I am trying to create a simple program to apply the statement of the Collatz Conjecture to an integer that the user can enter, I have:
def collatz(n):
print n,
if n % 2 ==0:
n = n / 2
elif n == 0:
Print "Collatz Conjecture true for" , 'n'
else:
n = n *3 + 1
input("\n\nInsert a positive integer:")
def collatz(n)
However it is saying there is a syntax error in the line:
Print "Collatz Conjecture true for" , 'n'
I can't see what mistake ther is in this line.
Also as I haven't been able to test it yet, does this look as though it will work ok?
Python is case sensitive. Use "print" not "Print".
Well, your syntax error is that python is case-sensitive, so you need print rather than Print.
But you've got more problems:
'n' prints the string n. I think what you want is n to print the value of the variable (or if not, then you can just make a single string "... true for n").
Finally (I think), in order to run the function collatz, you don't need the def; that's just for the definition.
More problems:
The stopping condition should be n == 1, not n == 0.
You have to recur or iterate, as is you're only making one step.
Check the input, make sure it really is a positive number.
def collatz_steps(n):
steps=0
if n==1:
return 0
else:
while n!=1:
if n%2==0:
n=n/2
steps+=1
else:
n = 3*n+1
steps+=1
return steps
Related
When I run this code and give input as 25 it should return me its not a prime num,
But when I debug the code the range values are not iterating into if condition, only the first value of the range is passed and if its not == 0 it moves to the else part.
def find(x):
if x > 1:
for i in range(2,x):
if x % i == 0:
return "its not a prime num"
else:
return "Its a prime num"
user = int(input("Enter your no: "))
print(find(user))
Please help me why its working like this , I am new to programming . TIA
As stated in a comment, this is an easy fix. Simply move the else statement's return to outside of the loop.
def find(x):
if x > 1:
for i in range(2,x):
if x % i == 0:
return "its not a prime num"
return "Its a prime num"
user = int(input("Enter your no: "))
print(find(user))
Using a return inside of a loop will break it and exit the function even if the iteration is still not finished. use print instead.
I discovered that for whatever reason for loops never run with the final value an easy fix is to just add 1 to the ending value.
I am trying to write code to find whether a given number is odd or even using recursion in Python.
When I execute my code, the recursive function descends down to 0 correctly but then doesn't halt, and keeps going with negative values. I expect it to stop when it reaches 0.
How to make sure the function returns after it reaches zero?
My code:
def odeven(n):
if n%2 == 0:
print("even no. : ",n)
elif n%2 != 0:
print("odd no. : ",n)
elif (n == 0):
return 0
return odeven(n-1)
result = odeven(10)
print("result odd ={}".format(result))
Fixing the most obvious mistake
Your main issue is that the branch elif (n == 0): will never be reached because 0 is even, so 0 gets caught in the first branch if n%2 == 0:
You can fix it by changing the order of the branches:
def odeven(n):
if (n == 0):
return 0
elif n%2 == 0:
print("even no. : ",n)
else: # n%2 != 0
print("odd no. : ",n)
return odeven(n-1)
result = odeven(10)
print("result odd ={}".format(result))
Output:
even no. : 10
odd no. : 9
even no. : 8
odd no. : 7
even no. : 6
odd no. : 5
even no. : 4
odd no. : 3
even no. : 2
odd no. : 1
result odd =0
Further improvements
You can figure out whether a number n is odd or even simply by checking the value of n % 2. There is no need for recursion here.
Presumably this is an exercise that was given by a teacher who wants you to use recursion to figure out whether n is odd or even without using %. This is terribly inefficient and has no use in practice, and is purely an exercise to learn about recursion. In that case, do not use %. Operator % solves the whole problem by itself so if you use it, you don't need to use recursion. It defeats the purpose of the exercise.
Compare the two following functions:
def oddeven_direct(n):
if (n % 2 == 0):
return 'even'
else:
return 'odd'
def oddeven_recursive(n):
if (n == 0):
return 'even'
elif (n == 1):
return 'odd'
elif (n > 1):
return oddeven_recursive(n-2)
elif (n < 0):
return oddeven_recursive(-n)
print(oddeven_direct(10))
print(oddeven_recursive(10))
Note that I did not call function print inside the function. The function returns a result, which is 'even' or 'odd', and doesn't print anything. This is more consistent. If the user calling the function wants to print something, they can call print themselves.
A variation
Notice how the recursive call jumped from n to n-2? This is because I know that n and n-2 will have the same parity (both odd or both even); and we have two base cases, 0 and 1, to which we're guaranteed to arrive when jumping 2 by 2.
If we had jumped from n to n-1 instead, we'd have run into an issue because n and n-1 have different parity (odd and even or even and odd), so we need some way to keep track of this change of parity during the recursion.
This can be achieved by implementing two distinct functions, is_odd and is_even, and using the recursion to express the facts "n is odd if n-1 is even" and "n is even if n-1 is odd".
Since those functions have names that sound like yes/no question, or true/false question, we will make them return True or False, which are called boolean values.
def is_odd(n):
if n == 0:
return False
else:
return is_even(n-1)
def is_even(n):
if n == 0:
return True
else:
return is_odd(n-1)
print('10 is even? {}'.format(is_even(10)))
print('10 is odd? {}'.format(is_odd(10)))
Note that python knows boolean values very well and we can make the code shorter using the two lazy logic operators and and or:
def is_odd(n):
return n != 0 and is_even(n-1)
def is_even(n):
return n == 0 or is_odd(n-1)
print('10 is even? {}'.format(is_even(10)))
print('10 is odd? {}'.format(is_odd(10)))
You get an error, because you call the function inside the function and never stop that function call so it recurses over and over again. Python allows recursion but only until a defined end is reached
To change this limit:
https://www.geeksforgeeks.org/python-handling-recursion-limit/
Also your recursion gets into minus, because you call your function attribute (= (n)) allways one number lower than the function call before:
Your function is now called with
odeven(9)
and so on until it reaches -985, which is in your case the maximum recursion depth which results in:
[Previous line repeated 992 more times]
File "/python/test.py", line 5, in odeven
print("odd no. : ",n)
I'm relatively new to Python and I decided to try and code a relatively simple collatz conjecture where the user enters a number (integer). The code is just a simple function that calls itself. i is a list that should have every number that the function calculates appended to it. I'm new to executing Python scripts and I have tried using the IDLE shell to run the code. It asks me what number I want but when I enter a number nothing is printed? I'm sure I just need to edit a small bit of this code (or maybe it's all wrong yikes) but does anybody have any idea why my script returns nothing? Sorry about this and thanks.
Here's the code:
l = input("Enter a number: ")
l = int(l)
i = []
def collatz(n):
if n==1:
return i
if n%2 == 0:
n = n/2
i.append(n)
return collatz(n)
else:
n = ((n*3) + 1) / 2
i.append(n)
return collatz(n)
print(i)
collatz(l)
There are three returns before your print and one of them is inside an else statement, which means that at least one of them will be executed, so your print won't even be reached to be executed, you should move it right after the function definition to see something:
def collatz(n):
print(i) # <= print here
if n==1:
....
See more about the return statement. A snippet:
return leaves the current function call with the expression list (or None) as return value.
As others have mentioned, all of the execution paths in your function end in a return statement, so that print call is unreachable. So if you want each value of n or i to be printed you need to move the call to somewhere that it will be reachable. ;)
Also, there's a little bit of redundancy in that code. You don't need
i.append(n)
return collatz(n)
in both the if and else branches, you can move them outside the if...else block.
Here's a modified version of your code. I've also changed the / operators to // so that the results of the divisions will be integers.
i = []
def collatz(n):
print(n)
if n==1:
return i
if n%2 == 0:
n = n // 2
else:
n = ((n*3) + 1) // 2
i.append(n)
return collatz(n)
# Test
print(collatz(12))
output
12
6
3
5
8
4
2
1
[6, 3, 5, 8, 4, 2, 1]
I'm new to python and I am trying to make a code to print all the square numbers until the square of the desired value entered by the user.
n = raw_input("Enter number")
a = 1
while a < n:
a = 1
print(a*a)
a += 1
if a > n:
break
When I run this code it infinitely prints "1" ... I'm guessing that the value of a does not increase by += so it's a=1 forever. How do I fix this?
There are some problems. First, your input (what raw_input() returns) is a string, so you must convert it to integer:
n = int(raw_input(...))
Second, you are setting a = 1 each iteration, so, since the loop condition is a < n, the loop will run forever ( if n > 1). You should delete the line
a = 1
Finally, it's not necesary to check if a > n, because the loop condition will handle it:
while a < n:
print a * a
a += 1
# 'if' is not necessary
There is a small error in your code:
while a < n:
a=1 # `a` is always 1 :)
print a*a
a += 1
if a > n:
break
You're setting the value of a back to 1 on every iteration of the loop, so every time it checks against n, the value is 2. Remove the a=1 line.
As others have noted, your specific problem is resetting a each time you loop. A much more Pythonic approach to this is the for loop:
for a in range(1, n):
print(a ** 2)
This means you don't have to manually increment a or decide when to break or otherwise exit a while loop, and is generally less prone to mistakes like resetting a.
Also, note that raw_input returns a string, you need to make it into an int:
n = int(raw_input("Enter number: "))
an even better idea is to make a simple function
def do_square(x):
return x*x
then just run a list comprehension on it
n = int(raw_input("Enter #:")) #this is your problem with the original code
#notice we made it into an integer
squares = [do_square(i) for i in range(1,n+1)]
this is a more pythonic way to do what you are trying to do
you really want to use functions to define functional blocks that are easy to digest and potentially can be reused
you can extend this concept and create a function to get input from the user and do some validation on it
def get_user_int():
#a function to make sure the user enters an integer > 0
while True:
try:
n = int(raw_input("Enter a number greater than zero:"))
except TypeError:
print "Error!! expecting a number!"
continue;
if n > 0:
return n
print "Error: Expecting a number greater than zero!"
and then you can build your input right into your list
squares = [do_square(i) for i in range(1,get_user_int()+1)]
and really do_square is such a simple function we could easily just do it in our loop
squares = [x*x for x in range(1,get_user_int())]
The first line in your loop sets's a to one on every iteration.
You assign a=1 inside the loop. That means it's overwriting the a+=1.
try this:
n = eval(raw_input("Enter number"))
a=1
while a < n:
print a*a
a += 1
The issue here is that the value of a gets overridden every time you enter in the loop
Problem is in the condition of the while loop, to print squares of numbers upto a limiting value, try this..
def powers(x):
n=1
while((n**2)<=x):
print(n**2, end =' ')
n +=1
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Simple Prime Generator in Python
First I will prompt user to input any number. Then my code will check whether does the number input by the user is a Prime number not.
Here is my codes:
num = int(raw_input("Input any of the number you like:"))
for x in range(2, int(num**0.5)+1):
if num % x == 0:
print "It is not a prime number"
else:
print "It is a prime number"
But question is I cant seem to get the output for 2 and 3. And when I randomly input any numbers like 134245, the system will output alot of sentences. And I do not know why?
Appreciate any kind souls to help me up :)
import urllib
tmpl = 'http://www.wolframalpha.com/input/?i=is+%d+a+prime+number'
def is_prime(n):
return ('is a prime number' in urllib.urlopen(tmpl % (n,)).read())
you should stop once num % x == 0 is true (no need for further testing) and print 'it is a prime number' only if the loop completed without anything printed before.
A number is prime if it only divides by 1 and itself. A pseudocode follows:
boolean prime = true;
for (int i = 2; i * i <= num; i++)
if (num % i == 0) {
prime = false;
break;
}
if (prime)
println("It is prime!");
else
println("It is not prime!");
Look at your code like this:
num = ...
for x in range(2, int(num**0.5)+1):
print something
The body of the loop is executed at every iteration. That means you're printing something at every iteration, i.e. for each x that you check to see if it's a factor of num, you print. That's not what you should be doing; in order to determine whether a number is prime, you check all possible factors first, then print your result. So you shouldn't be printing anything until after the loop.
But question is I cant seem to get the output for 2 and 3.
You're looping from 2 to ceil(sqrt(n)). For 2 and 3, this is an empty range, hence no iteration happens. Either special-case it or rewrite the code such that it assumes that n is prime and tries to disprove it in the loop.
the system will output alot of sentences.
You're printing on every iteration. Instead, use a boolean flag (or a premature return, if you factor it out into a function) to determine prime-ness and print once, after the loop, based on that prime.
Your code is not structured well-- the algorithm continues to loop all the way up to the top of your range, even if you already know that the number is composite, and it also prints some result on each iteration when it should not.
You could put the logic into a function and return True or False for prime-ness. Then you could just check the result of the function in your if statement.
def is_prime(num):
for x in range(2, int(num**0.5)+1):
if num % x == 0:
return False
return True
num = int(raw_input("Input any of the number you like:"))
if not is_prime(num):
print "It is not a prime number"
else:
print "It is a prime number"
Here are two Python routines for calculating primes. (Hint: Google for Sieve of Eratosthenese):
def pythonicSieve(maxValue):
"""
A Pythonic Sieve of Eratosthenes - this one seems to run slower than the other.
see http://love-python.blogspot.com/2008/02/find-prime-number-upto-100-nums-range2.html
"""
return [x for x in range(2,maxValue) if not [t for t in range(2,x) if not x%t]]
def sieveOfEratosthenes(maxValue):
"""
see http://hobershort.wordpress.com/2008/04/15/sieve-of-eratosthenes-in-python/
"""
primes = range(2, maxValue+1)
for n in primes:
p = 2
while n*p <= primes[-1]:
if n*p in primes:
primes.remove(n*p)
p += 1
return primes