I have to print nth prime palindrome number with the help of this program, where n is number given by the user but I have a problem in this program, as it is taking much time to print the answer.
n=int(input())
l=[]
for i in range(1,1000000):
y=True
if str(i)==str(i)[::-1]:
if i>=2:
for j in range(2,i):
if i%j==0:
y=False
break
if y:
l.append(i)
print("Your Prime Palindrome Number Is:",l[n-1])
How can I make this code time efficient?
The first part of this code is not specific to this question. It's a general purpose strategy for testing prime numbers. It's faster than sympy.isprime() for values lower than ~500,000 (Python 3.11.1 on Intel Xeon) after which the sympy version betters this implementation.
from math import sqrt, floor
def isprime(n):
if n < 2:
return False
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
for i in range(5, floor(sqrt(n))+1, 6):
if n % i == 0 or n % (i + 2) == 0:
return False
return True
Now you need something to test for a palindrome. This function will return True if the string representation of the object is palindromic.
def ispalindrome(o):
return (_ := str(o)) == _[::-1]
And this is the main part of the program. As 2 is the only even prime number, let's treat that as a special case. Otherwise start with 3 and just test subsequent odd numbers.
N = int(input('Enter value for N: '))
if N > 0:
if N == 1:
print(2)
else:
p = 3
while True:
if isprime(p) and ispalindrome(p):
if (N := N - 1) == 1:
print(p)
break
p += 2
Sample output:
Enter value for N: 11
313
Below is my code to make a prime number list from Fibonacci list, but the prime list is not working.
what am I missing? I need to make a prime number list from Fibonacci list.
a1 = []
a2 = []
count = 0
n = int(input())
def fib(n):
a = 1
b = 1
a1.append(a)
a1.append(b)
for i in range(n):
c = a + b
a = b
b = c
if len(a1) <= n - 1:
a1.append(c)
print(a1)
print(len(a1))
fib(n)
def pr(a1):
count = 0
for i in a1:
for j in range(2, i):
if i % j == 0:
count += 1
if count > 0:
a2.append()
print(a2)
why the prime number list not working?
I made some fixes to your code:
def fib(n):
a = 1
b = 2
a1.append(a)
a1.append(b)
for i in range(n):
c=a+b
a=b
b=c
if len(a1)<=n-1:
a1.append(c)
print(a1)
print(len(a1))
def pr(a1):
for i in a1:
isPrime = True
for j in range(2,i):
if i % j == 0:
isPrime = False
break
if isPrime:
a2.append(i)
print(a2)
a1=[]
a2=[]
n=int(input())
fib(n)
pr(a1)
You never called the function pr(). It seems a bit, like you are not quite sure how functions work, maybe read a bit into them (or just ask away :) )
Since you already append a and b anyway, why not start them of with the correct fibonacci sequence (1, 2,..) to avoid duplicates?
I changed the code to find prime numbers a bit, to make it a tiny bit faster. I'm sure this is by no means the optimized, but at least it immedately breaks the loop, if it is clear, that we don't have a prime number.
In pr() you used the append() method without an argument. You need to tell the program what you want to append to the list. As far as I can tell this is, where you got an exception.
Hope this helps a bit.
Try the below code. Edited your code and a2 will give prime numbers in the fibonacci list.
a1 = []
a2 = []
def recur_fibo(n):
if n <= 1:
return n
else:
return(recur_fibo(n-1) + recur_fibo(n-2))
n = 10
if n > 0:
for i in range(n):
a1.append(recur_fibo(i))
for num in a1:
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
a2.append(num)
print(a2)
I am attempting to find the 1000 prime number and am trying to do it without cheating memorization or other code. Can you please tell me whether I am on the right track with my code or completely off base?
primes = []
num = 3
while (len(primes)) < 1000:
if num / 2 == 0:
num = num + 1
else:
x = num - 1
for z in range(2,x):
if num % z == 0 :
num = num + 1
else:
primes.append(num)
num = num + 1
print primes[1000]
You've got a few problems (num / 2 == 0 should be num % 2 == 0)1, but really, using continue and break would really help. e.g.
for z in range(2,x): # xrange would be a performance win on python2.x
if num % z == 0 :
num = num + 1
else:
primes.append(num)
num = num + 1
In this loop, you could find yourself incrementing num a whole bunch of times. in reality, you only want to increment it once (to move on to the next number). So here you'd want to break after you increment it. Also, the else statement would need to be on the for loop:
for z in range(2, x):
if num % z == 0:
break # exit the for loop, don't execute `else` clause.
else:
# only executes if `break` was never encountered.
primes.append(num)
# This happens no matter what and should only happen once
# Might as well pull it out of the loop.
num += 1
1the checking for divisibility by 2 is sort of useless here -- It's the first iteration of your loop. If you used xrange (for python2.x) and break as I've described, then this optimization (likely) isn't worth it.
Aside: You don't really need to run the loop from 2 to N - 1, you can actually get away with running the loop from 2 to int(math.sqrt(N)) which makes this lots more efficient :-) although still not the best you can do... The Sieve of Eratosthenes is another algorithm for doing this more efficiently and there's probably even better methods than that (though I'm not an expert in this area)
I need to print out numbers between 1 and n(n is entered with keyboard) that do not divide by 2, 3 and 5.
I need to use while or for loops and the remainder is gotten with %.
I'm new here and I just don't understand the usage of %?
I tried something like this:
import math
print("Hey. Enter a number.")
entered_number = int(input())
for i in range(1, entered_number):
if i%2 != 0:
print("The number", i, "is ok.")
else:
pass
if i%3 != 0:
print("The number", i, "is ok.")
else:
pass
if i%5 != 0:
print("The number", i, "is ok.")
help?
You need to test for all 3 conditions in one statement, not in 3:
for i in range(1, entered_number):
if i % 2 != 0 and i % 3 != 0 and i % 5 != 0:
print("The number", i, "is ok.")
The and operators here make sure that all three conditions are met before printing.
You are testing each condition in isolation, which means that if the number is, say, 10, you are still printing The number 10 is ok. because it is not divisible by 3. For numbers that are okay, you were printing The number ... is ok. 3 times, as your code tests that it is not divisible by 3 different numbers separately, printing each time.
If something divides by 7 then:
something % 7 == 0
If something divides by 7 and 9 then:
something % 7 == 0 and something % 9 == 0
Conversely, if something divides by 7 or 9 then:
something % 7 == 0 or something % 9 == 0
Something that does not divide by 7 or 9 is given by the expression:
not (something % 7 == 0 or something % 9 == 0)
You don't require the else: pass bits from your code and one if statement with an if-expression that has three %, == bits in it should suffice.
You should probably check the three conditions at the same time:
if i%2 != 0 and i%3 != 0 and i%5 != 0:
print("The number", i, "is ok.")
Otherwise, you would print the same message several times for a single number.
Anyway, for your second question, the% operation is called modulo and it gives you the remainder of a division. For instance, 5%3 = 2 because 5 = 3*1 + 2. And when you check i%2 != 0, you actually check if i can be divided by 2.
print("Hey. Enter a number.")
entered_number = int(input())
for i in range(1, entered_number):
if i%2 != 0 and i%3 !=0 and i%5!=0:
print("The number", i, "is ok.")
a%b returns the remainder when a is divided by b. Example:
>> 5%3
2
What you are doing wrong here is that you are printing after checking a single condition so it will print even if i is divisible by other numbers. For example if i is 3, it will satisfy the first condition and therefore print that the number is ok but it is actually divisible by 3.
I saw you've solved your problem but my answer may worth reading.
This problem is actually doing filtering over a list of numbers 1..n. You can define a base function to test if number x is dividable by number y, and then use this base function to filter the list to get the result.
Here's my version.
import math
from functools import partial
print("Hey. Enter a number.")
entered_number = int(input())
def not_dividable_by(x, y):
return False if x % y == 0 else True
number_list = range(1, entered_number)
for i in [2, 3, 5]:
test_func = partial(not_dividable_by, y=i)
number_list = filter(test_func, number_list)
for number in number_list:
print("%d is OK" % (number,))
I'm truly a beginner at python so I apologise for the lack of knowledge, but the reason I'm asking is that reading the Python manual and tutorial (http://docs.python.org/2.7/tutorial) I'm not unable to totally grasp how loops work. I've written some simple programs so I think I get the basics but for whatever reason this program that is meant to list all primes less than or equal to n is not working:
n = int(raw_input("What number should I go up to? "))
p = 2
while p <= n:
for i in range(2, p):
if p%i == 0:
p=p+1
print "%s" % p,
p=p+1
print "Done"
The output when I enter 100 for example is:
2 3 5 7 11 13 17 19 23 27 29 31 35 37 41 43 47 53 59 61 67 71 73 79 83 87 89 95 97 101 Done
Which looks almost right but for some reason contains 27, 35, 95, which are composite of course. I've been trying to pick apart the way my loop works but I just don't see where it skips checking for divisibility suddenly. I figured that if someone had a look they could explain to me what about the syntax is causing this. Thanks a bunch!
I would actually restructure the program to look like this:
for p in range(2, n+1):
for i in range(2, p):
if p % i == 0:
break
else:
print p,
print 'Done'
This is perhaps a more idiomatic solution (using a for loop instead of a while loop), and works perfectly.
The outer for loop iterates through all the numbers from 2 to n.
The inner one iterates to all numbers from 2 to p. If it reaches a number that divides evenly into p, then it breaks out of the inner loop.
The else block executes every time the for loop isn't broken out of (printing the prime numbers).
Then the program prints 'Done' after it finishes.
As a side note, you only need to iterate through 2 to the square root of p, since each factor has a pair. If you don't get a match there won't be any other factors after the square root, and the number will be prime.
Your code has two loops, one inside another. It should help you figure out the code if you replace the inner loop with a function. Then make sure the function is correct and can stand on its own (separate from the outer loop).
Here is my rewrite of your original code. This rewrite works perfectly.
def is_prime(n):
i = 2
while i < n:
if n%i == 0:
return False
i += 1
return True
n = int(raw_input("What number should I go up to? "))
p = 2
while p <= n:
if is_prime(p):
print p,
p=p+1
print "Done"
Note that is_prime() doesn't touch the loop index of the outer loop. It is a stand-alone pure function. Incrementing p inside the inner loop was the problem, and this decomposed version doesn't have the problem.
Now we can easily rewrite using for loops and I think the code gets improved:
def is_prime(n):
for i in range(2, n):
if n%i == 0:
return False
return True
n = int(raw_input("What number should I go up to? "))
for p in range(2, n+1):
if is_prime(p):
print p,
print "Done"
Note that in Python, range() never includes the upper bound that you pass in. So the inner loop, which checks for < n, we can simply call range(2, n) but for the outer loop, where we want <= n, we need to add one to n so that n will be included: range(2, n+1)
Python has some built-in stuff that is fun. You don't need to learn all these tricks right away, but here is another way you can write is_prime():
def is_prime(n):
return not any(n%i == 0 for i in range(2, n))
This works just like the for loop version of is_prime(). It sets i to values from range(2, n) and checks each one, and if a test ever fails it stops checking and returns. If it checks n against every number in the range and not any of them divide n evenly, then the number is prime.
Again, you don't need to learn all these tricks right away, but I think they are kind of fun when you do learn them.
This should work and is bit more optimized
import math
for i in range(2, 99):
is_prime = True
for j in range(2, int(math.sqrt(i)+1)):
if i % j == 0:
is_prime = False
if is_prime:
print(i)
Please compare your snippet with the one pasted below and you will notice where you were wrong.
n = int(raw_input("What number should I go up to? "))
p = 2
while p <= n:
is_prime=True
for i in range(2, p):
if p%i == 0:
is_prime=False
break;
if is_prime==True:
print "%d is a Prime Number\n" % p
p=p+1
you do not re-start the i loop after you find a non-prime
p = i = 2
while p <= n:
i = 2
while i < p:
if p%i == 0:
p += 1
i = 1
i += 1
print p,
p += 1
print "Done"
A while loop executes the body, and then checks if the condition at the top is True, if it is true, it does the body again. A for loop executes the body once for each item in the iterator.
def is_prime(n):
if n>=2:
for i in range(2, n):
if n%i == 0:
return False
return True
else:
return False
To find PRIME NUMBER
Let's do a couple more improvements.
You know 2 is the only even prime number, so you add 2 in your list and start from 3 incrementing your number to be checked by 2.
Once you are past the half-way point (see above sqrt and * examples), you don't need to test for a prime number.
If you use a list to keep track of the prime numbers, all you need to do is to divide by those prime numbers.
I wrote my code and each of the above items would improve my code execution time by about 500%.
prime_list=[2]
def is_prime(a_num):
for i in prime_list:
div, rem = divmod(a_num, i)
if rem == 0:
return False
elif div < i:
break;
prime_list.append(a_num)
return True
This in my opinion is a more optimised way. This finds all the prime numbers up to 1,000,000 in less than 8 seconds on my setup.
It is also one of my very first attempts at python, so I stand to be corrected
class prime:
def finder (self):
import math
n = long(raw_input("What number should I go up to? "))
for i in range(2, n):
is_prime = True
if i % 2 == 0:
is_prime = False
for j in range(3, long(math.sqrt(i) + 1), 2):
if i % j == 0:
is_prime = False
break
if is_prime:
print(i)
prime().finder()
print('Enter a Number: ')
number=abs(int(input()))
my_List=[0,1]
def is_prime(n):
if n in my_List:
return True
elif n>=2:
for i in range(2, n):
if n%i == 0:
return False
return True
else:
return False
if is_prime(number):
print("%d is Prime!"%number)
else:
print(number,'is not prime')
for i in range(2, p):
if p%i == 0:
p=p+1
print "%s" % p,
p=p+1
I am going to tell your error only,in line 3 you are incrimenting p but actually what you are missing is your i if your i in previous case is let say 13 then it will check your loop after 13 but it is leaving 2,3,5,7,11 so its an error .that is what happening in case of 27 your i before 27 is 13 and now it will check from 14.and I don't think u need an solution.
def findprime(num):
count = 0
for i in range(1,num+1):
list1 = []
for ch in range(1,i+1):
if i%1==0 and i%ch==0:
list1.append(ch)
if len(list1)==2:
count += 1
print(i,end=", ")
print()
return count
num2 = int(input("enter a number: "))
result=findprime(num2)
print("prime numbers between 1 and",num2,"are",result)
Here's a more extensive example with optimization in mind for Python 3.
import sys
inner_loop_iterations: int = 0
def is_prime(n):
a: int = 2
global inner_loop_iterations
if n == 1:
return("Not prime")
elif n == 2:
return("Prime")
while a * a <= n + 1:
inner_loop_iterations += 1
# This if statement reduces the number of inner loop iterations by roughy 50%
# just weeding out the even numbers.
if a % 2 == 0:
a += 1
else:
a += 2
if n % 2 == 0 or n % a == 0:
return ("Not prime")
else:
return ("Prime")
while True:
sys.stdout.write("Enter number to see if it's prime ('q' to quit): ")
n = input()
if not n:
continue
if n == 'q':
break
try:
n = int(n)
except ValueError:
print("Please enter a valid number")
if n < 1:
print("Please enter a valid number")
continue
sys.stdout.write("{}\n".format(is_prime(n)))
sys.stderr.write("Inner loops: {}\n\n".format(inner_loop_iterations))
inner_loop_iterations=0
This program has two main optimizations, first it only iterates from 2 to the square root of n and it only iterates through odd numbers. Using these optimizations I was able to find out that the number 1000000007 is prime in only 15811 loop iterations.
My fast implementation returning the first 25 primes:
#!/usr/bin/env python3
from math import sqrt
def _is_prime(_num: int = None):
if _num < 2:
return False
if _num > 3 and not (_num % 2 and _num % 3):
return False
return not any(_num % _ == 0 for _ in range(3, int(sqrt(_num) + 1), 2))
_cnt = 0
for _ in range(1, 1000):
if _is_prime(_):
_cnt += 1
print(f"Prime N°: {_:,} | Count: {_cnt:,}")
Better use
for i in range(2, p//2 + 1):