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Sorrow2 wrote: ↑March 22nd, 2023, 9:37 am Is there any codebusters book that you guys can recommend? I love to read. Thank you in advance!

knightmoves wrote: ↑March 16th, 2023, 10:28 am

Mythic wrote: ↑March 15th, 2023, 5:48 pm Hello, I was wondering if someone could explain how you solve the Cryptanalysis of an Affine.

Here's a worked example for you. You are given the cyphertext
"ABOBRFLYBMLHGQBPSLYLSSRYBTODGWLYLQDFRF", and told that the first two letters of the plaintext are H and E.

So you know that under the affine encryption, E(4) -> B(1) and H(7) ->A (0).

So (4a + b) mod 26 = 1
and (7a + b) mod 26 = 0

Subtracting these, you get 3a mod 26 = -1 = 25.

So we're hunting for a value of a for which 3a mod 26 = 25.
3a = 25 is no good, 'cause a needs to be an integer.
3a = 51 gives a = 17, which is a viable solution.
3a = 77 doesn't have an integer solution, and then there aren't any other valid solutions, because you know a<26.

So with a=17, the first equation becomes
(68 + b) mod 26 = 1
or equivalently
(16 + b) mod 26 = 1

And so b = 11.

Now you have determined a and b, you can compute the mappings.
A(0) -> (b mod 26) = 11 = L
B(1) -> (a + b) mod 26 = 2 = C
C(2) -> (2a + b) mod 26 = 19 = T
D(3) -> (3a + b) mod 26 = 10 = K.

And so on.

You might prefer to compute the mappings for the high frequency letters (etaoin shrdlu...) rather than doing them in alphabetical order. Finishing this off is left as an exercise for the reader

Mythic wrote: ↑March 27th, 2023, 2:25 pm

knightmoves wrote: ↑March 16th, 2023, 10:28 am

Mythic wrote: ↑March 15th, 2023, 5:48 pm Hello, I was wondering if someone could explain how you solve the Cryptanalysis of an Affine.

Here's a worked example for you. You are given the cyphertext
"ABOBRFLYBMLHGQBPSLYLSSRYBTODGWLYLQDFRF", and told that the first two letters of the plaintext are H and E.

So you know that under the affine encryption, E(4) -> B(1) and H(7) ->A (0).

So (4a + b) mod 26 = 1
and (7a + b) mod 26 = 0

Subtracting these, you get 3a mod 26 = -1 = 25.

So we're hunting for a value of a for which 3a mod 26 = 25.
3a = 25 is no good, 'cause a needs to be an integer.
3a = 51 gives a = 17, which is a viable solution.
3a = 77 doesn't have an integer solution, and then there aren't any other valid solutions, because you know a<26.

So with a=17, the first equation becomes
(68 + b) mod 26 = 1
or equivalently
(16 + b) mod 26 = 1

And so b = 11.

Now you have determined a and b, you can compute the mappings.
A(0) -> (b mod 26) = 11 = L
B(1) -> (a + b) mod 26 = 2 = C
C(2) -> (2a + b) mod 26 = 19 = T
D(3) -> (3a + b) mod 26 = 10 = K.

And so on.

You might prefer to compute the mappings for the high frequency letters (etaoin shrdlu...) rather than doing them in alphabetical order. Finishing this off is left as an exercise for the reader

Thanks for the idea but I was wondering how you went from (68 + b) mod 26 = 1 --> (16 + b) mod 26 = 1

knightmoves wrote: ↑March 27th, 2023, 6:57 pm Same principle. You know T(19) -> M(12)and H(7) -> W (22)

So (19a + b) mod 26 = 12
and (7a + b) mod 26 = 22

Subtract to get

12a mod 26 = (-10) = 16 [-10 is 16 because we're doing arithmetic in mod 26]

Now hunt for a solution
12a = 16 is no good (16/12 has remainder 4)
12a = 42 is no good (42/12 has remainder 6)
12a = 68 is no good (68/12 has remainder 8)
you can see the pattern now, right? we're going to want
12a = 120 -> a = 10
Are there other possibilities?
12a = 146 is no good (146/12 has remainder 2)
12a = 172 is no good (172/12 has remainder 4)
...
12a = 276 -> a = 23

So either a = 10 or a = 23.
If a = 10, (70 + b) mod 26 = 22
so b=4
If a = 23, (161+b)mod26 = 22
so (5+b) = 22
b = 17.

So you have two possibilities = (a,b) = (10,4) or (23,17).
But a=10 isn't a viable solution (10 is not coprime with 26), so (a,b) = (23,17) is the only possible solution.

Now you have a and b, you can build the mapping table as before.

Clue: the answer begins "THESE DAYS, ALL WE HEAR ABOUT".

knightmoves wrote: ↑March 16th, 2023, 10:28 am

Mythic wrote: ↑March 15th, 2023, 5:48 pm Hello, I was wondering if someone could explain how you solve the Cryptanalysis of an Affine.

Here's a worked example for you. You are given the cyphertext
"ABOBRFLYBMLHGQBPSLYLSSRYBTODGWLYLQDFRF", and told that the first two letters of the plaintext are H and E.

So you know that under the affine encryption, E(4) -> B(1) and H(7) ->A (0).

So (4a + b) mod 26 = 1
and (7a + b) mod 26 = 0

Subtracting these, you get 3a mod 26 = -1 = 25.

So we're hunting for a value of a for which 3a mod 26 = 25.
3a = 25 is no good, 'cause a needs to be an integer.
3a = 51 gives a = 17, which is a viable solution.
3a = 77 doesn't have an integer solution, and then there aren't any other valid solutions, because you know a<26.

So with a=17, the first equation becomes
(68 + b) mod 26 = 1
or equivalently
(16 + b) mod 26 = 1

And so b = 11.

Now you have determined a and b, you can compute the mappings.
A(0) -> (b mod 26) = 11 = L
B(1) -> (a + b) mod 26 = 2 = C
C(2) -> (2a + b) mod 26 = 19 = T
D(3) -> (3a + b) mod 26 = 10 = K.

And so on.

You might prefer to compute the mappings for the high frequency letters (etaoin shrdlu...) rather than doing them in alphabetical order. Finishing this off is left as an exercise for the reader

Sorrow2 wrote: ↑March 28th, 2023, 6:03 am Affine Cipher looks confusing ;-;

Mythic wrote: ↑April 5th, 2023, 6:17 pm Hi, quick question on Fractionated Morse does it always end in X or XX or does it vary?