Question

The following is an English sentence encrypted by means of a Vigen`ere cipher. IYMEC GOBDO JBSNT VAQLN BIEAO YIOHV XZYZY LEEVI PWOBB OEIVZ HWUDE AQALL KROCU WSWRY SIUYB MAEIR DEFYY LKODK OGIKP HPRDE JIPWL LWPHR KYMBM AKNGM RELYD PHRNP ZHBYJ DPMMW BXEYO ZJMYX NYJDQ WYMEO GPYBC XSXXY HLBEL LEPRD EGWXL EPMNO CMRTG QQOUP PEDPS LZOJA EYWNM KRFBL PGIMQ AYTSH MRCKT UMVST VDBOE UEEVR GJGGP IATDR ARABL PGIMQ DBCFW XDFAW UWPPM RGJGN OETGD MCIIM EXTBE ENBNI CKYPW NQBLP GIMQO ELICM RCLAC MV (i) Compute its index of coincidence. (ii) Use this to estimate the length of the keyword. (iii) Use Kasiski’s method to estimate the length of the keyword. Find the Vigen`ere keyword (which need not be an English word) and find the plaintext.

Answer #1

(i)

The index of coincidence is the probability of two randomly selected letters to be the same and computed using the following formula:

Where, n_{i} is the number of occurrence of the alphabet
i in the given string and N is the total number of letters.

The value obtained from the above formula is around
**0.04285.**

(ii)

For the above value of IC = 0.04285, the estimated length of the
keyword is around **9** for the English language.

(iii)

Find the trigrams and bigrams in the ciphertext and the ciphertext and keep track of the distance between 2 successful encounters. The most common trigram is GIM which occurs at an interval of 45 between first and second occurrence and 99 between first and third occurrence.

The gcd of the numbers gives the estimate of the length. Gcd of
45 and 99 is 9. **Hence, the key length is equal to
9.**

To find the key, pick the letters in the interval of 9 starting with the first letter in the ciphertext and perform the frequency attack. This results in the first letter as A.

Then start with the second letter of the ciphertext and pick every other letter after an interval of 9. Again, perform the frequency attack on this string. The second letter of the key is expected to be N.

Repeat the above process 9 times to find each letter of the key.
The final key is equal to **ANIELTKYW**.

Use the above key to find the plain text. The plain text is as follows:

ILEAR NEDHO WTOCA LCULA TETHE AMOUN TOFPA PERNE EDEDF ORARO OMWHE NIWAS ATSCH OOLYO UMULT IPLYT HESQU AREFO OTAGE OFTHE WALLS BYTHE CUBIC CONTE NTSOF THEFL OORAN DCEIL INGCO MBINE DANDD OUBLE ITYOU THENA LLOWH ALFTH ETOTA LFORO PENIN GSSUC HASWI NDOWS ANDDO ORSTH ENYOU ALLOW THEOT HERHA LFFOR MATCH INGTH EPATT ERNTH ENYOU DOUBL ETHEW HOLET HINGA GAINT OGIVE AMARG INOFE RRORA NDTHE NYOUO RDERT HEPAP ER

The above plain text translates to the following English paragraph:

I LEARNED HOW TO CALCULATE THE AMOUNT OF PAPER NEEDED FOR A ROOM WHEN I WAS AT SCHOOL YOU MULTIPLY THE SQUARE FOOTAGE OF THE WALLS BY THE CUBIC CONTENTS OF THE FLOOR AND CEILING COMBINED AND DOUBLE IT YOU THEN ALLOW HALF THE TOTAL FOR OPENINGS SUCH AS WINDOWS AND DOORS THEN YOU ALLOW THE OTHER HALF FOR MATCHING THE PATTERN THEN YOU DOUBLE THE WHOLE THING AGAIN TO GIVE A MARGIN OF ERROR AND THEN YOU ORDER THE PAPER

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