Closest Prime Number in Python

问题

I need a user to enter a number and enter out the closest prime number to the value they put in. I am struggling on how to check the prime numbers before and after the number they put in. The last part is to print the smaller value of the two prime numbers if they are the same distance away from the inputted number.

n = int(input("Enter n: "))

holder1 = n
holder2 = n

prime = True

holder3 = 0
holder4 = 0

for i in range(2,n):
    if (n % i) == 0:
        prime = False


if(prime == True):
    print("The prime closest to " + str(n) + " is " + str(n))
else:
    while (prime == False):

        holder1 -= 1
        holder2 += 1

        for i in range(2,holder1):
            if (n % i) == 0:
                prime = False
            else:
                prime = True
                holder3 = holder1

        for i in range(2,holder2):
            if (n % i) == 0:
                prime = False
            else:
                prime = True
                holder4 = holder2


    if(abs(n - holder3) <= abs(n-holder4)):
        print("The prime closest to " + str(n) + " is " + str(holder3))
    elif (abs(n - holder3) > abs(n-holder4)):
        print("The prime closest to " + str(n) + " is " + str(holder4))

回答1:

Even though I didn't debug your code, the following piece of code should work to find the closest prime number :

n = int(input("Enter n: "))

def chk_prime(n):
    if n>1:
        for i in range(2, n//2+1):
            if n%i==0:
                return False
                break
        else:
            return True
    else:
        return False

if chk_prime(n):
    print(f"{n} is itself a prime.")
else:
    count = 1
    while count<n:
        holder1 = n-count
        holder2 = n+count
        holder1_chk = chk_prime(holder1)
        holder2_chk = chk_prime(holder2)
        if holder1_chk and holder2_chk:
            print(f"closest primes are {holder1}, {holder2}")
            break
        elif holder1_chk and not holder2_chk:
            print(f"closest prime is {holder1}")
            break
        elif holder2_chk and not holder1_chk:
            print(f"closest prime is {holder2}")
            break
        else:
            count = count + 1

First, we define a function specifically to check whether a number is prime or not. Next we initiate count = 1 and create two place-holder values by subtracting count from original number and adding count to the original number. If both of these place-holder values are prime numbers, then we print both of them as closest primes, else the closest one between them.

回答2:

If I understood your question correctly, you are trying to find a way of finding the closest number to the inputted number. If this is the case, the Sieve of Eratosthenes method to calculate all of the prime numbers up to a given range, and then find the prime to the number you entered

# Import math for the infinity functionality
import math

# The Sieve of Eratosthenes method of calculating the primes less than the limit
def getPrimes(limit):
    # The list of prime numbers
    primes = []
    # The boolean list of whether a number is prime
    numbers = [True] * limit
    # Loop all of the numbers in numbers starting from 2
    for i in range(2, limit):
        # If the number is prime
        if numbers[i]:
            # Add it onto the list of prime numbers
            primes.append(i)
            # Loop over all of the other factors in the list
            for n in range(i ** 2, limit, i):
                # Make them not prime
                numbers[n] = False

    # Return the list of prime numbers
    return primes

# The number to find the closest prime of
number = int(input("Enter a number: > "))
# The list of primes using the function declared above
primes = getPrimes(number + 100)

# The distance away from the closest prime
maxDist = math.inf
# The closest prime
numb = 0

# Loop all of the primes
for p in primes:
    # If the prime number is closer than maxDist
    if abs(number - p) < maxDist:
        # Set maxDist to the number
        maxDist = abs(number - p)
        # Set numb to the number
        numb = p

# Print the output
print(numb, "is the closest prime number to the number you entered!")

I hope this answers your question

***** EDIT *****

You said that you cannot use the python math library, so below is the slightly adjusted code that does not use it:


# The Sieve of Eratosthenes method of calculating the primes less than the limit
def getPrimes(limit):
    # The list of prime numbers
    primes = []
    # The boolean list of whether a number is prime
    numbers = [True] * limit
    # Loop all of the numbers in numbers starting from 2
    for i in range(2, limit):
        # If the number is prime
        if numbers[i]:
            # Add it onto the list of prime numbers
            primes.append(i)
            # Loop over all of the other factors in the list
            for n in range(i ** 2, limit, i):
                # Make them not prime
                numbers[n] = False

    # Return the list of prime numbers
    return primes

# The number to find the closest prime of
number = int(input("Enter a number: > "))
# The list of primes using the function declared above
primes = getPrimes(number + 100)

# The distance away from the closest prime
maxDist = 99999999
# The closest prime
numb = 0

# Loop all of the primes
for p in primes:
    # If the prime number is closer than maxDist
    if abs(number - p) < maxDist:
        # Set maxDist to the number
        maxDist = abs(number - p)
        # Set numb to the number
        numb = p

# Print the output
print(numb, "is the closest prime number to the number you entered!")

来源：https://stackoverflow.com/questions/58680930/closest-prime-number-in-python

标签

python

for-loop

while-loop

primes

closest