I'm trying to write a program in python that can either print 1 through nth numbers of the prime number sequence, or print just the nth number of the prime number sequence. Here is the code.
import math
P = 2
X = raw_input('Choose a number: ')
Y = 1
def prime(P, Y):
Choice = raw_input('Select 1 to print X numbers of the Prime sequence. \nSelect 2 to print the Xth number in the Prime sequence. \nWhat is your choice: ')
if Choice == "1":
while Y <= int(X):
isprime = True
for x in range(2, int(P) - 1):
if P % x == 0:
isprime = False
break
if isprime:
print P
Y += 1
P += 1
elif Choice == "2":
prime(P, Y)
Basically, i have the first part down, so that it prints 1 through nth numbers of the prime sequence. However, I'm quite lost on how to make it calculate just the nth prime, where the nth prime is the given through raw input. It must be possible to do this in python, however, how would it be done, and what would be the best way to do so, without having to add too many new variables that i don't have here, (though i would be fine doing so). Help would be appreciated.
Add a condition so that if either the user wants to print all numbers, or you have reached the final prime number of the sequence, the number will be printed. (I have also replaced some of the variable names with more descriptive ones, and altered it so that the function is passed the number_of_primes as its only parameter, which would seem to make more sense.)
def print_primes(X):
choice = raw_input('Select 1 to print X numbers of the Prime sequence. \nSelect 2 to print the Xth number in the Prime sequence. \nWhat is your choice: ')
count = 1
n = 2
while count <= X:
is_prime = True
for i in range(2, int(n) - 1):
if n % i == 0:
is_prime = False
break
if is_prime:
if choice == "1" or count == X:
print n
count += 1
n += 1
number_of_primes = int(raw_input('Choose a number: '))
print_primes(number_of_primes)
Just only print if it's the Yth number:
if isprime:
Y += 1
if Y == X:
print P
Related
Write a python script to print all Prime numbers between two given numbers (both values inclusive)
can anyone please tell what am I doing wrong here ?
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
for k in range(a,b):
for i in range(2,k):
if k%i!=0:
if k!=i:
continue
elif k==i:
print(k)
break
elif k!=i:
break
you are checking if a number is prime the wrong way there are many
unhandled cases in your code for example if a is larger than b
you will start looping from "a" anyway
and even if the values are sat up correctly the algorithm is not right
here is my optimal solution hope it will help
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
def is_prime(n):
# negative numbers cannot be primes 1 and 0 are also not primes
if n <= 1:
return False
# since 2 is the first prime we will start looping from it
# until n since you mentioned that n is included
for i in range(2, n + 1):
# if n is cleanly divisible by any number less than n
# that means that n is not prime
if n % i == 0 and n != i:
return False
return True
for k in range(a,b):
if a > b:
print ("a cannot be bigger than b")
if is_prime(k):
print(k)
Here's the solution. I added some comments to explain what the code does.
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
# Include b in the range by adding 1
for num in range(a, b + 1):
# Prime numbers are greater than 1
if num > 1:
for i in range(2, num):
# If the number is divisible, it is not prime
if num % i == 0:
# Check if the number is equal to the number itself
if num != i:
break
else:
# If the loop was not broken, the number isn't divisible
# by other numbers except itself and 1, so it's prime
print(num)
Well, a prime is "a number that is divisible only by itself and 1", so to actually first I would go only to range(2, k-1). This approach you have is one of the most straightforward ways of doing it, and is not computationally friendly. There are algorithms that specialize in this kind of prime number finding.
I have fixed the code, simplifying the expressions and adding like already mentioned +1 for inclusivity.
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
for k in range(a,b+1):
for i in range(2,k+1):
if k==i:
print(k)
elif k%i!=0:
continue
else: # is divisible and isn't 1 or the number
break
I encourage you to see this post, how they do it.
An other way of doing it would be:
#Check only 1 number at time:
def check_prime(check):
if check > 1:
for i in range(2, check):
if check % i == 0:
return False
return True
return False
#then check your numbers in a loop
list_prime = []
for i in range(50, 250):
if check_prime(i):
list_prime.append(i)
print(list_prime)
That way you can check 1 possible prime number at time.
If you need numbers in between just put it in loop.
Usually, when generating prime numbers utilizing some form of the "Sieve of Erotosthenes" algorithm, it is only necessary to check for denominator values up to the square root of the number being evaluated. With that in mind, here is one more possible take on your prime number loop test.
import math
a = int(input("Enter the value of a : "))
b = int(input("Enter the value of b : "))
for k in range(a,b):
if k == 2 or k == 3: # Pick up the prime numbers whose square root value is less than 2
print(k)
x = int(math.sqrt(k) + 1) # Need only check up to the square root of your test number
for i in range(2,x):
if k%i!=0:
if x-1 > i:
continue
else:
print(k)
else: # If the remainder is zero, this is not a prime number
break
Another version to try.
Write a python program to accept 10 numbers through command line arguments and calculate the sum of prime numbers among them.
This is my question.
I tried this
for Number in range (1, 101):
count = 0
for i in range(2, (Number//2 + 1)):
if(Number % i == 0):
count = count + 1
break
if (count == 0 and Number != 1):
print(" %d" %Number, end = ' ')
but this is naive and works only for given range. I want to input 10 numbers in command prompt and it should find the sum of prime numbers in my input. I tried using import sys and giving sys.args[input] but it is showing errors.
Can someone please help with this?I am just starting to learn coding.
Thanks in advance:)
A solution can be this:
import sys
nums = sys.argv
def with_loop():
total = 0 # to collect the prime numbers
count = 1 # a counter for the numbers that you entered
for i in range(10):
num = int(input("{}. Please enter a number: ".format(count)))
if num > 1: # if number is larger than 1, we need to check
for j in range(2, num):
if (num % j) == 0:
break
else:
total += num
elif num == 1: # 1 is a prime number
total += num
else: # otherwise the number is negative so we skip.
pass
count += 1
print("\nTotal : {}".format(total))
def with_argv(nums):
total = 0 # to collect the prime numbers
count = 1 # a counter for the numbers that you entered
for i in range(1, len(nums)):
if int(nums[i]) > 1: # if number is larger than 1, we need to check
for j in range(2, int(nums[i])):
if (int(nums[i]) % j) == 0:
break
else:
total += int(nums[i])
elif int(nums[i]) == 1: # 1 is a prime number
total += int(nums[i])
else: # otherwise the number is negative so we skip.
pass
count += 1
print("\nTotal : {}".format(total))
with_loop()
with_argv(nums)
A simple example:
sum = 0
for count in range(1,11):
sum += int(input(f'Enter number {count} : '))
print(f'Total sum : {sum}')
To solve this problem, I would first define a function that checks whether a number is prime or not. It would only return a Boolean and nothing else.
Then I would create two lists- one for storing all the numbers entered by a user and another for storing all the prime numbers among the numbers entered by the user. I would use my predefined function for this check.
When the list of prime numbers is ready, I would use the sum() method to find the sum of the items of the list viz. the primes numbers the user has entered.
Here is the code-
def isPrime(num): #defining a function to check a given number is prime
factors = []
if num == 2:
return True
else:
for i in range(1, num + 1):
if (num % i == 0):
factors.append(i)
if len(factors) == 2:
return True
else:
return False
numbers = [] #list for user entered numbers
primes = [] #list for prime numbers among them
print("Enter 10 numbers.\n")
for i in range(10):
numbers.append(int(input(f'Enter number #{i + 1}:')))
#adding user inputs to numbers list
for i in numbers: #checking if the entered numbers are prime
if isPrime(i):
primes.append(i)
print("Sum of the prime numbers enters by the user is " + str(sum(primes)))
#sum() method adds all elements in an iterable
This code will do.
Spyder uses IPython Console. That is the command line referred in the question. So, if you run the code in Spyder, the IPython console will run this program.
I am pretty new to python, so I don't fully understand how to use loops. I am currently working on a piece of code that I have to find the first N prime numbers.
The result that is desired is if you input 5, it outputs 2, 3, 5, 7, and 11, but no matter what I input for 'max', the output always ends up being 2 and 3. Is there a way to improve this?
max=int(input("How many prime numbers do you want: "))
min=2
while(min<=(max)):
for c in range(2, min):
if min%c==0:
break
else:
print min
min=min+1
You only increment min in the else block, i.e., if min % c is nonzero for all c, i.e., if min is prime. This means that the code won't be able to move past any composite numbers. You can fix this by unindenting min=min+1 one level so that it lines up with the for and else.
number = int(input("Prime numbers between 2 and "))
for num in range(2,number + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Solution: Get the nth prime number entry. Iterate through each natural numbers for prime number and append the prime number to a list. Terminate the program when length of a list satisfies the user nth prime number entry.
# Get the number of prime numbers entry.
try:
enterNumber = int(input("List of nth prime numbers: "))
except:
print("The entry MUST be an integer.")
exit()
startNumber = 1
primeList = []
while True:
# Check for the entry to greater than zero.
if enterNumber <= 0:
print("The entry MUST be greater than zero.")
break
# Check each number from 1 for prime unless prime number entry is satisfied.
if startNumber > 1:
for i in range(2,startNumber):
if (startNumber % i) == 0:
break
else:
primeList.append(startNumber)
if (len(primeList) == enterNumber):
print(primeList)
break
else:
startNumber = startNumber + 1
continue
Try that :
n = int(input("First N prime number, N ? "))
p = [2]
c = 2
while len(p) < n:
j = 0
c += 1
while j < len(p):
if c % p[j] == 0:
break
elif j == len(p) - 1:
p.append(c)
j += 1
print(p)
Its simple. Check the below code, am sure it works!
N = int(input('Enter the number: ')
i=1
count=0
while(count<N):
for x in range(i,i+1):
c=0
for y in range(1,x+1):
if(x%y==0):
c=c+1
if(c==2):
print(x)
count=count+1
i=i+1
The following code will give you prime numbers between 3 to N, where N is the input from user:
number = int(input("Prime numbers between 2, 3 and "))
for i in range(2,number):
for j in range(2,int(i/2)+1):
if i%j==0:
break
elif j==int(i/2):
print(i)
You can see to check a number i to be prime you only have to check its divisibility with numbers till n/2.
I am beginner in programming, So can you please tell me what's wrong with my code?
I want to print next palindrome number if the number entered by the user (n) is not palindrome
n = int(input("Enter any number :- "))
reverse = 0
temp = n
while (n!=0):
reverse = reverse * 10
reverse = reverse + n%10
n=n//10
if(temp==reverse):
print ("Already palindrome:: ")
if(temp != reverse):
new_temp = temp
new_reverse = 0
for i in range(new_temp,new_temp+10):
while(temp != 0):
new_reverse = new_reverse * 10
new_reverse = new_reverse + temp%10
temp = temp//10
if(new_temp==new_reverse):
print ("Next pallindrome is :- ",new_temp)
break
if(new_temp != new_reverse):
temp = new_temp+1
There are two problems with your code.
1) Your "for i in range" loop calculates the reverse of the temp variable, but you don't change the temp variable's value.
You do
new_temp = temp
for i in range(new_temp,new_temp+10):
[SNIP]
if(new_temp != new_reverse):
temp = new_temp+1 #this value never changes.
So you're making 10 iterations with one and the same value.
2) Ten iterations might not be enough to find a palindrome. Keep going until you find a palindrome.
Working code:
def reverse(num):
reverse= 0
while num:
reverse= reverse*10 + num%10
num= num//10
return reverse
num= int(input("Enter any number :- "))
if num==reverse(num):
print ("Already palindrome.")
else:
while True:
num+= 1
if num==reverse(num):
print ("Next palindrome is : %s"%num)
break
To check if a number is a palindrome, you don't need to convert it to a number. In fact, its a lot simpler if you just check the string equivalent of your number.
>>> i = '212'
>>> i == i[::-1]
True
>>> i = '210'
>>> i == i[::-1]
False
Use this to your advantage, and create a function:
def is_palindrome(foo):
return str(foo) == str(foo)[::-1]
Next, to find the next palindrome, simply increment the number till your palindrome check is true.
Combine all that, and you have:
def is_palindrome(n):
return str(n) == str(n)[::-1]
n = raw_input('Enter a number: ')
if is_palindrome(n):
print('Congratulations! {0} is a palindrome.'.format(n))
else:
n1 = n
while not is_palindrome(n1):
n1 = int(n1)+1
print('You entered {0}, but the next palindrome is {1}'.format(n, n1))
Here is how it works:
$ python t.py
Enter a number: 123
You entered 123, but the next palindrome is 131
$ python t.py
Enter a number: 121
Congratulations! 121 is a palindrome.
If it helps, I believe it's possible to solve this problem with n/2 iterations where n is the length of the input number. Here's my solution in Python:
def next_palin_number(number):
number+=1
# Convert the number to a list of its digits.
number = list(str(number))
# Initialize two indices for comparing symmetric digits.
i = 0
j = len(number) - 1
while i < j:
# If the digits are different:
if number[i] != number[j]:
# If the lower-power digit is greater than the higher-power digit:
if int(number[j]) > int(number[i]):
if number[j-1]!='9':
number[j - 1] = str(int(number[j - 1]) + 1)
number[j] = number[i]
else:
number = list(str(int(''.join(number[:j]))+1))+number[j:]
else:
number[j] = number[i]
i += 1
j -= 1
# Concatenate and return the result.
return "".join(number)
This problem has a wonderful number of ways to solve them.
One of them is
def nearest_palindrome(number):
#start writitng your code here
while True:
number+=1
if str(number) == str(number)[::-1]:
return number
number=12300
print(nearest_palindrome(number))
Thanks for your time to read my answer : )
I have written this for finding next pallindrome number given a pallindrome number.
def palindrome(num):
bol=False
#x=len(str(num))
num=num+1
while(bol==False):
if(check_palindrome(num)):
bol=True
else:
num=num+1
return num
def check_palindrome(n):
temp=n
rev=0
while(n>0):
dig=n%10
rev=rev*10+dig
n=n//10
if(temp==rev):
return True
b=palindrome(8)
print(b)
def next_palin_drome(n):
while True:
n+=1
if str(n) == str(n)[::-1]:
return n
n=12231
print(next_palin_drome(n))
output:12321
def nearest_palindrome(number):
for i in range(1,number):
number=number+1
tem=str(number)
tem1=tem[-1::-1]
if(tem==tem1):
return number
else:
continue
number=12997979797979797
print(nearest_palindrome(number))
def nearest_palindrome(number):
n = len(str(number))//2
if(len(str(number)) % 2 == 0):
#number like 1221
number_1 = int((str(number))[:n]) #12
number_2 = int((str(number))[n:]) #21
if(number_1 < number_2):
number_1 += 1
number_2 = int(str(number_1)[::-1])
else:
number_2 = int(str(number_1)[::-1])
# if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(number_1)[::-1]
#combining the both parts
ans = int(str(number_1) + str(number_2))
return ans
else:
#numer like 12510 n=2
nu = int((str(number))[:n+1]) #add in this number
number_1 = int((str(number))[:n]) # 12
number_2 = int((str(number))[n+1:]) # 21
if (number_1 < number_2):
nu += 1
number_2 = int((str(nu))[::-1][1:])
else:
number_2 = int((str(nu))[::-1][1:])
#if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(nu)[::-1]
number_2 = number_2[1:]
#combinning both parts
ans = int(str(nu) + str(number_2))
return ans
number=12331
print(nearest_palindrome(number))
If a definite range is given:
# function to check if the number is a palindrome
def palin(x):
s=str(x)
if s==s[::-1]:
return True
else:
return False
n=int(input("Enter the number"))
# Putting up range from the next number till 15 digits
for i in range(n+1,int(10e14)):
if palin(i) is True:
print(i)
break
A brute force method:
def math(n):
while not var:
n += 1
if str(n) == str(n)[::-1] : f = 'but next is : '+str(n); return f
n = int(input()); t = math(n); print('Yes',t) if str(n) == str(n)[::-1] else print('No',t); global var; var = False
This is a good fast solution. I saw that the other solutions were iterating and checking through every +1 they did, but this is really slow for big numbers.
This solution has O(n) time if you look at the length of the number
beginNumber = 123456789101112131415161718 #insert number here for next palidrome
string = str(beginNumber + 1)
length = len(string)
number= [int(x) for x in list(string)]
for i in range(length//2):
if (number[i] != number[length-1-i]):
if (number[i]<number[length-1-i]):
number[length-2-i] += 1
number[length-1-i] = number[i]
print("".join([str(x) for x in number]))
I have written this for finding next pallindrome number given a pallindrome number..
#given a pallindrome number ..find next pallindrome number
input=999
inputstr=str(input)
inputstr=inputstr
#append 0 in beginning and end of string ..in case like 99 or 9999
inputstr='0'+inputstr+'0'
length=len(inputstr)
halflength=length/2;
#if even length
if(length%2==0):
#take left part and reverse it(which is equal as the right part )
temp=inputstr[:length/2]
temp=temp[::-1]
#take right part of the string ,move towards lsb from msb..If msb is 9 turn it to zero and move ahead
for j,i in enumerate(temp):
#if number is not 9 then increment it and end loop
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now you have right hand side...mirror it and append left and right part
output=temp[::-1]+temp
#if the length is odd
if(length%2!=0 ):
#take the left part with the mid number(if length is 5 take 3 digits
temp=inputstr[:halflength+1]
#reverse it
temp=temp[::-1]
#apply same algoritm as in above
#if 9 then make it 0 and move on
#else increment number and break the loop
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now the msb is the middle element so skip it and copy the rest
temp2=temp[1:]
#this is the right part mirror it to get left part then left+middle+right isoutput
temp2=temp2[::-1]
output=temp2+temp
print(output)
similarly for this problem take the left part of given number ...reverse it..store it in temp
inputstr=str(number)
if(inputstr==inputstr[::-1])
print("Pallindrome")
else:
temp=inputstr[:length/2]
temp=temp[::-1]
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
now depending on length of your number odd or even generate the output..as in the code
if even then output=temp[::-1]+temp
if odd then temp2=temp1[1:]
output=temp2[::-1]+temp
I am not sure about this solution..but hope it helps
I wrote a code in python to find the nth prime number.
print("Finds the nth prime number")
def prime(n):
primes = 1
num = 2
while primes <= n:
mod = 1
while mod < (num - 1):
ptrue = 'true'
if num%(num-mod) == 0:
ptrue = 'false'
break
mod += 1
if ptrue == 'true':
primes += 1
return(num)
nth = int(input("Enter the value of n: "))
print(prime(nth)
The code looked fine to me, but it returns an error when I run it:
Traceback (most recent call last):
File "C:/Users/AV/Documents/Python/nth Prime.py", line 17, in <module>
print(prime(nth))
File "C:/Users/AV/Documents/Python/nth Prime.py", line 13, in prime
if ptrue == 'true':
UnboundLocalError: local variable 'ptrue' referenced before assignment
It appears to me as if it is trying to say that I am referring to ptrue in the last line even though I am not. What is the problem here... Can anyone help?
how about using Boolean ? and initalize ptrue out of while loop
print("Finds the nth prime number")
def prime(n):
primes = 1
num = 2
while primes <= n:
mod = 1
ptrue = True
while mod < (num - 1):
if num%(num-mod) == 0:
ptrue = False
break
mod += 1
if ptrue == True:
primes += 1
return(num)
nth = int(input("Enter the value of n: "))
print prime(nth)
ptrue is local to your while loop which goes out of scope as soon as the while loop ends. so declare ptrue before the start of your inner while loop
Get rid of ptrue entirely and use else with your inner loop. For example:
while mod < (num - 1):
if num % (num - mod) == 0:
break
mod += 1
else:
primes += 1 # only executes if loop terminates normally, without `break`
You can try this:
#This program finds nth prime number
import math
def is_prime(number):
if number < 2:
return False
if number % 2 == 0:
return False
else:
for i in range(3, number):
if not number % i:
return False
return True
n = input('Enter n: ')
#This array stores all the prime numbers found till n
primes = []
for i in range(100000):
if is_prime(i):
primes.append(i)
if len(primes) == n:
break
print("nth prime number is: " + str(primes[n-1]))
The first part is define a function that calculates the next prime number given any number.
import math
def is_prime(x): # function
for i in range(2,int(math.sqrt(x))+1):
if x%i == 0:
return is_prime(x+1)
return x
For example, is_prime(10) will return 11.
The next step is to write a generator that returns a list of prime numbers.
def get_prime(k): # generator
cnt = 1
n = 2
while cnt <= k:
yield(is_prime(n))
n = is_prime(n) + 1
cnt += 1
For example, get_prime(5) will return [2,3,5,7,11].
The code below can help you test the results.
a = get_prime(50)
lists = list(a)[:]
for idx, value in enumerate(lists):
print("The {idx}th value of prime is {value}.".format(idx = idx+1, value = value))
All answers depends on user input, but here is a simple code to give nth number, no matter how big the n is ....
def isprime(n): # First the primality test
if n<2:
return False
for i in range(2,n):
if n%i==0:
return False
break
else:
return True
def nthprime(n): # then generic code for nth prime number
x=[]
j=2
while len(x)<n:
if (isprime(j)) == True:
x.append(j)
j =j+1
print(x[n-1])
n=int(input('enter n'))
a=[2,3,5,7]
i=3
j=9
while i<n:
flag=0
j=j+2
for k in range(len(a)):
if (a[k]<=int(j**0.5) and j%a[k]==0):
flag=1
break
if flag==0:
a=a+[j]
i=i+1
print(a[n-1])
Try this.
n = int(input())
count=1
u=2
prime=[]
while(count<=n):
temp=0
for i in range(2,u):
if(u%i==0):
temp=1
if(temp==0):
count+=1
prime.append(u)
u+=1
print(prime[-1])
Program to find nth Prime Number.
def nth_Prime(num):
Semi = num*num
Res_1 = [True for i in range(Semi+1)]
prime = 2
while prime*prime <= Semi:
if Res_1[prime] == True:
for i in range(prime*prime, Semi+1, prime):
Res_1[i] = False
prime += 1
Res_2 = []
for i in range(2, Semi+1):
if Res_1[i]:
Res_2.append(i)
return Res_2[num-1]
if __name__ == "__main__":
num = int(input("Enter nth Number: "))
print(nth_Prime(num))
Try this out ,I just made few changes in yours.
Here I am checking for each prime number using all(num%i!=0 for i in range(2,num)) checking its remainder not equal to zero so if it is true for that range (starting from 2 and less than itself) it is a prime and for that all() function helps me later if its a prime I increment the 'p' count and check till 'p' is less than the 'n'(Input Number) so when it equates the condition its the nth prime we are looking for.
n=raw_input("enter the nth prime ")
num=4
p=2
while p <int(n):
if all(num%i!=0 for i in range(2,num)):
p=p+1
num=num+1
print "nTH prime number: ",num-1