1. The following describes the simple hash function:

    Choose p, q primes and compute N = pq.
    Choose g relatively prime to N and less than N.

    Then a number n is hashed as follows:

    H = gn mod N

    If there is an m that hashes to the same value as n, then

    gm ≡ gn mod N


    gm-n ≡ 1 mod N

    which implies that

    m –n ≡ 0 mod φ (N)

    So breaking this amounts to finding a multiple of φ (N), which is the hard problem in RSA.

    1. Write a function that takes a bit length n and generates a modulus N of bitlength n and g less than N and relatively prime to it.
    2. Show the output of your function from part (a) for a few outputs.
    3. Using N, g, n as arguments write a function to perform the hashing.
    4. For the following parts (a)-(d) compute the simple hash:
    5. N = 600107, g = 154835, n = 239715
    6. N = 548155966307, g = 189830397891, n = 44344313866
    7. N = 604766153, g = 12075635, n = 443096843
    8. Write a function that creates a collision given p and q.  Show that your function works for a couple of examples.
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