Reputation: 99
def gei(a, b):
'''
a, b: only positive integers! If you don't, we will make you. Max 1337
For extracting the juice of the numbers in the form of a common divider
'''
#Not Ruby, no to_i, but int()
smallest = int(abs(min(a, b)))
biggest = int(abs(max(a, b)))
print "You inputed: ", smallest, " and ", biggest, " their order doesn't matter."
print "Do you want to see guess numbers? Type 'yes', if you do! "
selection = raw_input()
print
print
#To evade infinite loops and too big numbers, we use count.
count = 0
ans = smallest
truth = str(selection) == str('yes')
def condition(ans, base):
ans1 = base % ans == 0
return ans1
while condition(ans, biggest) == False or condition(ans, smallest) == False:
ans -= 1
count += 1
if count >= 1337:
break
elif truth == True:
print ans
if truth == True:
print
print
print "After weeks of calculation, here is your greater common divider: "
return ans
So yeah, 8th grade informatics assignment to extract common greater dividers. I was wondering, maybe you guys know how can I make it less cumbersome? How to avoid using definition inside and naming so many variables?
Upvotes: 1
Views: 5879
Reputation: 7908
Use Euclid's algorithm
int GCD(int a,int b){
if(b==0){
return a;
}
return GCD(b,a%b);
}
Upvotes: 1
Reputation: 104712
A few things can be improved in your code.
First off, truth
is a truly poor variable name, since it doesn't really tell you what it's contents mean. I'd use something like show_work
instead.
Next, you have an internal function that tells you a variable divides another evenly (returning a Boolean value). I'd suggest just using the modulus value directly, rather than coverting it to a bool with == 0
, and you also don't need it to be in a function. Also, it's never necessary to compare a value to True
or False
. Just use the value itself (or use not
to invert it). Putting these together will make your while
loop's condition not biggest % ans and not smallest % ans
.
Finally, you can use a better algorithm than trying every value one by one. Euclid's algorithm is a great way to calculate the Greatest Common Divisor quickly, and it's very easy to implement in Python (but I'll leave it to you).
Upvotes: 0
Reputation: 80346
import fractions
print fractions.gcd(4,8)
>>> 4
But you can also look at the source:
def gcd(a, b):
"""Calculate the Greatest Common Divisor of a and b.
Unless b==0, the result will have the same sign as b (so that when
b is divided by it, the result comes out positive).
"""
while b:
a, b = b, a%b
return a
Reference: http://en.wikipedia.org/wiki/Euclidean_algorithm
Upvotes: 8
Reputation: 41813
This is nice an consise, and names no variable:
def gcd(a,b):
return max(d for d in xrange(1, min(a, b)+1) if a % d == b % d == 0)
print gcd(15, 25)
>>> 5
But please don't claim it as your own :)
Upvotes: -1