Iris CTF 2025 Writeup

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As with last year, I participated in Iris CTF as my first CTF of the new year.

I did not participate for the full event, but I wrote a brief writeup for a few of the challenges I solved.

Now this will run on my 486?(Rev)
deldeldel(Forensic)
No Shark?(Network)
Not Eelaborate(OSINT)
Summary

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!

Analyzing the program provided with the challenge shows that it first registers sig4_func with sigaction as the handler for SIGILL (an illegal instruction not supported by the processor).

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

After that, it allocates a 0x378-byte memory region, writes a hardcoded byte sequence into it, marks that region executable, and tries to run it.

However, the byte sequence written here does not satisfy the execution requirements for x86-64, so SIGILL is raised.

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

When SIGILL is raised, the sig4_func handler registered with sigaction is called.

Here, it appears to analyze the byte data written to that memory region and replace it with different data.

Looking at this hardcoded memory data, we can see that it starts with 0x17.

So, as a first step, I rewrote the part that handles 0x17 in Python.

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

Looking at the result, it became clear that the branch starting with 0x17 writes x86-64-executable code into memory. (It also seems to pad unnecessary parts with NOPs.)

Since the program was already writing the executable code into memory for me, there was no need to build a decoder from scratch, so I decided to use gdb to extract the expanded code from memory.

As shown below, the decoded executable code was laid out in memory with NOPs as separators.

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
・・・

Below is the code expanded in memory after removing the NOP sections.

Reading this assembly shows that it checks whether the sum of the 0x20-byte string received by read, added one character at a time, equals 0xcff.

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

Naturally, the information “a 0x20-character string whose characters sum to 0xcff” is not enough to uniquely determine the flag, so I supplied an arbitrary input satisfying that condition and debugged it again.

Doing so causes new executable code, which was not decoded before, to be written into memory.

Looking at this code, it verifies the result of splitting the input into 4-character chunks and XORing them with a hardcoded key, which means the correct flag can be recovered by reversing those values.

Using the same method, I recovered the string 4 characters at a time from the beginning and identified the correct flag.

# 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?

The challenge provides a pcap file capturing USB keyboard input.

Cryptoverse CTF 2023 Writeup - The Cyber Heist(Forensic) and similar ones are classic problems I have solved several times before.

I wrote about the details before, so this time I solved it quickly using the following script I found online.

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)

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

The output did not seem to handle deletions or arrow-key operations particularly well, but by guessing I was able to determine that the correct flag was irisctf{this_keylogger_is_too_hard_to_use}.

No Shark?(Network)

Where’s baby shark at?

The challenge provides the following text file.

Each line, such as 00000000000000000000000008004500003c7d1540004006bfa47f0000017f000001815e1a6f049536f500000000a002ffd7fe3000000204ffd70402080a0fddf1d40000000001030307, seems to correspond to a network packet.

00000000000000000000000008004500003c7d1540004006bfa47f0000017f000001815e1a6f049536f500000000a002ffd7fe3000000204ffd70402080a0fddf1d40000000001030307
00000000000000000000000008004500003c0000400040063cba7f0000017f0000011a6f815ee579bd34049536f6a012ffcbfe3000000204ffd70402080a0fddf1d40fddf1d401030307
0000000000000000000000000800450000347d1640004006bfab7f0000017f000001815e1a6f049536f6e579bd3580100200fe2800000101080a0fddf1d40fddf1d4
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
0000000000000000000000000800450000347d1740004006bfaa7f0000017f000001815e1a6f049536f6e579c13580100219fe2800000101080a0fddf1d50fddf1d5
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0000000000000000000000000800450000347d2a40004006bf977f0000017f000001815e1a6f049536f6e57a766480100322fe2800000101080a0fddf1d70fddf1d6
0000000000000000000000000800450000347d2b40004006bf967f0000017f000001815e1a6f049536f6e57a766480110322fe2800000101080a0fddf79e0fddf1d6
0000000000000000000000000800450000347d2c40004006bf957f0000017f000001815e1a6f049536f7e57a766480140322fe2800000101080a0fddf79e0fddf1d6

When I used a quick script generated by ChatGPT to parse the headers of some of the packet data, it suggested that these were most likely packet trace data from some TCP/IP communication over loopback addresses, as shown below.

Here is the header-parsing script I used at that time.

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)

So I decided to convert this text data into a pcap file with the following script.

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 is a Python packet manipulation library that can create packets for various protocols and save them as pcap files.

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

Here, I converted each line in the provided text file into a byte string, parsed it as an Ethernet packet with the Ether function, and saved the result as a pcap file.

Analyzing the pcap file dumped by this script in Wireshark showed that it was transmitting an image file, as shown below.

Saving this packet data as raw data and giving it a .jpg extension produced the following image, along with the correct flag irisctf{welcome_to_net_its_still_ez_to_read_caps_without_wireshark}.

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/

This challenge asked us to identify the restaurant shown below.

I first searched for places around Nara Park where I could get hitsumabushi, and quickly found a very similar image like the one below.

It appears to be a photo of “Unagi no Nedoko Edogawa Kintetsu Nara Ekimae Branch.”

However, entering the restaurant’s name did not yield the correct flag.

After thinking about it for a bit, I tried entering not the restaurant’s actual name but the name shown on Google Maps (Edogawa Kintetsu Nara), and found that the correct flag was irisctf{Edogawa_Kintetsu_Nara}.

Summary

Happy New Year.

Lately I have been doing little besides work and have not been able to find much time for CTFs, but I plan to keep at it this year as well.

Published Jan 9, 2025

Aspiring Reverse Engineer and CTF Player (Team: 0nePadding). Passionate about WinDbg and Anti-Virus internals. OSCP / CISSP. Working at Microsoft Japan, but all views expressed are my own.かしわば(@kash1064) on Twitter