Question

IN PYHTON CODE

Question #3 Produce a function prime_factors.dict that has one integer variable n that is assumed to be greater than 1. The function will return a dictionary with keys the integers from 2 to n, inclusive, and with values the set of prime factors of each key. So, the command prime_factors.dict (8) will return the dictionary 2 123,3: 3),4 2),5 (53,6 2,3),7:7),8 {2)) Note that even though the prime number 2 appears twice in the prime fac- torization of 4, it only appears once in the set of its prime divisors The function smallest_divisor is available to use in your code. This function has one integer argument n that is greater than 1. It will return the smallest integer divisor of n that is greater than 1. So smallest.divisor (35) will return the integer 5 and smallest divisor (23) will return 23. Note that in all cases, this function will return a prime number and that smallest divisor(n) will return the number n if and only if n is a prime num ber. The code for this function appears at the top of the question #3 cell. You may find it useful in solving this question, but you are not required to use it Hint If n is the product of the integers c and d, then the set of prime divisors of n is equal to the union of the set of prime divisors of c and the set of prime divisors of d

Answer 1

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Code:

def smallest_divisor(n):

a=[]

for i in range(2, n+1):

if(n%i==0):

a.append(i)

a.sort()

return a[0]

def prime_factors_dict(n):

pf_d = {}

for i in range(2, n+1):

pf_d[i] = []

for j in range(2, i+1):

if i%j==0 and smallest_divisor(j)==j:

pf_d[i].append(j)

pf_d[i] = set(pf_d[i])

return pf_d

print(prime_factors_dict(8))