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AMR-KELEGAMR-KELEG
committed
Mar 2019
Add solutions to practice problems of CyberTalents
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‎CyberTalents/cryptography/Hide Data.ipynb

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{
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"def update_str(s, off):\n",
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" orig_index = [ord(c)-ord('a') if not (c>='0' and c<='9') else c for c in s]\n",
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" modified_index = [chr(ord('a') + ((ind + off) %26)) if (isinstance(ind, int)) else ind for ind in orig_index]\n",
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" return ''.join(modified_index)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"def update_sentence(s, off):\n",
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" return ' '.join([update_str(st, off) for st in s.split()])"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1 hvs tzou wg 2x68mtvezn qh wg dfshhm sogm hc gss hvs tzou pih qob mci gss wh w hccy bsofzm 1 awbihs hc sbqcrs hvwg kwhv zwb13 uccr ziqy wb gczjwbu hvoh\n",
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"2 iwt uapv xh 2y68nuwfao ri xh egtiin tphn id htt iwt uapv qji rpc ndj htt xi x iddz ctpgan 1 bxcjit id tcrdst iwxh lxiw axc13 vdds ajrz xc hdakxcv iwpi\n",
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"3 jxu vbqw yi 2z68ovxgbp sj yi fhujjo uqio je iuu jxu vbqw rkj sqd oek iuu yj y jeea duqhbo 1 cydkju je udsetu jxyi myjx byd13 weet bksa yd ieblydw jxqj\n",
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"4 kyv wcrx zj 2a68pwyhcq tk zj givkkp vrjp kf jvv kyv wcrx slk tre pfl jvv zk z kffb evricp 1 dzelkv kf vetfuv kyzj nzky cze13 xffu cltb ze jfcmzex kyrk\n",
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"5 lzw xdsy ak 2b68qxzidr ul ak hjwllq wskq lg kww lzw xdsy tml usf qgm kww al a lggc fwsjdq 1 eafmlw lg wfugvw lzak oalz daf13 yggv dmuc af kgdnafy lzsl\n",
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"6 max yetz bl 2c68ryajes vm bl ikxmmr xtlr mh lxx max yetz unm vtg rhn lxx bm b mhhd gxtker 1 fbgnmx mh xgvhwx mabl pbma ebg13 zhhw envd bg lheobgz matm\n",
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"7 nby zfua cm 2d68szbkft wn cm jlynns yums ni myy nby zfua von wuh sio myy cn c niie hyulfs 1 gchony ni yhwixy nbcm qcnb fch13 aiix fowe ch mifpcha nbun\n",
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"8 ocz agvb dn 2e68taclgu xo dn kmzoot zvnt oj nzz ocz agvb wpo xvi tjp nzz do d ojjf izvmgt 1 hdipoz oj zixjyz ocdn rdoc gdi13 bjjy gpxf di njgqdib ocvo\n",
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"9 pda bhwc eo 2f68ubdmhv yp eo lnappu awou pk oaa pda bhwc xqp ywj ukq oaa ep e pkkg jawnhu 1 iejqpa pk ajykza pdeo sepd hej13 ckkz hqyg ej okhrejc pdwp\n",
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"10 qeb cixd fp 2g68vceniw zq fp mobqqv bxpv ql pbb qeb cixd yrq zxk vlr pbb fq f qllh kbxoiv 1 jfkrqb ql bkzlab qefp tfqe ifk13 dlla irzh fk plisfkd qexq\n",
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"11 rfc djye gq 2h68wdfojx ar gq npcrrw cyqw rm qcc rfc djye zsr ayl wms qcc gr g rmmi lcypjw 1 kglsrc rm clambc rfgq ugrf jgl13 emmb jsai gl qmjtgle rfyr\n",
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"12 sgd ekzf hr 2i68xegpky bs hr oqdssx dzrx sn rdd sgd ekzf ats bzm xnt rdd hs h snnj mdzqkx 1 lhmtsd sn dmbncd sghr vhsg khm13 fnnc ktbj hm rnkuhmf sgzs\n",
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"13 the flag is 2j68yfhqlz ct is pretty easy to see the flag but can you see it i took nearly 1 minute to encode this with lin13 good luck in solving that\n",
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"14 uif gmbh jt 2k68zgirma du jt qsfuuz fbtz up tff uif gmbh cvu dbo zpv tff ju j uppl ofbsmz 1 njovuf up fodpef uijt xjui mjo13 hppe mvdl jo tpmwjoh uibu\n",
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"15 vjg hnci ku 2l68ahjsnb ev ku rtgvva gcua vq ugg vjg hnci dwv ecp aqw ugg kv k vqqm pgctna 1 okpwvg vq gpeqfg vjku ykvj nkp13 iqqf nwem kp uqnxkpi vjcv\n",
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"16 wkh iodj lv 2m68biktoc fw lv suhwwb hdvb wr vhh wkh iodj exw fdq brx vhh lw l wrrn qhduob 1 plqxwh wr hqfrgh wklv zlwk olq13 jrrg oxfn lq vroylqj wkdw\n",
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"17 xli jpek mw 2n68cjlupd gx mw tvixxc iewc xs wii xli jpek fyx ger csy wii mx m xsso rievpc 1 qmryxi xs irgshi xlmw amxl pmr13 kssh pygo mr wspzmrk xlex\n",
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"18 ymj kqfl nx 2o68dkmvqe hy nx uwjyyd jfxd yt xjj ymj kqfl gzy hfs dtz xjj ny n yttp sjfwqd 1 rnszyj yt jshtij ymnx bnym qns13 ltti qzhp ns xtqansl ymfy\n",
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"19 znk lrgm oy 2p68elnwrf iz oy vxkzze kgye zu ykk znk lrgm haz igt eua ykk oz o zuuq tkgxre 1 sotazk zu ktiujk znoy cozn rot13 muuj raiq ot yurbotm zngz\n",
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"20 aol mshn pz 2q68fmoxsg ja pz wylaaf lhzf av zll aol mshn iba jhu fvb zll pa p avvr ulhysf 1 tpubal av lujvkl aopz dpao spu13 nvvk sbjr pu zvscpun aoha\n",
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"21 bpm ntio qa 2r68gnpyth kb qa xzmbbg miag bw amm bpm ntio jcb kiv gwc amm qb q bwws vmiztg 1 uqvcbm bw mvkwlm bpqa eqbp tqv13 owwl tcks qv awtdqvo bpib\n",
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"22 cqn oujp rb 2s68hoqzui lc rb yancch njbh cx bnn cqn oujp kdc ljw hxd bnn rc r cxxt wnjauh 1 vrwdcn cx nwlxmn cqrb frcq urw13 pxxm udlt rw bxuerwp cqjc\n",
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"23 dro pvkq sc 2t68ipravj md sc zboddi okci dy coo dro pvkq led mkx iye coo sd s dyyu xokbvi 1 wsxedo dy oxmyno drsc gsdr vsx13 qyyn vemu sx cyvfsxq drkd\n",
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"24 esp qwlr td 2u68jqsbwk ne td acpeej pldj ez dpp esp qwlr mfe nly jzf dpp te t ezzv yplcwj 1 xtyfep ez pynzop estd htes wty13 rzzo wfnv ty dzwgtyr esle\n",
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"25 ftq rxms ue 2v68krtcxl of ue bdqffk qmek fa eqq ftq rxms ngf omz kag eqq uf u faaw zqmdxk 1 yuzgfq fa qzoapq ftue iuft xuz13 saap xgow uz eaxhuzs ftmf\n",
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"26 gur synt vf 2w68lsudym pg vf cerggl rnfl gb frr gur synt ohg pna lbh frr vg v gbbx arneyl 1 zvahgr gb rapbqr guvf jvgu yva13 tbbq yhpx va fbyivat gung\n"
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]
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}
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],
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"source": [
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"s= 'gur synt vf 2w68lsudym Vg vf cerggl rnfl gb frr gur synt ohg pna lbh frr vg v gbbx arneyl 1 zvahgr gb rapbqr guvf jvgu EBG13 tbbq yhpx va fbyivat gung'\n",
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"for offset in range(1, 27):\n",
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" print(offset, update_sentence(s, offset))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"'the flag is 2j68yfhqlz ct is pretty easy to see the flag but can you see it i took nearly 1 minute to encode this with lin13 good luck in solving that'"
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]
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},
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"execution_count": 4,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"update_sentence(s, 13)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.6"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎CyberTalents/cryptography/assets/genfei/encrypt.py

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import sys
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from struct import pack, unpack
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def F(w):
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return ((w * 31337) ^ (w * 1337 >> 16)) % 2**32
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def encrypt(block):
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a, b, c, d = unpack("<4I", block)
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for rno in xrange(32):
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a, b, c, d = b ^ F(a | F(c ^ F(d)) ^ F(a | c) ^ d), c ^ F(a ^ F(d) ^ (a | d)), d ^ F(a | F(a) ^ a), a ^ 31337
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a, b, c, d = c ^ F(d | F(b ^ F(a)) ^ F(d | b) ^ a), b ^ F(d ^ F(a) ^ (d | a)), a ^ F(d | F(d) ^ d), d ^ 1337
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return pack("<4I", a, b, c, d)
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pt = open(sys.argv[1]).read()
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while len(pt) % 16: pt += "#"
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ct = "".join(encrypt(pt[i:i+16]) for i in xrange(0, len(pt), 16))
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open(sys.argv[1] + ".enc", "w").write(ct)

‎CyberTalents/cryptography/assets/genfei/flag.enc

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P��ي ܮ��hq�>ޒ�b ��M�S\��j

‎CyberTalents/cryptography/assets/genfei/genfei.zip

2.31 KB
Binary file not shown.

‎CyberTalents/cryptography/assets/lineq/lineq.txt

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15345857135052644158 * x[0] mod 15914389274045831441 = 10753153698913165324
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10107862342967460188 * x[1] mod 12471333000718257439 = 8412981602133999892
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15514951512896411 * x[2] mod 11189085043185963643 = 1482464683316586574
4+
4918093547848646552 * x[3] mod 14953553045254805869 = 2025900790652430240
5+
9458472078791077712 * x[4] mod 16969044464796096757 = 5398900907079313999
6+
10389810153032407159 * x[5] mod 12219369241978883401 = 7029759589492702570
7+
10317166141181737080 * x[6] mod 11051035063490494121 = 3468633932944308360
8+
4739071230119101568 * x[7] mod 12446419077417014895 = 8675623443003281737
9+
5602877978471499087 * x[8] mod 17033822019842116078 = 10322234074859615825
10+
6817739495547298027 * x[9] mod 9516363895804911673 = 4437950083761802261
11+
6427678547440599488 * x[10] mod 15079736889469193431 = 3279819731898553304
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9457480281190568299 * x[11] mod 10650632135951148921 = 9305609025967996118
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3811243068786264022 * x[12] mod 17812647846133398533 = 11526794456945961531
14+
775941643434169870 * x[13] mod 11412472367550805013 = 4738382726559211541
15+
7024610526778714279 * x[14] mod 9778393688778074575 = 5323545124401969226
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11651575732234344955 * x[15] mod 16630642531754496501 = 2873213552124115822
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6625087805180364133 * x[16] mod 9364964556418467329 = 2787565658998643270
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3669801280657684382 * x[17] mod 14097382369878612419 = 7567005815588497736
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5443984100737085739 * x[18] mod 10512797518085747507 = 7873334411288369657
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12346616364541216766 * x[19] mod 15943449689777265613 = 3904160665454908058
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8986565623928863916 * x[20] mod 18324134584075223407 = 188593124624644258
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944611705353292388 * x[21] mod 11593718893451959913 = 5953933343058555785
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9480419754515455118 * x[22] mod 14797616246348898583 = 13228205597299662061
24+
6357470584650383564 * x[23] mod 10888050221402213645 = 3331190950406564692
25+
4804105847411349449 * x[24] mod 15964574501738046999 = 9808705001033677355
26+
The flag is chr(x[0]%256)+chr(x[1]%256)+...

‎CyberTalents/cryptography/genfei.ipynb

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# The encryption algorithm"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"import sys\n",
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"from struct import pack, unpack\n",
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"\n",
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"def F(w):\n",
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" return ((w * 31337) ^ (w * 1337 >> 16)) % 2**32\n",
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"\n",
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"def encrypt(block):\n",
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" a, b, c, d = unpack(\"<4I\", block)\n",
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" for rno in xrange(32):\n",
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" a, b, c, d = b ^ F(a | F(c ^ F(d)) ^ F(a | c) ^ d), c ^ F(a ^ F(d) ^ (a | d)), d ^ F(a | F(a) ^ a), a ^ 31337\n",
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" a, b, c, d = c ^ F(d | F(b ^ F(a)) ^ F(d | b) ^ a), b ^ F(d ^ F(a) ^ (d | a)), a ^ F(d | F(d) ^ d), d ^ 1337\n",
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" return pack(\"<4I\", a, b, c, d)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Reverse engineering the encryption algorithm\n",
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"- First it's noticed that the algorithm works on blocks of 16 bytes\n",
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"- Each block is divided into four 4-byte ints (a,b,c,d)\n",
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"- 32 Rounds are performed on these ints\n",
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"- Each round performs two operations on these 4 ints\n",
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"- We will need to figure out a way to reverse these functions"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [],
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"source": [
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"def reverse_second_op(a1, b1, c1, d1):\n",
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" # Variables should be decoded in the given sequence d0, a0, b0, c0\n",
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" d0 = d1 ^ 1337\n",
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" \n",
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" fd0 = F(d0 | F(d0) ^ d0)\n",
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" a0 = c1 ^ fd0\n",
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"\n",
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" fad0 = F(d0 ^ F(a0) ^ (d0 | a0))\n",
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" b0 = b1 ^ fad0\n",
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"\n",
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" fabd0 = F(d0 | F(b0 ^ F(a0)) ^ F(d0 | b0) ^ a0)\n",
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"\n",
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" c0 = a1 ^ fabd0\n",
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" \n",
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" return a0, b0, c0, d0"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"def reverse_first_op(a1, b1, c1, d1):\n",
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" a0 = d1 ^ 31337\n",
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" \n",
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" fa0 = F(a0 | F(a0) ^ a0)\n",
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" d0 = c1 ^ fa0\n",
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" \n",
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" fad0 = F(a0 ^ F(d0) ^ (a0 | d0))\n",
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" c0 = b1 ^ fad0\n",
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" \n",
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" facd0 = F(a0 | F(c0 ^ F(d0)) ^ F(a0 | c0) ^ d0)\n",
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" b0 = a1 ^ facd0\n",
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" \n",
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" return a0, b0, c0, d0"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"metadata": {},
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"outputs": [],
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"source": [
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"def decrypt(block):\n",
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" a, b, c, d = unpack(\"<4I\", block)\n",
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" for rno in range(32):\n",
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" # Second operation is reversed first\n",
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" a, b, c, d = reverse_second_op(a, b, c, d)\n",
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" # First operation is reversed\n",
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" a, b, c, d = reverse_first_op(a, b, c, d)\n",
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" \n",
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" return pack(\"<4I\", a, b, c, d)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# How to encrypt a new message?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 5,
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"metadata": {},
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"outputs": [],
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"source": [
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"def encrypt_file(file_name):\n",
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" pt = open(file_name).read()\n",
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" while len(pt) % 16: pt += \"#\"\n",
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"\n",
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" ct = \"\".join(encrypt(pt[i:i+16]) for i in xrange(0, len(pt), 16))\n",
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" open(file_name + \".enc\", \"w\").write(ct)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# How to decrypt an encrypted message?"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 6,
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"metadata": {},
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"outputs": [],
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"source": [
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"def decrypt_file(file_name):\n",
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" pt = open(file_name, 'rb').read()\n",
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" # No need to add any padding\n",
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"\n",
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" # Decode the bytes stream using .decode() before joining the list\n",
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" ct = ''.join([decrypt(pt[i:i+16]).decode() for i in range(0, len(pt), 16)])\n",
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" print(ct)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 7,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"FLAG{G3N3R4L123D_F31573L_EZ!}###\n"
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]
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}
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],
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"source": [
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"decrypt_file('assets/genfei/flag.enc')"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.6"
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}
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},
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"nbformat": 4,
184+
"nbformat_minor": 2
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}

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