Question

Given the key and the ciptherext, find the plaintext. Do you recognize the plaintext? [ 22 17 171 A= 13 21 9 L 5 0 21 VBMEB JGYDW UJFRC RSZIO ENVSR RTWUJ ZFCDO YHOXR RYOJJ RQCKS IBAUM OLHHS XAHFS ZNBFZ TSPAS BDGTE PDVNR KAFCE SYBSS SFSZM QSOGU YWAZV YZLLY GQGHP ECRBL FSGAB HEDEX IUMYC JFVRF TACBL EYEQM QXZVF HMKGH FDJJG OMWGP HUXJH FJYOG UXOOU PJXOO UPJXO OUPJS MCBVH XOOUP JHOXH TOJDZ LBDET LUSWP FWJTI NQUJG OOGUT ZNTWZ BACLD DOLHT OGVQK KMSJG JPONJ GJSNJ AMDCE LAMDH ZIHBV VQKYG QGRRO BVFSN IUMUC BOGUD NTOPX CZJEM GOGUY KEYME XKYTH FATNV GVFUZ QRKRN MESNY MWOGU XOOUP JXOOU PJXOO UPJHO XHTOJ DZOGU WISMT FFCCE CPBRG YESSC XDWYV FQSOR OUUJT BTOGV QKKMS JGJPO NJGJS NJAMD CELAM DHZIH BVVQK PTJQR LSGCO ZZLBS UZVWL ZICYF FDDHG DVNRK AFCES YBYME XRZWC CRKAZ ISUZA GHQJJ RJVUM OLZDX RDSFS ZHOKN AASBL HXANK AJVUM OLIGU RSZHO XLKOC ZJMOL EYUGG CWCST ZNYGQ GHPHZ KZLLL BDLUL OGUFI CYTZI UWHAF ZFCSM KGGCA QOSMC FPKCL JHONI SHIKH UCBJI MLPDA THIGG ZTCJO VLTAJ OVLTA JOVKU LXNHG ESGTP USWPF WJTIN QUJGO OGUTZ NTWZB ACLDD QRLSG CHOXJ GJSNJ AMDCE LAMDH ZIHBV VQKTO GVQKK MSJGJ TGPOY AYUNA MDHZI HBVVQ KLBSS MUXZV FHMKG HFDJS MELBS GWYYU IUNVS MUXZV FHMKG HFDJL VZJBF IUCLM KTMAK EMNEJ NYAPR ANYAF VXODT PQVRX NHERH TSIXP ONXZM MVDXE MK

Answer 1

This is hill cipher with key of 3*3 matrix.

I have decoded first fer words and here they are->
Cipher = VBMEB JGYDW UJFRC RSZIO ENVSR RTWUJ ZFCDO YHOXR RYOJJ RQCKS IBAUM OLHHS XAHFS ZNBFZ

Plain = WHEN I GET TO THE BOTTOM I GO BACK TO THE TOP OF THE SLIDE WHERE IS TOP AND I TURN AND I GO FOR ARID..........

Steps below->

Encryption

We have to encrypt the message ‘ACT’ (n=3).The key is ‘GYBNQKURP’ (in ascii % 65) which can be written as the nxn matrix:

[6 24 11 | 13 16 10 20 17 15 |

The message ‘ACT’ is written as vector:

The enciphered vector is given as:

B [6 24 1] 13 16 10 12=6=-) ona (mod 26 6 20 17 15

which corresponds to ciphertext of ‘POH’

Decryption

To decrypt the message, we turn the ciphertext back into a vector, then simply multiply by the inverse matrix of the key matrix (IFKVIVVMI in letters).The inverse of the matrix used in the previous example is:

6 24 1 「s 5 10 13 16 10 |=|21 8 21 || 21 12 8 26 2017 15

For the previous Ciphertext ‘POH’:

5 260 ||||||) 574 218 21 21 128 539

which gives us back ‘ACT’.