Write a Python function squareprime(l) - python-3.x
Write a Python function squareprime(l) that takes a non-empty list of integers and returns True if the elements of l alternate between perfect squares and prime numbers, and returns False otherwise. Note that the alternating sequence of squares and primes may begin with a square or with a prime.
Here are some examples to show how your function should work.
>>> primesquare([4])
True
>>> primesquare([4,5,16,101,64])
True
>>> primesquare([5,16,101,36,27])
False
Function prime_checker checks if a number is prime or not.
Function is_square checks if a number is prefect_square or not.
If the number at index 0 is square, then the chain is like [square, prime, square...]
else the sequence goes like [prime, square, prime, square..]
If the input sequence is not either of the two, then it is not a valid sequence and it will return False.
import math
def prime_checker(num):
flag = True
if num == 2:
return True
elif num < 2:
return False
else:
for i in range(2, int(num/2)):
if num % i == 0:
flag = False
break
return flag
def is_square(integer):
if integer == 0:
return false
root = math.sqrt(integer)
if int(root + 0.5) ** 2 == integer:
return True
return False`
def primesquare(list_nums):
if len(list_nums) == 0:
return False
if len(list_nums) == 1:
if (is_square(list_nums[0]) or prime_checker(list_nums[0])):
return True
else:
return False
else:
flag = True
if is_square(list_nums[0]):
check_for = 'prime'
elif prime_checker(list_nums[0]):
check_for = 'square'
else:
return False
for i in range(1,len(list_nums)):
if (check_for == 'prime' and prime_checker(list_nums[i])):
check_for = 'square'
elif (check_for == 'square' and is_square(list_nums[i])):
check_for = 'prime'
else:
flag = False
break
if flag:
return True
else:
return False
Update:
As the element at the 0th index has already been checked, we are not concerned about that number anymore. Hence, if the 0th element is a prime number, then the sequence will be [prime, square, prime, square,...].
If it is a perfect square, then the sequence will be [square, prime, square, prime...].
If it is neither of the two, then it is not a valid sequence and hence, false is returned.
Now, if the first number was either of the two, and the length of the list is greater than 1, then we will iterate over the remaining elements and check if they are similar to what we expected, but changing the value of check_for variable.
If the value of check_for is prime and the value that we encountered is also prime, then we know that the next number of the sequence should be a square number for the sequence to be a valid sequence. The similar thing happens when a square number is encountered.
from math import sqrt
def square(n):
if(sqrt(n) % 1 == 0):
return True
else:
return False
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def primesquare(list_nums):
if len(list_nums) == 0:
return False
if len(list_nums) == 1:
if (square(list_nums[0]) or isprime(list_nums[0])):
return True
else:
return False
else:
flag = True
if square(list_nums[0]):
check_for = 'prime'
elif isprime(list_nums[0]):
check_for = 'square'
else:
return False
for i in range(1,len(list_nums)):
if (check_for == 'prime' and isprime(list_nums[i])):
check_for = 'square'
elif (check_for == 'square' and square(list_nums[i])):
check_for = 'prime'
else:
flag = False
break
if flag:
return True
else:
return False
from math import sqrt
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def primesquare(l):
flag=0
if len(l)==1:
n=l[0]
if(sqrt(n)%1==0):
return True
else:
for i in range(0,len(l)):
if(sqrt(l[i])%1==0):
if(i==0):
if(isprime(l[i+1])==True):
flag=1
else:
if(isprime(l[i-1])==True):
if(isprime[i+1]==True):
flag=1
else:
flag=0
else:
flag=0
if(flag==0):
return False
else:
return True
from math import sqrt
def square(n):
if(sqrt(n) % 1 == 0):
return True
else:
return False
def isprime(n):
if (n == 1):
return(False)
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def squareprime(l):
s=l[0]
if(isprime(s)):
for i in range(0,len(l),2):
if(isprime(l[i])==False):
return False
for i in range(1,len(l),2):
if(square(l[i])==False):
return False
return True
elif(square(s)):
for i in range(0,len(l),2):
if(square(l[i])==False):
return False
for i in range(1,len(l),2):
if(isprime(l[i])==False):
return False
return True
else:
return False
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Exercise in python is returning None
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I don't understand why you return False when the length of the list is equal to 1. I mean 2 in [2] => True despite len([2]) => 1 and: busca([2], 2) => True This makes more sense to me: def busca(lst, tgt): if len(lst) !=0: if lst[0] == tgt: return True else: return busca(lst[1:], tgt) else: return False Can you explain?
Recursive palindrome check - having issues when recalling the function
The problem is simple, check if palindrome or not using recursion. They also provided a template so I can't change that.
The template:
def isPalindrome(s): # Wrapper function
def isPalindromeRec(s,low,high):
""" Recursive function which checks if substring s[low ... high] is palindrome
returns a True/False value"""
n = len(s)
return isPalindromeRec(s,0,n-1)
I am nearly there but I think I am having trouble understanding how recursion exactly works. (especially how the values changes in the recursion)
My code:
def isPalindrome(s): # Wrapper function
def isPalindromeRec(s,low,high):
if len(s)<=1:
return True
else:
if s[0]==s[len(s)-1]:
return isPalindromeRec(s[low+1:high],low+1,high-1)
else:
return False
n = len(s)
return isPalindromeRec(s,0,n-1)
print(isPalindrome("anna"))
print(isPalindrome("civic"))
print(isPalindrome("a"))
print(isPalindrome("tx1aa1xt"))
print(isPalindrome(""))
print(isPalindrome("Civic"))
print(isPalindrome("ab"))
this is the output:
runfile('/Users/Rayan/Desktop/AUB Spring 2019/EECE 230 /HW/Homework 7/Problem2.py', wdir='/Users/Rayan/Desktop/AUB Spring 2019/EECE 230 /HW/Homework 7')
True
True
True
False
True
False
False
the first false should be True.
Thank you for your help!
Rewrote it:
def isPalindrome(s):
def isPalindromeRec(s,low,high):
if (low == high):
return True
if (s[low] != s[high]) :
return False
if (low < high + 1) :
return isPalindromeRec(s, low + 1, high - 1);
return True
n = len(s)
if (n == 0) :
return True
return isPalindromeRec(s, 0, n - 1);
print(isPalindrome("anna"))
print(isPalindrome("civic"))
print(isPalindrome("a"))
print(isPalindrome("tx1aa1xt"))
print(isPalindrome(""))
print(isPalindrome("Civic"))
print(isPalindrome("ab"))
output:
True
True
True
True
True
False
False
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