CRYPTOGRAPHY Course
Cryptography
Cryptography is the knowledge related to the encryption and decryption of messages. This course is more a mathematical course than a programming course, at least for now (sorry 😖).
Mathematical references
A summary of what you need to know.
Euclidean division (Division euclidienne
)
Dividing $a$ by $b$, mean solving $a = b * q + r$. You need to find quotient $q$ and the remainder (reste
) $r$, with $r \lt b$. Both are unique.
We are saying that $b$ is a divisor of $a$ if $r = 0$, written $a\ \ b$. It would also mean that $a$ is a multiple of $b$.
 $25/5$: $25 = 5 * 5 + 0$ so $525$
 $25/4$: $25 = 4 * 6 + 1$
Greatest common divisor (Plus grand diviseur commun
)
$D(a,b)$ is the set of the common divisors between $a$ and $b$. We are calling greatest common divisor (GCD or PGCD
), the greatest value of $D(a,b)$, written $GCD(a,b)$ or $a \wedge b$.
Example: what's the GCD of $(27, 15)$
\begin{split} D(27, 15) = D(2715{\color{grey}*1}, 15)\\ = D(12, 15) = D(12, 1512{\color{grey}*1})\\ = D(12, 3) = D(123{\color{grey}*4}, 3)\\ = D(0, 3) = 3 = 27 \wedge 15 \end{split}
Formula: $a \wedge b = (a  b * q) \wedge b$
Pro tip: $a \wedge b = c \wedge (\frac{b}{c} \wedge \frac{a}{c})$, so $27 \wedge 15 = 3 \wedge (9 \wedge 5) = 3 \wedge 0 = 3 $.
Bézout's identity (Thèorème de Bezout
)
The formula is $a \wedge b = a * u + b * v$. Bézout coefficients $u$ and $v$ are not unique. Sometimes, this is easy to find the coefficients, but if this isn't, use a table (tip: read the example first)
k  $r_k$  $q_k$  $u_k$  $v_k$  Bézout 

0  a  ❌  1  0  
1  b  the q in a/b  0  1  
...  the r in a/b  ...  $u_{k2}q_{k1}*u_{k1}$  $v_{k2}q_{k1}*v_{k1}$  $r_k=a*u_k+b*v_k$ 
Example: $a=98$ and $b=77$.
k  $r_k$  $q_k$  $x_k$  $y_k$  Bézout 

0  98  ❌  ${\color{red}1}$  0  $98 = 98 * 1 + 77 * 0$ 
1  77  ${\color{green}1}$  ${\color{blue}0}$  1  $77 = 98 * 0 + 77 * 1$ 
2  21  3  ${\color{red}1}{\color{green}1}*{\color{blue}0}=1$  $0  1*1=1$  $21 = 98 * 1 + 77 * 1$ 
3  14  1  $0  3 * 1 = 3$  $1  3 * 1=4$  $14 = 98 * 3 + 77 * 4$ 
3  7  2  $1  1 * 3 = 4$  $1  1 * 4=5$  $7 = 98 * 4 + 77 * 5$ (solution ✅) 
3  0  ❌  $3  2 * 4 = 11$  $4  2 * 5=14$  $0 = 98 * 11 + 77 * 14$ 
Of course, we are doing this because this one was hard, but if you got $5 * a + 7 * b = 5 \wedge 7 = 1$, then you could find almost immediately that you can use $a=3$ and $b=2$ giving us $15  14 = 1$ (you got 10 and 9 too, etc.).
Note: Bézout's identity is a diophantine equation (
Équation diophantienne
, wiki).
Prime numbers (Nombres premiers
)
📚 Definition 📚
A prime number is a number greater or equals to 2, which is only divisible by 1 and itself.
 ✅: 2, 3, 5, 7, 11, 13, etc.
 ❌: 4 (→2), 6 (→2, →3), 9 (→3), 10 (→2, →5), etc.
Note: If $p$ and $q$ are prime numbers, and $p \neq q$, then $p \wedge q = 1$.
Note: $\phi(n)$ is called Euler's totient function (indicatrice d'Euler
) and is the number of prime numbers with $n$.
🧺 Prime Factorization 🧺

French:
Décomposition en produit de facteurs premiers
 Definition: Every number ($\ge 2$) can be expressed as a product of prime numbers
 Examples
 $27 = 3 * 9 = 3^3$
 $60 = 6 * 10 = 2 * 3 * 5 * 2 = 2^2 * 3 * 5$
We are calling $\xi_p(n)$, the exponent of $p$ in the factorization of a number $n$ with prime numbers (puissance de p dans n
, ex: $\xi_3(27) = 3$ or $\xi_5(60)=1$).
🚀 GCP with prime numbers 🚀
You can calculate the GCD easily. Simply express each number with prime numbers. Then, take each unique number in both factorizations, they will be in the GCD. Their exponent is the lowest exponent that we have for each number in the two factorizations.
\begin{split} GCP(a, b) = \prod_{i\ \in\ unique\ prime\ numbers} i^{\min(\xi_i(a),\ \xi_i(b))} \\ 98 = 2 * 49 = 2 * 7^2 \\ 77 = 7 * 11 \\ GCP(98, 77) = 2^{min(1, 0)} * 7^{min(2, 1)} * 11^{min(0, 1)} = 1 * 7 * 1 = 7\\ \end{split}
Note: if you replace min with max, you will have the least common multiple (
Plus petit commun multiple/PPCM
).
Congruence
Let a and b two numbers. We are saying that $a$ is congruent (congru
, $\equiv$) with $b$ modulus (modulo
) $m$, if we can find a $q$ giving us
@ a = b + m * q @
We are using one of the notations below
Ex: $27 \equiv 3\ (mod\ 12)$ as we have $12*2 + 3$
Formula: Given $a = m * q + r$, then $a \equiv r\ (mod\ m)$
🐱🏍 Find $r$ for a complex $a$ 🐱🏍
If you need to evaluate the value $r$ given modulus $m$ of a complex value $a$, then simply split rewrite your value $a$ as a product of factors, and evaluate the modulus on each one.
 Ex: $256\ \text{mod}\ 7 = 2^8 = 2^3 * 2^3 * 2^2$
 $2^3 = 8 \equiv 1\ (\text{mod}\ 7)$
 $2^2 = 4 \equiv 4\ (\text{mod}\ 7)$
 Giving us $256 \equiv 1 * 4 \equiv 4 \ (\text{mod}\ 7)$
 Ex: $2021\ \text{mod}\ 3 = 2 * 10^3 + 2 * 10^2 + 1$
 $10 \equiv 1\ \text{mod}\ 3$
 $10^2 = 10 * 10 \equiv 1 \ (\text{mod}\ 3)$
 $10^3 = 10 * 10 * 10 \equiv 1 \ (\text{mod}\ 3)$
 $2021= 2 * 1 + 2 * 1 + 1 = 5 \equiv 2 \ (\text{mod}\ 3)$
This is called Euler theorem (wiki).
🧐 Invertible numbers 🧐
A number $a$ is invertible modulus $m$ if, $\exists a^{1}$ giving us $a * a^{1} \equiv 1\ (mod\ m)$. To find $a^{1}$, you can use Bézout: $a u + m v = 1 \Leftrightarrow a u = 1\ (mod\ m) \Leftrightarrow a^{1} \equiv u\ (mod\ m)$ with $u \gt 0$.
 Condition: $a \wedge m\ \ 1$

Ex: in Bézout example, we had $98 * 4 + 77 * 5 = 7$
 ❌: not invertible

Ex: we found that $5 * 3 + 7 * 2 = 1$
 $m = 7$, and $a = 5$
 $5^{1} = 3\ (\text{mod}\ 7)$
 check: $5 * 3 = 15 = 1\ (\text{mod}\ 7)$
 $7^{1} = 2 = 3\ (\text{mod}\ 5)$
 check: $3 * 7 = 21 = 1\ (\text{mod}\ 5)$
Main idea
You will encrypt your message with a key using an algorithm, generating a cipher (message encrypté
): $E(\text{key}_1, \text{message})=^{algo}c$. The receiver will receive your cipher, decrypt it using a key and an algorithm: $D(\text{key}_2, c)=^{algo}\text{message}$.
 Note: $k_1$ should be different from $k_2$ (otherwise, it will be broken)
 Note: The algorithm is wellknown (ex: AES, DES, RSA, SSL, ...)
We will use the terms
 public key: a key used to encrypt a message that is shared with everyone. Only the one generating the key is supposed to be able to decrypt a message.
 private key: a key associated with the public key, not shared, used to decrypt messages encrypted with the public key
Playing with binaries 0️⃣1️⃣
A binary is a number in base 2, so we are using only 0 and 1. You will need to do a lot of operations on binaries.

Addition: note that $1+0=1$, while $0+0=0$, and $1+1=0$
 Generalization: $1+1+1=3\equiv1\ (mod\ 2)$
 Ex: $(10101)_2 + (11000)_2 = (01101)_2$

Conversion: convert a number from base 10 to base 2
 We know that $2^0={\color{blue}1},\ 2^1={\color{blue}2},\ 2^2={\color{blue}4},\ 2^3={\color{blue}8}, ...$ (16, 32, 64, 128, 256, 512, etc.)
 Ex: $56 = 32 + 16 + 8$
 Ex: $56 = 2^5 + 2^4 + 2^3 = {\color{grey} 0*2^0 + 0*2^1 + 0*2^2} + {\color{grey} 1*} 2^3 + {\color{grey} 1*} 2^4 + {\color{grey} 1*} 2^5 $
 Ex: $(56)_{10} = (000111)_2$

Conversion: from base 2 to base 10
 Ex: $(000111)_2$
 $(000111)_2 = {\color{grey} 0 * 2^0 + 0 * 2^1 + 0 * 2^2} + {\color{grey} 1*} 2^3 + {\color{grey} 1*} 2^4 + {\color{grey} 1*} 2^5$
 $(000111)_2 = (8+16+32)_{10} = (56)_{10}$
Some algorithms
Most of them aren't used, either because they are inefficient or because they were broken. You can move to the next section if you are not interested 🧐.
Caesar cipher (Code de César
)
Caesar was replacing letters like this $a \to c,\ b \to d,\ etc.$. On the same idea, the substitution cipher is replacing each letter by another like $a \to w,\ b \to e,\ etc.$.
Unfortunately, we can try to replace the most common letters in the ciphertext, or the most common groups of letters, with the most common letters (or group of letters) in a Language. And break the cipher.
Vigenère cipher (Cryptage en bloc de Vigenère
)
This is the same as the Caesar cipher/Substitution cipher. We are working on letters. But, this time we are splitting letters into groups. The problem is that if we got the key, we can easily get back the message.

message:
memorize

key:
vgn

cipher text:
memorize+vgn=memorize+vgnvgnv=hkzjxvuk
 $M=12$, $V=21$, $12+21=33\equiv 7\ (mod\ 26)=H$
 $E=4$, $G=6$, $4+6=10\equiv 10\ (mod\ 26)=K$
 $M=12$, $N=13$, $12+13=25\equiv 25\ (mod\ 26)=Z$
 ...

decrypt:
hkzjxvuk+vgnvgnv=memorize
 $H=7$, $V=21$, $7  21 + 26 \equiv 12\ (mod\ 26)=M$
 $K=10$, $G=6$, $10  6 + 26 \equiv 4\ (mod\ 26)=E$
 $Z=25$, $N=13$, $25  13 + 26 \equiv 12\ (mod\ 26)=M$
 $J=9$, $V=21$, $9  21 + 26 \equiv 14\ (mod\ 26)=O$
 ...
Onetime pad
This algorithm is creating a key as long or longer than the message, so even though it is secure, we can't use it. For each message, both user are agreeing on a key (ex: $01010$), and the Algorithm is XOR (ou exclusif
, $0+0=1+1=0$ and $1+0=0+1=1$).
 Message: $10011$
 Key (encrypt): $01010$
 Cipher text: $11001$
 Key (decrypt): $01010$
 Message: $10011$
Knapsack problem (sacàdos
) of MerkleHellman
You got a "bag/Knapsack" of numbers (ex: 23, 5, 11, 2, 55). Your message is made of 0
and 1
, and using this method 1
means you picked something from the bag, 0
means you didn't. Then, make the sum of the numbers you picked in the bag to create the cipher. Note that you will have to split the message into groups having the size of the knapsack.
 Knapsack (private key): $2, 5, 11, 23, 55$ (up to you, size=6)
 Message: $1100111001 = 11001\ 10001$ (group of 6)

Cipher
 $2 + 5 + 0 + 0 + 55 = 62$ (first group)
 $2 + 0 + 0 + 0 + 55 = 57$ (second group)
 Cipher text: $62, 57$
But, this is too easy to find what generated this cipher. So we are using the superincreasing knapsack problem to generate a public key, and this easy key as the private key. We will pick a value $N$ greater than the sum of the values in the Knapsack, and a number $W$, so that $N \wedge W1$ (=no common divisor aside 1).
 We are picking $N=113$, $W=27$
 We got $W^{1} = 67\ (\text{mod}\ 113)$

Knapsack (public key)
 We are multiplying the private key by $W$, modulus $N$
 $54, 22, 71, 56, 16$ (ex: $27 * 2 = 54\ (\text{mod}\ 113)$)

private cipher
 $54 + 22 + 0 + 0 + 16 = 92$ (first group)
 $54 + 0 + 0 + 0 + 16 = 70$ (second group)
 Cipher text: $92, 70$

decrypt (given N and W)
 We are multiplying the public key by $W$, modulus $N$
 we get back the private key: $2, 5, 11, 23, 55$
 we can easily find that: $2 + 5 + 0 + 0 + 55 = 62$ and ...
 so the message was: $1100110001$
Diffie–Hellman key exchange
A is sending a message to B. Both are agreeing on two numbers $p$ and $g$ with $(p1) \wedge g\ \ 1$ (=the only divisor is 1), and they are picking a private number (resp. $a$ and $b$, lesser than $p$).
A is sending $g^a\ (mod\ p)$ to B (resp. b for B to A). The common key for A is $(g^a)^b\ (mod\ p)$ and for B, it is $(g^b)^a\ (mod\ p)$. They both have the same number, but they do not know about the exponent of the other.
Rivest–Shamir–Adleman (RSA)
We are picking two prime numbers $p$ and $q$, and deducting both $n = p * q$ and $\phi(n) = (p1) * (q1)$. The, use Bézout to find the coefficients that we will call $d$ and $e$.
 Public key: $(n, e)$ (used by others to encrypt)
 Private key: $(n, d)$ (used by only me, to decrypt)
To encrypt a message $T$, simply do $S = T^e\ (mod\ n)$, while the message is supposed to be a number $\lt n$.
To decrypt a message $S$, simply do $T = S^d\ (mod\ n)$.
This is working, because an attacker would have a lot of prime numbers to test ($10^{497}$ for $n \approx 10^{1000}$) to find back $\phi(n) = (p1) * (q1)$ from $n$.