substitution0, 1 and 2 - picoCTF 2022

Uncomment cipher, plain and crypt to solve either the first, second or third substitution challenge.

Always start with the assumption that the last few words are "the flag is picoCTF{<something>}" and you already have the substitution for some of the letters. With these, you can try to understand what is missing looking at the English text and add letters to the substitution until you get a fully formed text (and the flag).

#cipher = '''QWITJSYHXCNDFERMUKGOPVALBZ 
#Hjkjpmre Djykqet qkrgj, axoh q ykqvj qet goqojdb qxk, qet wkrpyho fj ohj wjjodj
#skrf q ydqgg iqgj xe ahxih xo aqg jeidrgjt. Xo aqg q wjqpoxspd giqkqwqjpg, qet, qo
#ohqo oxfj, penerae or eqopkqdxgog—rs irpkgj q ykjqo mkxzj xe q gixjeoxsxi mrxeo
#rs vxja. Ohjkj ajkj oar krpet wdqin gmrog ejqk rej jlokjfxob rs ohj wqin, qet q
#drey rej ejqk ohj rohjk. Ohj giqdjg ajkj jlijjtxeydb hqkt qet ydrggb, axoh qdd ohj
#qmmjqkqeij rs wpkexghjt yrdt. Ohj ajxyho rs ohj xegjio aqg vjkb kjfqknqwdj, qet,
#oqnxey qdd ohxeyg xeor iregxtjkqoxre, X irpdt hqktdb wdqfj Cpmxojk srk hxg rmxexre
#kjgmjioxey xo.
#Ohj sdqy xg: mxirIOS{5PW5717P710E3V0DP710E03055505}'''
#plain = 'theflagispcornudywmbkx' these were good for the first substitute
#cryp = 'ohjsdqyxgmirkeptbafwnl'
#cipher = '''WYHg (gzray hra wimybas yzs hvij) ias i yums rh wrombysa gswbakyu wromsykykrl. Wrlysgyilyg ias masgslysn dkyz i gsy rh wzivvsljsg dzkwz ysgy yzska wasiykxkyu, yswzlkwiv (iln jrrjvklj) gckvvg, iln marqvso-grvxklj iqkvkyu. Wzivvsljsg bgbivvu wrxsa i lboqsa rh wiysjraksg, iln dzsl grvxsn, siwz uksvng i gyaklj (wivvsn i hvij) dzkwz kg gbqokyysn yr il rlvkls gwraklj gsaxkws. WYHg ias i jasiy diu yr vsial i dkns iaaiu rh wrombysa gswbakyu gckvvg kl i gihs, vsjiv slxkarlosly, iln ias zrgysn iln mviusn qu oilu gswbakyu jarbmg iarbln yzs dravn hra hbl iln maiwykws. Hra yzkg marqvso, yzs hvij kg: mkwrWYH{HA3FB3LWU4774WC54A3W0017II384QW}
#
#'''
#plain = 'isflagthepcorumyndwvkbq'
#cryp = 'kghvijyzsmwraboulndxcqf'

cipher = '''xidcddhsgxgdqdcjvwxidczdvvdgxjyvsgidrisoigtiwwvtwufbxdcgdtbcsxltwufdxsxswpgsptvbrspotlydcfjxcswxjprbgtlydctijvvdpodxidgdtwufdxsxswpgawtbgfcsujcsvlwpglgxdugjruspsgxcjxswpabprjudpxjvgzistijcdqdclbgdabvjprujcmdxjyvdgmsvvgiwzdqdczdydvsdqdxidfcwfdcfbcfwgdwajisoigtiwwvtwufbxdcgdtbcsxltwufdxsxswpsgpwxwpvlxwxdjtiqjvbjyvdgmsvvgybxjvgwxwodxgxbrdpxgspxdcdgxdrspjprdhtsxdrjywbxtwufbxdcgtsdptdrdadpgsqdtwufdxsxswpgjcdwaxdpvjywcswbgjaajscgjprtwudrwzpxwcbppspotidtmvsgxgjprdhdtbxspotwpasogtcsfxgwaadpgdwpxidwxidcijprsgidjqsvlawtbgdrwpdhfvwcjxswpjprsufcwqsgjxswpjprwaxdpijgdvdudpxgwafvjlzdydvsdqdjtwufdxsxswpxwbtispowpxidwaadpgsqddvdudpxgwatwufbxdcgdtbcsxlsgxidcdawcdjydxxdcqdistvdawcxdtidqjpodvsguxwgxbrdpxgspjudcstjpisoigtiwwvgabcxidczdydvsdqdxijxjpbprdcgxjprspowawaadpgsqdxdtipsnbdgsgdggdpxsjvawcuwbpxspojpdaadtxsqdrdadpgdjprxijxxidxwwvgjprtwpasobcjxswpawtbgdptwbpxdcdrsprdadpgsqdtwufdxsxswpgrwdgpwxvdjrgxbrdpxgxwmpwzxidscdpduljgdaadtxsqdvljgxdjtispoxiduxwjtxsqdvlxispmvsmdjpjxxjtmdcfstwtxasgjpwaadpgsqdvlwcsdpxdrisoigtiwwvtwufbxdcgdtbcsxltwufdxsxswpxijxgddmgxwodpdcjxdspxdcdgxsptwufbxdcgtsdptdjuwpoisoigtiwwvdcgxdjtispoxidudpwboijywbxtwufbxdcgdtbcsxlxwfsnbdxidsctbcswgsxluwxsqjxspoxiduxwdhfvwcdwpxidscwzpjprdpjyvspoxiduxwydxxdcrdadprxidscujtispdgxidavjosgfstwTXA{P6C4U4P41L5151573R10B5702A03AT}'''
plain = 'picotfslaghenmurydwvbkx'
cryp = 'fstwxagvjoidpubclrzqymh'
result = ''
cipher = cipher.lower()
for letter in cipher:
    if letter in cryp:
        position = cryp.find(letter)
        result += plain[position]
    else:
        result += letter
print(result)