libsnark_related
近日因为种种原因又拾起了零知识证明方向的相关工作,于是在此简略做一下libsnark相关的记录
https://github.com/scipr-lab/libsnark/tree/2af440246fa2c3d0b1b0a425fb6abd8cc8b9c54d
基于zk-SNARKs实现非交互式零知识证明应用的开发顺序可以概括如下:
创建一个r1cs_constraint_system(l ...
Zerocash_Report
Report
ReferenceSurvey-SNARKs.pdf
Zerocash_ Decentralized anonymous payments from bitcoin(extended).pdf.pdf)
zkSNARK-explained-III.pdf
Appendix零知识证明 - 从QSP到QAP
参考自V神($R1CS\Rightarrow QAP$ ...
Issue Research for Blockchain(Chaitin)
Password required.
5ab08714a2c6ce6a221d23e7b0eacf85644656658ec2ec7ca211478449cab1ef73702a27b916e9f4b253659eee730a9cf3f78bf0d0405a7b40f35a0f3aa4353fe73ffc7872761c2ad0d770884d2b918c ...
Learning of NIZK
R1CS of zk-SNARKS需提前将待证明的命题表达为 R1CS (Rank One Constraint System)
e.g. 给定等式$x_{1}^{2}x_{2}+x_{1}+1=22\ (x_{1}=3,x_{2}=2)$,将其化简如下(单元仅包含加/减/乘)
$y=x_{1}+1\quad\leftarrow(1)$$z=x_{1}\cdot x_{1}\quad\leftar ...
金融密码杯2020-挑战赛
Prefacesl和coin带我躺了ojzzzz,这里仅作dec-01的记录,其他题目全是SM国密系列
dec - 01[题目文件]Click Here to Download
[题解分析]111111 round的feistel,但每轮使用的轮密钥均相同(思维固化一直以为是差分一直卡着,111111轮想🍑
且给出的明密文对中,给出的前65536组pt前四个字节fixed(‘aaaa’),后65 ...
WriteUp(GACTF2020) - Crypto
da Vinci after rsa[题解分析]12factor(0x1d42aea2879f2e44dea5a13ae3465277b06749ce9059fd8b7b4b560cd861f99144d0775ffffffffffff)# 9749 * 11237753507624591 * 9127680453986244150392840833873266696712898279308227 ...
WriteUp(强网杯2020) - Crypto
modestudy[题解分析]qwb经典study系列(套娃
Challenge-1: AES/CBC字节翻转
Challenge-2: AES/CBC Encrypt Oracle(iv fixed),发送b'\x00'*32,即可还原iv
Challenge-3: ECB,将第三组cipher换为第五组cipher即可
Challenge-4: ECB选择明文攻击 ( ...
CTF-Training-Record-2
GXYCTF2019 - CommonModulusAttack[题目考点]
Java nextInt() using Truncated LCG
[题目文件]Click Here to Download
[题解分析]并不知道和题目名字有什么关系- -
123456789101112131415161718public static void main(String[] paramArrayOf ...
WriteUp(WMCTF2020) - Crypto
Prefaceidiot box ===> hellman拿了一血,祥哥拿了二血,顺利落幕(草草草草草草才知道hellman博士都毕业了dbq
Piece of Cake[题解分析]考察Wiener’s Attack & NTRU正常解密 & RSA common private key attack
在本题中,RSA cryptosystem的ph=(p-1)(q+1), ...
WriteUp(中国电信2020天翼杯) - Crypto
EasyRSA[题解分析]Cry签到题,逐字节的RSA,且assert(e < 20000),爆破即可
[exp]123456789101112131415n = c = [...]e = Nonefor _ in range(3, 20000): if pow(ord('f'), _, n) == c[0]: e = _ break# e = 11299f ...