Alice’s warm up
描述
Welcome to the XCTF-*CTF2022, I'm Alice and interested in AI security.
I prepared a easy warm-up for you before you enjoy those pure AI security challenges.
Like humans, AI also needs to warm up before running. Can you find something strange in these initialized parameters?
题面大部分是废话，不过还是有两点有用的信息的：第一点是需要选手关注模型的初始参数(initialized parameters)，第二点是提示这题不是纯粹的AI问题(pure AI security challenges)。
Challenge 1
我在这题中设置的第一个挑战是如何正确载入模型。这个子任务的出发点是希望传达一个观念——AI模型文件和当前的AI模型不同，它是有着完整且清晰结构的文件，而后者更多时候是一个黑漆漆的黑盒。
从比赛网站下载的文件为xxxx.zip，解压后会发现里面有一个archive文件夹，文件夹里面有一个hint.py和若干AI模型文件。这里很多师傅试图直接使用torch.load()导入data.pkl时会发现报错，实际如果熟悉pytorch模型保存的文件结构或者自己尝试保存一个模型，就会发现其实模型本身就是zip格式的，而且结构和得到的zip一样(除了多了个hint.py)。所以最方便的解法就是直接torch.load载入zip。
savepath='./0bdb74e42cdf4a42923ccf40d2a66313.zip'
net=torch.load(savepath)
载入模型之后还需要恢复模型。首先保存信息会提示不存在AliceNet1，其实只需要补上一个空的类就行了。
classAliceNet1(nn.Module):pass
当然如果你想知道内部的结构，直接打印就可以知道了。
savepath='./0bdb74e42cdf4a42923ccf40d2a66313.zip'
net=torch.load(savepath)print(net)############################################################################################################
AliceNet1((fc): Sequential((0): Linear(in_features=47, out_features=47, bias=True)(1): Linear(in_features=47, out_features=10, bias=True)(2): Linear(in_features=10, out_features=1, bias=True)))############################################################################################################
Challenge 2
开始挑战之前首先让我们回顾一下到目前为止，还有多少提示没有用上。“关注模型的初始参数”和hint.py这些提示没有用上，也没有用上提示“这题不是纯粹的AI问题”。
首先让我们查看一下net中的参数是个什么样的情况。net的结构是一个47X10X1的全连接网络，除了第0层是方阵而其他都是一般的矩阵。通过使用pytorch的state_dict()函数，我们可以观察每层矩阵的参数情况
for name in net.state_dict():print(name)############################################################################################################
fc.0.weight
fc.0.bias
fc.1.weight
fc.1.bias
fc.2.weight
fc.2.bias
############################################################################################################
for name in net.state_dict():print(net.state_dict()[name])############################################################################################################
fc.0.weight
tensor([[0.,0.,1.,...,0.,0.,0.],[0.,0.,0.,...,0.,0.,0.],[0.,0.,0.,...,0.,0.,0.],...,[0.,0.,0.,...,0.,0.,0.],[0.,0.,0.,...,0.,0.,0.],[0.,0.,0.,...,0.,0.,0.]])
fc.0.bias
tensor([-0.0481,0.0083,0.0302,0.0301,-0.0125,0.1012,0.0500,0.0477,0.0310,0.0276,-0.0967,0.0889,-0.0678,-0.1428,0.1141,-0.1254,0.1439,-0.0778,0.0316,-0.0729,0.0046,-0.1126,-0.0916,0.1256,0.0641,-0.0317,0.0422,-0.1378,-0.0489,0.0612,0.0500,0.0190,-0.0649,-0.1008,-0.1050,-0.0399,0.0906,0.0275,-0.0825,-0.1431,0.1126,-0.0183,0.1168,-0.0400,-0.0637,0.0806,-0.0296])
fc.1.weight
tensor([[-0.0512,0.1195,-0.0837,-0.0519,-0.0278,0.0530,0.1385,-0.0872,-0.1022,-0.0910,-0.0098,0.0995,0.0531,-0.0997,-0.0123,-0.0347,-0.0872,0.0987,-0.0472,0.0851,-0.1073,-0.0153,-0.0942,0.0949,0.0522,0.0521,0.1063,0.1335,0.0305,0.1082,0.0114,-0.1429,-0.1264,0.1127,-0.1318,0.0350,0.1166,0.1224,0.0600,-0.0837,-0.0425,-0.0854,0.0214,-0.1391,-0.0359,-0.0529,-0.0379],...[-0.0641,-0.1137,-0.0556,0.1383,-0.0967,0.0524,-0.0661,0.0510,-0.1030,-0.0732,0.1109,0.1101,-0.1164,-0.0505,-0.0610,-0.0219,-0.1451,-0.0486,-0.0898,-0.1229,0.1050,-0.0934,-0.0408,0.0432,0.0159,0.0220,0.0875,-0.0512,-0.0437,0.0833,0.0277,0.0892,-0.1136,-0.1330,-0.0778,0.0363,0.0043,0.1038,-0.1079,-0.0030,-0.0358,-0.1302,0.0822,0.0155,0.0218,0.0417,0.1241]])
fc.1.bias
tensor([0.0427,0.1376,-0.0805,0.1418,0.1263,0.0791,-0.0008,0.0997,0.0499,-0.0104])
fc.2.weight
tensor([[-0.2986,-0.2412,0.2308,-0.0410,0.1583,-0.0777,-0.2521,0.0848,-0.0169,-0.0387]])
fc.2.bias
tensor([-0.1993])############################################################################################################
可以发现除了第一个方阵为0-1矩阵，其他参数都在[-1,1]之间。第一个方阵是比较特别的。
现在再看看hint.py，flagset告知了flag的内容选择范围以及flag格式，同时也告知了flag的长度。可以发现，flagset中的字母数量正好为47。
import string
assertlen(flag)==16assert flagset=string.printable[0:36]+"*CTF{ALIZE}"
首先第一步是找出flagset与0-1矩阵之间的关系，这里可以大胆猜测flagset的字母和0-1矩阵的行列是一一对应的，也就是矩阵是一个邻接矩阵。这可以通过分析字母之间的相关性得到——CTF{是连在一起的且没有其他分支，}只和字母r相连。不过进一步的分析会发现，整个邻接矩阵代表的是一个带环的有向图。对于破解flag，flag的长度是需要知道的。我们可以使用dfs来完成这个工作，最大深度设置为16，初始位置为""。完整exp的如下
import torch
from torch import nn
import string
classAliceNet1(nn.Module):passdefchar2num(ch):
tmpset = string.printable[0:36]+'*CTF{ALIZE}'
tmplen=len(tmpset)for i inrange(tmplen):if(ch==tmpset[i]):return i
defdfs(ch,depth,ans):
ans+=ch
if(len(ans)==flaglen and ans[-1]=='}'):print('flag is:',ans)elif(len(ans)==flaglen):returnelse:
tmpi=char2num(ch)for i inrange(setlen):if(flagset[i]==ch):continue
tmpj=char2num(flagset[i])if(mymat[tmpi][tmpj]==1.0and used[tmpj]==False):
used[tmpj]=True
dfs(flagset[i],depth+1,ans)
used[tmpj]=False
savepath='./0bdb74e42cdf4a42923ccf40d2a66313.zip'
net=torch.load(savepath)print(net)
mymat=net.state_dict()['fc.0.weight'].tolist()
flagset=string.printable[0:36]+'*CTF{ALIZE}'
setlen=len(flagset)
flaglen=10+6
used=[False]*setlen
flag=''
used[char2num('*')]=True
dfs('*',0,flag)############################################################################################################
flag is:*CTF{qx1jukznmr}############################################################################################################
可以看出第二部分的解法其实是和AI无关的，也就是不是纯的AI题目。然后就是这题比较难分析出的点在于0-1矩阵的含义，这一点实际上当初出题的时候有考虑到，也准备了hint dfs。
不过由于实际解题情况，最终就没有放了。给各位师傅带来困扰了，我在这里真诚的道歉QAQ。
Alice’s challenge
这是一题就是一道纯的AI安全的题目了，出这题的目的在于和大家分享一种隐私攻击方法——梯度泄露攻击(毕竟大家做FGSM也做腻了对吧)。题目的来源是NeurlPS 2019的论文《Deep leakage from gradient》中的开源代码。
https://github.com/mit-han-lab/dlg
由于这题的出题目的是分享算法，所以解题只需要通过搜索诸如"从梯度恢复图像"这种短句就可以找到很多恢复算法以及对应的代码。之后稍微理解一下修改对应的代码即可获得答案。一些介绍这种算法以及对应代码的链接如下
https://zhuanlan.zhihu.com/p/476936278
https://github.com/JonasGeiping/breaching
https://www.anquanke.com/post/id/259060#h3-4
简单修改了dlg-master中的代码得到exp
from PIL import Image
import matplotlib.pyplot as plt
import torch
import torch.nn as nn
import torch.nn.functional as F
from torchvision import transforms
classAliceNet2(nn.Module):def__init__(self):super(AliceNet2, self).__init__()
self.conv = \
nn.Sequential(
nn.Conv2d(3,12,kernel_size=5,padding=2,stride=2),
nn.Sigmoid(),
nn.Conv2d(12,12,kernel_size=5,padding=2, stride=2),
nn.Sigmoid(),
nn.Conv2d(12,12,kernel_size=5,padding=2,stride=1),
nn.Sigmoid(),
nn.Conv2d(12,12,kernel_size=5,padding=2,stride=1),
nn.Sigmoid(),)
self.fc = \
nn.Sequential(
nn.Linear(768,200))defforward(self, x):
x = self.conv(x)
x = x.view(x.size(0),-1)
x = self.fc(x)return x
defcriterion(pred_y, grand_y):# This is the Cross entropy loss function
tmptensor=torch.mean(
torch.sum(- grand_y * F.log_softmax(pred_y, dim=-1),1))return tmptensor
ts1 = transforms.Compose([transforms.Resize(32),transforms.CenterCrop(32),transforms.ToTensor()])
ts2 = transforms.ToPILImage()
my_device ="cpu"if torch.cuda.is_available():
my_device ="cuda"
Net = torch.load('./Net.model').to(my_device)
outpath='./grad/'
torch.manual_seed(0)for i inrange(25):
original_dy_dx=dy_dx=torch.load(outpath+str(i)+'.tensor')
dummy_data = torch.randn(1,3,32,32).to(my_device).requires_grad_(True)
dummy_label = torch.randn(1,200).to(my_device).requires_grad_(True)
optimizer = torch.optim.LBFGS([dummy_data, dummy_label])
history =[]for iters inrange(300):defclosure():
optimizer.zero_grad()
pred = Net(dummy_data)
dummy_onehot_label = F.softmax(dummy_label, dim=-1)
dummy_loss = criterion(pred,
dummy_onehot_label)
dummy_dy_dx = torch.autograd.grad(dummy_loss, Net.parameters(), create_graph=True)
grad_diff =0
grad_count =0for gx, gy inzip(dummy_dy_dx, original_dy_dx):
grad_diff +=((gx - gy)**2).sum()
grad_count += gx.nelement()
grad_diff.backward()return grad_diff
optimizer.step(closure)if iters %10==0:
current_loss = closure()print(iters,"%.4f"% current_loss.item())
history.append(ts2(dummy_data[0].cpu()))
plt.figure(figsize=(12,8))for i inrange(30):
plt.subplot(3,10, i +1)
plt.imshow(history[i *10])
plt.title("iter=%d"%(i *10))
plt.axis('off')print("Dummy label is %d."% torch.argmax(dummy_label, dim=-1).item())
plt.show()
效果示例如下
原始flag图片如下
flag:*CTF{PZEXHBEARAK8MQT5NAliceE}
本文转载自: https://blog.csdn.net/qq_42940521/article/details/124242259
版权归原作者 AliceNCsyuk 所有， 如有侵权，请联系我们删除。
版权归原作者 AliceNCsyuk 所有， 如有侵权，请联系我们删除。