分类： Forensic

一丢丢的 FingerPrint Analysis

好久没有更新技术文章了，写写最近研究的小玩意。

​image​​aa0f388fc1dd3ddb01864eb95513d4a​

（下边就是我自己做的指纹膜，实操后是可以绕过一些签到机和手机指纹锁的）

我们的指纹为什么能用于解密手机？因为每个人的指纹都是独一无二、终生不变的，手机通过识别模块收集指纹信息，与之前存储在手机中的指纹信息进行对比，匹配成功即可解锁。

原理很简单，但是实现起来并不轻松。

指纹采集技术获取的指纹图像通常为二维灰度图像，其中脊线是暗的，而谷线是亮的。虽然指纹图像并不是深度图像，但是通过将灰度视为高度，可以将指纹显示为曲面（越黑越高），近似反映了实际手指皮肤上的高低起伏。成人脊线的宽度从0.1毫米到0.3毫米不等，脊线的周期约为0.5毫米。手指的轻微损伤，如表皮烧伤、擦伤或割伤，不会影响真皮层的脊线结构，新长出的皮肤还会恢复为原来的脊线结果，这就是指纹的终生不变性。

​image​

指纹增强

但是指纹图像常常会受到噪声、光照变化、模糊等影响，因此我们需要对目标指纹图片进行图片增强

​image​

CLAHE （Contrast Limited Adaptive Histogram Equalization）， 一种增强图像对比度的方法，特别适用于局部图像区域的对比度调整，基于直方图均衡化，但对比度的增强是自适应的，并通过一个限制因子 （clipLimit） 来避免过度增强噪声。

自适应阈值化， 一种将图像转换为二值图像的方法，通过在每个局部区域内计算阈值来进行分割，而不是使用全局固定阈值，与传统的全局阈值化方法不同，自适应阈值化能够根据局部区域的不同亮度特征自动调整阈值，在图像中亮度不均匀或噪声较多的情况下，能够较好地分割出指纹区域。

def preprocess_fingerprint(image):
    # 检查图像格式并转换为灰度
    if len(image.shape) == 3:
        gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
    else:
        gray = image

    # CLAHE 增强对比度
    clahe = cv2.createCLAHE(clipLimit=2.0, tileGridSize=(8, 8))
    enhanced = clahe.apply(gray)

    # 自适应阈值化
    binary = cv2.adaptiveThreshold(
        enhanced, 255,
        cv2.ADAPTIVE_THRESH_GAUSSIAN_C,
        cv2.THRESH_BINARY,
        blockSize=11,
        C=2
    )
    return binary

指纹关键点提取

SIFT（Scale-Invariant Feature Transform，尺度不变特征变换）是一种计算机视觉算法，用于从图像中提取局部特征点，并对它们进行描述和匹配，具体分为五步：

1. 尺度空间构造

为了找到图像的特征点，SIFT 构造了一组尺度空间来检测关键点。

  • 高斯模糊 (Gaussian Blur) ：通过逐渐增加高斯核的标准差 $\sigma$，对图像进行多次模糊，得到多尺度图像。

  • 高斯差分 (DoG) ：用相邻的模糊图像相减构造差分图像（Difference of Gaussian, DoG），公式为：

    $D(x, y, \sigma) = L(x, y, k\sigma) - L(x, y, \sigma)$

  • 其中 $L(x, y, \sigma)$ 是高斯模糊图像。

2. 关键点检测

在 DoG 图像中，关键点通过极值检测找到：

  • 每个像素点在其当前尺度的 $3 \times 3$ 邻域，以及上下相邻尺度的 $3 \times 3$ 邻域中，寻找局部极值点。

3. 关键点过滤

为了确保关键点的稳定性和准确性，SIFT 对检测到的关键点进行了进一步优化：

  • 去掉低对比度点：如果关键点的 DoG 值低于某个阈值，丢弃。
  • 去掉边缘响应点：通过计算 Hessian 矩阵，去除对边缘敏感的关键点。

4. 关键点方向分配

为实现旋转不变性，SIFT 为每个关键点分配一个主方向：

  • 以关键点为中心，计算邻域内像素的梯度幅值和方向。
  • 构建方向直方图（36个bin，覆盖 $0^\circ$ 到 $360^\circ$）。
  • 主方向是直方图中幅值最大的方向，必要时添加次方向。

5. 生成特征描述符

根据关键点的尺度和方向，计算关键点周围区域的描述。

但是当我们尝试利用 SIFT 算法进行关键点信息提取识别时，就会发现本算法会对关键点进行全局匹配，这可能导致对一些局部区域的错配，尤其是在旋转、位移或纹理局部损坏的情况下。

因此为减小误差，我们还要限制一下特征匹配的范围，即：

  • 提取指纹图像的核心区域（ROI，Region of Interest），只在ROI内进行匹配。
  • ROI可以通过二值化后提取连通区域，定位指纹的主要部分。

我们还可以添加几何约束，比如：

  • 角度一致性：匹配点之间的相对角度。
  • 距离一致性：匹配点之间的距离。
# === 提取 SIFT 关键点与描述符 ===
def extract_keypoints_sift(image):
    sift = cv2.SIFT_create()
    keypoints, descriptors = sift.detectAndCompute(image, None)
    return keypoints, descriptors

# === 匹配 SIFT 关键点 ===
def match_sift_keypoints(desc1, desc2, kp1, kp2):
    bf = cv2.BFMatcher(cv2.NORM_L2, crossCheck=True)
    matches = bf.match(desc1, desc2)

    filtered_matches = []
    for match in matches:
        pt1 = kp1[match.queryIdx].pt
        pt2 = kp2[match.trainIdx].pt
        distance = np.linalg.norm(np.array(pt1) - np.array(pt2))
        if distance < 100:  # 距离阈值
            filtered_matches.append(match)

    return sorted(filtered_matches, key=lambda x: x.distance)
# === ROI 提取函数 ===
def extract_roi(binary_image):
    contours, _ = cv2.findContours(binary_image, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
    if contours:
        largest_contour = max(contours, key=cv2.contourArea)
        x, y, w, h = cv2.boundingRect(largest_contour)
        roi = binary_image[y:y + h, x:x + w]
        return roi, (x, y, w, h)
    return binary_image, None

骨架化（脊线提取）

骨架化（脊线提取）是指在指纹图像中提取脊线（即指纹的主要纹理结构）的一种过程，在骨架化中腐蚀操作可以逐步去除图像边缘，膨胀则恢复图像的区域，结合腐蚀和膨胀，逐渐提取出指纹的细节；我们用到了 skimage 库中的 skeletonize​ 函数来进行骨架化。

# === 骨架化（脊线提取） ===
def extract_ridges(image):
    inverted = cv2.bitwise_not(image)
    skeleton = skeletonize(inverted // 255)
    skeleton = (skeleton * 255).astype(np.uint8)
    return skeleton

而在指纹形成脊线的同时，也产生了重要的关键点，也是指纹的独特标识符：

在脊线分裂成两条脊线的地方，我们称为分叉点，即指纹的脊线在某一点发生了分叉，形成两个方向。

当脊线到达某个点后终止的位置，我们称为端点，即脊线的尽头，没有继续延伸下去。

# === 提取分叉点和端点 ===
def extract_minutiae(skeleton):
    minutiae = []
    for y in range(1, skeleton.shape[0] - 1):
        for x in range(1, skeleton.shape[1] - 1):
            if skeleton[y, x] == 255:
                neighbors = skeleton[y - 1:y + 2, x - 1:x + 2].sum() // 255
                if neighbors == 2:  # 端点
                    minutiae.append((x, y, 'ending'))
                elif neighbors > 3:  # 分叉点
                    minutiae.append((x, y, 'bifurcation'))
    return minutiae
# 如果一个脊线像素周围只有一个相邻的脊线像素，那么该像素是一个端点。
# 如果一个脊线像素周围有三个或更多的相邻脊线像素，那么该像素是一个分叉点。

统计与计算

这个地方我们综合分析，两种算法的结果都要尊重，因此最后我们进行加权：

# === 计算匹配率 ===
def calculate_match_ratio(kp1, kp2, matches):
    return len(matches) / max(len(kp1), len(kp2))

# === 综合特征匹配函数 ===
def compare_fingerprints(img1, img2):
    binary1 = preprocess_fingerprint(img1)
    binary2 = preprocess_fingerprint(img2)

    roi1, _ = extract_roi(binary1)
    roi2, _ = extract_roi(binary2)

    # SIFT 匹配
    kp1, desc1 = extract_keypoints_sift(roi1)
    kp2, desc2 = extract_keypoints_sift(roi2)
    matches = match_sift_keypoints(desc1, desc2, kp1, kp2)
    sift_ratio = calculate_match_ratio(kp1, kp2, matches)

    # 脊线特征匹配
    ridge1 = extract_ridges(roi1)
    ridge2 = extract_ridges(roi2)
    minutiae1 = extract_minutiae(ridge1)
    minutiae2 = extract_minutiae(ridge2)
    minutiae_matches = len(set(minutiae1) & set(minutiae2))
    ridge_ratio = minutiae_matches / max(len(minutiae1), len(minutiae2))

    # 加权融合得分
    final_score = 0.7 * sift_ratio + 0.3 * ridge_ratio

    return final_score, matches, sift_ratio, ridge_ratio

处理结果

  1. 100.png：这个是原图，两个对比率均为100%

​image​

  1. 100_0.png：结果是对的，这两个确实来自同一个人，图片来源：Andrey_Kuzmin/Shutterstock

​image​

​image​

改进？

改进肯定是有的，例如在局部脊线方向和频率估计上，我们采用二维傅里叶变换检测局部区域的多个候选正弦波，然后利用相邻区域正弦波的连续性来确定正确的脊线方向和频率。

指纹识别中的傅里叶变换_matlab中如何计算指纹图像脊线的方向场与频率场-CSDN博客

还有背景纹理去除、指纹残缺等问题亟需解决，另外如何加入LLM，利用AI识别也是一个新的发展方向......指纹识别的道路也是任重道远，就算是上面写的这个我也是更迭了10个版本，不停地优化信息提取算法以及可视化分析，也只能算是小打小闹，上不得台面，希望大家还是多多批评指正🙏。

2024 HuaWeiCup misc partly WriteUp

Draw_what_you_like

flag1

在桌面上有flag.txt，直接vol提取

​image​

​image​

flag2

搜索flag2，发现有flag2.zip​

​image​

提取出来，里面有一个sqlite文件

​image​

发现有密码Digital5211314​

桌面上有一个draw.zip​文件，提取出来，密码同上

有流量包，是数位板流量，用脚本进行处理：

import os
import matplotlib.pyplot as plt
os.system("tshark -r draw.pcap -T fields -e usbhid.data| sed '/^\s*$/d' > 1.txt")
data=[]
with open('1.txt',"r") as f:
    for line in f.readlines():
        if line[16:18] !="00":
            data.append(line)
X = []
Y = []
for line in data:
        x0=int(line[4:6],16)
        x1=int(line[6:8],16)
        x=x0+x1*256
        y0=int(line[8:10],16)
        y1=int(line[10:12],16)
        y=y0+y1*256
        X.append(x)
        Y.append(-y)
fig = plt.figure()
ax1 = fig.add_subplot(111)
ax1.set_title("result")
ax1.scatter(X, Y, c='b', marker='o')
plt.show()

​image​

flag3

附件给了secret.zip​，观察大小刚好是50mb，怀疑是VC容器

在查看文件时发现有一个打什么CTF.jpg​文件

​image​

提取出来，winhex打开删除多余的空字符，作为密钥文件加载secret.zip​

​image​

flag03:Verakey_graph}

Secret of the Varied Gif

binwalk分离出一个decode，是SVG路径数据，写脚本处理：

import matplotlib.pyplot as plt
from svg.path import parse_path, Line, CubicBezier, QuadraticBezier, Arc
# 定义SVG路径数据，每个子路径作为一个列表元素
svg_paths = [['m320.66772,62.66697c0,0 0,0.59068 0,1.77203c0,2.36269 0,5.90674 0,8.26943c0,3.54405 0,8.26944 0,10.63214c0,2.95337 0,5.90674 0,10.04146c0,3.54405 0,5.31607 0,7.67877c0,1.18135 0,3.54404 0,5.31606c0,1.77203 0,3.54405 0,4.7254c0,1.18135 0,2.95337 0,4.13471c0,1.18135 0,2.3627 0,3.54405c0,0.59067 0,2.36269 0,3.54404c0,0.59067 0,2.36269 0,2.95337c0,1.77203 0,2.95338 0,4.13472c0,1.77202 0,3.54404 0,4.72539c0,1.18135 0,2.95338 0,4.13472c0,1.77202 0,2.95337 0,4.13471c0,1.18135 0,2.3627 0,3.54405c0,1.18135 0,2.36269 0,3.54404c0,1.18135 0,2.36269 0,3.54404c0,1.18136 0,1.18136 0,2.3627c0,1.18135 0,1.77202 0,2.36269c0,1.18135 0,1.77202 0,2.36269c0,1.18135 0,1.77202 0,2.95337c0,0.59068 0,1.18136 0,1.77203c0,0.59067 0,1.18135 0,1.77202c0,0.59067 0,0.59067 0,2.36269c0,0.59067 0,1.18135 0,1.77202c0,0 0,0.59067 0,1.18135c0,0 -0.34444,0.48657 -0.70833,1.77203c-0.16275,0.57487 -0.54214,0.6806 0,1.77202c0.38335,0.77175 0.70833,1.18135 0.70833,1.18135c-1.41667,0 -1.41667,0 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-2.83333,0c-0.70833,0 -1.41667,0 -1.41667,0c-0.70833,0 -1.41668,0 -2.12501,0c-0.70833,0 -1.41667,0 -2.125,0l0,0'], ['m518.66791,60.66697c0,0 0,0.59067 0,1.77203c0,2.36269 0,5.90674 0,8.26943c0,3.54405 0,8.26944 0,10.63214c0,2.95337 0,5.90674 0,10.04146c0,3.54405 0,5.31607 0,7.67877c0,1.18135 0,3.54404 0,5.31606c0,1.77203 0,3.54405 0,4.72539c0,1.18135 0,2.95337 0,4.13472c0,1.18135 0,2.36271 0,3.54405c0,0.59067 0,2.36269 0,3.54404c0,0.59067 0,2.36269 0,2.95337c0,1.77203 0,2.95337 0,4.13472c0,1.77202 0,3.54404 0,4.72539c0,1.18135 0,2.95338 0,4.13473c0,1.77202 0,2.95337 0,4.13472c0,1.18135 0,2.3627 0,3.54405c0,1.18135 0,2.36269 0,3.54404c0,1.18135 0,2.36269 0,3.54404c0,1.18136 0,1.18136 0,2.36271c0,1.18135 0,1.77202 0,2.36269c0,1.18135 0,1.77202 0,2.36269c0,1.18135 0,1.77202 0,2.95337c0,0.59068 0,1.18135 0,1.77203c0,0.59067 0,1.18135 0,1.77202c0,0.59067 0,0.59067 0,2.36269c0,0.59067 0,1.18135 0,1.77202c0,0 0,0.59067 0,1.18135c0,0 -0.29581,0.48658 -0.60833,1.77203c-0.13978,0.57487 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# 创建图表
fig, ax = plt.subplots()
ax.set_aspect('equal')

# 存储所有点的列表
all_x_points = []
all_y_points = []

# 遍历每个SVG路径
for path_str in svg_paths:
    # 解析SVG路径
    path_string = path_str[0]   
    path = parse_path(path_string)   
    # 存储当前路径的点
    current_x_points = []
    current_y_points = []

    # 遍历路径中的每个段落
    for segment in path:
        if isinstance(segment, Line):
            # 处理直线段
            current_x_points.append(segment.start.real)
            current_y_points.append(segment.start.imag)
            current_x_points.append(segment.end.real)
            current_y_points.append(segment.end.imag)
        elif isinstance(segment, CubicBezier):
            # 处理三次贝塞尔曲线段
            current_x_points.append(segment.start.real)
            current_y_points.append(segment.start.imag)
            for t in [0.0, 0.5, 1.0]:
                point = segment.point(t)
                current_x_points.append(point.real)
                current_y_points.append(point.imag)
        elif isinstance(segment, QuadraticBezier):
            # 处理二次贝塞尔曲线段
            current_x_points.append(segment.start.real)
            current_y_points.append(segment.start.imag)
            for t in [0.0, 0.5, 1.0]:
                point = segment.point(t)
                current_x_points.append(point.real)
                current_y_points.append(point.imag)
        elif isinstance(segment, Arc):
            # 处理圆弧段
            current_x_points.append(segment.start.real)
            current_y_points.append(segment.start.imag)
            for t in [0.0, 0.5, 1.0]:
                point = segment.point(t)
                current_x_points.append(point.real)
                current_y_points.append(point.imag)

    # 将当前路径的点添加到总列表中
    all_x_points.extend(current_x_points)
    all_y_points.extend(current_y_points)

    # 在每个子路径结束时插入NaN以断开线条
    all_x_points.append(float('nan'))
    all_y_points.append(float('nan'))

# 绘制路径
ax.plot(all_x_points, all_y_points)

# 设置坐标轴范围
x_min, x_max = min(all_x_points), max(all_x_points)
y_min, y_max = min(all_y_points), max(all_y_points)
ax.set_xlim(x_min - 10, x_max + 10)
ax.set_ylim(y_min - 10, y_max + 10)

# 设置图表标题和轴标签
plt.title('SVG-title')
plt.xlabel('X-axis')
plt.ylabel('Y-axis')

# 显示网格线
plt.grid()

# 显示图表
plt.show()

​image​

猪圈密码，但是画反了，上下反转一下

​image​

但是最后一个对不上，猜了一下是g，acadesvg​

​image​

2024第一届“长城杯”信息安全铁人三项线下决赛 取证溯源Writeup

关卡描述：黑客攻击此服务器所使用的2个IP分别是什么（ascii码从小到大排列，空格分隔）

​202.1.1.1 202.1.1.129​

​image​

​image​

关卡描述：存在安全问题的apk中使用的登录密码是什么?

password663399

​image​

关卡描述：黑客尝试上传一个文件但显示无上传权限的文件名是什么？

​image​

​image​

关卡描述：黑客利用的漏洞接口的api地址是什么?（http://xxxx/xx)

​image​

关卡描述：黑客上传的webshell绝对路径是什么？

​image​

连上服务器可以找到在/usr/local/tomcat/webapps/ROOT/static/s74e7vwmzs21d5x6.jsp​

关卡描述：黑客上传的webshell的密码是什么？

​bing_pass​

​image​

关卡描述：黑客通过webshell执行的第一条命令是什么？

​pwd​

看一下这个🐎，和冰蝎差不多，是获取随机的uuid之后，将-​替换为空，取前16位做密钥然后输出出来，所以后面相应包里前面的16位字符其实就是冰蝎密钥，解密即可。

​image​

​image​

导入jadx​反编译一下

​image​

关卡描述：黑客获取webshell时查询当前shell的权限是什么？

同样的步骤

​image​

​image​

​image​

关卡描述：利用webshell查询服务器Linux系统发行版本是什么？

​image​

​image​

​image​

关卡描述：黑客从服务器上下载的秘密文件的绝对路径是什么？

连服务器找一下就有/usr/local/tomcat/webapps/ROOT/static/secert.file​

关卡描述：黑客通过反连执行的第一条命令是什么？

​image​

关卡描述：黑客通过什么文件修改的root密码（绝对路径）

​image​

关卡描述：黑客设置的root密码是多少？

​image​image​

关卡描述：黑客留下后门的反连的ip和port是什么？（ip:port)

​202.1.1.129:9999​

好像是这个目录/var/spool/mail/root​

​image​

或者直接看计划任务里面有写

关卡描述：黑客通过后门反连执行的第一条命令是什么？

​image​

关卡描述：黑客通过什么文件留下了后门？

查一下服务器的时间

​image​

​pam_unix.so​

关卡描述：黑客设置的后门密码是什么？

​image​

​ssh_back_pwd​

关卡描述：黑客的后门将root密码记录在哪个文件中？（绝对路径）

​image​

​/tmp/.sshlog​

这个地方从flag.sh​题目自己的更新脚本里找到了.sshlog​，存着之前的密码123456​和Come.1234​算非预期了

The Last Diary of Forensic

MemLabs Lab 4 | Obsession

下载链接：MemLabs_Lab4

Challenge Descryption

My system was recently compromised. The Hacker stole a lot of information but he also deleted a very important file of mine. I have no idea on how to recover it. The only evidence we have, at this point of time is this memory dump. Please help me.

Note : This challenge is composed of only 1 flag.

The flag format for this lab is: inctf{s0me_l33t_Str1ng}

我的系统最近遭到入侵。黑客窃取了很多信息，但他还删除了我的一个非常重要的文件。我不知道如何恢复它。目前我们拥有的唯一证据就是这个内存转储。请帮我。

Progress

Flag

不多谈了好吧：

image​

image​

嗨嗨嗨，运气~

image​

结合描述，文件被删除了，尝试恢复一下。

image​

说一下 MFT表：

  • NTFS文件系统包含一个叫主文件表（Master File Table）的文件，简称为MFT。对于在 NTFS 文件系统卷上的每个文件，在 MFT 中都至少会有一个条目。 MFT 条目会存储文件所有的信息，包括名称、大小、时间、时间戳、权限和数据内容，或者会存储在 MFT 条目所描述的 MFT 之外的空间。
  • 随着文件被添加到 NTFS 文件系统卷，会有更多的条目添加到 MFT ，并且 MFT 大小也会随之增加。但是当从 NTFS 卷中删除文件时，它们的 MFT 条目会被重新标记为空闲状态，并且可以重复使用。但是已为这些条目分配的磁盘空间是不会再重新分配的，并且 MFT 的空间不会减小。
  • 文件大小 小于等于 1024字节的文件，会直接存储在 MFT 表中（称为 驻留文件），如果超过1024字节，MFT 表就会包含其位置信息，不会存储文件。（称为 非驻留文件）

在volatility中提供了mftparser插件来查看系统的 MFT表：

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字符串分散开了：inctf{1_is_n0t_EQu4l_7o_2_bUt_th1s_d0s3nt_m4ke_s3ns3}

MemLabs Lab 5 | Black Tuesday

下载链接：MemLabs Lab 5

Challenge Description

We received this memory dump from our client recently. Someone accessed his system when he was not there and he found some rather strange files being accessed. Find those files and they might be useful. I quote his exact statement,

The names were not readable. They were composed of alphabets and numbers but I wasn't able to make out what exactly it was.

Also, he noticed his most loved application that he always used crashed every time he ran it. Was it a virus?

Note-1 : This challenge is composed of 3 flags. If you think 2nd flag is the end, it isn't!! 😛

Note-2 : There was a small mistake when making this challenge. If you find any string which has the string " L4B_3_D0n3 !! " in it, please change it to " L4B_5_D0n3 !! " and then proceed.

Note-3 : You'll get the stage 2 flag only when you have the stage 1 flag.

最近我们从客户那里收到了这个内存转储。有人趁他不在时访问了他的系统，客户发现一些相当奇怪的文件正在被访问。找到这些文件，它们可能很有用。客户的原话是这样：

名字不可读。它们由字母和数字组成，但我不清楚它到底是什么。

注 1 ：此挑战由 3 个flag组成。如果您认为第二个标志是结束，它不是！:P、

注 2：挑战时有一个小错误。如果您发现任何包含字符串“ L4B_3_D0n3 !! ”的字符串，请将其更改为“ L4B_5_D0n3 !! ”然后继续。

注意 3 ：只有当您拥有flag1时，您才会获得flag2。

Progress

Flag 1

不想说了：

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pslist

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看到了特殊的进程，查看了命令行历史：

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确实不可读🤔，提取出来：

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emm，Stage2.png 看来是第二部分了，还得去找第一部分。

这个地方用到了iehistory（想不到吧:P）

iehistory插件可以恢复IE浏览器的历史 index.dat 缓存文件的片段。iehistory可以提取基本的访问协议（如http、ftp等）链接、重定向链接（-REDR）和已删除条目（-LEAK）。此外，不仅仅是IE浏览器，它适用于任何加载和使用的 winnet.dll库 的进程，通常包括 Windows 资源管理器 甚至恶意软件样本。

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运气不错，熟悉的base64：

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flag{!!_w3LL_d0n3_St4g3-1_0f_L4B_5D0n3!!}

Flag 2

有了第一个flag，去解密压缩包：

Stage2​

直接出了

flag{W1thth1s$taGe_2_1sc0mPL3T3!!}

Flag 3

前面看到了 notepad.exe，提取文件，转储可执行文件，丢入IDA：

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JO8DJR0SR06JOJUUH​

XFEMYOO44F8AMYCGF57J​

flag3：bi0s{M3m_l4b5OVeR!}

MemLabs Lab 6 | The Reckoning

下载链接：MemLabs Lab 6

Challenge Description

We received this memory dump from the Intelligence Bureau Department. They say this evidence might hold some secrets of the underworld gangster David Benjamin. This memory dump was taken from one of his workers whom the FBI busted earlier this week. Your job is to go through the memory dump and see if you can figure something out. FBI also says that David communicated with his workers via the internet so that might be a good place to start.

Note : This challenge is composed of 1 flag split into 2 parts.

The flag format for this lab is: inctf{s0me_l33t_Str1ng}

我们从情报局收到了这个内存转储。他们说这个证据可能包含黑帮 大卫·本杰明 的一些秘密。这个内存转储是从本周早些时候被 FBI 逮捕的他的一名手下那里获取的。你的工作是通过内存转储，看看你是否能找出一些东西。联邦调查局还表示，大卫通过互联网与他的手下交流，因此这个内存可能是一个很好的案件突破口。

注意 ：此挑战由 1 个flag 组成，分为 2 个部分。

本实验的flag格式为：inctf{s0me_l33t_Str1ng}

Progress

The first part of flag

。。。

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排查一下可疑进程：

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先看WinRAR.exe吧

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提取一下：

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经典，又是加密。。。

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🤔emmm，有点生硬：

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flag2​

First Part：aNAm4zINg!_igU3Ss???}

The second part of flag

还有浏览器历史，之前安装过了插件：https://github.com/superponible/volatility-plugins

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向下翻，有这么一条：

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有一条回收站：

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看一下回收站的链接：

Important - Google 文档，google文档

额，全是拉丁语，不过幸好，有Google 翻译

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有个网盘链接：Mega网盘

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emm又有加密

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靠运气找Key果然还是行不通吗呜呜呜

直接 strings 全局搜：

strings Lab6.raw | grep "Mega Drive Key"

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直接看是打不开的，拖进Winhex看看

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这个地方要大写的IHDR，修复一下，16进制从69改成49

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flag_​

Second part：inctf{thi5cH4LL3Ng3!sg0nn4b3?

综上，flag为：inctf{thi5cH4LL3Ng3!s_g0nn4b3?_aNAm4zINg!_igU3Ss???}

The Second Diary of Forensic

MemLabs Lab_2 | A New World

下载链接：MemLabs Lab_2

Challenge description

One of the clients of our company, lost the access to his system due to an unknown error. He is supposedly a very popular "environmental" activist. As a part of the investigation, he told us that his go to applications are browsers, his password managers etc. We hope that you can dig into this memory dump and find his important stuff and give it back to us.

Note : This challenge is composed of 3 flags.

我们公司的一位客户由于未知错误而失去了对其系统的访问权限。据推测，他是一位非常受欢迎的“环保”主义者。作为调查的一部分，他告诉我们他的应用程序是浏览器、他的密码管理器等。我们希望你能深入这个内存转储并找到他的重要资料并将其还给我们。

注意：这个挑战由3个flag组成

Progress

Flag 1

老规矩：

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根据题目描述，查看进程，重点查看浏览器和密码管理相关进程：

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此外，上面还提到了环境变量，envars查看一下：

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啊！这串熟悉的base64开头

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flag{w3lc0m3T0$T4g3_!_Of_L4B_2}

Flag 2

回到浏览器，提取浏览器历史记录，volatility是不自带这个插件的

https://github.com/superponible/volatility-plugins

(255条消息) volatility2各类外部插件使用简介_Blus.King的博客-CSDN博客_volatility插件

注意： --plugins后写清插件位置，比如这样：

┌──(root㉿SanDieg0)-[/mnt/d/volatility-master]
└─# python2 vol.py  --plugins=./volatility/plugins/ -f "/mnt/f/Memlabs/lab2/Lab2.raw" --profile=Win7SP1x64 chromehistory

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发现了一个下载链接，

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上个实验第三部分flag：flag{w3ll_3rd_stage_was_easy}

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flag{oK_So_Now_St4g3_3_is_DoNE!!}

Flag 3

还有一个密码管理器进程KeePass.exe没有用到

KeePass会存储密码在以.kdbx为后缀的数据库中，并用主密码（master password）进行管理

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filescan并进行筛选：

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将Hidden.kdbx转储出来后，找密码，文件里面有一张叫Password.png的图片

Password​

密码右下角：P4SSw0rd_123

有了密码后，在KeePass里面打开这个数据库：

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右键直接复制出来密码：flag{w0w_th1s_1s_Th3_SeC0nDST4g3!!}

（咦？这个才是第二个flag吗？没事，我懒得改了：）

MemLabs Lab 3 | The Evil's Den

下载链接：MemLabs Lab 3

Challenge Descryption

A malicious script encrypted a very secret piece of information I had on my system. Can you recover the information for me please?

Note-1 : This challenge is composed of only 1 flag. The flag split into 2 parts.

Note-2 : You'll need the first half of the flag to get the second.

You will need this additional tool to solve the challenge,

sudo apt install steghide

The flag format for this lab is: inctf{s0me_l33t_Str1ng}

恶意脚本加密了我系统上的一条非常机密的信息。你能为我恢复信息吗？

注意-1：本次挑战只有一个flag，但被分为两个部分。

注意-2：你需要得到第一部分的flag才能得到第二部分flag。

Progress

The first part of the flag

老样子：

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题目描述说有恶意脚本，看一下cmd的记录：

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确实有一个叫恶意脚本的py脚本🤔还有一个vip.txt

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evilscript.py.py：

import sys
import string

def xor(s):

    a = ''.join(chr(ord(i)^3) for i in s)
    return a

def encoder(x):

    return x.encode("base64")

if __name__ == "__main__":

    f = open("C:\\Users\\hello\\Desktop\\vip.txt", "w")

    arr = sys.argv[1]

    arr = encoder(xor(arr))

    f.write(arr)

    f.close()

vip.txt：

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呃。。。

看一下脚本过程比较简单，先用一个字符将vip.txt的内容进行异或，然后base64加密一遍，解密也很简单，把过程逆过来就好：

s = 'am1gd2V4M20wXGs3b2U='
d = s.decode('base64')
a = ''.join(chr(ord(i)^3) for i in d)

print a

执行结果：inctf{0n3_h4lf，这是第一部分

The second part of the flag

按照题目描述，还会用到steghide，扫一下图片文件：

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.jpg都是些临时文件，.jpeg这个可能性最大，而且名字就很可疑🤔导出来看看：

suspision1image上面说，有了第一部分的flag才能获取到第二部分，那提示很明显了，密码应该就是第一部分flag

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_1s_n0t_3n0ugh}

综上，flag为：inctf{0n3_h4lf_1s_n0t_3n0ugh}