Proper Carmichael Function

Question

I'm creating all the necessary functions for RSA algorithm. Unfortunately i can't seem to a make proper Carmichael function.

These are the functions that i've written:

def gcd(a, b):  # Greatest Common Divisor Generator (Euclidean Algorithm)
    while b != 0:  # While remainder exists
        t = b  # Initially r[k-1]
        b = a % t  # Initially r[k] = r[k-2] mod r[k-1] (where r[k-2] is a)
        a = t  # Predecessor of remainder (b)
    return a

def phi(n):  # Leonard Euler's Totient Function
    y = 0
    for k in range(1, n + 1):  # Phi(+n) is the number of integers k in the range (1 <= k >= n)...
        if gcd(n, k) == 1:  # for which gcd(n, k) = 1
            y += 1
    return y

def carmichael(n):  # Robert Daniel Carmichael's Function
    y = (phi(n) * 1/2) if (n > 4 and ((n & (n - 1)) == 0)) else phi(n)  # phi(n) * 1/2 if 2^x = n, else phi(n) * 1
    return y

I'm using totient function for number generation. From my knowledge there is a simple rule, If number is power of 2 and it's greater than 4, Amount of it's prime numbers shall be halved, otherwise it's equal to phi(n).

The rule above is perfectly working in my code, For example, if the input value is 8, these are the results:

phi(8) = 4
carmichael(8) = 2

But the problem is, Carmichael function is also halving other numbers for some reason, for example if input is 12, this is what my functions return:

phi(12) = 4
carmichael(12) = 4

But this is how it should look like:

phi(12) = 4
carmichael(12) = 2

Why is this happening? Perhaps non-prime odd numbers should be treated differently? Is there something that i need to add to my function?

Thank you!

There's a [codegolf answer with a Python Carmichael function implementation](https://codegolf.stackexchange.com/questions/93739/compute-the-carmichael-function) and detailed explanation that you might find helpful. — import random, Dec 12 '17 at 01:25
Because the stated connection to Euler's totient only exists [for powers or twice the power of an odd prime and powers of 2](http://mathworld.wolfram.com/CarmichaelFunction.html)? Apart from that - isn't it easier to calculate Euler's totient with `n (1-1/p1)(1-1/p2)...(1-1/pm)` — Mr. T, Dec 12 '17 at 01:38
@Piinthesky Thanks! Euler's product formula seems to be easier but i think gcd rule is quicker. — ShellRox, Dec 12 '17 at 12:30
@ShellRox Depends. For one number the gcd approach is probably faster. For repetitive calls, it is probably faster to create a prime number list - composite numbers usually break down pretty fast. There are [prime sieve](https://github.com/kimwalisch/primesieve) generators [written in C](https://github.com/hickford/primesieve-python), who probably outpace a Python script any time. — Mr. T, Dec 12 '17 at 13:26
@Piinthesky Thanks a lot! Definitely going to try both methods! — ShellRox, Dec 12 '17 at 14:42

user9073637 · Answer 1 · 2019-07-28T15:19:19.323

First we create the gcd function to calculate greatest common divisor of 2 numbers, we will need it later in lambda function.

 def gcd(a,b):
        while (a>0):
            b=b%a
            (a,b)=(b,a)
        return b

Then we look at how carmichael function works.

Let n be a positive integer. Then λ(n) is defined to be the smallest positive integer k such that
a^k≡1(mod n)
for all a such that gcd(a,n)=1.

Note that we are looking for k, the values of a is determined once we have n.

Now we initialize the function with default condition

n=int(n)
k=2
a=1
alist=[]

To find all a values we use gcd(a,n)=1 to test whether a and n have the greatest common divisor as 1, which means they are coprime.
If not, a++
if gcd(a,n)==1, we store this value to the list of a and test next a until we test all a<=n

        while not ((gcd(a,n))==1):
            a=a+1

        while ((gcd(a,n))==1) & (a<=n) :
            alist.append(a)
            a=a+1
            while not ((gcd(a,n))==1):
                a=a+1

Ok now we have all a in the list alist, look back at definition

the smallest positive integer k such that
a^k≡1(mod n)

First we count the number of a, which is the length of alist
timer=len(alist)
Then we use
if (a**k)%n==1:
to test whether this k makes a^k≡1(mod n) for all a value in alist. We construct a loop

for a in alist:
      if (a**k)%n==1:
           timer=timer-1
              if timer <0:
                  break
              pass
      else:
           timer=len(alist)
           k=k+1

to test all k number from 2, if it doesnot meet requirement, we do k=k+1

Now we have the whole function as following

    def carmichael(n):
        n=int(n)
        k=2
        a=1
        alist=[]

        while not ((gcd(a,n))==1):
            a=a+1

        while ((gcd(a,n))==1) & (a<=n) :
            alist.append(a)
            a=a+1
            while not ((gcd(a,n))==1):
                a=a+1

        timer=len(alist)
        while timer>=0:
            for a in alist:
                if (a**k)%n==1:
                    timer=timer-1
                    if timer <0:
                        break
                    pass
                else:
                    timer=len(alist)
                    k=k+1
        return k

I'm sorry if this is a little late, but I need to say that although your solution is good, your code is completely unpythonic. — Gderu, Feb 28 '20 at 08:52
This is how I would write the function instead: `def carmichael(n): coprimes = [] for i in range(n): if gcd(i, n) == 1: coprimes.append(i) k = 0 while True: for coprime in coprimes: if (coprime ** k) % n != 1: break else: return k k += 1` — Gderu, Feb 28 '20 at 09:05

Proper Carmichael Function

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