Vigenére Cipher

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)/((0.065-index)+(lc*(index-0.0385))));

Friedman Example Ciphertext: OAKKADKMFCRBCKBOMTTITMIF Frequencies: Length = 24 letters Sum of frequencies = Σ letterCount*(letterCount-1) Key: Cat Plaintext: mariakimmariakimmariakim

Program Demonstration Ciphertext:KSEUOBHCSABMBSGBCLHCSATYLSLXHNJGRNRMWGMVSMCVCADJJQCOCPRTGCKCASUQSPNWRRGLXHWXIKRUFCRGBZGTSWXIPWCQNOJXHGVOERBCRTWXIPYFCLWMDGLXGCFSPNGYWRURQFNRGWPCCKCNBWXIPCCCBHFJHAUSYAZWFCSUR8WCRKSYCFCJHDXFWXIBKSDXFANRRXGKNZJHCSATCNHWXIPWCQNKMDZBKSYBCSAQCXTBASYMKCASGCORCOAQSBJHMYMMDFFNOBRHQXCLFCSURBAWTNMMDHMMSQYOGATMASTNFRRQIUSBKMWXIPQOGAKGCVGWMMDFCJFWXIPWCQNKMDZBKSYWOZBCJDHCLORJGRACNQSDXFUQSLHCSFSPNCZUWENRRXGLNSXNMMDFZAOGWKMDZBAORCZCOFMVHFNPPNSXNMMDFLXGCRBQCSYMHFACSPVRQWATOLMHFRBPNAYRBQKSRFSCWMMDFCHSQJBBLVGWBMCDYBHCMCLBCKNCRQSPYZYLSZNUJJRWXIPWCQNWQXBWXIPOOAN Key: joy

Deciphered Plaintext: BEGLADYOURNOSEISONYOURFACENOTPASTEDONSOMEOTHERPLACEFORIFITWEREWHEREITISNOTYOUMIGHTDISLIKEYOURNOSEALOTIMAGINEIFYOURPRECIOUSNOSEWERESANDWICHEDINBETWEENYOURTOESTHATCLEARLYWOULDTNOTBEATREATFORYOUDBEFORCEDTOSMELLYOURFEETYOURNOSEWOULDBEASOURCEOFDREADWEREITATTACHEDATOPYOURHEADITSOONWOULDDRIVEYOUTODESPAIRFOREVERTICKLEDBYYOURHAIRWITHINYOUREARYOURNOSEWOULDBEANABSOLUTECATASTROPHEFORWHENYOUWEREOBLIGEDTOSNEEZEYOURBRAINWOULDRATTLEFROMTHEBREEZEYOURNOSEINSTEADTHROUGHTHICKANDTHINREMAINSBETWEENYOUREYESANDCHINNOTPASTEDONSOMEOTHERPLACEBEGLADYOURNOSEISONYOURFACE

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 QNAHY YEDBL WOSHM AUCER CELVE LXSSU ZLQWBSVYXA RRMJFIAWFNAHBZO UQNAH ULKHG YLWQI STBHW LJCYZ EIYWV YJPFN TQQYY IRNPH SHZOR WBSVY XAEEM JFIAW F -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.

Vigenére Cipher