Vigenére Cipher
Kimberly Chiffens
&
Maria Jannelli
Progress Report
•Implementation of Vigenere
encryption and decryption
•Applet design
•Website design
•Implementation of Friedman
attack which one way to find the
key length in cryptanalysis
Future Plans
•December: Finish introduction of the
paper, Finish Kasiski implementation
•January: Start to implement the
Cryptanalysis and finish draft of the
paper
•February: Implement cryptanalysis
•March: Implement cryptanalysis
•April: Finish final project and paper
Cryptanalysis
•Given only the Ciphertext, we must find
the Plaintext and the key
•The first step in doing this is finding the
KEY LENGTH
•Two ways to find this key length is:
•Friedman Attack implementation
•Kasiski Attack implementation
Friedman Attack
• The goal is to find the key length
• Uses frequencies to count the amount of each
letter in the ciphertext
• We multiply the count of each letter by count
minus 1 and then add up the sum.
• It computes the sum of the frequencies as
follows:
• for(int k = 0; k < 26; k++)
• sum = sum + Fcount[k]*(Fcount[k]-1);
• }
Friedman Attack
• We then find the index of coincidence
• To find this we divide the entire sum of the
frequencies with the length of the cipher
times the length minus 1.
• index= sum/(lc*(lc-1)); , where lc is the
length
• Then we must calculate the key length. To do
so we use this equation:
• keyword2= ((0.0265*lc)/((0.065index)+(lc*(index-0.0385))));
Friedman Example
Ciphertext:
OAKKADKMFCRBCKBOMTTITMIF
Frequencies:
A
B C D E F
G H I
J
K L
M N O P Q R S T U V W X Y Z
2 2 2 1 0 2 0 0 2 0 4 0 3 0 2 0 0 1 0 3 0 0 0 0 0 0
Length = 24 letters
Sum of frequencies = Σ letterCount*(letterCount-1)
Key: Cat
Plaintext: mariakimmariakimmariakim
Program Demonstration
Ciphertext:
KSEUOBHCSABMBSGBCLHCSATYLSLXHNJGRNRMWGMVSMCVCADJJQCOCPRTGCKC
ASUQSPNWRRGLXHWXIKRUFCRGBZGTSWXIPWCQNOJXHGVOERBCRTWXIPYFCLW
MDGLXGCFSPNGYWRURQFNRGWPCCKCNBWXIPCCCBHFJHAUSYAZWFCSUR8WCRK
SYCFCJHDXFWXIBKSDXFANRRXGKNZJHCSATCNHWXIPWCQNKMDZBKSYBCSAQCXTB
ASYMKCASGCORCOAQSBJHMYMMDFFNOBRHQXCLFCSURBAWTNMMDHMMSQYO
GATMASTNFRRQIUSBKMWXIPQOGAKGCVGWMMDFCJFWXIPWCQNKMDZBKSYWO
ZBCJDHCLORJGRACNQSDXFUQSLHCSFSPNCZUWENRRXGLNSXNMMDFZAOGWKMD
ZBAORCZCOFMVHFNPPNSXNMMDFLXGCRBQCSYMHFACSPVRQWATOLMHFRBPNAY
RBQKSRFSCWMMDFCHSQJBBLVGWBMCDYBHCMCLBCKNCRQSPYZYLSZNUJJRWXIP
WCQNWQXBWXIPOOAN
Key:
joy
Deciphered
Plaintext:
BEGLADYOURNOSEISONYOURFACENOTPASTEDONSOMEOTHERP
LACEFORIFITWEREWHEREITISNOTYOUMIGHTDISLIKEYOURNOSE
ALOTIMAGINEIFYOURPRECIOUSNOSEWERESANDWICHEDINBET
WEENYOURTOESTHATCLEARLYWOULDTNOTBEATREATFORYOUDB
EFORCEDTOSMELLYOURFEETYOURNOSEWOULDBEASOURCEOFD
READWEREITATTACHEDATOPYOURHEADITSOONWOULDDRIVEYO
UTODESPAIRFOREVERTICKLEDBYYOURHAIRWITHINYOUREARYOU
RNOSEWOULDBEANABSOLUTECATASTROPHEFORWHENYOUWER
EOBLIGEDTOSNEEZEYOURBRAINWOULDRATTLEFROMTHEBREEZE
YOURNOSEINSTEADTHROUGHTHICKANDTHINREMAINSBETWEEN
YOUREYESANDCHINNOTPASTEDONSOMEOTHERPLACEBEGLADYO
URNOSEISONYOURFACE
Kasiski Attack
• The goal is to find the key length
• Count the gaps between repeated trigrams in
the ciphertext
• The key is a divisor of the GCDs of the gaps
Kasiski Attack Example
Ciphertext:
WCZOU
LXSSU
UQNAH
YJPFN
JFIAW
QNAHY YEDBL
ZLQWB SVYXA
ULKHG YLWQI
TQQYY IRNPH
F
WOSHM AUCER CELVE
RRMJF IAWFN AHBZO
STBHW LJCYZ EIYWV
SHZOR WBSVY XAEEM
-NAH repeated three times
-There are 48 letters from the first occurrence of NAH to the
second and 8 letters from the second to the third
occurrence.
-WBSVYZAEEMJFIAWF repeated twice
-There are 72 letters between the long groups
Kasiski Attack Example
· The greatest common divisor of 72,48, and 8
is 8
· The keyword length is likely to be 8, 4, or 2
· A 2 or 4 letter keyword offers little security so it
is likely to have the keyword length of 8
References
•Principles of Operating Systems: Design and
Application, Brian L. Stuart. Course Technology.
2009.
•Invitation to Cryptology, Thomas H. Barr.
Prentice Hall. 2002.