I've started the book "Automate The Boring Stuff" by Al Sweigart.
At the end of Chapter 3, the author suggests creating a Collatz Sequence in Python as a practice exercise. (the practice exercise suggests I use a the print function and return statement)
When I use a print() function in my code, it works great and I get all the evaluated values I want to see on the screen:
print("This is The Collatz Sequence")
user = int(input("Enter a number: "))
def collatz(n):
print(n)
while n != 1:
if n % 2 == 0:
n = n // 2
print(n)
else:
n = n * 3 + 1
print(n)
collatz(user)
Question:
How come when I want to use the return statement, the while loop only runs once?
For example, passing the integer 3 into my function with the return statement only gives me the return value of 3 and 10:
print("This is The Collatz Sequence")
user = int(input("Enter a number: "))
def collatz(n):
print(n)
while n != 1:
if n % 2 == 0:
n = n // 2
return n
else:
n = n * 3 + 1
return n
result = collatz(user)
print(result)
return exits the function and, therefore terminates your while loop.
Perhaps you meant to use yield instead:
print("This is The Collatz Sequence")
user = int(input("Enter a number: "))
def collatz(n):
print(n)
while n != 1:
if n % 2 == 0:
n = n // 2
yield(n)
else:
n = n * 3 + 1
yield(n)
print(list(collatz(user)))
Output:
This is The Collatz Sequence
Enter a number: 3
3
[10, 5, 16, 8, 4, 2, 1]
Yield is logically similar to a return but the function is not terminated until a defined return or the end of the function is reached. When the yield statement is executed, the generator function is suspended and the value of the yield expression is returned to the caller. Once the caller finishes (and assumably uses the value that was sent) execution returns to the generator function right after the yield statement.
In your code you don't re-feed the new value back into your equation. Try separating your while loop from the collatz module. I have an example of this below:
def collatz(number):
if number % 2 == 0:
return number // 2
elif number % 2 == 1:
return 3 * number + 1
chosenInt = int(input('Enter an integer greater than 1: '))
print(chosenInt)
while chosenInt != 1:
chosenInt = collatz(chosenInt)
print(chosenInt)
def collatz(number):
if (number%2 == 0):
return print(number//2);
else:
return (print(number*3+1));
inputNumber = input("Enter a number greater than 1:");
result = collatz(int(inputNumber));
while result != 1:
result = collatz(result);
I am getting a typeError with it! Don't know why?
Related
When I run the following code:
def collatz(number):
if number % 2 == 0:
return number // 2
result = number
print(number)
elif number % 2 == 1:
return 3 * number + 1
result = number
print(number)
n = input('Enter a number: ')
while n != 1:
n = collatz(int(n))
the code runs to get the number to 1, but it always seems to skip the lines
result = number
print(number)
because it does not show any output. But when I visualize it, it is running the number to one. Could someone help explain why that is happening? Thanks so much.
You are using a recursive loop that bypasses those two statements.
def collatz(number):
if number % 2 == 0:
return number // 2 <---- exits function here
result = number
print(number)
elif number % 2 == 1:
return 3 * number + 1 <---- exits function here
result = number
print(number)
The same concept applies if you have the following function.
def add(x,y):
return x + y
print(x)
You would never get to the print(x) statement as you are exiting the function before that line of code!
You need to remove the return statement. Return exits the collatz function immediatly.
this is because it returns the value and it doesn't continue after the return, so you could do:
def collatz(number):
if number % 2 == 0:
result = number / 2
print(result)
return result
elif number % 2 == 1:
result = 3 * number + 1
print(result)
return result
n = input('Enter a number: ')
while n != 1:
n = collatz(int(n))
When a function sees return it ends immediately and outputs the value after the word. So to get the lines to run, they need to be before the word return:
def collatz(number):
if number % 2 == 0:
result = number
print(number)
return number // 2
elif number % 2 == 1:
result = number
print(number)
return 3 * number + 1
n = input('Enter a number: ')
while n != 1:
n = collatz(int(n))
i have to write a hailstone program in python
you pick a number, if it's even then half it, and if it's odd then multiply it by 3 and add 1 to it. it says to continue this pattern until the number becomes 1.
the program will need methods for the following:
accepting user input
when printing the sequence, the program should loop until the number 1.
print a count for the number of times the loop had to run to make the sequence.
here's a sample run:
prompt (input)
Enter a positive integer (1-1000). To quit, enter -1: 20
20 10 5 16 8 4 2 1
The loop executed 8 times.
Enter a positive integer (1-1000). To quit, enter -1: 30
30 15 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
The loop executed 19 times.
Enter a positive integer (1-1000). To quit, enter -1: -1
Thank you for playing Hailstone.
right now i have this:
count = 0
def hailstone(n):
if n > 0
print(n)
if n > 1:
if n % 2 == 0:
hailstone(n / 2)
else:
hailstone((n * 3) + 1)
count = count + 1
i don't know what to do after this
Try to think in a modular way, make two functions: check_number() and user_call(). Check_number will verify if the current number in the loop is odd or even and the user_call() just wraps it to count how many times the loop did iterate.
I found the exercise in a great book called Automate Boring Stuff with Python, you have to check it out, if you don't know it already.
Here's my code. Try to use what serves you the best.
from sys import exit
def check_number(number):
if number % 2 ==0:
print(number // 2)
return(number // 2)
else:
print(number*3+1)
return number*3+1
def user_call(number):
count = 0
while number != 1:
count += 1
number = check_number(number)
return count
if __name__ == "__main__":
try:
number = int(input('Give a number \n'))
count = user_call(number)
print('count ',count)
except Exception as e:
exit()
you can use global
visit https://www.programiz.com/python-programming/global-keyword to learn more
import sys
res = []
def hailstone(number):
global res
if number > 1:
if number % 2 == 0:
res.append( number // 2 )
hailstone(res[len(res)-1])
else:
res.append(number * 3 + 1)
hailstone(res[len(res)-1])
return res
number = int(input('Enter a positive integer. To quit, enter -1: '))
if number <= 0 or number == 0:
print('Thank you for playing Hailstone.')
sys.exit()
else:
answers = hailstone(number)
for answer in answers:
print(answer)
print('The loop executed {} times.'.format(len(answers) + 1))
I used recursion to solve the problem.
Heres my code:
Edit: All criteria met
count = 0
list_num = []
def input_check():
number = int(input("Enter a positive integer (1-1000). To quit, enter -1: "))
if number >= 1 and number <= 1000:
hailstone_game(number)
elif number == -1:
return
else:
print("Please type in a number between 1-1000")
input_check()
def hailstone_game(number):
global count
while number != 1:
count += 1
list_num.append(number)
if number % 2 == 0:
return hailstone_game(int(number/2))
else:
return hailstone_game(int(number*3+1))
list_num.append(1) # cheap uncreative way to add the one
print(*list_num, sep=" ")
print(f"The loop executed {count} times.")
return
input_check()
Additional stuff that could be done:
- Catching non-integer inputs using try / except
Keep in mind when programming it is a good habit to keep different functions of your code separate, by defining functions for each set of 'commands'. This leads to more readable and easier to maintain code. Of course in this situation it doesn't matter as the code is short.
Your recursive function is missing a base/terminating condition so it goes into an infinite loop.
resultArray = [] #list
def hailstone(n):
if n <= 0: # Base Condition
return
if n > 0:
resultArray.append(n)
if n > 1:
if n % 2 == 0:
hailstone(int(n/2))
else:
hailstone((n * 3) + 1)
# function call
hailstone(20)
print(len(resultArray), resultArray)
Output
8 [20, 10, 5, 16, 8, 4, 2, 1]
Here's a recursive approach for the problem.
count=0
def hailstone(n):
global count
count+=1
if n==1:
print(n)
else:
if n%2==0:
print(n)
hailstone(int(n/2))
else:
print(n)
hailstone(3*n+1)
hailstone(21)
print(f"Loop executed {count} times")
Beginner question here.
I have just attempted an exercise from Automate the boring stuff. I've completed the question in the format suggested by first defining a function as below:
"Write a function named collatz() that has one parameter named number. If number is even, then collatz() should print number // 2 and return this value. If number is odd, then collatz() should print and return 3 * number + 1."
and then using that same function, meeting those minimal constraints, to write a programme that meets the following requirements:
Then write a program that lets the user type in an integer and that keeps calling collatz() on that number until the function returns the value 1.
I've managed to generate a sequence of numbers ending with one, following the above rules, but my program prints each number in the sequence three times. Is anyone able to explain why this might be?
Thanks so much for your help
def collatz(number):
if int(number) % 2 == 0:
print(int(number)//2)
return int(number)//2
else:
print(3 * int(number) + 1)
return 3 * int(number) + 1
collatz(5)
print('Enter a number')
entry = input()
while collatz(entry) != 1:
collatz(entry)
entry = collatz(entry)
Your loop should look like this:
entry = input()
while entry != 1:
entry = collatz(entry)
You are calling the function 3 times and you have a print call in the function.
Only call the function once and I would remove the print statements from the collatz method and just print in the calling loop, e.g.:
In []:
def collatz(number):
if number % 2 == 0:
return number//2
return 3*number + 1
entry = int(input("Enter a number: "))
print(entry)
while entry != 1:
entry = collatz(entry)
print(entry)
Out[]:
Enter a number: 10
10
5
16
8
4
2
1
You can try:
def collatz(number):
if number == 0:
return 'Try again with an integer other than 0'
elif number == 1:
return 1
elif number % 2 == 0:
n = number // 2
print(n)
elif number % 2 == 1:
n = 3 * number + 1
print(n)
while n != 1:
n = collatz(n)
return n
return n
The last statement return n in line 15 is optional.
I'm trying to write a program that finds if a given number is in the Fibonacci sequence and I keep getting recursion that doesn't terminate and I'm not sure why. Line 17 seems to be the big problem. When I input 0 or 1 I get the desired answers. Just looking for help getting to an answer, I'm trying to learn so just telling me the answer doesn't help me much.
number = int(input("Enter your number:"))
def fib(n):
if n == 0 or n == 1:
return 1
else:
return (fib(n-1) + fib(n-2))
def isfib(number):
n = 0
if number < 0:
print("Invalid input")
elif number == fib(n):
print("Number is in the sequence")
elif number < fib(n):
print("Number is not in the sequence")
elif number > fib(n):
n = n +1
isfib(number) #where the problem occurs
isfib(number)
There are many little mistakes so i corrected those(I have added a better implementation of Fibonacci code with simple addition too):
number = int(input("Enter your number:"))
def fib(n):
if n == 0 or n == 1: return 1
else:
temp1=1
temp=2
temp3=0
for z in range(n-2):
temp3=temp
temp+=temp1
temp1=temp3
return temp
def isfib(number): #it is ok not to return anything unless you need to stop the function in between
done=0
n=0
while done!=1:
if number < 0:
print("Invalid input")
done=1
elif number == fib(n):
print("Number is in the sequence")
done=1
elif number < fib(n):
print("Number is not in the sequence")
done=1
elif number > fib(n):
n = n +1
#i have used done instead of return to show the function can exit even if you dont return a value
#you can just 'return' instead of changing done variable and making the loop infinite
isfib(number)
Since you have used lot of recursions, i am guessing you want to do it only by using recursions. So, the code will be like this:
number = int(input("Enter your number:"))
def fib(n):
if n == 0 or n == 1: return 1
else: return (fib(n-1) + fib(n-2))
def isfib(number,n=0):
if number < 0: print("Invalid input")
elif number == fib(n): print("Number is in the sequence")
elif number < fib(n): print("Number is not in the sequence")
elif number > fib(n):
n = n +1
isfib(number,n)
isfib(number)
Tested of course, it works(but again i wouldn't recommend it this way :D)
Your fib function is wrong (one of the terms should be n-2)
The last elif of the isfib() function is not returning a call of itself,i.e it is calling itself and not doing anything with that result.
Put simply, you are calling isfib(number) with the very same number you were given originally. Your "n = n + 1" just before that suggests that maybe you wanted to call isfib(n) not isfib(number).
Basically isfib(number) is just an infinite loop.
I don't really know why you want to use recursive to solve this question. When you get a number, you don't know what are n-1 and n-2. Therefore, you cannot recursive on n-1 and n-2. The following example can easily check the number is in Fibonacci sequence or not:
number = int(input("Enter your number:"))
def fib(n):
ni = 1
ni_minus_1 = 0
while ni + ni_minus_1 < n:
temp = ni_minus_1
ni_minus_1 = ni
ni = ni + temp
if ni + ni_minus_1 == n:
return True
return False
print(fib(number))
update
I guess you want to build the sequence by recursive. For example, given number is 5, you want to return the 5th element in Fibonacci sequence. Then the code would be like the following:
def build_fib(n):
if n == 0:
return 0
if n == 1:
return 1
return build_fib(n-1) + build_fib(n-2)
print(build_fib(6))
I am currently a noob learning Python, and I am trying to complete an exercise. The exercise requires me to:
Input an integer.
Depending on whether that integer is odd or even, do a specific calculation and print the answer.
Take the given answer, and repeat specific calculations again until answer is equal to 1.
The code I have so far completes the first 2 actions, but I am struggling to implement the loop which will continue to rerun the calculations until the answer is 1. Here is my code so far:
def collatz(getNumber):
if getNumber % 2 == 0:
print(getNumber // 2)
elif getNumber % 2 == 1:
print(3 * getNumber + 1)
print('Please write a number')
number = collatz(int(input()))
Use a while loop:
def collatz(number):
print(number)
while number != 1:
if number % 2 == 0:
number //= 2
else:
number = number * 3 + 1
print(number)
Alternatively, you could use recursion:
def collatz(number):
print(number)
if number == 1:
return
collatz(number // 2 if number % 2 == 0 else number * 3 + 1)
def collatz(n):
print n
if n == 1:
return
if n % 2 == 0:
n2 = (n / 2)
elif n % 2 == 1:
n2 = (3 * n + 1)
collatz(n2)
print('Please write a number')
number = collatz(int(input()))