问题
I have to find out whether number(N) is a prime or not using recursion, no loops are allowed. I've tried converting the usual code that uses a for loop to a recursive one, but it's not behaving the same. This function is included in another function, which is part of another function. only parameters a and N should be used and passed Here is my function.
a=2
def is_prime(a,N):
prime = True
if N <=1:
return
else:
if a >= N:
return
else:
if N == 2:
prime = True
print(N)
return
elif (N % a) == 0:
prime = False
return is_prime(a+1,N)
else:
prime = True
print(N)
return
I believe the bug is somewhere here.
elif (N % a) == 0:
prime = False
return is_prime(a+1,N)
else:
prime = True
print(N)
Here is the code I tried to convert.
if num > 1:
for i in range(2,num):
if (num % i) == 0:
print(num,"is not a prime number")
print(i,"times",num//i,"is",num)
break
else:
print(num,"is a prime number")
else:
print(num,"is not a prime number")
回答1:
Your solution is close, with just a few changes needed to make it work.
def is_prime(a,N):
print(a, N)
if N <= 1:
return
else:
if a >= N:
print(N)
else:
if N == 2:
print(N)
elif (N % a) == 0:
return False
else:
return is_prime(a+1,N)
return False
You didn't give any examples of calling this function, but I assume it's always called with a being 2, since any other value wouldn't make sense. So if you run the above function like so, you should get the right output:
print(is_prime(2, 7)) => True
print(is_prime(2, 4)) => False
print(is_prime(2, 37)) => True
I think you have a misunderstanding of how recursion works, you're assigning this prime variable in the body of the function, but never doing anything with it. Maybe your confusion comes from a misunderstanding of scopes in Python. That prime variable will not be 'shared' across invocations, it will just create a new prime every time.
EDIT: Didn't realize you wanted the function to just print out the prime if it's a prime, changed the code accordingly.
回答2:
Your function sometimes returns something and sometimes returns nothing -- it should be either all one or the other, not both. In this case is_prime() looks like a boolean function so it should return True or False. We'll leave the printing to the caller:
def is_prime(N, a=3):
if N == 2: # special case
prime = True
elif N <= 1 or N % 2 == 0: # too small or even
prime = False
elif a * a > N: # tried all divisors to sqrt, must be prime
prime = True
elif (N % a) == 0: # divides evenly, not a prime
prime = False
else: # can't tell yet, recursively try the next (odd) divisor
prime = is_prime(N, a+2)
return prime
for x in range(100):
if is_prime(x):
print(x)
Keep it simple. Think through each possible case. Avoid increasing the indention depth unnecessarily, it makes your code more complicated.
The above solution tries to speed up prime detection by avoiding even numbers (both divisor and number) and limiting the divisor to the square root of the number. This can matter as without these optimizations, a recursive solution will likely run out of call stack space at around N=1,000 whereas the above should go to N=1,000,000 without expanding the call stack.
回答3:
def prime(n,j):
if(n<2):
return False
if(j==n):
return True
if(n%j==0):
return False
return prime(n,j+1)
print(prime(n,2))
A number is called prime if it is only divisible by itself and 1.
So iterate from 2 to n-1, if n is divisible by any of (2,3,4,..n-1) return False.
If j == n then there is no such number from (2,3,4...n-1) divisible by n, Hence it's Prime.
回答4:
Since the goal is to print the number in case it's prime let's do that part first. You've already got a condition for it in your code but there was no print:
if a >= N:
print(N)
return
Next we need to handle all the cases where N > 1:
if N == 2:
prime = True
print(N)
return
elif (N % a) == 0:
prime = False
return is_prime(a+1,N)
else:
prime = True
print(N)
First check, if N == 2 is unnecessary since there's already a block before that handles all the cases where N is prime so it can be removed. That said having it there doesn't cause any harm.
The next block that checks if N is divisible by a should terminate the recursion. Since you know that N isn't prime you should just stop there.
Final block that gets executed when N is not divisible by a should do the recursion instead. As it stands now the recursion stops as soon as N % a != 0 which is clearly wrong.
Here's a working sample with above modifications and cleanup:
def is_prime(N, a=2):
if N <= 1:
return
elif a >= N:
print(N)
elif N % a != 0:
is_prime(N, a + 1)
回答5:
to print the list of prime numbers between a given range
l=[]
def primenum(x,y):
global l
if x==y:
print(l)
else:
m=0
for i in range(1,x+1):
if x%i==0:
m+=1
if m==2 or x==1:
l+=[x,]
return primenum(x+1,y)
else:
primenum(x+1,y)
来源：https://stackoverflow.com/questions/37095508/how-do-i-find-a-prime-number-using-recursion-in-python