Description
Internetwache CTF 2016 : SPIM
Category: Reversing Points: 50 Solves: 208 Description:Description: My friend keeps telling me, that real hackers speak assembly fluently. Are you a real hacker? Decode this string: “IVyN5U3X)ZUMYCs”
Attachment: rev50.zip
Resolution
We are given a message that was crypted using this piece of asm:
User Text Segment [00400000]..[00440000] [00400000] 8fa40000 lw $4, 0($29) ; 183: lw $a0 0($sp) # argc [00400004] 27a50004 addiu $5, $29, 4 ; 184: addiu $a1 $sp 4 # argv [00400008] 24a60004 addiu $6, $5, 4 ; 185: addiu $a2 $a1 4 # envp [0040000c] 00041080 sll $2, $4, 2 ; 186: sll $v0 $a0 2 [00400010] 00c23021 addu $6, $6, $2 ; 187: addu $a2 $a2 $v0 [00400014] 0c100009 jal 0x00400024 [main] ; 188: jal main [00400018] 00000000 nop ; 189: nop [0040001c] 3402000a ori $2, $0, 10 ; 191: li $v0 10 [00400020] 0000000c syscall ; 192: syscall # syscall 10 (exit) [00400024] 3c081001 lui $8, 4097 [flag] ; 7: la $t0, flag [00400028] 00004821 addu $9, $0, $0 ; 8: move $t1, $0 [0040002c] 3401000f ori $1, $0, 15 ; 11: sgt $t2, $t1, 15 [00400030] 0029502a slt $10, $1, $9 [00400034] 34010001 ori $1, $0, 1 ; 12: beq $t2, 1, exit [00400038] 102a0007 beq $1, $10, 28 [exit-0x00400038] [0040003c] 01095020 add $10, $8, $9 ; 14: add $t2, $t0, $t1 [00400040] 81440000 lb $4, 0($10) ; 15: lb $a0, ($t2) [00400044] 00892026 xor $4, $4, $9 ; 16: xor $a0, $a0, $t1 [00400048] a1440000 sb $4, 0($10) ; 17: sb $a0, 0($t2) [0040004c] 21290001 addi $9, $9, 1 ; 19: add $t1, $t1, 1 [00400050] 0810000b j 0x0040002c [for] ; 20: j for [00400054] 00082021 addu $4, $0, $8 ; 24: move $a0, $t0 [00400058] 0c100019 jal 0x00400064 [printstring]; 25: jal printstring [0040005c] 3402000a ori $2, $0, 10 ; 26: li $v0, 10 [00400060] 0000000c syscall ; 27: syscall [00400064] 34020004 ori $2, $0, 4 ; 30: li $v0, 4 [00400068] 0000000c syscall ; 31: syscall [0040006c] 03e00008 jr $31 ; 32: jr $ra
The important thing to note here is that we have a xor inside this.
After trying some tools to bruteforce the decryption of the message, we try to find the key because we can think the message is “IW{..}”
And when we do that, we find that the first three bytes are 0,1,2. So we try to decipher the message with a xor key being [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14] and we find the flag.
Here is the C# code used for that:
using System; using static System.Console; namespace SPIM { class Program { static void Main() { string cipher = "IVyN5U3X)ZUMYCs"; byte b = 0; for (int i = 0; i < cipher.Length; i++) { Write(Convert.ToChar(cipher[i] ^ b)); b++; } WriteLine(); ReadKey(); } } }
Flag is: IW{M1P5_!S_FUN}