Rating:
# IWCTF 2016 - Reversing - SPIM - 50 pts
> 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](rev50.zip)
# Write-up
The attachment contains a MIPS disassembly of the decoding algorithm:
```shell
$ cat README.txt
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 decoding part starts at 0x00400024, so let's check what it does:
```
[00400024] 3c081001 lui $8, 4097 [flag] ; 7: la $t0, flag (t0 <= flag)
[00400028] 00004821 addu $9, $0, $0 ; 8: move $t1, $0 (t1 = 0)
[0040002c] 3401000f ori $1, $0, 15 ; 11: sgt $t2, $t1, 15 (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 (t2 = t0 + t1)
[00400040] 81440000 lb $4, 0($10) ; 15: lb $a0, ($t2) (a0 = [t2])
[00400044] 00892026 xor $4, $4, $9 ; 16: xor $a0, $a0, $t1 (a0 = a0 ^ t1)
[00400048] a1440000 sb $4, 0($10) ; 17: sb $a0, 0($t2) (t2 = a0)
[0040004c] 21290001 addi $9, $9, 1 ; 19: add $t1, $t1, 1 (t1 = t1 + 1)
[00400050] 0810000b j 0x0040002c [for] ; 20: j for
```
`t1` is a counter variable, and the encoding stops when it reaches 15. `t2` is a pointer to the current char in the loop, and the char at that position get's xored with the current index.
This boils down to a simple xor encryption:
```python
#!/usr/bin/python
import sys
encoded = "IVyN5U3X)ZUMYCs"
for i in range(0,len(encoded)):
sys.stdout.write(str(chr(ord(encoded[i])^i)))
```
This should do the decoding for us:
```shell
$ python dec.py
IW{M1P5_!S_FUN}
```
