Diffie-Hellman Key Exchange - Problem

Implement the Diffie-Hellman Key Exchange algorithm to simulate secure key exchange between two parties (Alice and Bob). This cryptographic protocol allows two parties to establish a shared secret key over an insecure communication channel without prior knowledge of each other.

Given a prime number p and a primitive root g, along with Alice's private key a and Bob's private key b, your task is to:

1. Calculate Alice's public key: A = g^a mod p

2. Calculate Bob's public key: B = g^b mod p

3. Calculate the shared secret from Alice's perspective: secret_alice = B^a mod p

4. Calculate the shared secret from Bob's perspective: secret_bob = A^b mod p

5. Return an object containing all intermediate values and verify that both parties arrive at the same shared secret.

The beauty of this algorithm is that even if an eavesdropper intercepts the public keys A and B, they cannot easily compute the shared secret without knowing the private keys a or b (assuming the discrete logarithm problem is hard).

Input & Output

Example 1 — Basic Diffie-Hellman Exchange

$ Input: p = 23, g = 5, a = 6, b = 15

› Output: {"alice_public":8,"bob_public":19,"alice_secret":2,"bob_secret":2,"shared_secret":2,"verification":true}

💡 Note: Alice computes A = 5^6 mod 23 = 8, Bob computes B = 5^15 mod 23 = 19. Then Alice calculates 19^6 mod 23 = 2 and Bob calculates 8^15 mod 23 = 2. Both arrive at the same shared secret: 2

Example 2 — Different Private Keys

$ Input: p = 11, g = 2, a = 3, b = 7

› Output: {"alice_public":8,"bob_public":7,"alice_secret":2,"bob_secret":2,"shared_secret":2,"verification":true}

💡 Note: With smaller prime p=11: Alice gets A = 2^3 mod 11 = 8, Bob gets B = 2^7 mod 11 = 7. Alice calculates 7^3 mod 11 = 2, Bob calculates 8^7 mod 11 = 2. Shared secret is 2

Example 3 — Edge Case with Small Values

$ Input: p = 7, g = 3, a = 2, b = 4

› Output: {"alice_public":2,"bob_public":4,"alice_secret":4,"bob_secret":4,"shared_secret":4,"verification":true}

💡 Note: Alice: A = 3^2 mod 7 = 2, Bob: B = 3^4 mod 7 = 4. Alice: secret = 4^2 mod 7 = 2, Bob: secret = 2^4 mod 7 = 2. Wait - this should give 2, not 4. Let me recalculate: 4^2 mod 7 = 16 mod 7 = 2, and 2^4 mod 7 = 16 mod 7 = 2. The shared secret should be 2

Constraints

2 ≤ p ≤ 10⁶ (p is prime)
1 ≤ g < p (g is primitive root mod p)
1 ≤ a, b < p (private keys)
All computations done modulo p

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 22 f Meta 15 🍎 Apple 12

The Diffie-Hellman Key Exchange enables secure key sharing over insecure channels. The key insight is using modular exponentiation to create a mathematical 'trapdoor' - easy to compute forward, hard to reverse. Best approach uses fast binary exponentiation for efficiency. Time: O(log n), Space: O(1).

Common Approaches

✓ Direct Calculation

⏱️ Time: O(max(a,b)) Space: O(1)

Directly compute each modular exponentiation step by step using repeated multiplication. Calculate public keys, then shared secrets, and verify they match.

Fast Modular Exponentiation

⏱️ Time: O(log(max(a,b))) Space: O(1)

Implement fast modular exponentiation using the binary method (exponentiation by squaring). This reduces the time complexity from O(n) to O(log n) for each power calculation.

Direct Calculation — Algorithm Steps

Calculate Alice's public key: A = g^a mod p
Calculate Bob's public key: B = g^b mod p
Calculate shared secret from Alice's view: secret_alice = B^a mod p
Calculate shared secret from Bob's view: secret_bob = A^b mod p
Verify both secrets are equal and return all values

Visualization

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Step-by-Step Walkthrough

Calculate Public Keys

A = g^a mod p, B = g^b mod p

Exchange Public Keys

Alice gets B, Bob gets A

Calculate Shared Secrets

Alice: B^a mod p, Bob: A^b mod p

Verify Match

Both secrets are equal

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

long long modPow(long long base, long long exp, long long mod) {
    long long result = 1;
    base = base % mod;
    while (exp > 0) {
        if (exp % 2 == 1) {
            result = (result * base) % mod;
        }
        exp = exp >> 1;
        base = (base * base) % mod;
    }
    return result;
}

void solution(int p, int g, int a, int b) {
    // Calculate Alice's public key: A = g^a mod p
    long long A = modPow(g, a, p);
    
    // Calculate Bob's public key: B = g^b mod p
    long long B = modPow(g, b, p);
    
    // Calculate shared secrets
    long long secret_alice = modPow(B, a, p);
    long long secret_bob = modPow(A, b, p);
    
    bool verification = (secret_alice == secret_bob);
    
    printf("{\"alice_public\":%lld,\"bob_public\":%lld,\"alice_secret\":%lld,\"bob_secret\":%lld,\"shared_secret\":%lld,\"verification\":%s}\n",
           A, B, secret_alice, secret_bob, secret_alice, verification ? "true" : "false");
}

int main() {
    int p, g, a, b;
    scanf("%d", &p);
    scanf("%d", &g);
    scanf("%d", &a);
    scanf("%d", &b);
    
    solution(p, g, a, b);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(max(a,b))

Each exponentiation requires up to max(a,b) multiplications

⚡ Linearithmic

Space Complexity

O(1)

Only stores a few integer variables for calculations

✓ Linear Space

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Medium Frequency

~35 min Avg. Time

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Diffie-Hellman Key Exchange - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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