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2017::HITCON-qual::Sakura

CTF Reverse HITCON
Word count: 849Reading time: 4 min
2017/11/07 Share

要求輸入 400 個字元
然後會把輸入當作20*20的陣列丟入function做驗證
有 150 個迴圈
在進入迴圈會把需要的 local 變數設好
流程如下

取一個本地變數出來
每個迴圈做的次數不等

迴圈
    根據剛剛取出的數為準從使用者輸入中拿字元出來
    檢查範圍是否在1~9
    不能有重複的數字
    都沒問題就把它累加進本回圈的sum並且放入sha256要用的陣列中
    再取一次本地變數出來
迴圈結束

檢查sum是否符合本迴圈要求的數值
通過150個迴圈的話則會印出flag

我看有人用angr解，但是我覺得塞位置給 angr avoid 有點麻煩

不如把 psudocode dump 出來用正規表達取變數直接生一段 z3 code 解限制式

方便等等 parse 先把一些奇怪的 function name 修改一下

大多數只是單純回傳，我命名為 nothing

有做加法再回傳的命名為 plus_數字

把 psudocode 放進 sakura_de.c

然後前面 local variable 的指定式長得太大眾臉了，我把它移到 variable 文字檔

solver.py

#! /usr/bin/env python
# -*- coding: utf-8 -*-
import re

f = open("./sakura_de.c", "r").read()

#get the variable and constraint of every loop
idx     = re.findall("v[0-9]* = \(__int64 \*\)nothing_[0-9]*\(\(__int64\)\&v([0-9]*)\)", f)
loop_n  = re.findall("v[0-9]* = plus_([0-9]*)", f)
con     = re.findall("\( v[0-9]* \!= ([0-9]*) \)", f)

f = open("./variables", "r").read()

#get the local variable on stack
equ   = re.findall("v([0-9]*) = ([0-9]*);", f) 
stack = [0]*1944
for ( x , y ) in equ:
    stack[int(x)] = int(y)

get_stack = lambda a,b: (stack[int(idx[a])+b*2] + (stack[int(idx[a])+1+b*2] <<32))


z3_code = """
#! /usr/bin/env python
# -*- coding: utf-8 -*-
from z3 import *

s = Solver()
"""

#add constraint to z3
for row in range(20):
    for col in range(20):
        name = chr(ord('A')+row) + str(col)
        z3_code += "{} = Int('{}')\n".format(name,name)
        z3_code += "s.add( 0 < {} , {} < 10)\n".format(name,name)

for j in range(150):
    loop = int(int(loop_n[j])/8)
    this_loop = []
    for i in range(loop):
        xx = get_stack(j,i)
        row =  xx&0xffffffff
        col = xx>>32
        this_loop.append((row, col))

    for i in range(0,len(this_loop)-1):
        (row, col) = this_loop[i]
        a = chr(ord('A')+row) + str(col)
        for k in range(i+1,len(this_loop)):
            (row, col) = this_loop[k]
            b = chr(ord('A')+row) + str(col)
            z3_code += "s.add( {} != {})\n".format(a, b)
    z3_code += "s.add("
    for i in range(len(this_loop)):
        (row, col) = this_loop[i]
        a = chr(ord('A')+row) + str(col)
        z3_code += " {} + ".format(a)
    z3_code += " 0 == {})\n".format(con[j])

z3_code += """
s.check()
m = s.model()
sakura = [0]*400

for i in m:
    row = ord(str(i)[0]) - ord('A')
    col = int(str(i)[1:])
    sakura[ 20*row + col ] = str(m[i])

print ''.join(sakura)
"""

exec z3_code

sakura_de.c (partial)

__int64 __fastcall sub_850(__int64 UserInput)
{
  v604 = (__int64 *)nothing_16((__int64)&v904);
  v754 = plus_16((__int64)&v904);
  while ( v604 != (__int64 *)v754 )
  {
    v1 = *v604;
    *(&OutPutArray[20 * (signed int)v1] + SHIDWORD(v1)) = *(_BYTE *)(UserInput + 20LL * (signed int)v1 + SHIDWORD(v1));
    v454 = *(_BYTE *)(UserInput + 20LL * (signed int)v1 + SHIDWORD(v1)) - '0';
    if ( v454 <= 0 || v454 > 9 )
      Success = 0;
    if ( (v304 >> v454) & 1 )
      Success = 0;
    v304 |= 1 << v454;
    v154 += v454;
    ++v604;
  }
  if ( v154 != 17 )
    Success = 0;
  v155 = 0;
  v305 = 0;
  v605 = (__int64 *)nothing_16((__int64)&v908);
  v755 = plus_16((__int64)&v908);
  while ( v605 != (__int64 *)v755 )
  {
    v2 = *v605;
    *(&OutPutArray[20 * (signed int)v2] + SHIDWORD(v2)) = *(_BYTE *)(UserInput + 20LL * (signed int)v2 + SHIDWORD(v2));
    v455 = *(_BYTE *)(UserInput + 20LL * (signed int)v2 + SHIDWORD(v2)) - 48;
    if ( v455 <= 0 || v455 > 9 )
      Success = 0;
    if ( (v305 >> v455) & 1 )
      Success = 0;
    v305 |= 1 << v455;
    v155 += v455;
    ++v605;
  }
  if ( v155 != 3 )
    Success = 0;

variables (partial)

v904 = 1;
v905 = 6;
v906 = 2;
v907 = 6;
v908 = 1;
v909 = 7;
v910 = 2;
v911 = 7;
v912 = 2;
v913 = 1;
v914 = 3;
v915 = 1;
v1312 = 2;
v1313 = 2;
v1314 = 3;
v1315 = 2;

Original Author: Terrynini

Original link: http://blog.terrynini.tw/tw/2017-HITCON-qual-Sakura/

Publish at: November 7th 2017, 4:39:57

Copyright: This article is licensed under CC BY-NC 4.0

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