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Cipher decryption (vigenere, substitution etc) Watch

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    Hi,

    I was wondering if anyone knew how I could decrypt a piece of encrypted text that may have been vigenere encrypted twice and then encrypted via substitution?

    This is the cipher text:


    Spoiler:
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    PUF ZHV JON IRL KGS CIY INZ FSO IPW THM LHR
    YBN NIU GSE NWG QUC QFR MYB PYK WRV ODP IDX
    LZQ PRZ FZN XMV JMU CTS EET ROI QML EXL TFE
    TOP NMG HVE MIF TIN QAG KUJ ZJH JXA UVN TIH
    AOZ PDE VYJ XZA DRF WVF IQK TON OKP BSN SUO
    NAP PTS GGU MTZ ZJZ QZM NRC HLS CMC UUB PWS
    BMW DYX XSW EDQ QBH GRE ILU XJK SII HZE HUN
    NEP MGJ YZS HPY XFN FRZ KDW CBN XVP VVJ XGC
    VHY MCQ PUU TQL JKH HTB PQP GEL AXY OCU QYX
    UZN LQI MCK IVH SSA KGY HXW QNU NRZ KJG XBH
    UKJ YRQ TUV SJJ IQJ DAO QZT RDT ULC KLB DHY
    PJH QKS WUW WNJ HTV FLP VKI KTD QID KGN JDL
    AOL DZE KIS KJA UTW TTX OTC IUD OAQ MKX PNL
    NSK DVT PQX TWD FFC RNH AJD SRL FEU NKY LGN
    ELX YZK KJL IXR TAS RJU NRN EKB HPO HEQ YIL
    KRF ALF ZXA PPK YRG NSS LCW MQP VJM NBS FLY
    XYP JZW CFO CWN RBE PHP UJB MMI KPC OBR BQF
    QIK GRW SLF IMH QJY XGI GAC AJR KHL EEM BAP
    HDX IUA BDF GTF TCH UKY UXX WTI UPG AQP FIV
    MKO DOX AER TIW NRP NAA WAM WLT YVR WQV UOS
    GIR ZJZ FVE YPX PBX NRQ KBY ZJV INP RXM XAX
    HMP EEM TDP DGL IRL QUX UNE GOE UXO MUK IRZ


    This the plaintext:

    Spoiler:
    Show
    Right now I am happy as I am visiting London zoo on a trip with my Lampton
    School biology and psychology pupils. Firstly, I am going to visit a pack of
    Troglodytes gorillas and Orangutons. Our day has just got awful as it is now
    raining cats and dogs and I am not wearing any boots. On top of that I do not
    own a raincoat. My lunch is the only thing I am now looking forward to as I am
    starving. My dusty pink lunchbox contains a potato salad, ham sandwich, a kiwi
    along with a pack of chilli crisps and four tropical flavour drinks. If only my
    mum would put in an additional two drinks and a dual kitkat chunky with an
    apricot yogurt. Hang on! I can go buy a bunch of snacks from two candy shops
    that I saw on my way, using six pounds that I took from my mums Louis Vuitton
    clutchbag. If only I was not so hungry I could have brought a raincoat for
    myself. This day is such an oxymoron, I was happy but now I am sad.


    Thank you.
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    Well after doing what you should have done and hitting up Professor Google who directed me to Dr. Wikipedia - I found out what a Vigenere cipher is.

    So, what you need to do is try and work out or estimate the length of the key (try looking for repetitions of two or three letter groupings in the cipher text and look at the space between them and work on the hypothesis that the key length is a divisor)

    Then, you match up with the plain text in the relevant columns and see if it works. You could just try and see if it is a key word of length 3 or 6 straight away because of the grouping of the cipher text.
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    (Original post by Mark85)
    Well after doing what you should have done and hitting up Professor Google who directed me to Dr. Wikipedia - I found out what a Vigenere cipher is.
    You obviously didn't participate in the National Cipher Challenge when you were at school ...
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    (Original post by Mr M)
    You obviously didn't participate in the National Cipher Challenge when you were at school ...
    Looks like it only started after I left school. To be fair, at that age - I had no interest in maths so I probably wouldn't have had anything to do with it anyway.
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    (Original post by Khodu)
    Hi,

    I was wondering if anyone knew how I could decrypt a piece of encrypted text that may have been vigenere encrypted twice and then encrypted via substitution?

    Thank you.
    Ok so you've plaintext + vig + vig + subs = ciphertext.

    Working backwards, we know that substitution doesn't mask digrams (and trigrams), so we can skip straight to Kasiski's method, i.e. guessing the keyword length of the Vigenere cipher by finding the gcd of the distance between common digrams. The fact that it's a double vigenere just increases the length of the keyword. Divide into substrings and apply frequency analysis to each (remember double substitution is just as easy to crack as single substitution).

    I think that's right

    To be honest, if I were you, I'd crack a few Caesars on the website and keep upto speed with the course; cracking stream ciphers is time consuming, especially since most students go crazy with encryption or their plaintext is a load of nonsense. Things get crazy from Week 8, and Public Key Cryptography is amaaaaaaazing.
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    (Original post by Topmanfaz)
    Ok so you've plaintext + vig + vig + subs = ciphertext.

    Working backwards, we know that substitution doesn't mask digrams (and trigrams), so we can skip straight to Kasiski's method, i.e. guessing the keyword length of the Vigenere cipher by finding the gcd of the distance between common digrams. The fact that it's a double vigenere just increases the length of the keyword. Divide into substrings and apply frequency analysis to each (remember double substitution is just as easy to crack as single substitution).

    I think that's right

    To be honest, if I were you, I'd crack a few Caesars on the website and keep upto speed with the course; cracking stream ciphers is time consuming, especially since most students go crazy with encryption or their plaintext is a load of nonsense. Things get crazy from Week 8, and Public Key Cryptography is amaaaaaaazing.


    Thanks dude,

    I've cracked a couple of easy ciphers, (single vig or single sub with the method you mentioned) but nothing with more than one encryption method. But yeah, just getting my head down now, cant waste any more time trying to crack hard ciphers.
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    (Original post by Khodu)
    Thanks dude,

    I've cracked a couple of easy ciphers, (single vig or single sub with the method you mentioned) but nothing with more than one encryption method. But yeah, just getting my head down now, cant waste any more time trying to crack hard ciphers.
    Nice. I couldn't be bothered last year, ended up getting 0.
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    (Original post by Topmanfaz)
    Nice. I couldn't be bothered last year, ended up getting 0.
    :eek: You always amaze me, and yet you did fab last year from I can gather, mA.
 
 
 
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