when i try to print this line:
print(perfect_square(0))
i should get True but instead i get a time limit exceeded error and i dont know how to fix it.
i tried chaging it to an elif statment instead of 2 separate if statements but i still get that error
This is my current code:
def perfect_square(n):
s = 1
while s != n:
if s*s == n:
return True
elif s == 0:
return True
else:
s +=1
return False
def perfect_cube(n):
s = 1
while s != n:
if s*s * s == n:
return True
elif s == 0:
return True
else:
s +=1
return False
Seems quite clear to me why the perfect_square(0) and perfect_cube(0) cases cause an infinite loop. You start s=1 and always increment it s+=1. It will never be equal to n=0 so you get an infinitely running program. Maybe try making checks for invalid values of n?
def perfect_cube(n):
if n < 1: return False
# ...
Related
Hi I'm a beginner and I'm stuck on this question that wants me to use only while loop to solve. The question wants me to write a function that returns True when the given number is a prime number and it returns False if the given number is not a prime number.
My code so far:
def is_prime(n):
i = 2
while i <= n//2:
if n%i != 0:
return True
else:
return False
i+=1
The problem I have is I think my code displays the correct output for numbers 4 and above and it returns 'None' for 1, 2, and 3. I've debugged it and I think the problem is the while loop condition. But I don't know how to fix it. I would appreciate it if any of you pros can help me out!
edit:
I changed the while condition but 1 still returns None.. and 2 returns False when it's supposed to return True
def is_prime(n):
i = 2
while i <= n:
if n%i != 0:
return True
else:
return False
i+=1
import math;
def is_prime(n):
i = 2
while i < max(math.sqrt(n),2):
if n%i != 0:
return True
else:
return False
if i == 2:
i+=1
else
i+=2
You could hard-code these 3 cases, in case you dont want to use sqrt:
def is_prime(n):
i = 2
if n in (1,3):
return True
elif n == 2:
return False
while i <= n//2:
if n%i != 0:
return True
else:
return False
i+=1
for x in range(1, 5):
print(x, '=', is_prime(x))
Output:
(1, '=', True)
(2, '=', False)
(3, '=', True)
(4, '=', False)
Want to get really fancy? Make a Sieve of Eratosthenes:
def is_prime(n):
a = list()
# Assume all are prime
a[0:n+1] = (n+1)*[1]
# Start with removing even numbers
i = 2
while i*i <= n:
print ("I: ", i)
# Set all divisible by i to 0
a[0:n+1:i] = len(a[0:n+1:i])*[0]
# If a[n] is zero, return False
if a[n] == 0:
return False
# Increment i until we have a prime number
while a[i] == 0:
i+=1
if a[n] == 0:
return False
else:
return True
If you want to impress your lesson teacher you can show him a fast probabilistic prime number isprime for numbers larger than 2**50. I haven't found any errors in it after weeks of cpu time stress testing it on a 6 core AMD:
import random
import math
def lars_last_modulus_powers_of_two(hm):
return math.gcd(hm, 1<<hm.bit_length())
def fast_probabilistic_isprime(hm):
if hm < 2**50:
return "This is to only be used on numbers greater than 2**50"
if lars_last_modulus_powers_of_two(hm+hm) != 2:
return False
if pow(2, hm-1, hm) == 1:
return True
else:
return False
def fast_probabilistic_next_prime(hm):
if hm < 2**50:
return "This is to only be used on numbers greater than 2**50"
if hm % 2 == 0:
hm = hm + 1
hm += 2
while fast_probabilistic_isprime(hm) == False:
hm += 2
return hm
""" hm here is bitlength, which must be larger than 50.
usage is create_probabilistic_prime(1000)
"""
def create_probabilistic_prime(hm):
if 2**hm < 2**50:
return "This is to only be used on numbers greater than 2**50"
num = random.randint(2**hm,2**(hm+1))
return fast_probabilistic_next_prime(num)
A friend of mine told me that she needs help with some homework, I owe her a favor so I said fine, why not. she needed help with a program that checks a sequence, if the sequence is made of the same 2 chars one after the other it will print "yes" (for example "ABABABAB" or "3$3$3$3:)
The program works fine with even length strings (for example "abab") but not with odd length one ("ububu")
I made the code messy and "bad" in purpose, computers is her worst subject so I don't want it to look obvious that someone else wrote the code
the code -
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = StringInput[0]
L2 = StringInput[1]
i = 0
while i <= len(StringInput):
if i % 2 == 0:
if StringInput[i] == L1:
i = i + 1
else:
GoodOrBad = False
break
if i % 2 != 0:
if StringInput[i] == L2:
i = i + 1
else:
GoodOrBad = False
break
if GoodOrBad == True:
print("yes")
elif GoodOrBad != True:
print("no")
main()
I hope someone will spot the problem, thanks you if you read everything :)
How about (assuming s is your string):
len(set(s[::2]))==1 & len(set(s[1::2]))==1
It checks that there is 1 char in the even locations, and 1 char in the odd locations.
a) Showing your friend bad and messy code makes her hardly a better programmer. I suggest that you explain to her in a way that she can improve her programming skills.
b) If you check for the character at the even position and find that it is good, you increment i. After that, you check if i is odd (which it is, since you found a valid character at the even position), you check if the character is valid. Instead of checking for odd position, an else should do the trick.
You can do this using two methods->
O(n)-
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = StringInput[0]
L2 = StringInput[1]
i = 2
while i < len(StringInput):
l=StringInput[i]
if(l==StringInput[i-2]):
GoodOrBad=True
else:
GoodOrBad=False
i+=1
if GoodOrBad == True:
print("yes")
elif GoodOrBad == False:
print("no")
main()
Another method->
O(1)-
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = set(StringInput[0::2])
L2 = set(StringInput[1::2])
if len(L1)==len(L2):
print("yes")
else:
print("no")
main()
There is a lot in this that I would change, but I am just showing the minimal changes to get it to work. There are 2 issues.
You have an off by one error in the code:
i = 0
while i <= len(StringInput):
# in the loop you index into StringInput
StringInput[i]
Say you have 5 characters in StringInput. Because your while loop is going from i = 0 to i < = len(StringInput), it is going to go through the values [0, 1, 2, 3, 4, 5]. That last index is a problem since it is off the end off StringInput.
It will throw a 'string index out of range' exception.
You need to use:
while i < len(StringInput)
You also need to change the second if to an elif (actually it could just be an else, but...) so you do not try to test both in the same pass of the loop. If you go into the second if after the last char has been tested in the first if it will go out of range again.
elif i % 2 != 0:
So the corrected code would be:
def main():
StringInput = input('your string here - ')
GoodOrBad = True
L1 = StringInput[0]
L2 = StringInput[1]
i = 0
while i < len(StringInput):
if i % 2 == 0:
if StringInput[i] == L1:
i = i + 1
else:
GoodOrBad = False
break
elif i % 2 != 0:
if StringInput[i] == L2:
i = i + 1
else:
GoodOrBad = False
break
if GoodOrBad == True:
print("yes")
elif GoodOrBad != True:
print("no")
main()
def main():
StringInput = input('your string here - ')
MaxLength = len(StringInput) // 2 + (len(StringInput) % 2 > 0)
start = StringInput[:2]
chained = start * MaxLength
GoodOrBad = chained[:len(StringInput)] == StringInput
if GoodOrBad == True:
print("yes")
elif GoodOrBad != True:
print("no")
I believe this does what you want. You can make it messier if this isn't bad enough.
In the console, my program prints the first question, and once the input is entered, prints the second one and terminates. It appears to skip the function. Obviously I've done something(s) wrong, any help would be appreciated. That while-loop still feels wrong.
def Prime(n):
i = n - 1
while i > 0:
if n % i == 0:
return False
print("This number is not prime.")
else:
i = i - 1
return True
print("This number is prime.")
def Main():
n = int(input("What is the number you'd like to check?"))
Prime(n)
answer2 = input("Thank you for using the prime program.")
Main()
Your function returns before printing output, so nothing ever gets to the console. Consider printing before returning:
def Prime(n):
i = n - 1
while i > 0:
if n % i == 0:
print("This number is not prime.") # Here
return False
else:
i = i - 1
print("This number is prime.") # And here
return True
In this primality test program, 'None' is printed when I input a prime, instead of 'True'. How can I get it to print 'True'?.
def main():
import math
def check_n(n):
n_s = int(math.sqrt(n))
for i in range(2, n_s):
if (n_s % i) == 0:
return False
break
else:
return True
def msg():
n = int(input('Enter a number, I will return True if it is a prime'))
return n
print(check_n(msg()))
main()
You need to change int(math.sqrt(n)) to int(math.sqrt(n)+1), because range runs until n_s-1. So if the input is 5, range(2,int(math.sqrt(5))) is just range(2,2), which is empty.
In addition, you need to take the return True outside of the for loop, otherwise your code may stop in a too early stage. You also don't need the break statement after return False (the function will never arrive to that line, as it will return False if it enters to that if statement).
Finally, change if (n_s % i) == 0: to if (n % i) == 0:, as you need to check if n is divisible by i (and not its square root).
Here is a more clean version:
import math
def check_n(n):
n_s = int(math.sqrt(n)+1)
for i in range(2, n_s):
if (n % i) == 0:
return False
return True
def msg():
n = int(input('Enter a number, I will return True if it is a prime'))
return n
print(check_n(msg()))
First: Your break statement is redundant.
Second: For values such as 3 the for loop is never executing because value
n_s is less than 2 and since the for loop isn't executing the
python is returning the default value None(which is returned when
no value is specified).
Hence your check_n(n) function has to be
def check_n(n):
n_s = int(math.sqrt(n))
for i in range(2, n_s + 1):
if (n_s % i) == 0:
return False
return True
one liner :
check_n = lambda n : sum([i for i in range(2, int(math.sqrt(n)+1)) if n % i == 0]) == 0
don't overcomplicate things ..
Your range is (2,2) or None when you choose anything less than 9.
So to solve your first problem: add 2 to n_s (for input 3)
You also have a problem with your logic.
Your for loop should be checking that n mod i is 0, not n_s.
This should work:
def main():
import math
def check_n(n):
n_s = int(math.sqrt(n)+1)
for i in range(2, n_s):
if (n % i) == 0:
return False
return True
def msg():
n = int(input('Enter a number, I will return True if it is a prime'))
return n
print(check_n(msg()))
main()
Hello I'm very new to python and was wondering if you could help me with something.
I've been playing around with this code and can't seem to get it to work.
import math
def main():
if isPrime(2,7):
print("Yes")
else:
print("No")
def isPrime(i,n):
if ((n % i == 0) and (i <= math.sqrt(n))):
return False
if (i >= math.sqrt(n)):
print ("is Prime: ",n)
return True
else:
isPrime(i+1,n)
main()
Now the output for the isPrime method is as follows:
is Prime: 7
No
I'm sure the function should return true then it should print "Yes".
Am I missing something?
You are discarding the return value for the recursive call:
def isPrime(i,n):
if ((n % i == 0) and (i <= math.sqrt(n))):
return False
if (i >= math.sqrt(n)):
print ("is Prime: ",n)
return True
else:
# No return here
isPrime(i+1,n)
You want to propagate the value of the recursive call too, include a return statement:
else:
return isPrime(i+1,n)
Now your code prints:
>>> isPrime(2,7)
is Prime: 7
True