7CCSMCIS Cryptography and Information Security

Rail fence cipher: exercises

Decrypt the following ciphertext that was generated using a rail fence cipher with 2 rails.

TEETN WRTRA HNWSE EOEBA TUSHR ISHBS KONOO MCIEA DVLPD YRHRC EBU

1. Count characters and divide by 2 to get split point. 53/2 = 26.5, so split at H.

T E E T N W R T R A H N W S E E O E B A T U S H R I S
 H B S K O N O O M C I E A D V L P D Y R H R C E B U

2. Gives: THE BEST KNOWN ROTOR MACHINE WAS DEVELOPED BY ARTHUR SCHERIBIUS

Use a rail fence cipher with 3 rails to encipher this message: alan turing the enigma.

A  N  R  G  E  I  A
 L  T  I  T  E  G
  A  U  N  H  N  M

Gives:
ANRGEIALTITEGAUNHNM

Rotating Grille

A more complex transposition cipher: exercise

As in the slides of the lecture, write the plaintext message:
The enigma cipher machine had the confidence of German forces who depended on its security
in a rectangle, row by row, and read the message off column by column but permute the order of the columns using the key 3571426

Columns:
3. tmrecnrceoc
5. hamhocmepnu
7. ecaaneaseir
1. eicdfonwnti
4. nphtiffhdst
2. ihihdgooesy
6. geneeerddex

1. eicdfonwnti
2. ihihdgooesy
3. tmrecnrceoc
4. nphtiffhdst
5. hamhocmepnu
6. geneeerddex
7. ecaaneaseir

Encrypted with 3571426:

eicdfonwntiihihdgooesytmrecnrceocnphtiffhdsthamhocmepnugeneeerddexecaaneaseir

Second transposition with the same key

Columns:
3. ewhyrhscedn
5. indtcttmnee
7. ctgmeiheexa
1. diorofapees
4. fioecfmnece
2. ohecnhhurai
6. nisnpdogdar

1. diorofapees
2. ohecnhhurai
3. ewhyrhscedn
4. fioecfmnece
5. indtcttmnee
6. nisnpdogdar
7. ctgmeiheexa

Encrypted with 3571426:

diorofapeesohecnhhuraiewhyrhscednfioecfmneceindtcttmneenisnpdogdarctgmeiheexa

The ADFGVX cipher: exercise

Consider again the arrangement:

and encrypt: The quick brown fox jumps over the lazy dog first with the key MARK and then with the key TURING.

Stage 1:

T  H  E  Q  U  I  C  K  B  R  O  W  N  F  O  X  J  U  M  P  S  O  V  E  R  T  H  E  L  A  Z  Y  D  O  G
DD DX XD XX GD VG FG FD FF VV DG GG AX XV DG VA GA GD GX AD VD DG VF XD VV DD DX XD DA DV FX XF AG DG GV

Stage 2:

M A R K
-------
D D D X
X D X X
G D V G
F G F D
F F V V
D G G G
A X X V
D G V A
G A G D
G X A D
V D D G
V F X D
V V D D
D X X D
D A D V
F X X F
A G D G
G V

A K M R
-------
D X D D
D X X X
D G G V
G D F F
F V F V
G G D G
X V A X
G A D V
A D G G
X D G A
D G V D
F D V X
V D V D
X D D X
A V D D
X F F X
G G A D
V   G

DDDGFGXGAXDFVXAXGV XXGDVGVADDGDDDVFG DXGFFDADGGVVVDDFAG DXVFVGXVGADXDXDXD

Feistel Encryption/Decryption: exercise

Show that output of decryption round 13 is equal to 32-bit swap of input to encryption round … (first identify the corresponding encryption round and then prove the equivalence).

Given:

• $LD_{16-i} \parallel RD_{16-i} = RE_i \parallel LE_i$ .
• Encryption: $LE_i = RE_{i-1}$ and $RE_i = LE_{i-1} \oplus F(RE_{i-1}, K_i)$ .
• Decryption: $LD_i = RD_{i-1}$ and $RD_i = LD_{i-1} \oplus F(RD_{i-1}, K_i)$ .

We want to show that $LD_{13} \parallel RD_{13} = RE_3 \parallel LE_3$ .

Encryption:

• $LE_3 = RE_2$
• $RE_3 = LE_2 \oplus F(RE_2, K_3)$

Decryption:

• $LD_{13} = RD_{12} = LE_4 = RE_3$
• $RD_{13} = LD_{12} \oplus F(RD_{12}, K_{13})$
• $ = RE_4 \oplus F(LE_4, K_{13})$
• $ = RE_4 \oplus F(RE_3, K_{13})$
• $ = [LE_3 \oplus F(RE_3, K_{13})] \oplus F(RE_3, K_{13})$
• $ = LE_3$

Hence: $LD_{13} \parallel RD_{13} = RE_3 \parallel LE_3$ .