EXERCISE 2 — Diffie-Hellman with real (small) numbers
======================================================

The real DH formula uses modular exponentiation:
    public  =  g^private  mod p
    shared  =  (other side's public)^(my private)  mod p

PUBLIC parameters (agreed openly, an eavesdropper sees these):
    p = 23     (a small prime)
    g = 5      (the base / generator)

PRIVATE choices (each side keeps secret):
    client private a = 6
    server private b = 15

STEP 1 — each computes their PUBLIC value:
    client A = g^a mod p = 5^6  mod 23 = 15625 mod 23 = 8
    server B = g^b mod p = 5^15 mod 23 = 30517578125 mod 23 = 19
    -> client sends 8, server sends 19  (both travel in the open)

STEP 2 — each computes the SHARED secret from the other's public value:
    client: B^a mod p = 19^6  mod 23 = 47045881 mod 23  = 2
    server: A^b mod p = 8^15  mod 23 = 35184372088832 mod 23 = 2

    BOTH GET 2 -> the shared secret is 2, and it was NEVER sent.

QUICK CHECK with command-line bc (optional):
    echo "5^6  % 23" | bc     -> 8
    echo "5^15 % 23" | bc     -> 19
    echo "19^6 % 23" | bc     -> 2
    echo "8^15 % 23" | bc     -> 2

WHAT THE EAVESDROPPER KNOWS vs DOESN'T:
    KNOWS:        p=23, g=5, A=8, B=19   (everything transmitted)
    DOESN'T KNOW: a=6, b=15, and the shared secret 2

  To find the secret, the eavesdropper must solve g^a mod p = A for a
  (i.e. 5^a mod 23 = 8) — the DISCRETE LOGARITHM problem. With p=23 you
  could brute-force it, but with the ~2048-bit primes (or 256-bit elliptic
  curves) used in real TLS, it is computationally infeasible. THAT is the
  "un-mixing the paint is hard" property made concrete.

MAPPING TO THE PAINT ANALOGY:
    p, g          = the shared base colour (yellow)
    a, b          = each side's secret colour (red, blue)
    A, B          = the mixed colours they send (orange, green)
    shared secret = the final brown both arrive at, never transmitted