Affine Cipher Find And B Calculator

Find Affine Cipher Keys ‘a’ and ‘b’

Enter two plaintext-ciphertext character pairs (A-Z) and the modulus (m) to find the affine cipher keys ‘a’ and ‘b’.

Character to Number Mapping (A=0 to Z=25)

Char	Num	Char	Num	Char	Num	Char	Num	Char	Num	Char	Num	Char	Num
A	0	B	1	C	2	D	3	E	4	F	5	G	6
H	7	I	8	J	9	K	10	L	11	M	12	N	13
O	14	P	15	Q	16	R	17	S	18	T	19	U	20
V	21	W	22	X	23	Y	24	Z	25

What is an Affine Cipher and Finding ‘a’ and ‘b’?

The Affine cipher is a type of monoalphabetic substitution cipher, where each letter in an alphabet is mapped to its numeric equivalent, encrypted using a simple mathematical function, and converted back to a letter. The encryption function is `E(x) = (ax + b) mod m`, where `x` is the numerical equivalent of the plaintext letter, `m` is the size of the alphabet, and ‘a’ and ‘b’ are the keys of the cipher.

To encrypt a message, you need ‘a’, ‘b’, and ‘m’. To decrypt, you also need them. The affine cipher find a and b calculator is used when you have a couple of plaintext-ciphertext pairs and you want to determine the keys ‘a’ and ‘b’ that were used, assuming you know the modulus ‘m’ (usually 26 for the English alphabet).

This calculator is useful for cryptanalysis or for understanding how the affine cipher works by reverse-engineering the keys from known pairs. For ‘a’ to be a valid key, it must be coprime with ‘m’ (i.e., `gcd(a, m) = 1`).

Who should use it?

Students learning cryptography, hobbyists interested in ciphers, or anyone trying to solve an affine cipher puzzle with some known plaintext-ciphertext mappings can use this affine cipher find a and b calculator.

Common Misconceptions

A common misconception is that any pair of ‘a’ and ‘b’ will work. However, ‘a’ must be coprime to ‘m’ for decryption to be possible and unique. If `gcd(a, m) != 1`, then multiple plaintext letters could map to the same ciphertext letter, making decryption ambiguous.

Affine Cipher ‘a’ and ‘b’ Formula and Mathematical Explanation

Given two plaintext-ciphertext pairs (P1, C1) and (P2, C2), and their numerical equivalents (p1, c1) and (p2, c2) where A=0, B=1, …, Z=25, we have:

1. `c1 = (a * p1 + b) mod m`
2. `c2 = (a * p2 + b) mod m`

Subtracting the second equation from the first (modulo m):

`(c1 – c2) mod m = ((a * p1 + b) – (a * p2 + b)) mod m`

`(c1 – c2) mod m = (a * p1 – a * p2) mod m`

`(c1 – c2) mod m = a * (p1 – p2) mod m`

Let `diffC = (c1 – c2 + m) mod m` and `diffP = (p1 – p2 + m) mod m`. So, `diffC ≡ a * diffP (mod m)`.

To solve for ‘a’, we need to find the modular multiplicative inverse of `diffP` modulo `m`, let’s call it `invDiffP`. This inverse exists if and only if `gcd(diffP, m) = 1`.

If the inverse exists, then `a = (diffC * invDiffP) mod m`.

Once ‘a’ is found, we can find ‘b’ using the first equation: `b = (c1 – a * p1) mod m`. To ensure ‘b’ is non-negative, we calculate `b = (c1 – a * p1 + m * m) mod m`.

Variables Table

Variable	Meaning	Unit	Typical Range
P1, P2	Plaintext characters	Character	A-Z
C1, C2	Ciphertext characters	Character	A-Z
p1, p2	Numerical value of P1, P2	Integer	0-25 (for m=26)
c1, c2	Numerical value of C1, C2	Integer	0-25 (for m=26)
m	Modulus (alphabet size)	Integer	26 (English), >1
a	Multiplicative key	Integer	1 to m-1, gcd(a,m)=1
b	Additive key	Integer	0 to m-1
diffP	(p1-p2) mod m	Integer	0 to m-1
diffC	(c1-c2) mod m	Integer	0 to m-1
invDiffP	Modular inverse of diffP mod m	Integer	1 to m-1 (if exists)

Practical Examples

Example 1: Finding ‘a’ and ‘b’

Suppose we know ‘H’ maps to ‘P’ and ‘E’ maps to ‘F’, with m=26.

• P1=’H’ (p1=7), C1=’P’ (c1=15)
• P2=’E’ (p2=4), C2=’F’ (c2=5)
• m=26

diffP = (7 – 4) mod 26 = 3

diffC = (15 – 5) mod 26 = 10

We need to solve `3a ≡ 10 (mod 26)`. First, find inverse of 3 mod 26. `gcd(3, 26) = 1`. Inverse of 3 mod 26 is 9 (since 3 * 9 = 27 ≡ 1 mod 26).

a = (10 * 9) mod 26 = 90 mod 26 = 12. Hmm, `gcd(12, 26) = 2 != 1`. Let’s recheck. 3 * 9 = 27 = 1 mod 26. a = 10 * 9 = 90 = 3 * 26 + 12. So a=12. If a=12, gcd(12, 26)=2, not valid. Maybe my initial H->P, E->F was bad for a valid ‘a’.

Let’s use the calculator with H->P and E->F. It gives a=12. gcd(12,26)=2. The calculator should report this.

If we use P1=’B'(1), C1=’F'(5) and P2=’C'(2), C2=’K'(10):

• p1=1, c1=5, p2=2, c2=10, m=26

diffP = (1-2) mod 26 = -1 mod 26 = 25

diffC = (5-10) mod 26 = -5 mod 26 = 21

Solve `25a ≡ 21 (mod 26)` or `-a ≡ 21 (mod 26)` or `a ≡ -21 ≡ 5 (mod 26)`. So a=5. `gcd(5, 26)=1`. Valid ‘a’.

b = (c1 – a * p1) mod 26 = (5 – 5 * 1) mod 26 = 0 mod 26 = 0.

So, a=5, b=0.

Example 2: No Unique Solution

Suppose P1=’A'(0), C1=’C'(2) and P2=’N'(13), C2=’C'(2), m=26.

• p1=0, c1=2, p2=13, c2=2, m=26

diffP = (0-13) mod 26 = 13

diffC = (2-2) mod 26 = 0

Solve `13a ≡ 0 (mod 26)`. `gcd(13, 26) = 13 != 1`. Inverse of 13 mod 26 does not exist. However, `13a` is a multiple of 26. So `13a = 26k`, `a=2k`. ‘a’ could be 0, 2, 4, …, 24. There isn’t a unique valid ‘a’ (coprime to 26). The calculator should indicate no unique ‘a’ or that `gcd(diffP, m) != 1`.

How to Use This Affine Cipher Find a and b Calculator

1. Enter Plaintext 1 Character: Input the first known plaintext character (A-Z).
2. Enter Ciphertext 1 Character: Input the corresponding ciphertext character for the first plaintext.
3. Enter Plaintext 2 Character: Input the second known plaintext character (A-Z).
4. Enter Ciphertext 2 Character: Input the corresponding ciphertext character for the second plaintext.
5. Enter Modulus (m): The default is 26 for the English alphabet. Change it if you are using a different alphabet size.
6. Calculate: The calculator automatically updates as you type or you can click “Calculate Keys”.
7. Read Results: The primary result will show the values of ‘a’ and ‘b’ if a valid, unique ‘a’ is found. It will also state if `gcd(a,m) != 1` or if `gcd(diffP, m) != 1`.
8. Intermediate Values: Check the numerical values, differences, GCD, and inverse used in the calculation.
9. Chart: The chart visually represents your two plaintext-ciphertext number pairs.
10. Reset: Use the “Reset” button to clear inputs to default values.
11. Copy: Use “Copy Results” to copy the main results and intermediate steps.

If the calculator indicates `gcd(diffP, m) != 1`, it means either no solution for ‘a’ exists or multiple solutions exist, but no unique modular inverse was found for `diffP`.

Key Factors That Affect Affine Cipher ‘a’ and ‘b’ Results

• Choice of Plaintext-Ciphertext Pairs: The difference between the numerical values of the plaintexts (p1-p2) is crucial. If `gcd(p1-p2, m) != 1`, you won’t find a unique ‘a’.
• Modulus (m): The size of the alphabet determines the modulo base. All calculations are done modulo m.
• Coprimality of ‘a’ and ‘m’: For a valid affine cipher key, ‘a’ must be coprime to ‘m’ (`gcd(a, m) = 1`). If the calculated ‘a’ is not coprime, decryption might not be unique or possible in the standard way.
• Coprimality of (p1-p2) and m: If `gcd(p1-p2, m) != 1`, we cannot find a unique modular inverse of (p1-p2), and thus cannot find a unique ‘a’ using this method.
• Correct Mapping: Ensure you have the correct plaintext-ciphertext pairs. Incorrect pairs will lead to incorrect ‘a’ and ‘b’.
• Alphabet Used: The calculator assumes A=0, B=1, etc. If a different numerical mapping was used, the results for ‘a’ and ‘b’ would differ.

Frequently Asked Questions (FAQ)

What if gcd(a, m) is not 1?

If the calculated ‘a’ is not coprime to ‘m’, the encryption function is not a one-to-one mapping over the alphabet modulo m, meaning decryption would be ambiguous for some ciphertext characters. The affine cipher is typically defined with `gcd(a,m)=1`.

What if gcd(p1-p2, m) is not 1?

If `gcd(p1-p2, m) = d > 1`, then `d * a * (p1-p2)/d ≡ (c1-c2) (mod m)`. A solution for ‘a’ exists only if ‘d’ also divides (c1-c2). If it does, there are ‘d’ possible values for ‘a’ modulo m, but none might be coprime to ‘m’. Our affine cipher find a and b calculator will highlight when `gcd(p1-p2, m) != 1`.

Can ‘a’ be 0?

No, ‘a’ cannot be 0 because it needs to be coprime with m (for m>1), and gcd(0,m)=m. Also, if a=0, E(x)=b, which is a trivial and insecure shift cipher where all letters map to one letter.

Can I use this calculator for other alphabets?

Yes, if you know the modulus ‘m’ for that alphabet size and your characters can be mapped to 0 to m-1. However, the current calculator is hardcoded for A-Z input.

What if the two plaintext characters are the same (P1=P2)?

If P1=P2, then p1=p2, and p1-p2=0. `gcd(0, m) = m`. You won’t find a unique ‘a’. You need two different plaintext characters.

What if the two ciphertext characters are the same (C1=C2)?

If C1=C2 but P1!=P2, then c1-c2=0. This means `a * (p1-p2) ≡ 0 (mod m)`. If `gcd(p1-p2, m)=1`, then `a` must be 0 or m, which isn’t valid. If `gcd(p1-p2, m) = d > 1`, then `a * (p1-p2)/d ≡ 0 (mod m/d)`, meaning ‘a’ is a multiple of m/d.

Is the affine cipher secure?

No, it is very insecure. With only two plaintext-ciphertext pairs, we can often find the keys ‘a’ and ‘b’ using a tool like this affine cipher find a and b calculator. It’s easily broken with frequency analysis as well.

How many possible keys are there for an affine cipher with m=26?

The number of possible ‘a’ values is the number of integers between 1 and 25 coprime to 26 (phi(26) = phi(2)*phi(13) = 1*12 = 12). There are 26 possible ‘b’ values. So, 12 * 26 = 312 possible keys.