Reputation: 228
I made a code that measure number of steps it takes to return to 1 in a Collatz Conjecture. Here Is my code
counter = 0
def collatz(n):
global counter
counter += 1
if n <= 0 :
return "Invalid Number"
elif n == 1 :
return counter
elif n % 2 == 1 :
n = 3*n + 1
return collatz(n)
elif n % 2 == 0 :
n = n/2
return collatz(n)
print(collatz(9921615699))
print(collatz(9921615699))
I expect the Last two print commands to print 311 and 311. Instead, they print 311 and 622. I guess that was easy enough to see in the code what is wrong. How can i fix that? how can counter reset each time a command is completed and not when the function is run.
Upvotes: 1
Views: 109
Reputation: 51847
You're using recursion, so just use it properly:
def collatz(n, counter=0):
counter += 1
if n <= 0 :
return "Invalid Number"
elif n == 1 :
return counter
elif n % 2 == 1 :
n = 3*n + 1
return collatz(n, counter)
elif n % 2 == 0 :
n = n/2
return collatz(n, counter)
print(collatz(9921615699))
print(collatz(9921615699))
You were using global
in your original version which is usually not what you want to do. You got to see first-hand why.
You could have reset the counter, e.g.
result = counter
counter = 0
return result
But that's pretty nasty, let's not do that. When you're implementing a recursive algorithm there's probably no good reason to have a global variable.
Upvotes: 1
Reputation: 17263
Instead of using global variable you could make the counter a parameter with default value:
def collatz(n, counter=0):
counter += 1
if n <= 0 :
return "Invalid Number"
elif n == 1 :
return counter
elif n % 2 == 1 :
n = 3*n + 1
return collatz(n, counter)
elif n % 2 == 0 :
n = n/2
return collatz(n, counter)
Upvotes: 4