Examining the cycle structure and order of columnar transposition ciphers as elements of the symmetric group on L elements (L = length of message). Talk given at Ball State University Faculty Mathematics Colloquium, 2 April 2009.

1. DECONSTRUCTING
COLUMNAR TRANSPOSITION
CIPHERS
Robert Talbert, PhD
Associate Professor of Mathematics and Computing Science
Franklin College, Franklin, IN
Ball State University Mathematics Faculty Colloquium
2 April 2009

2. How encryption/decryption works

3. How encryption/decryption works

4. How encryption/decryption works

5. How encryption/decryption works
Message (plaintext)

6. How encryption/decryption works
Message (plaintext)
Key

7. How encryption/decryption works
Message (plaintext)
Encrypted
message
(ciphertext)
Key

8. How encryption/decryption works
Message (plaintext)
Encrypted
message
(ciphertext)
Key

9. How encryption/decryption works
Message (plaintext)
Encrypted
message
(ciphertext)
Key
Key

10. How encryption/decryption works
Message (plaintext) Message (plaintext)
Encrypted
message
(ciphertext)
Key
Key

11. How encryption/decryption works
Message (plaintext) Message (plaintext)
Encrypted
message
(ciphertext)
Key
Key
Alice and Bob share the same key

12. How encryption/decryption works
Message (plaintext) Message (plaintext)
Encrypted
message
(ciphertext)
Key
Key
Alice and Bob share the same key
Should be easy to decrypt with the key

13. How encryption/decryption works
Message (plaintext) Message (plaintext)
Encrypted
message
(ciphertext)
Key
Key
Alice and Bob share the same key
Should be easy to decrypt with the key
Should be very difficult to decrypt without the key

14. CLASSICAL
CIPHER
SYSTEMS
SUBSTITUTION TRANSPOSITION

15. CLASSICAL
CIPHER
SYSTEMS
SUBSTITUTION TRANSPOSITION
Replace plaintext symbols
by other symbols.

16. CLASSICAL
CIPHER
SYSTEMS
SUBSTITUTION TRANSPOSITION
Rearrange plaintext
Replace plaintext symbols
according to a well-
by other symbols.
defined rule.

17. Columnar transposition cipher

18. Columnar transposition cipher
: Agree upon a positive integer, C

19. Columnar transposition cipher
: Agree upon a positive integer, C
C
•••
•••
•••
•••
•••
•••
•••
•
•
•
•••

20. Columnar transposition cipher
: Agree upon a positive integer, C
C Enter plaintext into the
grid one row at a time;
•••
wrap to first column.
•••
•••
•••
•••
•••
•••
•
•
•
•••

21. Columnar transposition cipher
: Agree upon a positive integer, C
C Enter plaintext into the
grid one row at a time;
•••
wrap to first column.
•••
Read text off starting in top-left
position and going down first
•••
column; wrap to first row.
•••
•••
•••
•••
•
•
•
•••

22. Columnar transposition cipher
: Agree upon a positive integer, C
C Enter plaintext into the
grid one row at a time;
•••
wrap to first column.
•••
Read text off starting in top-left
position and going down first
•••
column; wrap to first row.
•••
•••
•••
•••
•
•
Enter ciphertext into the
•
grid one column at a time;
•••
wrap to first row & read off.

23. THE ENEMY ADVANCES AT DAWN
(USING C=5)

24. THE ENEMY ADVANCES AT DAWN
(USING C=5)
T H E E N
E M Y A D
V A N C E
S A T D A
W N

25. THE ENEMY ADVANCES AT DAWN
(USING C=5)
T H E E N
E M Y A D
V A N C E
S A T D A
W N
TEVSWHMAANEYNTEACDNDEA

26. Double encryption = Double security?
Multiple encryption using CTC with C = 4:

27. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY

28. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY
CTAROPYGHPRY

29. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY
CTAROPYGHPRY
COHTPPAYRRGY

30. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY
CTAROPYGHPRY
COHTPPAYRRGY
CPROPRHAGTYY

31. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY
CTAROPYGHPRY
COHTPPAYRRGY
CPROPRHAGTYY
CPGPRTRHYOAY

32. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY
CTAROPYGHPRY
COHTPPAYRRGY
CPROPRHAGTYY
CPGPRTRHYOAY
CRYPTOGRAPHY

33. Double encryption = Double security?
Multiple encryption using CTC with C = 4:
CRYPTOGRAPHY
CTAROPYGHPRY
COHTPPAYRRGY Columnar transposition
on 12 characters using 4
columns has order = 5.
CPROPRHAGTYY
CPGPRTRHYOAY
CRYPTOGRAPHY

34. AGENDA FOR TALK

35. AGENDA FOR TALK
• Address: What is the order of a columnar transposition cipher?
• Explicit formula for underlying permutation
• Specialize to C = 2, the “rail fence cipher”
• Analyze cycle structure when C = 2
• Determine order when C = 2
• Unanswered questions

36. A FORMULA FOR THE
COLUMNAR TRANSPOSITION
CIPHER PERMUTATION

37. π C, L = Permutation implementing C.T.C.
C = Number of columns being used
L = Length of plaintext (= length of ciphertext) (an element of SL )

38. π C, L = Permutation implementing C.T.C.
C = Number of columns being used
L = Length of plaintext (= length of ciphertext) (an element of SL )
C A1 R
π 3,9 : CARDINALS CDA2A1ILRNS
D I N
A2 L S

39. π C, L = Permutation implementing C.T.C.
C = Number of columns being used
L = Length of plaintext (= length of ciphertext) (an element of SL )
C A1 R
π 3,9 : CARDINALS CDA2A1ILRNS
D I N
A2 L S
0 1 2
012345678 036147258
3 4 5
6 7 8

40. π C, L = Permutation implementing C.T.C.
C = Number of columns being used
L = Length of plaintext (= length of ciphertext) (an element of SL )
C A1 R
π 3,9 : CARDINALS CDA2A1ILRNS
D I N
A2 L S
0 1 2
012345678 036147258
3 4 5
6 7 8
π 3,9 = (1 3)(2 6)(5 7)

41. π 4,13 :
t0 t1 t2 t3
t4 t5 t6 t7
t 0t 4 t 8t12t1t 5t 9t 2t 6t10t 3t 7t11
t8 t9 t10 t11
t12
π 4,13 = (1, 4)(2, 7, 11, 12, 3, 10, 9, 6, 8)
0 is fixed (always); 5 is fixed
Where does the character in position n end up?

42. C
•••
•••
n •••
•••
•••
•••
•••
•
•
•
•••
π C, L (n) = (# of preceding rows) + (# of positions in preceding columns)
A B

43. THE ENEMY ADVANCES AT DAWN
(USING C=5)
T H E E N
E M Y A D
V A N C E
S A T D A
W N
TEVSWHMAANEYNTEACDNDEA

44. 0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
15 16 17 18 19
20 21
π 5,22 (2) = 10 (# char's in preceding columns)
π 5,22 (5) = 1 (# of preceding rows)
π 5,22 (11) = 7 (5 in prec column + 2 prec rows)
π 5,22 (13) = 10 + 4 + 2 = 16

45. }
•••
q
A •••
n •••
•••
•••
•••
•••
•
•
•
•••
n’ = n mod C
n = Cq + n′
n − n′
q=
C

46. C
•••
B
•••
n
L/C, round up •••
•••
•••
•••
•••
•
•
•
a •••
If a column preceding n’s column is
not full, fill it with a “dummy”.
# characters in any quot;fullquot; column:
# dummies:
L 
 C
0 if L ′ = 0, or if L ′ ≠ 0 and 0 ≤ n′ ≤ L ′
# full columns: C, or L’ n′ − L ′ if L ′ ≠ 0 and n′ > L ′

47. Theorem 1
Let C be the number of columns used in a CTC and let L be the length of
the message. Also let n be one of the character position indices (0 ≤ n < L)
and let n’ = n mod C and L’ = L mod C. Then:
 n − n′   L 
 + n′     if L ′ = 0, or if L ′ ≠ 0 and 0 ≤ n′ ≤ L ′
C   C 

π C, L (n) = 
 n − n ′ + n ′   L   − (n ′ − L ′ ) if L ′ ≠ 0 and n′ > L ′
  C 
C   


48. π5,12
01234
56789
10 11

49. π5,12
1−1  12 
π 5,12 (1) = + 1⋅   = 3
5 5
01234
3− 3   12 
π 5,12 (3) = + 3    − (3 − 2) = 0 + 3(3) − 1 = 8
5   5 
56789
8−3   12 
π 5,12 (8) = + 3    − (3 − 2) = 1 + 8 = 9
5   5 
10 11

50. π5,12
1−1  12 
π 5,12 (1) = + 1⋅   = 3
5 5
01234
3− 3   12 
π 5,12 (3) = + 3    − (3 − 2) = 0 + 3(3) − 1 = 8
5   5 
56789
8−3   12 
π 5,12 (8) = + 3    − (3 − 2) = 1 + 8 = 9
5   5 
10 11
7−2   12 
π 5,12 (7) = + 2 ⋅    = 1 + 2(3) = 7
5   5 

51. π5,12
1−1  12 
π 5,12 (1) = + 1⋅   = 3
5 5
01234
3− 3   12 
π 5,12 (3) = + 3    − (3 − 2) = 0 + 3(3) − 1 = 8
5   5 
56789
8−3   12 
π 5,12 (8) = + 3    − (3 − 2) = 1 + 8 = 9
5   5 
10 11
7−2   12 
π 5,12 (7) = + 2 ⋅    = 1 + 2(3) = 7
5   5 
π 5,12 = (1, 3, 8, 9,11, 5)

52. THE RAIL FENCE CIPHER

53. C Y T G A H
R P O R P Y

54. C Y T G A H
R P O R P Y
CYTGAHRPORPY

55. C Y T G A H
R P O R P Y
CYTGAHRPORPY
Rail fence cipher = π 2, L

56. C Y T G A H
R P O R P Y
CYTGAHRPORPY
Rail fence cipher = π 2, L
C R
Y P
T O
CYTGAHRPORPY
G R
A P
H Y

57.  n − n′   L 
 + n′     if L ′ = 0, or if L ′ ≠ 0 and 0 ≤ n′ ≤ L ′
C   C 

π C, L (n) = 
 n − n ′ + n ′   L   − (n ′ − L ′ ) if L ′ ≠ 0 and n′ > L ′
  C 
C   


58.  n − n′   L 
 + n′     if L ′ = 0, or if L ′ ≠ 0 and 0 ≤ n′ ≤ L ′
C   C 

π C, L (n) = 
 n − n ′ + n ′   L   − (n ′ − L ′ ) if L ′ ≠ 0 and n′ > L ′
  C 
C   

n’ = 0 (n even) or 1 (n odd)

59.  n − n′   L 
 + n′     if L ′ = 0, or if L ′ ≠ 0 and 0 ≤ n′ ≤ L ′
C   C 

π C, L (n) = 
 n − n ′ + n ′   L   − (n ′ − L ′ ) if L ′ ≠ 0 and n′ > L ′
  C 
C   

n’ = 0 (n even) or 1 (n odd)
 n
n even

2

π 2, L (n) = 
 n − 1 +  L  n odd
2
2 


60.  n − n′   L 
 + n′     if L ′ = 0, or if L ′ ≠ 0 and 0 ≤ n′ ≤ L ′
C   C 

π C, L (n) = 
 n − n ′ + n ′   L   − (n ′ − L ′ ) if L ′ ≠ 0 and n′ > L ′
  C 
C   

n’ = 0 (n even) or 1 (n odd)
L
 n L even

2
n even
 =
2
  L + 1 L odd
π 2, L (n) =  2
 n − 1 +  L  n odd 
2
2 


61. Corollary 2
Let L be the length of a message enciphered with the rail fence cipher. Also
let n be one of the character position indices (0 ≤ n < L). Then:
 n
n even

2

 n+L
π 2, L (n) =  n odd, L odd
2

 n + L −1
n odd, L even
 2


62. What character positions are fixed by the RFC?
C R
Y P
T O
CYTGAHRPORPY
G R
A P
H Y

63. What character positions are fixed by the RFC?
C R
Y P
T O
CYTGAHRPORPY
G R
A P
H Y

64. What character positions are fixed by the RFC?
C R
Y P
T O
CYTGAHRPORPY
G R
A P
H Y
Corollary 3
The first character in the message is always fixed by the
RFC. The last character is fixed if and only if L is even.
There are no other fixed points.

65. π 2, L (n) = n

66. π 2, L (n) = n
L odd:
L even:

67. π 2, L (n) = n
L odd:
L even:
n even: n odd:

68. π 2, L (n) = n
L odd:
L even:
n even: n odd:
n
=n⇔n=0
2

69. π 2, L (n) = n
L odd:
L even:
n even: n odd:
n n + L −1
=n⇔n=0 =n
2 2
n = L −1

70. π 2, L (n) = n
L odd:
L even:
n odd:
n even: n odd:
n n + L −1
=n⇔n=0 =n
2 2
n = L −1

71. π 2, L (n) = n
L odd:
L even:
n odd:
n even: n odd:
n n+L
n + L −1
=n⇔n=0 =n
=n
2 2
2
n=L ⊗
n = L −1
(0 ≤ n < L)

72. π 2, L (n) = n
L odd:
L even:
n odd:
n even: n odd:
n n+L
n + L −1
=n⇔n=0 =n
=n
2 2
2
n=L ⊗
n = L −1
(0 ≤ n < L)
Corollary 4
If L is even, then π2,L = π2,L+1. So we may assume for what
follows that L is odd.

73. THE INITIAL CYCLE AND THE
STRUCTURE OF THE RAIL
FENCE CIPHER

75. How does π2,L factor into a product of disjoint cycles?

76. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.

77. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.
Cycle containing 1 = initial cycle

78. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.
Cycle containing 1 = initial cycle
Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)

79. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.
Cycle containing 1 = initial cycle
Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)

80. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.
Cycle containing 1 = initial cycle
Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)
Initial cycle of π2,33:
(1, 17, 25, 29, 31, 32, 16, 8, 4, 2)

81. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.
Cycle containing 1 = initial cycle
Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)
Initial cycle of π2,33:
(1, 17, 25, 29, 31, 32, 16, 8, 4, 2)

82. How does π2,L factor into a product of disjoint cycles?
0th position always fixed; position 1 is first one that moves.
Cycle containing 1 = initial cycle
Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)
Initial cycle of π2,33:
(1, 17, 25, 29, 31, 32, 16, 8, 4, 2)
Theorem 5
the initial cycle of π2,L is
k-1, then
If L = 2
k −1 k−2
(1, 2 ,2 ,K , 8, 4, 2)

83. Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)

84. Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)
6 5 4
2 mod11 2 mod11 2 mod11

85. Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)
6 5 4
2 mod11 2 mod11 2 mod11
Theorem 6
Let l1 be the length of the initial cycle of π2,L. Then
k l1 − k
π 2, L (1) = 2 mod L

86. Initial cycle of π2,11:
(1, 6, 3, 7, 9, 10, 5, 8, 4, 2)
6 5 4
2 mod11 2 mod11 2 mod11
Theorem 6
Let l1 be the length of the initial cycle of π2,L. Then
k l1 − k
π 2, L (1) = 2 mod L
Corollary 7
l1 > log 2 L

87. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6

88. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6

89. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6

90. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6

91. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6
3x

92. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6
3x

93. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6
3x mod
3x
17

94. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6
3x mod
3x
17

95. What about the other cycles?
π 2,17
(1, 9, 13, 15, 16, 8, 4, 2)(3, 10, 5, 11, 14, 7, 12, 6)
1 9 13 15 16 8 4 2
3 10 5 11 14 7 12 6
3x mod
3x
Theorem 8
17
π 2, L (n) = ( n ⋅ π 2, L (1)) mod L
I.e.: Every cycle is determined by
the initial cycle.

96. Proof of Theorem 8

97. Proof of Theorem 8
L +1
π 2, L (1) =
2

98. Proof of Theorem 8
L +1
π 2, L (1) =
2
n nL + n  n
n even: π 2, L (n) − n ⋅ π 2, L (1) = − = L  −  ∈¢
 2
2 2
n + L nL + n 1− n
n odd: π 2, L (n) − n ⋅ π 2, L (1) = = L
−  ∈¢
2
2 2

99. Proof of Theorem 8
L +1
π 2, L (1) =
2
n nL + n  n
n even: π 2, L (n) − n ⋅ π 2, L (1) = − = L  −  ∈¢
 2
2 2
n + L nL + n 1− n
n odd: π 2, L (n) − n ⋅ π 2, L (1) = = L
−  ∈¢
2
2 2
In all cases, L divides difference.

100. Proof of Theorem 8
L +1
π 2, L (1) =
2
n nL + n  n
n even: π 2, L (n) − n ⋅ π 2, L (1) = − = L  −  ∈¢
 2
2 2
n + L nL + n 1− n
n odd: π 2, L (n) − n ⋅ π 2, L (1) = = L
−  ∈¢
2
2 2
In all cases, L divides difference.
Corollary 9
( ) mod L
k l1 − k
(n) = n ⋅ 2
π 2, L

101. THE ORDER OF THE RAIL
FENCE CIPHER

102. Proposition (basic group theory)
If a permutation in Sn is written as a product of disjoint
cycles, then the order of the permutation is the least
common multiple of the cycle lengths.

103. Proposition (basic group theory)
If a permutation in Sn is written as a product of disjoint
cycles, then the order of the permutation is the least
common multiple of the cycle lengths.
Theorem 10
The order of the rail fence cipher is the length of its initial
cycle.

104. Proposition (basic group theory)
If a permutation in Sn is written as a product of disjoint
cycles, then the order of the permutation is the least
common multiple of the cycle lengths.
Theorem 10
The order of the rail fence cipher is the length of its initial
cycle.
Proof outline:
Show that the length of each cycle in the disjoint cycle factorization
divides the length of the initial cycle.

106. G = π 2, L ⊆ SL

107. G = π 2, L ⊆ SL
{ }
k
orbG (n) = y : y = π (n) for some k = Cycle containing n
2, L

108. G = π 2, L ⊆ SL
{ }
k
orbG (n) = y : y = π (n) for some k = Cycle containing n
2, L
orbG (1) = Initial cycle

109. G = π 2, L ⊆ SL
{ }
k
orbG (n) = y : y = π (n) for some k = Cycle containing n
2, L
orbG (1) = Initial cycle
Define binary operation * on orbG(1):
a b a +b
π 2, L (1) ∗ π 2, L (1) = π 2, L (1)

110. G = π 2, L ⊆ SL
{ }
k
orbG (n) = y : y = π (n) for some k = Cycle containing n
2, L
orbG (1) = Initial cycle
Define binary operation * on orbG(1):
a b a +b
π 2, L (1) ∗ π 2, L (1) = π 2, L (1)
Claim: orbG(1) forms an abelian group under *.

111. G = π 2, L ⊆ SL
{ }
k
orbG (n) = y : y = π (n) for some k = Cycle containing n
2, L
orbG (1) = Initial cycle
Define binary operation * on orbG(1):
a b a +b
π 2, L (1) ∗ π 2, L (1) = π 2, L (1)
Claim: orbG(1) forms an abelian group under *.
−1
(π )
a
= π 2,− a (n)
l1
(1)
2, L L

112. Let x be the smallest element of its cycle, so cycle = orbG(x).
orbG(1) acts on orbG(x):

113. Let x be the smallest element of its cycle, so cycle = orbG(x).
orbG(1) acts on orbG(x):
( )(
π i2, L (1), ( x ⋅ π 2, L (1)) mod L a )
j
x ⋅ π 2, Lj (1) mod L
i+

114. Let x be the smallest element of its cycle, so cycle = orbG(x).
orbG(1) acts on orbG(x):
( )(
π i2, L (1), ( x ⋅ π 2, L (1)) mod L a )
j
x ⋅ π 2, Lj (1) mod L
i+
{ }
k k
Fx = π (1) ∈orbG (1) : x ⋅ π (1) = x mod L = Stabilizer of x
2, L 2, L

115. Let x be the smallest element of its cycle, so cycle = orbG(x).
orbG(1) acts on orbG(x):
( )(
π i2, L (1), ( x ⋅ π 2, L (1)) mod L a )
j
x ⋅ π 2, Lj (1) mod L
i+
{ }
k k
Fx = π (1) ∈orbG (1) : x ⋅ π (1) = x mod L = Stabilizer of x
2, L 2, L
Classical group theory:
Fx is a subgroup of orbG(1)
The following mapping is a bijection:
orbG (1)
→ orbG (x)
FX
π 2, L (1) ⋅ FX a π 2, L (x)
k k

116. π2,35 = (1, 18, 9, 22, 11, 23, 29, 32, 16, 8, 4, 2)(3, 19, 27, 31, 33,
34, 17, 26, 13, 24, 12, 6)(5, 20, 10)(7, 21, 28, 14)(15, 25, 30)

117. π2,35 = (1, 18, 9, 22, 11, 23, 29, 32, 16, 8, 4, 2)(3, 19, 27, 31, 33,
34, 17, 26, 13, 24, 12, 6)(5, 20, 10)(7, 21, 28, 14)(15, 25, 30)
(1, 18, 9, 22, 11, 23, 29, 32, 16, 8, 4, 2)
{ }
F7 = {n ∈orbG (1) : 7n = 7 mod 35} = {1,11,16} = π 0 35 (1),π 2, 35 (1),π 8 35 (1)
4
2, 2,

118. π2,35 = (1, 18, 9, 22, 11, 23, 29, 32, 16, 8, 4, 2)(3, 19, 27, 31, 33,
34, 17, 26, 13, 24, 12, 6)(5, 20, 10)(7, 21, 28, 14)(15, 25, 30)
(1, 18, 9, 22, 11, 23, 29, 32, 16, 8, 4, 2)
{ }
F7 = {n ∈orbG (1) : 7n = 7 mod 35} = {1,11,16} = π 0 35 (1),π 2, 35 (1),π 8 35 (1)
4
2, 2,
orbG (1)
= {1⋅ F7 , 18 ⋅ F7 , 9 ⋅ F7 , 22 ⋅ F7 }
F7
orbG (7) = {7, 21, 28,14}
= {1mod 35,(18 ⋅ 7)mod 35,(9 ⋅ 7)mod 35,(22 ⋅ 7)mod 35}

119. orbG (1)
orbG (1)
= orbG (x)
=
FX Fx
∴ orbG (1) = Fx ⋅ orbG (x)
Therefore the length of the cycle containing x divides the
length of the initial cycle.

120. Theorem 11
orbG (1) ≅ 2 ⊆ ¢ ∗
L
By Theorem 6, π 2, L (1) = 2 l1 − k mod L
k
Corollary 12
The order of the rail fence cipher on a text of length L (odd)
is the order of the integer 2 in ¢ L
∗
Corollary 13
π 2, L divides φ(L).

121. UNANSWERED QUESTIONS

122. UNANSWERED QUESTIONS
• Simple way to calculate length of initial cycle?

123. UNANSWERED QUESTIONS
• Simple way to calculate length of initial cycle?
• How much of this still works if C > 2?

124. UNANSWERED QUESTIONS
• Simple way to calculate length of initial cycle?
• How much of this still works if C > 2?
• What are the fixed points in a general CTC?

125. UNANSWERED QUESTIONS
• Simple way to calculate length of initial cycle?
• How much of this still works if C > 2?
• What are the fixed points in a general CTC?
• Can we tell when the RFC or general CTC has a k-cycle?

126. UNANSWERED QUESTIONS
• Simple way to calculate length of initial cycle?
• How much of this still works if C > 2?
• What are the fixed points in a general CTC?
• Can we tell when the RFC or general CTC has a k-cycle?
• When is the RFC or general CTC a single (L-1)-cycle?

127. THANK YOU
Contact:
rtalbert@franklincollege.edu
Slides/PDFs for this talk:
http://www.slideshare.net/rtalbert/deconstructing-
columnar-transposition-ciphers
http://www.box.net/shared/2ye298vm3g
Paper:
“The cycle structure and order of the rail fence cipher”.
Cryptologia, 30(2):159-172, 2006.