{"componentChunkName":"component---src-templates-post-template-js","path":"/ctf-angstromctf-2023-en","result":{"data":{"markdownRemark":{"id":"2884244a-195d-5aaa-8c62-32fd5d83b06e","html":"
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This page has been machine-translated from the original page.

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I participated in ångstrom CTF 2023 in April with team 0nePadding.

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We scored 920 points and finished in 151st place out of 1429 teams.

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I solved several Rev challenges this time, so I’ll write a brief writeup for each.

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Table of Contents

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zaza (Rev)

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Decompiling with Ghidra produced the following simple code.

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void main(void)\n\n{\n  int iVar1;\n  size_t sVar2;\n  long in_FS_OFFSET;\n  int local_60;\n  uint local_5c;\n  char local_58 [72];\n  long local_10;\n  \n  local_10 = *(long *)(in_FS_OFFSET + 0x28);\n  setbuf(stdout,(char *)0x0);\n  local_60 = 0;\n  local_5c = 0;\n  printf(\"I\\'m going to sleep. Count me some sheep: \");\n  __isoc99_scanf(&DAT_00102092,&local_60);\n  if (local_60 != 0x1337) {\n    puts(\"That\\'s not enough sheep!\");\n                    /* WARNING: Subroutine does not return */\n    exit(1);\n  }\n  printf(\"Nice, now reset it. Bet you can\\'t: \");\n  __isoc99_scanf(&DAT_00102092,&local_5c);\n  if (local_5c * local_60 == 1) {\n    printf(\"%d %d\",(ulong)local_5c,(ulong)(local_60 + local_5c));\n    puts(\"Not good enough for me.\");\n                    /* WARNING: Subroutine does not return */\n    exit(1);\n  }\n  puts(\"Okay, what\\'s the magic word?\");\n  getchar();\n  fgets(local_58,0x40,stdin);\n  sVar2 = strcspn(local_58,\"\\n\");\n  local_58[sVar2] = '\\0';\n  xor_(local_58);\n  iVar1 = strncmp(local_58,\"2& =$!-( <*+*( ?!&$$6,. )\\' $19 , #9=!1 <*=6 <6;66#\",0x32);\n  if (iVar1 != 0) {\n    puts(\"Nope\");\n                    /* WARNING: Subroutine does not return */\n    exit(1);\n  }\n  win();\n  if (local_10 != *(long *)(in_FS_OFFSET + 0x28)) {\n                    /* WARNING: Subroutine does not return */\n    __stack_chk_fail();\n  }\n  return;\n}
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The key section is this:

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It reads up to 0x40 bytes of user input, passes it to the xor_ function, and if the result matches 2& =$!-( <*+*( ?!&$$6,. )\\' $19 , #9=!1 <*=6 <6;66#, the flag is displayed.

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puts(\"Okay, what\\'s the magic word?\");\ngetchar();\nfgets(local_58,0x40,stdin);\nsVar2 = strcspn(local_58,\"\\n\");\nlocal_58[sVar2] = '\\0';\nxor_(local_58);\niVar1 = strncmp(local_58,\"2& =$!-( <*+*( ?!&$$6,. )\\' $19 , #9=!1 <*=6 <6;66#\",0x32);\nif (iVar1 != 0) {\nputs(\"Nope\");\n                /* WARNING: Subroutine does not return */\nexit(1);\n}\nwin();
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The xor_ function is implemented as follows.

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It checks that the input string length matches that of anextremelycomplicatedkeythatisdefinitelyuselessss, then XORs each input character with the corresponding key character.

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void xor_(long param_1)\n{\n  size_t sVar1;\n  int local_24;\n  \n  local_24 = 0;\n  while( true ) {\n    sVar1 = strlen(\"anextremelycomplicatedkeythatisdefinitelyuselessss\");\n    if (sVar1 <= (ulong)(long)local_24) break;\n    *(byte *)(param_1 + local_24) =\n         *(byte *)(param_1 + local_24) ^\n         \"anextremelycomplicatedkeythatisdefinitelyuselessss\"[local_24];\n    local_24 = local_24 + 1;\n  }\n  return;\n}
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So XOR-ing anextremelycomplicatedkeythatisdefinitelyuselessss with 2& =$!-( <*+*( ?!&$$6,. )\\' $19 , #9=!1 <*=6 <6;66# reveals that the magic word needed is SHEEPSHEEPSHEEPSHEEPSHEEP(OAPXJDIFJWTUTLE_NSLYEHEEB.

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needed is also 8 bytes × 1293 entries and has values hardcoded into it.

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Input is accepted as up to 1293 digits; each digit’s value determines how many times the corresponding function is called.

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(If the input is 123…, func0 is called once, func1 twice, func2 three times, and so on.)

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The task is to find an input that makes check exactly equal to needed after calling all functions the right number of times.

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I identified all of this but couldn’t come up with a solver algorithm — thinking too rigidly.

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In hindsight, if you let x, y, z, … denote the number of times each function runs, you get a system of 1293 equations in 1293 unknowns — solvable by simultaneous equations.

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It’s easy to get stuck in rigid thinking, so I need to be more flexible.

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I used the following reference for the final solver.

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Reference: ångstromCTF 2023 - Reverse Engineering Writeups - FazeCT Blogs

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First, use pwntools to extract the 1293 hardcoded needed values.

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from pwn import *\ne = ELF('./moon',checksec=False)\n\nneeded = []\nfor i in range(1293):\n    data = e.read(e.sym['needed']+i*8, 8)\n    data = int.from_bytes(data,'little')\n    needed.append(data)\n\nprint(needed)
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Then extract the addend from each func function to build the system of equations.

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This step uses pwntools’ ELF module and takes a fair amount of time.

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from sage.all import *\n\nmat = Matrix(check)\nmat = mat.T\nneeded = vector(needed)\nprint(mat.solve_right(needed))
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Solving the system with the extracted values yields the flag.

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Physics HW (Misc)

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Analyzing the provided pcap file, I found data that appeared to be a corrupted ZIP file being transmitted.

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Exporting the file from Wireshark and extracting it with 7-Zip let me examine the contents.

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The flag was recovered from the extracted file.

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Summary

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I was close to solving moon but couldn’t come up with the solver because I was thinking too rigidly.

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I’ve been saying this for about 3 years now — I really need to keep practicing.

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