这题很坑,刚开始我拿到就分析不出来了(/无奈),关键是不知道是什么加密算法,后来看题目描述的bubble,猜测是bubble

这种算法(听都没听说过。。。)

上图

这串编码

xinik-samak-luvag-hutaf-fysil-notok-mepek-vanyh-zipef-hilok-detok-damif-cusol-fezyx

百度也没找到哪里有,后面看到有大佬写的writeup,于是就借鉴一下下,233

这里是这种加密算法的解释(为了方便大家,加密算法我复制到文末),链接传送:http://wiki.yak.net/589/Bubble_Babble_Encoding.txt

附上解题代码,楼主用的python3.7,另外bubblepy这个库需要导入一下,网址附上https://pypi.python.org/pypi/bubblepy/

from bubblepy import BubbleBabble
#导入包bubblepy
str='xinik-samak-luvag-hutaf-fysil-notok-mepek-vanyh-zipef-hilok-detok-damif-cusol-fezyx'
#str是待解密字符
Str=BubbleBabble()
print(Str.decode(str))

最后,get Flag

这里是BubbleBabble加密算法

authors==Huima
status==Experimental
title==The Bubble Babble Binary Data Encoding
number==Internet Draft
date==April 2000
Network Working Group Antti Huima
Internet Draft SSH Communications Security
draft-huima-babble-01.txt April 2000 The Bubble Babble Binary Data Encoding Status of this Memo This memo provides information for the Internet community. It does
not specify an Internet standard of any kind. Distribution of this
memo is unlimited. Copyright Notice Copyright (C) The Internet Society (2000). All Rights Reserved. Abstract This document describes a new encoding method for binary data that is
intended to be used in conjunction with fingerprints of
security-critical data. 1. Introduction Hash values of certificates and public keys, known as fingerprints
or thumbprints, are commonly used for verifying that a received
security-critical datum has been received correctly. Fingerprints
are binary data and typically encoded as series of hexadecimal
digits. However, long strings hexadecimal digits are difficult for
comprehend and cumbersome to translate reliably e.g. over phone. The Bubble Babble Encoding encodes arbitrary binary data into
pseudowords that are more natural to humans and that can be
pronounced relatively easily. The encoding consumes asymptotically
the same amount of space as an encoding of the form HH HH HH HH ... where `H' is a hexadecimal digit, i.e. carries 16 bits in six
characters. However, the Bubble Babble Encoding includes a
checksumming method that can sometimes detect invalid encodings.
The method does not increase the length of the encoded data. 2. Encoding Below, _|X|_ denotes the largest integer not greater than X. Let the data to be encoded be D[1] ... D[K] where K is the length
of the data in bytes; every D[i] is an integer from 0 to 2^8 - 1.
First define the checksum series C[1] ... C[_|K/2|_] where C[1] = 1 C[n] = (C[n - 1] * 5 + (D[n * 2 - 3] * 7 + D[n * 2 - 2])) mod 36 The data is then transformed into _|K/2|_ `tuples'
T[1] ... T[_|K/2|_] and one `partial tuple' P so that T[i] = <a, b, c, d, e> where a = (((D[i * 2 - 3] >> 6) & 3) + C[i]) mod 6
b = (D[i * 2 - 3] >> 2) & 15
c = (((D[i * 2 - 3]) & 3) + _|C[i] / 6|_) mod 6
d = (D[i * 2 - 2] >> 4) & 15; and
e = (D[i * 2 - 3]) & 15. The partial tuple P is P = <a, b, c> where if K is even then a = (C[i]) mod 6
b = 16
c = _|C[i] / 6|_ but if it is odd then a = (((D[K] >> 6) & 3) + C[i]) mod 6
b = (D[K] >> 2) & 15
c = (((D[K]) & 3) + _|C[i] / 6|_) mod 6 The `vowel table' V maps integers between 0 and 5 to vowels as 0 - a
1 - e
2 - i
3 - o
4 - u
5 - y and the `consonant table' C maps integers between 0 and 16 to
consonants as 0 - b
1 - c
2 - d
3 - f
4 - g
5 - h
6 - k
7 - l
8 - m
9 - n
10 - p
11 - r
12 - s
13 - t
14 - v
15 - z
16 - x The encoding E(T) of a tuple T = <a, b, c, d, e> is then the string V[a] C[b] V[c] C[d] `-' C[e] where there are five characters, and `-' is the literal hyphen. The encoding E(P) of a partial tuple P = <a, b, c> is the
three-character string V[a] C[b] V[c]. Finally, the encoding of the whole input data D is obtained as `x' E(T[1]) E(T[2]) ... E(T[_|K/2|_]) E(P) `x' where `x's are literal characters. 3. Decoding Decoding is obviously the process of encoding reversed. To check the checksums, when a tuple <a, b, c, d, e> or partial
tuple <a, b, c> has been recovered from the encoded string, an
implementation should check that ((a - C[i]) mod 6) < 4 and that
((c - C[i]) mod 6) < 4. Otherwise the encoded string is not a valid
encoding of any data and should be rejected. 4. Checksum Strength Every vowel in an encoded string carries 0.58 bits redundancy; thus
the length of the `checksum' in the encoding of an input string
containing K bytes is 0.58 * K bits. 5. Test Vectors ASCII Input Encoding
------------------------------------------------------------------
`' (empty string) `xexax'
`1234567890' `xesef-disof-gytuf-katof-movif-baxux'
`Pineapple' `xigak-nyryk-humil-bosek-sonax' 6. Author's Address Antti Huima
SSH Communications Security, Ltd.
[XXX] 7. Full Copyright Statement Copyright (C) The Internet Society (2000). All Rights Reserved. This document and translations of it may be copied and furnished to
others, and derivative works that comment on or otherwise explain
it or assist in its implementation may be prepared, copied,
published and distributed, in whole or in part, without restriction
of any kind, provided that the above copyright notice and this
paragraph are included on all such copies and derivative works.
However, this document itself may not be modified in any way, such
as by removing the copyright notice or references to the Internet
Society or other Internet organizations, except as needed for the
purpose of developing Internet standards in which case the
procedures for copyrights defined in the Internet Standards process
must be followed, or as required to translate it into languages
other than English. The limited permissions granted above are perpetual and will not be
revoked by the Internet Society or its successors or assigns. This document and the information contained herein is provided on
an "AS IS" basis and THE INTERNET SOCIETY AND THE INTERNET
ENGINEERING TASK FORCE DISCLAIMS ALL WARRANTIES, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO ANY WARRANTY THAT THE USE OF
THE INFORMATION HEREIN WILL NOT INFRINGE ANY RIGHTS OR ANY IMPLIED
WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE.

CTF杂项之BubbleBabble加密算法的更多相关文章

  1. CTF杂项思路工具分享————2019/5/30

    分享碰到的一些奇奇怪怪的杂项解题方式: 键盘坐标密码: 题目给出一段字符串:11 21 31 18 27 33 34 对照上面的表格,就可以很清晰的看出来密文为:QAZIJCV 猪圈码: 题目为: 一 ...

  2. CTF杂项之音频隐写

    题目来自bugku 二话不说,直接上图 由题目可以看出,这题需要用到一个KEY,加上又是一段音频,很容易联想到一个著名的音频隐写解密软件Mp3stego 直接上工具 ok,成功Get Flag

  3. 【CTF杂项】常见文件文件头文件尾格式总结及各类文件头

    文件头文件尾总结 JPEG (jpg), 文件头:FFD8FF 文件尾:FF D9PNG (png), 文件头:89504E47 文件尾:AE 42 60 82GIF (gif), 文件头:47494 ...

  4. CTF杂项题解题思路

    下载压缩包解压 如果是图片就先查看图片信息 没有有用信息查看图片看是否是一个图片 如果不是图片就将文件进行还原 从还原文件中查找有用信息 例:这是一张单纯的图片 http://123.206.87.2 ...

  5. ctf杂项之easy_nbt

    下载附件查看 除了几个文件之外,没有思路 搜索nbt可知,可以使用nbtexplorer工具 果断下载,然后打开题目下载的目录 crrl+f搜索flag 猜测kflag{Do_u_kN0w_nbt?} ...

  6. BugKu CTF(杂项篇MISC)-贝斯手

    打开是以下内容 先看一下给了哪些提示 1.介绍 没了?不,拉到最底下还有 2.女神剧照 密码我4不会告诉你的,除非你知道我的女神是哪一年出生的(细品) 大致已经明白了,四位数密码,出生年份 文件是以下 ...

  7. bugku ctf 杂项 旋转跳跃 (熟悉的声音中貌似又隐藏着啥,key:syclovergeek)

    做这道题之前先给出工具   MP3Stego 下载地址 链接:https://pan.baidu.com/s/1W2mmGJcrm570EdJ6o7jD7g  提取码:1h1b 题目下载加压后 是一个 ...

  8. CTF 介绍及杂项

    CTF(Capture The Flag)中文一般译作夺旗赛,在网络安全领域中指的是网络安全技术人员之间进行技术竞技的一种比赛形式.CTF起源于1996年DEFCON全球黑客大会,以代替之前黑客们通过 ...

  9. 实验吧CTF练习题---安全杂项---异性相吸解析

    ---恢复内容开始--- 实验吧安全杂项之异性相吸   地址:http://www.shiyanbar.com/ctf/1855 flag值:nctf{xor_xor_xor_biubiubiu}   ...

随机推荐

  1. React 项目生产版本迭代页面不刷新问题

    React 页面缓存 react 打包项目进行服务端部署后,会发现index.html被浏览器缓存,访问项目url指向的还是上个版本的内容.原理是index.html被缓存后,路由指向和跳转都是旧版的 ...

  2. appium+python搭建自动化测试框架_Tools安装(一)

    作者的配置环境和版本: win10 + python3.6 +  Appium v1.4.16 1.下载node   https://nodejs.org/en/download/, 下载node.j ...

  3. Windows Server 2016-OU组织单位日常操作

    技术无所谓贵贱,既然曾经做过就总该是要留下点什么,毕竟做技术这些年给我们留下太多太多的成长经历,总有人问这些已经很皮毛了为什么还要写,其实没那么多花哨理由,就是想着做或者不做这一块总是要对过往做个简单 ...

  4. windows根据端口号杀进程

    有时候eclipse会被卡死, 结束进程后重新启动项目会出现端口已经被占用 这时候需要杀掉进程 1, cmd打开dos窗口 2, netstat -ano | findstr "端口号&qu ...

  5. Java相关资料分享(视频+电子书籍)

    正所谓“授人以鱼不如授人以渔”,你们想要的Java学习资料来啦!不管你是学生,还是已经步入职场的同行,希望你们都要珍惜眼前的学习机会,奋斗没有终点,知识永不过时. 关注底下的公众号,获取百度网盘提取码 ...

  6. Android 开发学习资源汇总

    下面这些资源对Android开发来说是很有帮助的! 1. 在线代码运行工具 地址:https://tool.lu/coderunner/ 说明:此工具站能在线运行C.C++.Java,基本能满足相关基 ...

  7. Appium在Android7.0及以上系统运行时报错的解决方案

    背景:在使用Samsung S系列手机进行自动化测试时,发现同样脚本的情况下华为荣耀系列可以正常运行,最终发现差异在于Android7.0及以上系统和appium版本不匹配,需要升级appium.但需 ...

  8. 使用 Moq 测试.NET Core 应用 -- Mock 属性

    第一篇文章, 关于Mock的概念介绍: https://www.cnblogs.com/cgzl/p/9294431.html 第二篇文章, 关于方法Mock的介绍: https://www.cnbl ...

  9. Mybatis之旅第四篇-输入输出映射

    一.引言 在日常开发用到mybatis时,因为实际的开发业务场景很复杂,不论是输入的查询条件,还是返回的结果,经常是需要根据业务来定制,这个时候我们就需要自己来定义一些输入和输出映射 二.parame ...

  10. springboot~mockMvc和asciidoctor生成基于TDD的API文档

    API文档是前端与后端快速开发,减少沟通成本的必要条件,有一份完善的文档是很必要的,由通过测试来生成文档的好处就是:测试数据有了,测试返回结果有了,而且可以对这些字段进行说明,很清晰,在springb ...