Least Common Multiple (LCM) – Python

Question

This is not much about how do i do it and more about whats wrong with this method. I managed to solve this using other methods but i dont know why i cant with this one. what am i missing here?

Example input: 4,6
Expected output: 12
Actual output: 4

n1, n2 = map(int, input("n1 and n2: ").split(','))

def lcmCalc (n1,n2):
    i = 2
    lcm = 1
    while (n1 != 1) and (n2 != 1):
        if n1 % i == 0 and n2 % i == 0:
            lcm *= i
            n1 = n1/i
            n2 = n2/i
        elif n1 % i != 0 and n2 % i == 0:
            lcm *= i
            n2 = n2/i
        elif n1 % i == 0 and n2 % i != 0:
            lcm *= i
            n1 = n1/i
        else:
            i += 1
    return lcm

print(lcmCalc(n1,n2))

Asked By: AleGPZ

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Source

Answer 1

I would be tempted to use the relationship :

lcm(a,b) = |a.b| / gcd(a,b)

And of course gcd(a,b) = gcd(b, a%b) & gcd(a,0) = a

So my code :

def gcd(a,b):
    if b ==0:
       return a
    else:
       return gcd(b, a % b)

def lcm(a,b):
    return int(abs(a*b) / gcd(a,b))

or – if you don’t object to a little bit of help from the standard library:

 from math import gcd

 def lcm(a,b):
    return int(abs(a*b) / gcd(a,b))

Answered By: Tony Suffolk 66

Answer 2

You were close. Here are the edits:

def lcmCalc(n1, n2):
    i = 2
    lcm = 1
    while (n1 != 1) and (n2 != 1):
        if n1 % i == 0 and n2 % i == 0:
            lcm *= i
            n1 = n1 // i   # <== use floor division operator
            n2 = n2 // i
        elif n2 % i == 0:  # <== remove unneeded 2nd test
            lcm *= i
            n2 = n2 // i
        elif n1 % i == 0:  # <== remove unneeded 2nd test
            lcm *= i
            n1 = n1 // i
        else:
            i += 1
    return lcm * n1 * n2    # <== need to include residuals

When the outer loop terminates, either of n1 or n2 may still be above 1. That residual needs to be included in the result.

Answered By: Raymond Hettinger

Answer 3

def lcm(num1,num2):
    for x in range(1,max(num1,num2)):
        if (num1 % x) == 0 and (num2 % x) == 0:
            e=x
    lcm = (num1 * num2) / e
    return lcm

Answered By: Job Ko

Least Common Multiple (LCM) – Python

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