[Crypto] Jack's Quest for the Flag of Unknown

-👤Aukro

Step.1

已知這個題目是Affine cipher

Fw j lpuko aju uhdpsho aupd gmh oftfgjk uhjkd, Yjvr apzwo mfdnhka pw jw hegujpuofwjux ypzuwhx gmupztm gmh lpwohun pa wjgzuh. Fg ljn j ojx kfrh jwx pgmhu, lfgm gmh nzw vjngfwt fgn ljud hdcujvh zipw gmh kjwonvjih, jn mh shwgzuho fwgp gmh mhjug pa gmh lfkohuwhnn, kpwtfwt apu joshwgzuh jwo gmh chjzgx pa gmh tuhjg pzgoppun.
Yjvr'n ypzuwhx chtjw fw j ohwnh jwo jwvfhwg apuhng, lmhuh gmh uzngkfwt khjshn lmfnihuho nhvuhgn pa j lpuko zwgpzvmho cx mzdjw mjwon. Jn mh ljkrho ohhihu fwgp gmh lppon, mh hwvpzwghuho j ajdfkx pa vzufpzn ujvvppwn, gmhfu djnrho ajvhn ihhrfwt pzg aupd chmfwo gmh guhhn. Gmhx vmjgghuho jdpwt gmhdnhkshn fw j kjwtzjth rwplw pwkx gp gmh apuhng, lhkvpdfwt Yjvr gp gmhfu lppokjwo mpdh.      
Vpwgfwzfwt pw mfn ijgm, Yjvr vjdh zipw j dhjwohufwt ufshu, fgn ljghun uhakhvgfwt gmh sfsfo ckzh pa gmh nrx jcpsh. Pw gmh ufshucjwr, j tupzi pa ikjxazk pgghun aupkfvrho fw gmh nmjkkpln, gmhfu nkhhr cpofhn tkfofwt haapugkhnnkx gmupztm gmh ljghu. Yjvr ljgvmho fw lpwohu jn gmhx nifujkho jwo npdhunjzkgho, gmhfu ypxpzn jwgfvn j uhdfwohu pa gmh nfdikh ikhjnzuhn pa kfah.
Jn gmh ojx lpuh pw, Yjvr apzwo mfdnhka fw j sjng pihw dhjopl, vjuihgho lfgm j rjkhfopnvpih pa lfkoakplhun. Gmh jfu ljn jkfsh lfgm gmh mzd pa chhn jwo gmh thwgkh npwtn pa cfuon ihuvmho pw nljxfwt cujwvmhn. Czgghuakfhn pa hshux mzh ojwvho tujvhazkkx jdpwt gmh ckppdn, gmhfu ohkfvjgh lfwtn ijfwgho lfgm gmh vpkpun pa j gmpznjwo nzwnhgn. Fg ljn j nxdimpwx pa wjgzuh'n afwhng vuhjgfpwn.   
Jn gmh nzw chtjw fgn ohnvhwg gpljuo gmh mpufqpw, Yjvr uhjvmho gmh nmpuhn pa j gujwbzfk ipwo. Gmh ljghu ljn jn vkhju jn vuxngjk, jkkplfwt mfd gp nhh gmh gjoipkhn ojugfwt jdpwt gmh kfkx ijon. Whjucx, j djyhngfv mhupw ngppo fw nfkhwg vpwghdikjgfpw, fgn kpwt whvr ipfnho kfrh jw hkhtjwg ojwvhu. Fg ljn j dpdhwg pa nhuhwfgx jwo vpwwhvgfpw lfgm gmh wjgzujk lpuko.
Czg gmh dpng hwvmjwgfwt ijug pa Yjvr'n ypzuwhx ljn xhg gp vpdh. Jn gmh dppw upnh fw gmh wftmg nrx, vjngfwt j nfkshux tkpl pshu gmh kjwonvjih, mh ngzdckho zipw j dxnghufpzn akjt akzgghufwt fw gmh thwgkh cuhhqh. Gmh akjt cpuh gmh vuxigfv dhnnjth "vrvnvVGA{jaafwh_v1im3u_vjw_nkpsh_cx_hxh5}." Fg ljn jw hwftdj lujiiho fw wjgzuh'n hdcujvh, j izqqkh ljfgfwt gp ch npksho.
Yjvr vpzkow'g uhnfng gmh vmjkkhwth. Lfgm ohghudfwjgfpw fw mfn mhjug jwo gmh lfnopd pa gmh wjgzujk lpuko jn mfn tzfoh, mh chtjw gp ohvfimhu gmh akjt'n dhnnjth. Hjvm vmjujvghu mhko j nhvuhg, j ifhvh pa gmh izqqkh gmjg lpzko uhshjk gmh akjt'n ngpux.
Jn mh ohvpoho gmh akjt, mh ahkg j ohhi vpwwhvgfpw lfgm gmh lpuko jupzwo mfd. Gmh apuhng, gmh ufshu, gmh dhjopl, jwo gmh ipwo mjo jkk ikjxho j ijug fw gmfn hegujpuofwjux joshwgzuh. Fg ljn jn fa wjgzuh fgnhka mjo vpwnifuho gp khjo mfd gp gmfn dpdhwg pa ofnvpshux.
Fw gmh hwo, Yjvr nzvvhnnazkkx zwujshkho gmh akjt'n dhnnjth: "vrvnvVGA{jaafwh_v1im3u_vjw_nkpsh_cx_hxh5}." Fg ljn j gufzdim pa cpgm dfwo jwo nifufg, j ghngjdhwg gp gmh dxnghufhn pa gmh wjgzujk lpuko jwo gmh mzdjw bzhng apu rwplkhoth.
Lfgm gmh akjt'n ngpux uhshjkho, Yjvr ahkg j iupapzwo nhwnh pa azkafkkdhwg. Mh uhjkfqho gmjg gmh chjzgx pa wjgzuh jwo gmh gmufkk pa ofnvpshux lhuh fwghuglfwho, tzfofwt mfd pw j ypzuwhx pa lpwohu jwo hwkftmghwdhwg. Jn mh kjx chwhjgm gmh ngju-ngzooho nrx, mh rwhl gmjg mfn joshwgzuhn fw gmh mhjug pa wjgzuh lpzko vpwgfwzh, hjvm nghi khjofwt mfd gp whl dxnghufhn jwo whl ngpufhn ljfgfwt gp ch zwvpshuho.

嘗試找一些已知的字元

$vrvnvVGA\stackrel{\mathrm{解}\mathrm{密}}{⟹}ckcscCTF$

Step.2

$\begin{array}{rrrrrrrrr}cipher& r& r& v& n& v& V& G& A\\ x& 21& 17& 21& 13& 21& 21& 6& 0\\ ax+b& 2& 10& 2& 18& 2& 2& 19& 5\\ plain& c& k& c& s& c& C& T& F\end{array}$

可以列出以下式子

$\begin{array}{}2a+b\equiv 21(\mathrm{mod}26)& (1)\\ 10a+b\equiv 17(\mathrm{mod}26)& (2)\\ 18a+b\equiv 13(\mathrm{mod}26)& (3)\\ 19a+b\equiv 6(\mathrm{mod}26)& (4)\\ 5a+b\equiv 0(\mathrm{mod}26)& (5)\end{array}$

由(3)(4)得

$a\equiv -7\equiv 19(\mathrm{mod}26)$

由(5)得

$b\equiv -5a(\mathrm{mod}26)$

$$\begin{gathered} b\equiv (-5)19(\mathrm{mod}26) \\ b\equiv 9(\mathrm{mod}26) \end{gathered}$$
求出b為9

• $a=19,b=9$

Step.3

加密方程式為

$C\equiv 19x+9(\mathrm{mod}26)$
解密方程式為

$P\equiv {19}^{-1}(x-9)(\mathrm{mod}26)$

Slove Script

import math
def base(ch:str):
    if ch.islower():
        return 97
    else :
        return 65
def decrypt(text:str,a:int,b:int):
    if math.gcd(a,26) != 1:
        return "invalid a"
    else :
        inv = pow(a,-1,26)
    final = ''
    for i in text:
        if i.isalpha():
            final += chr((inv*(ord(i)-base(i)-b))%26 + base(i))
        else:
            final += i
    return final
print(decrypt(FLAG_CIPHER,19,9))

EXTRA

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