CODEWITHSUNDEEP
pythonalgorithmgreatest-common-divisor
Heart Killer
Heart Killer

Reputation: 9

Python: Finding GCD using Middle School Procedure

Middle School Procedure GCD

Thus, for the numbers 60 and 24, we get

60 = 2 . 2 . 3 . 5

24 = 2 . 2 . 2 . 3

gcd(60, 24) = 2 . 2 . 3 = 12.

So using the above instructions, this is what I got so far:

import numpy as np

#find prime factors of m and output it to list fm
def Middle(m,n):
    c = 2
    fm = [ ]  
    while m > 1:
      if m % c == 0:
        fm.append(c)
        m = m/c
      else:
        c = c + 1             
    return fm

#find prime factors of n and output it to list fn
    d = 2
    fn = [ ]  
    while n > 1:
      if n % d == 0:
        fn.append(d)
        n = n/d
      else:
        d = d + 1 
    return fn

#compare fm and fn and multiply common items
#this is the part where I got wrong       
    cf = []
    for f in fm:
        if f in fn:
            cf.append(f) 
    return (np.prod(cf))

I know the last part is wrong but I have no idea how to fix it. The instructions said something about repeating the f to a minimum but I'm clueless. Please help.

Upvotes: 0

Views: 3910

Answers (2)

karthik reddy
karthik reddy

Reputation: 479

snap shot

This is one way to get your required output:

import functools
def gcd(a,b):
    def factArr(x):
        list = []
        i=2
        while i <= x:
            if (x % i) == 0:
                list.append(i)
                x = x/i
                i = 2
            else:
                i = i+1
        return list
    aArr = factArr(a);
    bArr = factArr(b);
    print("aArr",aArr,"bArr",bArr)
    cArr = []
    for v in aArr:
        if v in bArr:
            cArr.append(v)
            bArr.remove(v)
    print("cArr",cArr)
    return functools.reduce(lambda x, y: x*y, cArr)
gcd(60,24)`

Upvotes: 2

Adirio
Adirio

Reputation: 5286

import numpy as np
from collections import Counter

# Find the prime factors of a integer
def prime_factors(n):
    factors = []
    i = 2
    while n > 1:
        if n % i == 0:
            factors.append(i)
            n /= i
        else:
            i += 1
    return Counter(factors)

# Find prime factors of m and n and multiply their common ones
def Middle(m, n):
    fm = prime_factors(m)
    fn = prime_factors(n)
    cf = fm & fn
    return np.prod(list(cf.elements()))

Or you can also do it in a one liner:

def Middle(m, n):
    return np.prod(list((prime_factors(m) & prime_factors(n)).elements()))

Upvotes: 1

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