Cryptanalysis of substituttion ciphers.

Cryptanalysis of substituttion ciphers.

In this question, you are required to determine the plaintext and the keyword associated to the given cryptogram. Note that brute force attack (i.e., searching all possible keys) in order to find the keyword is not efficient. However, letter frequency attack and redundancy in natural languages are efficient tools for breaking substitution ciphers.

The cryptogram:

sjksjk jm snu sjsxlbd wdrsberkjms rz snu bpiuksxdum rz sjksjk, snu hrcjhm mudjum at aulejbk hbdsrrkjms nudeÉ. nu jm b duwrdsud bkp bpiuksxdud fnr sdbiulm bdrxkp snu frdlp fjsn njm pre mkrft. snu hnbdbhsud fbm hdubsup jk 1929 bkp jksdrpxhup jk lu wusjs ijkesjucu, b fuuvlt trxsn mxwwlucuks sr snu aulejbk kufmwbwud lu ijkesjucu mjuhlu. nu bwwubdm bm b trxke cbk, bdrxkp 20 sr 30 tubdm rlp (axs nudeu kuiud ejiu njc bk uobhs beu) fjsn b drxkp zbhu bkp gxjzz nbjdmstlu. sjksjk nbm b mnbdw jksulluhs, hbk puzukp njcmulz, bkp jm nrkums, puhuks, hrcwbmmjrkbsu, bkp vjkp. sndrxen njm jkiumsjebsjiu duwrdsjke, gxjhv-snjkvjke, bkp bll-bdrxkp errp kbsxdu, sjksjk jm blfbtm balu sr mrliu snu ctmsudt bkp hrcwlusu snu bpiuksxdu. xkljvu crdu hrlrxdzxl hnbdbhsudm snbs nu ukhrxksudm, sjksjk’m wudmrkbljst jm kuxsdbl, fnjhn bllrfm snu dubpud krs cudult sr zrllrf snu bpiuksxdum axs bmmxcu sjksjk’m wrmjsjrk fjsnjk snu msrdt. hrcajkup fjsn nudeÉ’m mjekbsxdu “hlubd ljku” mstlu, snjm nulwm snu dubpud “mbzult uksud b mukmxbllt msjcxlbsjke frdlp”.

Complete Answer:

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