tweak = int(input("Input an integer"))
def collatz(number):
while number != 1:
if number % 2 == 0:
return int(number)
elif number % 2 != 0:
return int((3 * number) + 1)
print(number)
collatz(tweak)
If you are trying to implement a Collatz conjecture script, then your first if is wrong - you should divide the number by 2. Also, your return causes the function to end, so you get only one while loop, doesn't matter what happens inside - so you basically return the number you inputed or that number * 3 + 1.
Here is the correct code with slight modifications:
tweak = int(input("Input an integer"))
def collatz(number):
steps = 0
num = number
while number != 1:
if number % 2 == 0:
number = number / 2
else:
number = int(3 * number + 1)
steps +=1
print("Reached 1 in {} iterations for number {}.".format(steps, num))
collatz(tweak)
You also don't need the elif, because number can only be dividable by 2 or not.
Example outputs:
collatz(22)
collatz(55)
collatz(234)
Reached 1 in 15 iterations for number 22.
Reached 1 in 112 iterations for number 55.
Reached 1 in 21 iterations for number 234.
Your question has only code, but I think this is what you might be looking for:
def collatz(tweak):
while tweak != 1:
if tweak % 2 == 0:
return int(tweak)
elif tweak % 2 != 0:
return int((3 * tweak) + 1)
print(tweak)
tweak = int(input("Input an integer:"))
result = collatz(tweak)
print(result)
Your returning values in if and else so it wont print beyond..
instead assign it in variable and do print..
tweak = input("Input an integer")
def collatz(number):
while number != 1:
if number % 2 == 0:
return int(number)
elif number % 2 != 0:
return int((3 * number) + 1)
print number # this wont work.. your returned already
print "output:%s"%collatz(tweak)
output:
me#dev-007:~/Desktop$ python test.py
Input an integer10
number 10
output:10
I'm trying to write my own formula to find a prime number, but it does not completely work and I cannot find the flaw in my logic. Bare in mind I have taken a look around but cannot find an algorithm that I find similar to mine.
My code:
#Challenge 7
prime = []
num = 0
found = False
while found == False:
if num == 0 or num == 1:
num+=1
else:
for value in range(2, num+1):
if len(prime) == 50:
print('Found all')
found = True
break
if num % value == 0:
num+=1
else:
if num not in prime:
prime.append(num)
else:
pass
print(prime)
This code works for first few primes (3, 5, 7...)
but it also gives incorrect values like 10, and I don't understand why. If someone could explain it to me so that I can understand where the logical mistake is, I'd appreciate it.
The error comes from this part
if num % value == 0:
num+=1
else:
if num not in prime:
prime.append(num)
else:
pass
You assume that the integer is a prime as soon as we find the first occurence of a non-divisor. But the def for primes is that every integer in the interval [2..prime] is a non-divisor. How do we check if any number does not have any divisors?
def isPrime(x):
for v in range(2, x):
if (x % v == 0):
return False;
return True;
Something like this would work to check if any given number is a prime or not. And since we now have taken the isPrime part out of the main loop, we no longer need a for loop inside the while. Something like this would do
def isPrime(x):
for v in range(2, x):
if (x % v == 0):
return False;
return True;
prime = [}
num = 2
found = False
while found == False:
if len(prime) == 50:
print("found all")
found = True
break
if(isPrime(num)):
print(num)
prime.append(num)
num+=1
else:
num+=1
If you set a breakpoint for when num == 10 you will see the problem clearly.
When you start doing you division check inside of for value in range(2, num + 1): the second number is 3, so num (10) modulo value (3) is 1, which is your test for determining a prime. What your test should be is that it not divisible by any number less than it (less than half is actually sufficient since you check with 2 anyway).
So, consider instead:
else:
is_indivisible = True
# loop through all numbers less than it not including itself
# (because x % x == 0)
for value in range(2, num - 1):
# it is only indivisible if it was previously indivisible
# And the check is same as before, modulo != 0
is_indivisible = is_indivisible and (num % value != 0)
if not is_indivisible:
break
# if it is indivisible and it doesn't exist in prime list yet
if is_indivisible and num not in prime:
prime.append(num)
# move on to the next number
num += 1
Have tried searching for this, but can't find exactly what I'm looking for.
I want to make a function that will recursively find the factors of a number; for example, the factors of 12 are 1, 2, 3, 4, 6 & 12.
I can write this fairly simply using a for loop with an if statement:
#a function to find the factors of a given number
def print_factors(x):
print ("The factors of %s are:" % number)
for i in range(1, x + 1):
if number % i == 0: #if the number divided by i is zero, then i is a factor of that number
print (i)
number = int(input("Enter a number: "))
print (print_factors(number))
However, when I try to change it to a recursive function, I am getting just a loop of the "The factors of x are:" statement. This is what I currently have:
#uses recursive function to print all the letters of an integer
def print_factors(x): #function to print factors of the number with the argument n
print ("The factors of %s are:" % number)
while print_factors(x) != 0: #to break the recursion loop
for i in range(1,x + 1):
if x % i == 0:
print (i)
number = int(input("Enter a number: "))
print_factors(number)
The error must be coming in either when I am calling the function again, or to do with the while loop (as far as I understand, you need a while loop in a recursive function, in order to break it?)
There are quite many problems with your recursive approach. In fact its not recursive at all.
1) Your function doesn't return anything but your while loop has a comparision while print_factors(x) != 0:
2) Even if your function was returning a value, it would never get to the point of evaluating it and comparing due to the way you have coded.
You are constantly calling your function with the same parameter over and over which is why you are getting a loop of print statements.
In a recursive approach, you define a problem in terms of a simpler version of itself.
And you need a base case to break out of recursive function, not a while loop.
Here is a very naive recursive approach.
def factors(x,i):
if i==0:
return
if x%i == 0:
print(i)
return factors (x,i-1) #simpler version of the problem
factors(12,12)
I think we do using below method:
def findfactor(n):
factorizeDict
def factorize(acc, x):
if(n%x == 0 and n/x >= x):
if(n/x > x):
acc += [x, n//x]
return factorize(acc, x+1)
else:
acc += [x]
return acc
elif(n%x != 0):
return factorize(acc, x+1)
else:
return acc
return factorize(list(), 1)
def factors(x,i=None) :
if i is None :
print('the factors of %s are : ' %x)
print(x,end=' ')
i = int(x/2)
if i == 0 :
return
if x % i == 0 :
print(i,end=' ')
return factors(x,i-1)
num1 = int(input('enter number : '))
print(factors(num1))
Recursion is a functional heritage and so using it with functional style yields the best results. This means avoiding things like mutations, variable reassignments, and other side effects. That said, here's how I'd write factors -
def factors(n, m = 2):
if m >= n:
return
if n % m == 0:
yield m
yield from factors(n, m + 1)
print(list(factors(10))) # [2,5]
print(list(factors(24))) # [2,3,4,6,8,12]
print(list(factors(99))) # [3,9,11,33]
And here's prime_factors -
def prime_factors(n, m = 2):
if m > n:
return
elif n % m == 0:
yield m
yield from prime_factors(n // m, m)
else:
yield from prime_factors(n, m + 1)
print(list(prime_factors(10))) # [2,5]
print(list(prime_factors(24))) # [2,2,2,3]
print(list(prime_factors(99))) # [3,3,11]
def fact (n , a = 2):
if n <= a :
return n
elif n % a != 0 :
return fact(n , a + 1 )
elif n % a == 0:
return str(a) + f" * {str(fact(n / a , a ))}"
Here is another way. The 'x' is the number you want to find the factors of. The 'c = 1' is used as a counter, using it we'll divide your number by 1, then by 2, all the way up to and including your nubmer, and if the modular returns a 0, then we know that number is a factor, so we print it out.
def factors (x,c=1):
if c == x: return x
else:
if x%c == 0: print(c)
return factors(x,c+1)
def get_nearest_multiple(minnum, factor):
"""
function get_nearest_multiple will calculate the nearest multiple that is greater than the min. value,
Parameters are the minimum value and factor,
Will return the ans - the nearest multiple
"""
ans = 0
x = 1
while ans < minnum:
if minnum == 0:
ans = 0
else:
ans = x * factor
x += 1
return ans
get_nearest_multiple(0, 1)
if __name__ == '__main__':
get_nearest_multiple(0, 1)
Can't seem to figure out why my function doesn't print out anything. The output doesn't even show up as an error. Just blank.
Nowhere in your code do you have a print() statement which is required to produce an output in the console
I am doing an assignment and the answers are coming back correctly but I would need them to say 5! = 120 instead of just = 120. How would I go about that?
def getInt():
getInt = int
done = False
while not done:
print("This program calcultes N!")
# get input for "N
N = int(input("Please enter a non-negative value for N: "))
if N < 0:
print("Non-Negative integers, please!")
else:
done = True
return N
def main():
n = getInt()
for i in range(n-1):
n = n * (i+1)
print("=" ,n)
main()
I hope this code will help.
print('Enter a positive integer')
a = int(input())
def factorial(n):
if n == 0:
return(1)
if n == 1:
return(1)
if n > 1:
return(n * factorial(n-1))
if a < 0:
print('Non-Negative integers, please!')
if a >= 0:
print(str(a) + '! = ' + str(factorial(a)))
In the for i in range(n-1)you could use another integer instead of n just to be sure things don't mess up and you can print like joel said print(i,"!=", n) but instead of n the integer you will use.
can you show me your homework instructions?
i'm not sure what the first value is in your example.. the current iteration or the original number entered?
# declare getInt()
def getInt():
getInt = int
done = False
while not done:
# write "this program calculates N!"
print("This program calcultes N!")
# get input for "N
N = int(input("Please enter a non-negative value for N: "))
# if N < 0 then
if N < 0:
print("Non-Negative integers, please!")
# else
else:
# done = true
done = True
# return N
return N
# main
def main():
n = entry = getInt()
for i in range(n-1):
n = n * (i+1)
print("{0}! = {1}".format(entry, n))
main()
results:
/*
This program calcultes N!
Please enter a non-negative value for N: 5
5! = 120
*/