber Heist(Forensic) とか、他にも何度か解いたことがある典型問題です。

詳細については以前にも書いたので、今回はインターネットに落ちてる以下のスクリプトを使用して雑に解きました。

import subprocess,sys,os
import shlex,string
usb_codes = {
    "0x04":['a','A'],"0x05":['b','B'], "0x06":['c','C'], "0x07":['d','D'], "0x08":['e','E'], "0x09":['f','F'],"0x0A":['g','G'],"0x0B":['h','H'], "0x0C":['i','I'], "0x0D":['j','J'], "0x0E":['k','K'], "0x0F":['l','L'],"0x10":['m','M'], "0x11":['n','N'], "0x12":['o','O'], "0x13":['p','P'], "0x14":['q','Q'], "0x15":['r','R'],"0x16":['s','S'], "0x17":['t','T'], "0x18":['u','U'], "0x19":['v','V'], "0x1A":['w','W'], "0x1B":['x','X'],"0x1C":['y','Y'], "0x1D":['z','Z'], "0x1E":['1','!'], "0x1F":['2','@'], "0x20":['3','#'], "0x21":['4','$'],"0x22":['5','%'], "0x23":['6','^'], "0x24":['7','&'], "0x25":['8','*'], "0x26":['9','('], "0x27":['0',')'],"0x28":['\n','\n'], "0x29":['[ESC]','[ESC]'], "0x2A":['[BACKSPACE]','[BACKSPACE]'], "0x2B":['\t','\t'],"0x2C":[' ',' '], "0x2D":['-','_'], "0x2E":['=','+'], "0x2F":['[','{'], "0x30":[']','}'], "0x31":['\',"|'],"0x32":['#','~'], "0x33":";:", "0x34":"'\"", "0x36":",<",  "0x37":".>", "0x38":"/?","0x39":['[CAPSLOCK]','[CAPSLOCK]'], "0x3A":['F1'], "0x3B":['F2'], "0x3C":['F3'], "0x3D":['F4'], "0x3E":['F5'], "0x3F":['F6'], "0x41":['F7'], "0x42":['F8'], "0x43":['F9'], "0x44":['F10'], "0x45":['F11'],"0x46":['F12'], "0x4F":[u'→',u'→'], "0x50":[u'←',u'←'], "0x51":[u'↓',u'↓'], "0x52":[u'↑',u'↑']
   }
data = "usb.capdata"
filepath = sys.argv[1]

def keystroke_decoder(filepath,data):
    out = subprocess.run(shlex.split("tshark -r  %s -Y \"%s\" -T fields -e %s"%(filepath,data,data)),capture_output=True)
    output = out.stdout.split() # Last 8 bytes of URB_INTERPRUT_IN
    message = []
    modifier =0
    count =0
    for i in range(len(output)):
        buffer = str(output[i])[2:-1]
        if (buffer)[:2] == "02" or (buffer)[:2] == "20":
            for j in range(1):
                count +=1 
                m ="0x" + buffer[4:6].upper()
                if m in usb_codes and m == "0x2A": message.pop(len(message)-1)
                elif m in usb_codes: message.append(usb_codes.get(m)[1])
                else: break
        else:
            if buffer[:2] == "01": 
                modifier +=1
                continue   
            for j in range(1):
                count +=1 
                m  = "0x" + buffer[4:6].upper()
                if m in usb_codes and m == "0x2A": message.pop(len(message)-1)
                elif m in usb_codes : message.append(usb_codes.get(m)[0])
                else: break

    if modifier != 0:
        print(f'[-] Found Modifier in {modifier} packets [-]')
    return message

if len(sys.argv) != 2 or os.path.exists(filepath) != 1:
    print("\nUsage : ")
    print("\npython Usb_Keyboard_Parser.py <filepath>")
    print("Created by \t\t\t Sabhya <sabhrajmeh05@gmail.com\n")
    print("Must Install tshark & subprocess first to use it\n")
    print("To install run \"sudo apt install tshark\"")
    print("To install run \"pip install subprocess.run\"")
    exit(1)

function_call = keystroke_decoder(filepath,data)
hid_data =''

for _ in range(len(function_call)): hid_data += function_call[_]

if(hid_data == ''):
    function_call = keystroke_decoder(filepath, "usbhid.data")
    print("\n[+] Using filter \"usbhid.data\" Retrived HID Data is : \n")
    for _ in range(len(function_call)): print(function_call[_],end='')
    print("\n")
else:
    print("\n[+] Using filter \"usb.capdata\" Retrived HID Data is : \n")
    print(hid_data)

参考：CTF-UsbKeyboardParser/UsbKeyboardParser.py at main · 5h4rrk/CTF-UsbKeyboardParser

削除や方向キーの操作がいまいち考慮されてなさそうな実行結果になりましたが、guess で正しい Flag が irisctf{this_keylogger_is_too_hard_to_use} であると特定できました。

image-20250105165629590

No Shark?(Network)

Where’s baby shark at?

問題バイナリとして以下のテキストファイルが与えられます。

00000000000000000000000008004500003c7d1540004006bfa47f0000017f000001815e1a6f049536f500000000a002ffd7fe3000000204ffd70402080a0fddf1d40000000001030307 のような各行が通信パケットと対応していそうです。

00000000000000000000000008004500003c7d1540004006bfa47f0000017f000001815e1a6f049536f500000000a002ffd7fe3000000204ffd70402080a0fddf1d40000000001030307
00000000000000000000000008004500003c0000400040063cba7f0000017f0000011a6f815ee579bd34049536f6a012ffcbfe3000000204ffd70402080a0fddf1d40fddf1d401030307
0000000000000000000000000800450000347d1640004006bfab7f0000017f000001815e1a6f049536f6e579bd3580100200fe2800000101080a0fddf1d40fddf1d4
000000000000000000000000080045000434505f40004006e8627f0000017f0000011a6f815ee579bd35049536f680180200022900000101080a0fddf1d50fddf1d4ffd8ffe000104a46494600010100000100010000ffdb004300030202020202030202020303030304060404040404080606050609080a0a090809090a0c0f0c0a0b0e0b09090d110d0e0f101011100a0c12131210130f101010ffdb00430103030304030408040408100b090b1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010ffc200110801f401f403011100021101031101ffc4001d000001050101010100000000000000000002010304050600070809ffc400190101010101010100000000000000000000000102030405ffda000c03010002100310000001f9b87aa4d8fea49a91a3c865cd4a13706e5eb106106c2b983945cecac5444819dda6f13f78899d5373dbd64cdc852b9628f590f1a9bbcb72c5cd91a48d660674f82aed95dcf563ac586e14d14a48535caa8ee752f9e9fc6a473d3243d487646a871471a93cef3a7ea4d8fe9229eb0336459a2dc7e9bd664d8d58dd88cb94965f78fbfbb73df9914f5e867ae4be1773e1bebf36dfcde8fa9f1baa3e52d67eb8cebc612eabcfacfa373af9f759ba97d9a5a44f98753ebccebc212fcadaf6e97c96cf9e7d9e6b3d960c39a3851cc6a5e35231a3e7a0b227490ec8f94220cba73cc71a915229fa7105452d358d0f59da3f0b647b95b1f64da8f73a4f1fa3ebec6fc94f384fa857e7c3dc4fcfcf4f0fa5bcfdafa5a93cc2cfb1a5f102cccca7b72f912432c57d80f97d3eac5c51f3659ebd18a3e9d5fcebf4f0d2f7e6814ae4af4d392398dccc69cce8650b98da9191acd6b4892dde6f99e2bda481cb5cc82c6d2f3a62e7ac9092889a8170f92a691616b9e8fc7e8faf31b64f9b13e9a5f9e8f783f3f3d3e7d5f2e9f65f3e98c4f9a6bee3978f224754cb33049ed8bc7c6c9f6e2f1f9fdd31e8f9b759bf4fafe767a7cfa5f4721ae94e5919d4bcd771b3cd7553363dc3363206686b50f36e737cc39d2a72d55747b596a4d274e76fda4d1b46359792742ce8ce90af2d178fd1f40e37bb3e633e9d3e7a3dd8fcfcf4f9fd87875b7cefe8a3e7d3e923c10bd36234684f102ccf603e614faad72e7cd567b2cbc636e7c1bd9e6b6eb9e448269ecd9d8dccc740cb9388ecb56356a668b75f9b739be65c7454e53839a376759acebc6dba9c18b934b1cd7a56b5b8bacc5bcf49e2f47ae637a33c553db96c4c9a7ca5ebf37a679fbfd1d8ddd1f2c27d64be0e5f9ea67851ef679010cf5d3e644faa57c2cb51c3c23a6331ecf3ccb95a2160a6dccd952cee7d13306806596f401726674aecdbacbcd78ecc3aeb041d9dd6761df959e80c2b33e6a76762dc6da174e6cde757c76de34015312de62d274e779d79d6674f71eac233d31271b8a1c3f513525e6b00549216a4bcd8563f2a05d7137ae5294587735fc69d9a22666a64
0000000000000000000000000800450000347d1740004006bfaa7f0000017f000001815e1a6f049536f6e579c13580100219fe2800000101080a0fddf1d50fddf1d5
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0000000000000000000000000800450000347d1840004006bfa97f0000017f000001815e1a6f049536f6e579c53580100219fe2800000101080a0fddf1d50fddf1d5
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0000000000000000000000000800450000347d2b40004006bf967f0000017f000001815e1a6f049536f6e57a766480110322fe2800000101080a0fddf79e0fddf1d6
0000000000000000000000000800450000347d2c40004006bf957f0000017f000001815e1a6f049536f7e57a766480140322fe2800000101080a0fddf79e0fddf1d6

適当に ChatGPT に生成させたスクリプトでいくつかのパケットデータのヘッダの解析をさせてみると、以下のようにループバックアドレス間で何らかの TCP/IP 通信を行った際のパケットトレースデータである可能性が高いことがわかります。

image-20250105181142530

この時使用したヘッダの解析スクリプトは以下です。

import struct
from pprint import pprint

# Hex data (the given data string)
hex_data = "000000000000000000000000080045000434506040004006e8617f0000017f0000011a6f815ee579c135049536f680180200022900000101080a0fddf1d50fddf1d5cd9c0237a26459b1674aec6aed3cbf8edc085d1282c91bcedfbf299aca313b367e767371b5a89bc43e98445854e15550996a4e5e41450a84412142a110e1292840b44438454154b2772939dcdce868730f468e0446432716bb1da073ddec9e5f8d4a88b482d05cd974cedfbf176e6665638e925a66abf721ef31fa64e0a1e9a544399e6524e38010e41144045a538e12901a1b46041ae39795cc9cc9fceacf1b7220ee132d272f1c9c393549cbbc4c6af24f29e7ab32003495ccddf5c6d7d1e7929698e93b9f54d2b7a483bc47dc6ac38761e9a71519165333a95110013a40384a516b8e14ea1028445181b14e5e53cd755ccd978d4a922e888082ab229d6399de6f8f7673abd4f24e7a990dd76880a69baf2d8f7f3d9e75658ea6d41dcadde6374cb68215aec8f4b29a0b0664333ace4e050044e10e5e3a4e3ab843b40044057a90e055ecd7e69ccd391f41015b102450995c76caf0ee39b7e9e479a50e874a329b2f4f0d9ef16f8e8e67707ae6b7798fbcb7608228e8fcb31a4a060738438e391b1138597a14445068280ed10411504b46b8595dce9fc1d821c451b56440d1587d60f3f466b876ecafee7c93356c998d3f34476b3b1f4f9f6b4e370f483bcb3b816356250da30f663e939b346d816044979144053828294c5cc112869ba6e868146929639795657257b35f9150c51a469385089719ee3e9a5e1d78bfb9f25cd5b2cf1a953416cad63d0fd5e79baace91b46b506c6ec440a1ba087a4766673a480591731d418ec9410514e0a5765772214eb1ba6d1918b42834e8e0a529a595e9a7d1c64811965b4e3ad7a4959e989f37aa273d125fa79229c584afa997ddb96c7d3c1bd0740d5151b96ece81d015a872244cc86e75ab20b03ac035d27660880a752e69c1c3f2ba388b608d91ea32852094b29cd14d38d3d63b31c803360082a4981e5d709e5f5a66ad682e7cad6666cac969dd676bebe561df801c0da9741a828234421a89f1261c96c75a36559164754512440611044838395d8910f84c9083551ad614116de68e539b2571936059119b5952a7b39953543c3be5fcde928ead0dcf9b4afe6b991d48ebcf77ede05db8866f2a5a96b751ec872c08afcc72b41536173a5dad182679135a0b3902033793ac4858394c722448f0598470deab2aca204a4dacd951b24c832d8cdad51cafa4cceb03e3f640e5b2a52fee7ceb15dcd347495db1b5f6f987a606135a695858453c435615339b7b9d16f3226e49989ad46f16172b20eaf0323683089d2a870a28b0f43d23d29c2896b6808369aab4abc722a0b31c66b8709446e3df03e2f5764ba117f73e7736b93a8f41f6cd87abceef5c9ab7519a88906234a92bb6d85e7a4b9b3d416e5b39b9bac369ac4eb82d5e0646e64744c89542c88585a58764707a5385041b155c94a6b81a1a4465962d6c"

# Convert hex string to bytes
binary_data = bytes.fromhex(hex_data)

# Parse Ethernet header (14 bytes)
eth_header = binary_data[:14]
eth = struct.unpack("!6s6sH", eth_header)
dest_mac = ":".join(f"{b:02x}" for b in eth[0])
src_mac = ":".join(f"{b:02x}" for b in eth[1])
eth_type = eth[2]

# Parse IP header (20 bytes starting from byte 14)
ip_header = binary_data[14:34]
ip = struct.unpack("!BBHHHBBH4s4s", ip_header)
version_ihl = ip[0]
version = version_ihl >> 4
ihl = version_ihl & 0xF
tos = ip[1]
total_length = ip[2]
identification = ip[3]
flags_offset = ip[4]
ttl = ip[5]
protocol = ip[6]
checksum = ip[7]
src_ip = ".".join(map(str, ip[8]))
dest_ip = ".".join(map(str, ip[9]))

# Parse TCP header (20 bytes starting from byte 34)
tcp_header = binary_data[34:54]
tcp = struct.unpack("!HHLLBBHHH", tcp_header)
src_port = tcp[0]
dest_port = tcp[1]
sequence = tcp[2]
acknowledgment = tcp[3]
offset_reserved = tcp[4]
data_offset = offset_reserved >> 4
flags = tcp[5]
window = tcp[6]
checksum_tcp = tcp[7]
urgent_pointer = tcp[8]

# Display parsed information
parsed_data = {
    "Ethernet": {
        "Destination MAC": dest_mac,
        "Source MAC": src_mac,
        "Type": eth_type,
    },
    "IP": {
        "Version": version,
        "Header Length": ihl,
        "Type of Service": tos,
        "Total Length": total_length,
        "Identification": identification,
        "Flags and Offset": flags_offset,
        "TTL": ttl,
        "Protocol": protocol,
        "Checksum": checksum,
        "Source IP": src_ip,
        "Destination IP": dest_ip,
    },
    "TCP": {
        "Source Port": src_port,
        "Destination Port": dest_port,
        "Sequence Number": sequence,
        "Acknowledgment Number": acknowledgment,
        "Data Offset": data_offset,
        "Flags": flags,
        "Window Size": window,
        "Checksum": checksum_tcp,
        "Urgent Pointer": urgent_pointer,
    },
}

pprint(parsed_data)

そこで、このテキストデータを以下のスクリプトで pcap ファイルに変換することにしました。

from scapy.all import *

with open("./noshark.txt", "r") as f:
    packet_data_list = f.readlines()

pcap_file = "output.pcap"

packets = []
for hex_data in packet_data_list:
    binary_data = bytes.fromhex(hex_data[:-1])
    packet = Ether(binary_data)
    packet_summary = packet.show(dump=True)
    packets.append(packet)
    print(packet)

wrpcap(pcap_file, packets)

scapy は Python によるパケット操作ライブラリであり、様々なプロトコルの通信パケットを作成したり pcap ファイルとして保存したりできる機能があります。

参考：secdev/scapy: Scapy: the Python-based interactive packet manipulation program & library.

今回は、問題バイナリとして与えられたテキストファイル内の各行をバイト文字列に変換した上で、Ether 関数でイーサネットパケットとして解析し、その結果を pcap ファイルとして保存しています。

このスクリプトでダンプした pcap ファイルを Wireshark で解析すると、以下の通り画像ファイルを送信していることがわかりました。

image-20250105173824423

このパケットデータを raw データとして保存して jpg 拡張子を付与すると以下の画像と共に正解の Flag である irisctf{welcome_to_net_its_still_ez_to_read_caps_without_wireshark} を取得することができました。

image-20250105174133841

Not Eelaborate(OSINT)

CHALLENGE’S DISCLAIMER: Please do not include any accent markings or special characters into the flag. Apply proper capitalization to the flag. Here’s an example of the flag irisctf{Restaurant_Name}

After my long train ride, I visited a deer park and got to feed the wildlife. There were so many restaurants to choose from but I was craving eel. I really like the soup mixed in with the rice and fish. The wasabi threw me off since I don’t normally have it served this way.

I would recommend this place if you want to find a quiet restaurant to eat at, and wouldn’t mind finding a few small fish bones. Eels are known to carry lots of tiny bones it’s inevitable that you’ll find it in a lot of places.

Question: What is the full name of the restaurant?

https://osint-food-blog-web.chal.irisc.tf/

以下のお店を特定せよという問題でした。

img

とりあえず奈良公園周辺でひつまぶしを食べられるお店を検索してみると、すぐに以下のようなよく似た画像が見つかります。

image-20250105180513549

これは「うなぎのねどこ えどがわ近鉄奈良駅前店」の写真らしいです。

image-20250105180333877

しかし、この店の名前を入力しても Flag の正解を得られません。

少し悩んだ結果、店の名前ではなく Google Map 上で表示されている名前(Edogawa Kintetsu Nara)を入力してみたところ、irisctf{Edogawa_Kintetsu_Nara} が正しい Flag であることがわかりました。

まとめ

新年あけましておめでとうございます。

最近仕事しかしてなくて CTF やる時間を中々確保できてないですが、今年も精進していこうと思います。

Published Jan 9, 2025

リバースエンジニアになりたいCTF Player(Team: 0nePadding)。WinDbgとYARAが好き。OSCP / CISSP / AtCoder 緑 Microsoft Japanに所属してます。発言はすべて個人の見解。かしわば on Twitter